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HOLT ®

ChemFile ®

Problem-Solving Workbook

Name Class Date

Skills Worksheet

Problem Solving

Conversions

One of the aims of chemistry is to describe changes—to tell what changed, how it

changed, and what it changed into. Another aim of chemistry is to look at matter

and its changes and to ask questions such as how much, how big, how hot, how

many, how hard, and how long did it take.

For example, chemistry asks the following:

• How much energy is needed to start a reaction?

• How much will the volume of a gas increase if you heat it?

• How long will a reaction take?

• How much can a reaction produce?

• How much of the reactant is needed to produce a required amount of product?

• How much energy does a reaction release?

• How high will the temperature of the solution get as a reaction occurs?

To answer these questions, chemists must make measurements. Measurements in

science can never be treated as just numbers; they must always involve both a

number and a unit. When you use measurements to calculate any quantity, the

unit must always accompany the number in the calculation. Sometimes the unit

given is not the most appropriate unit for the situation or calculation. In this case,

a conversion can change the impractical unit into a more useful one. For

instance, you would not want to measure the distance from New York City to San

Francisco in inches. A simple conversion can transform the number of inches

between the two cities to the much more practical number of miles.

1

4

Quantity

and unit

given

sought

General Plan for Converting Measurements

State the relationship

between the unit given

and the unit sought

as an equality.

Apply each conversion factor

successively so that the units

cancel properly.

Use the

equality

to write possible

conversion factors.

Relationship between

units given and

units sought

Conversion

factor

Holt ChemFile: Problem-Solving Workbook 1 Conversions

2

Problem Solving continued

CONVERTING SIMPLE SI UNITS

Sample Problem 1

A small bottle contains 45.5 g of calcium chloride. What is the mass of

calcium chloride in milligrams?

Solution

ANALYZE

What is given in the problem? the mass of calcium chloride in grams

What are you asked to find? the mass of calcium chloride in milligrams

A table showing what you know and what you do not know can help you organize

the data. Being organized is a key to developing good problem solving skills.

Items Data

Quantity given 45.5 g calcium chloride

Units of quantity given grams

Units of quantity sought milligrams

Relationship between units 1 g � 1000 mg

Conversion factor ?

Quantity sought ? mg calcium chloride

PLAN

What steps are needed to convert grams to milligrams?

Determine a conversion factor that relates grams and milligrams. Multiply the

number of grams by that factor. Arrange the factor so that units cancel to give

the units sought.

Mass of calcium

chloride in g

chloride in mg

write the relationship

between g and mg

Relationship between units: 1 g � 1000 mg

1 g 1000 mg

Possible conversion factors: or

1000 mg 1 g

possible conversion factors:

1 g

or

1000 mg

multiply by conversion factor

1 g � 1000 mg

3

Holt ChemFile: Problem-Solving Workbook 2 Conversions

The correct conversion factor is the one that when multiplied by the given quantity

causes the units to cancel.

COMPUTE

45.5 g� calcium chloride � � 1000mg

� � 45 500 mg calcium chloride

1 g�

EVALUATE

Are the units correct?

Yes; milligrams are the desired units. Grams cancel to give milligrams.

Is the answer reasonable?

Yes; the number of milligrams is 1000 times the number of grams.

Practice

g calcium chloride � 1 g

1. State the following measured quantities in the units indicated:

a. 5.2 cm of magnesium ribbon in millimeters ans: 52 mm

b. 0.049 kg of sulfur in grams ans: 49 g

c. 1.60 mL of ethanol in microliters ans: 1600 �L

d. 0.0025 g of vitamin A in micrograms ans: 2500 �g

e. 0.020 kg of tin in milligrams ans: 20 000 mg

f. 3 kL of saline solution in liters ans: 3000 L

conversion factor

g 3 mg

1000 mg quantity sought

� mg calcium chloride

Holt ChemFile: Problem-Solving Workbook 3 Conversions

2. State the following measured quantities in the units indicated:

a. 150 mg of aspirin in grams ans: 0.15 g

b. 2500 mL of hydrochloric acid in liters ans: 2.5 L

c. 0.5 g of sodium in kilograms ans: 0.0005 kg

d. 55 L of carbon dioxide gas in kiloliters ans: 0.055 kL

e. 35 mm in centimeters ans: 3.5 cm

h. 8740 m in kilometers ans: 8.74 km

i. 209 nm in millimeters ans: 0.000 209 mm

j. 500 000 �g in kilograms ans: 0.0005 khhtg

3. The greatest distance between Earth and the sun during Earth’s revolution is

152 million kilometers. What is this distance in megameters? ans: 152 000 Mm

Holt ChemFile: Problem-Solving Workbook 4 Conversions

Sample Problem 2

A metallurgist is going to make an experimental alloy that requires

adding 325 g of bismuth to 2.500 kg of molten lead. What is the total

mass of the mixture in kilograms?

What is given in the problem? the mass of bismuth in grams, the mass of lead

in kg

What are you asked to find? the total mass of the mixture

Quantity given 325 g of bismuth

Units of quantity sought kilograms

Relationship between units 1000 g � 1 kg

Quantity sought ? kg of bismuth

Mass of lead 2.500 kg

Total mass ? kg of mixture

To be added, the quantities must be expressed in the same units—in this case,

kilograms. Therefore, 325 g of bismuth must be converted to kilograms of bismuth.

What steps are needed to convert grams to kilograms?

Determine a conversion factor that relates grams to kilograms. Apply that

conversion factor to obtain the quantity sought.

What steps are needed to find the total mass of the mixture in kilograms?

Add the mass of the lead in kilograms to the mass of the bismuth in kilograms.

Holt ChemFile: Problem-Solving Workbook 5 Conversions

Mass of bismuth

between g and kg

1000 g � 1 kg

in g

1 kg

1000 g

1000 mg 1 kg

add the two

masses to obtain

the mass of the

mixture

Mass of mixture

Relationship between units: 1000 g � 1 kg

1000 g 1 kg

1 kg 1000 g

g 3 kg

given 1 kg quantity sought

g bismuth � � kg bismuth

calculated above

kg bismuth

325 g� bismuth � �� 0.325 kg bismuth

1000

g�

0.325 kg bismuth � 2.500 kg lead � 2.825 kg mixture

Yes; kilograms are the units sought.

Yes; the value, 0.325, is one-thousandth the given value, 325.

�

kg lead

� kg mixture

Holt ChemFile: Problem-Solving Workbook 6 Conversions

1. How many milliliters of water will it take to fill a 2 L bottle that already

contains 1.87 L of water? ans: 130 mL

2. A piece of copper wire is 150 cm long. How long is the wire in millimeters?

How many 50 mm segments of wire can be cut from the length? ans: 1500

mm; 30 pieces

3. The ladle at an iron foundry can hold 8500 kg of molten iron. 646 metric tons

of iron are needed to make rails. How many ladlefuls of iron will it take to

make 646 metric tons of iron? (1 metric ton � 1000 kg) ans: 76 ladlefuls

Holt ChemFile: Problem-Solving Workbook 7 Conversions

CONVERTING DERIVED SI UNITS

Sample Problem 3

A balloon contains 0.5 m 3 of neon gas. What is the volume of gas in cubic

centimeters?

What is given in the problem? the volume of neon in cubic meters

What are you asked to find? volume of neon in cubic centimeters

Quantity given 0.5 m 3 of neon

Units of quantity given cubic meters

Units of quantity sought cubic centimeters

Relationship between units 1 m � 100 cm

Quantity sought ? cm 3 neon

What steps are needed to convert cubic meters to cubic centimeters?

Rewrite the quantity in simple SI units. Determine the relationship between

meters and centimeters. Write a conversion factor for each of the units in the

given quantity. Multiply that quantity by the conversion factors. Arrange the

factors so that units will cancel to give the units of the quantity sought.

Volume of neon

in m3 1

in cm3 4

Volume of neon in

(m � m � m)

apply conversion factor three

times to convert from cubic

meters to cubic centimeters

1 m � 100 cm

possible

conversion

factors:

100 cm

1 m

1 m or

Holt ChemFile: Problem-Solving Workbook 8 Conversions

Rewrite the given quantity in simple units as follows.

m 3 � m � m � m

0.50 m 3 neon � 0.50 (m� � m� � m�) � � 100

cm

� � �

1 m�

100

� � 100

� � 500 000 cm

3 neon

Yes; cubic centimeters were the units sought.

Yes; 500 000 cm 3 is half the number of cm 3 in 1 m 3 .

Relationship between units: 1 m � 100 cm

1 m 100 cm

100 cm 1 m

conversion factors cm 3 m,

applied three times

m3 100 cm 100 cm

neon � �

1 m 1 m

1. State the following measured quantities in the units indicated.

a. 310 000 cm 3 of concrete in cubic meters ans: 0.31 m 3

b. 6.5 m 2 of steel sheet in square centimeters ans: 65 000 cm 2

c. 0.035 m 3 of chlorine gas in cubic centimeters ans: 35 000 cm 3

d. 0.49 cm 2 of copper in square millimeters ans: 49 mm 2

quantity sought

� cm 3 neon

Holt ChemFile: Problem-Solving Workbook 9 Conversions

e. 1200 dm 3 of acetic acid solution in cubic meters ans: 1.2 m 3

f. 87.5 mm 3 of actinium in cubic centimeters ans: 0.0875 cm 3

g. 250 000 cm 2 of polyethylene sheet in square meters ans: 25 m 2

2. How many palisade cells from plant leaves would fit in a volume of 1.0 cm 3 of

cells if the average volume of a palisade cell is 0.0147 mm 3 ? ans: 68 027 cells

Holt ChemFile: Problem-Solving Workbook 10 Conversions

Additional Problems

1. Convert each of the following quantities to the required unit.

a. 12.75 Mm to kilometers

b. 277 cm to meters

c. 30 560 m 2 to hectares (1 ha � 10 000 m 2 )

d. 81.9 cm 2 to square meters

e. 300 000 km to megameters

2. Convert each of the following quantities to the required unit.

a. 0.62 km to meters

b. 3857 g to milligrams

c. 0.0036 mL to microliters

d. 0.342 metric tons to kilograms (1 metric ton � 1000 kg)

e. 68.71 kL to liters

3. Convert each of the following quantities to the required unit:

a. 856 mg to kilograms

b. 1 210 000 �g to kilograms

c. 6598 �L to cubic centimeters (1 mL � 1 cm 3 )

d. 80 600 nm to millimeters

e. 10.74 cm 3 to liters

4. Convert each of the following quantities to the required unit:

a. 7.93 L to cubic centimeters

b. 0.0059 km to centimeters

c. 4.19 L to cubic decimeters

d. 7.48 m 2 to square centimeters

e. 0.197 m 3 to liters

5. An automobile uses 0.05 mL of oil for each kilometer it is driven. How much

oil in liters is consumed if the automobile is driven 20 000 km?

6. How many microliters are there in a volume of 370 mm 3 of cobra venom?

7. A baker uses 1.5 tsp of vanilla extract in each cake. How much vanilla extract

in liters should the baker order to make 800 cakes? (1 tsp � 5 mL)

8. A person drinks eight glasses of water each day, and each glass contains 300

mL. How many liters of water will that person consume in a year? What is the

mass of this volume of water in kilograms? (Assume one year has 365 days

and the density of water is 1.00 kg/L.)

Holt ChemFile: Problem-Solving Workbook 11 Conversions

9. At the equator Earth rotates with a velocity of about 465 m/s.

a. What is this velocity in kilometers per hour?

b. What is this velocity in kilometers per day?

10. A chemistry teacher needs to determine what quantity of sodium hydroxide to

order. If each student will use 130 g and there are 60 students, how many

kilograms of sodium hydroxide should the teacher order?

11. The teacher in item 10 also needs to order plastic tubing. If each of the 60

students needs 750 mm of tubing, what length of tubing in meters should the

teacher order?

12. Convert the following to the required units.

a. 550 �L/h to milliliters per day

b. 9.00 metric tons/h to kilograms per minute

c. 3.72 L/h to cubic centimeters per minute

d. 6.12 km/h to meters per second

13. Express the following in the units indicated.

a. 2.97 kg/L as grams per cubic centimeter

b. 4128 g/dm 2 as kilograms per square centimeter

c. 5.27 g/cm 3 as kilograms per cubic decimeter

d. 6.91 kg/m 3 as milligrams per cubic millimeter

14. A gas has a density of 5.56 g/L.

a. What volume in milliliters would 4.17 g of this gas occupy?

b. What would be the mass in kilograms of 1 m 3 of this gas?

15. The average density of living matter on Earth’s land areas is 0.10 g/cm 2 . What

mass of living matter in kilograms would occupy an area of 0.125 ha?

16. A textbook measures 250. mm long, 224 mm wide, and 50.0 mm thick. It has a

mass of 2.94 kg.

a. What is the volume of the book in cubic meters?

b. What is the density of the book in grams per cubic centimeter?

c. What is the area of one cover in square meters?

17. A glass dropper delivers liquid so that 25 drops equal 1.00 mL.

a. What is the volume of one drop in milliliters?

b. How many milliliters are in 37 drops?

c. How many drops would be required to get 0.68 L?

Holt ChemFile: Problem-Solving Workbook 12 Conversions

18. Express each of the following in kilograms and grams:

a. 504 700 mg

b. 9 200 000 �g

c. 122 mg

d. 7195 cg

19. Express each of the following in liters and milliliters:

a. 582 cm 3

b. 0.0025 m 3

c. 1.18 dm 3

d. 32 900 �L

20. Express each of the following in grams per liter and kilograms per cubic

meter.

a. 1.37 g/cm 3

b. 0.692 kg/dm 3

c. 5.2 kg/L

d. 38 000 g/m 3

e. 5.79 mg/mm 3

f. 1.1 �g/mL

21. An industrial chemical reaction is run for 30.0 h and produces 648.0 kg of

product. What is the average rate of product production in the stated units?

a. grams per minute

b. kilograms per day

c. milligrams per millisecond

22. What is the speed of a car in meters per second when it is moving at 100.

km/h?

23. A heater gives off energy as heat at a rate of 330 kJ/min. What is the rate of

energy output in kilocalories per hour? (1 cal � 4.184 J)

24. The instructions on a package of fertilizer tell you to apply it at the rate of 62

g/m 2 . How much fertilizer in kilograms would you need to apply to 1.0 ha? (1

ha � 10 000 m 2 )

25. A water tank leaks water at the rate of 3.9 mL/h. If the tank is not repaired,

what volume of water in liters will it leak in a year? Show your setup for

solving this. Hint: Use one conversion factor to convert hours to days and

another to convert days to years, and assume that one year has 365 days.

26. A nurse plans to give flu injections of 50 �L each from a bottle containing 2.0

mL of vaccine. How many doses are in the bottle?

Holt ChemFile: Problem-Solving Workbook 13 Conversions

Significant Figures

A lever balance used to weigh a truckload of stone may be accurate to the nearest

100 kg, giving a reading of 15 200 kg, for instance. The measurement should

be written in such a way that a person looking at it will understand that it represents

the mass of the truck to the nearest 100 kg, that is, that the mass is somewhere

between 15 100 kg and 15 300 kg.

Some laboratory balances are sensitive to differences of 0.001 g. Suppose you

use such a balance to weigh 0.206 g of aluminum foil. A person looking at your

data table should be able to see that the measurement was made on a balance

that measures mass to the nearest 0.001 g. You should not state the measurement

from the laboratory balance as 0.2060 g instead of 0.206 g because the balance

was not sensitive enough to measure 0.0001 g.

To convey the accuracy of measurements, all people working in science use

significant figures. A significant figure is a digit that represents an actual

measurement. The mass of the truck was stated as 15 200 kg. The 1, 5, and 2 are

significant figures because the balance was able to measure ten-thousands, thousands,

and hundreds of kilograms. The truck balance was not sensitive enough to

measure tens of kilograms or single kilograms. Therefore, the two zeros are not

significant and the measurement has three significant figures. The mass of the foil

was correctly stated as 0.206 g. There are three decimal places in this measurement

that are known with some certainty. Therefore, this measurement has three

significant figures. Had the mass been stated as 0.2060 g, a fourth significant figure

would have been incorrectly implied.

A.

B.

Rules for Determining Significant Figures

All digits that are not zeros are significant.

All are nonzero digits.

222

3 2 5 mL of ethanol

The measurement

has three

significant figures.

Zeros may or may not be significant. To determine whether a zero is

significant, use the following rules:

22 22

1.3 2 5 g of zinc

has four

Holt ChemFile: Problem-Solving Workbook 14 Significant Figures

1.

Zeros appearing between nonzero digits are significant.

2.

3.

4.

Nonzero digits

2 2

22 2

4 0. 7 L of ammonia

3 2 0 0 6 m of wire

1 11

Zero between

nonzero digits

Zeros appearing in front of nonzero digits are not significant.

0.0 5 7 2 m

2 222

of foil

Zeros to the right

of a number and after

a decimal point

0.000 2 g of RNA

has one

significant figure.

Zeros at the end of a number and to the right of a decimal are

significant figures. Zeros between nonzero digits and significant

zeros are also significant. This is a restatement of Rule 1.

22

9 7. 0 0 kg of tungsten

11

has five

1 2 0 0.0 0 cm3 22

of lead

1111

of a number and

after a decimal point

has six

Zeros at the end of a number but to the left of a decimal may or

may not be significant. If such a zero has been measured or is the

first estimated digit, it is significant. On the other hand, if the zero

has not been measured or estimated but is just a place holder, it is

not significant. A decimal placed after the zeros indicates that they

are significant.

Holt ChemFile: Problem-Solving Workbook 15 Significant Figures

3 4 0 0 g of sulfur

has two

The rules are summarized in the following flowchart:

4 0 0 0. mL of oxygen

Decimal point is present,

so these zeros are significant.

General Plan for Determining Significant Figures

There is a

decimal point.

Look for zeros at the end of

the number—they are

significant.

Look for nonzero digits—

all of them are significant.

Look for zeros between

nonzero digits—they are

Look for a decimal point.

There is no

Total number of significant

figures

Holt ChemFile: Problem-Solving Workbook 16 Significant Figures

Determine the number of significant figures in the following measurements:

a. 30 040 g

b. 0.663 kg

c. 20.05 mL

d. 1500. mg

e. 0.0008 m

What is given in the problem? five measurements

What are you asked to find? the number of significant figures in each measurement

a b c d e

Measured

quantity

30 040 g 0.663 kg 20.05 L 1500. mg 0.0008 g

What steps are needed to determine the number of significant figures in each

measurement?

Apply the steps in the flowchart to determine the number of significant figures.

Apply the following steps from the flowchart. Eliminate the steps that are not

applicable to the measurement in question.

How many nonzero digits are there? ?

How many zeros are there between nonzero digits? ?

Is there a decimal point? ?

How many significant zeros are at the end of the number? ?

Total number of significant figures ?

Holt ChemFile: Problem-Solving Workbook 17 Significant Figures

SOLVE

The final zero is not significant.

The zero only locates the decimal point and is not significant.

c. 20.05 L

How many nonzero digits are there? 2

How many zeros are there between nonzero digits? 2

Is there a decimal point? no

How many significant zeros are at the end of the number? NA

Total number of significant figures 4

How many nonzero digits are there? 3

How many zeros are there between nonzero digits? NA

Is there a decimal point? yes

Total number of significant figures 3

How many significant zeros are at the end of the number? 2

There is a decimal following the final two zeros, so all digits are significant.

Holt ChemFile: Problem-Solving Workbook 18 Significant Figures

e. 0.0008 g

The zeros are only place holders. They are not significant.

Are the answers reasonable?

Yes; all answers are in agreement with the rules for determining significant

figures.

How many nonzero digits are there? 1

Total number of significant figures 1

1. Determine the number of significant figures in the following measurements:

ans: 2 f. 20.900 cm ___________________

ans: 5

a. 640 cm 3 _____________________

b. 200.0 mL ____________________

ans: 4 g. 0.000 000 56 g/L ______________

ans: 2

c. 0.5200 g _____________________

ans: 4 ans: 4

d. 1.005 kg _____________________

ans: 4 ans: 6

e. 10 000 L _____________________

ans: 1 ans: 6

DETERMINING SIGNIFICANT FIGURES IN CALCULATIONS

Suppose you want to determine the density of an ethanol-water solution. You first

measure the volume in a graduated cylinder that is accurate to the nearest 0.1

mL. You then determine the mass of the solution on a balance that can measure

mass to the nearest 0.001 g. You have read each measuring device as accurately

as you can, and you record the following data:

Measurement Data

Mass of solution, m 11.079 g

Volume of solution, V 12.7 mL

Density of solution in g/mL, D ?

h. 0.040 02 kg/m 3 _______________

i. 790 001 cm 2 __________________

j. 665.000 kg�m/s 2 ______________

Holt ChemFile: Problem-Solving Workbook 19 Significant Figures

You can determine density on your calculator and get the following result:

D � � m 1.

079

g

� � �1 � � 0.872 362 204 g/mL

V 12.

7 mL

Although the numbers divide out to give the result shown, it is not correct to say

that this quantity is the density of the solution. Remember that you are dealing

with measurements, not just numbers. Consider the fact that you measured the

mass of the solution with a balance that gave a reading with five significant figures:

11.079 g. In addition, you measured the volume of the solution with a graduated

cylinder that was readable only to three significant figures: 12.7 mL. It seems

odd to claim that you now know the density with an accuracy of nine significant

You can calculate the density—or any measurement—only as accurately as

the least accurate measurement that was used in the calculation. In this case the

least accurate measurement was the volume because the measuring device you

used was capable of giving you a measurement with only three significant figures.

Therefore, you can state the density to only three significant figures.

Rules for Calculating with Measured Quantities

Operation Rule

Multiplication and division • Round off the calculated result to the same number

of significant figures as the measurement having the

fewest significant figures.

Addition and subtraction • Round off the calculated result to the same number

of decimal places as the measurement with the

fewest decimal places. If there is no decimal point,

round the result back to the digit that is in the same

position as the leftmost uncertain digit in the quantities

being added or subtracted.

In the example given above, you must round off your calculator reading to a

value that contains three significant figures. In this case, you would say:

m

D � �

V

11.079 g

12.7 mL

� 0.872 362 204 g/mL � 0.872 g/mL

Holt ChemFile: Problem-Solving Workbook 20 Significant Figures

In an experiment to identify an unknown gas, it is found that 1.82 L of

the gas has a mass of 5.430 g. What is the density of the gas in g/L?

What is given in the problem? the measured mass and volume of the gas

What are you asked to find? the density of the gas

What step is needed to calculate the density of the gas?

Divide the mass measurement by the volume measurement.

What steps are necessary to round the calculated value to the correct number of

significant figures?

Determine which measurement has the fewest significant figures. Round the

calculated result to that number of significant figures.

Mass of the gas, m gas

Volume of the gas, V gas

D gas �

D gas � m gas

V gas

four significant figures

5.430 g

1.82 L

three significant figures

Yes; density is given in units of mass per unit volume.

Are the significant figures correct?

Yes; the mass had only three significant figures, so the answer was rounded to

three significant figures.

round to correct

significant figures

mgas � numerical result rounded result

Vgas round to three significant figures

2.983 516 484

Density of the gas, D gas (numerical result) ? g/L

Least number of significant figures in measurements 3 (in 1.82 L)

Density of the gas, Dgas (rounded) ? g/L

� 2.98 g/L

the digit following the 8 is

less than 5, so the 8 remains

unchanged

Holt ChemFile: Problem-Solving Workbook 21 Significant Figures

Yes; the mass/volume ratio is roughly 3/1, so the density is approximately 3 g/L.

1. Perform the following calculations, and express the result in the correct units

and number of significant figures.

a. 47.0 � 2.2 s ans: 21 m/s

b. 140 cm � 35 cm ans: 4900 cm 2

c. 5.88 kg � 200 m 3 ans: 0.03 kg/m 3

d. 0.00 50 m 2 � 0.042 m ans: 0.000 21 m 3

e. 300.3 L � 180. s ans: 1.67 L/s

f. 33.00 cm 2 � 2.70 cm ans: 89.1 cm 3

g. 35 000 kJ � 0.250 min ans: 140 000 kJ/min

Holt ChemFile: Problem-Solving Workbook 22 Significant Figures

Three students measure volumes of water with three different devices.

They report the following volumes:

Device Volume measured

Large graduated cylinder 164 mL

Small graduated cylinder 39.7 mL

Calibrated buret 18.16 mL

If the students pour all of the water into a single container, what is the

total volume of water in the container?

What is given in the problem? three measured volumes of water

What are you asked to find? the total volume of water

First volume of water 164 mL

Second volume of water 39.7 mL

Third volume of water 18.16 mL

Total volume of water ?

What step is needed to calculate the total volume of the water?

Add the separate volumes.

Determine which measurement has the fewest decimal places. Round the calculated

result to that number of decimal places.

Vtotal � V1 � V2 � V3 � 164 mL � 39.7 mL � 18.16 mL

164 mL

� 39.7 mL

� 18.16 mL

221.86 mL

Round the sum to the same number of decimal places as the measurement with

the fewest decimal places (164 mL).

Vtotal � 221.86 mL � 222 mL

the digit following the 1 is

greater than 5, so the 1 is

rounded up to 2

Holt ChemFile: Problem-Solving Workbook 23 Significant Figures

Yes; the given values have units of mL.

Yes; three significant figures is correct.

Yes; estimating the values as 160, 40, and 20 gives a sum of 220, which is very

near the answer.

1. Perform the following calculations and express the results in the correct units

and number of significant figures:

a. 22.0 m � 5.28 m � 15.5 m ans: 42.8 m

b. 0.042 kg � 1.229 kg � 0.502 kg ans: 1.773 kg

c. 170 cm 2 � 3.5 cm 2 � 28 cm 2 ans: 150 cm 2

d. 0.003 L � 0.0048 L � 0.100 L ans: 0.108 L

e. 24.50 dL � 4.30 dL � 10.2 dL ans: 39.0 dL

f. 3200 mg � 325 mg � 688 mg ans: 2800 mg

g. 14 000 kg � 8000 kg � 590 kg ans: 23 000 kg

Holt ChemFile: Problem-Solving Workbook 24 Significant Figures

a. 0.0120 m

b. 100.5 mL

c. 101 g

d. 350 cm 2

e. 0.97 km

2. Round the following quantities to the specified number of significant figures:

a. 5 487 129 m to three significant figures

b. 0.013 479 265 mL to six significant figures

c. 31 947.972 cm 2 to four significant figures

d. 192.6739 m 2 to five significant figures

e. 786.9164 cm to two significant figures

f. 389 277 600 J to six significant figures

g. 225 834.762 cm 3 to seven significant figures

3. Perform the following calculations, and express the answer in the correct

units and number of significant figures.

a. 651 cm � 75 cm

d. 360 cm � 51 cm � 9.07 cm

b. 7.835 kg � 2.5 L

c. 14.75 L � 1.20 s

4. Perform the following calculations, and express the answer in the correct

a. 7.945 J � 82.3 J � 0.02 J

b. 0.0012 m � 0.000 45 m � 0.000 11 m

c. 500 g � 432 g � 2 g

d. 31.2 kPa � 0.0035 kPa � 0.147 kPa

e. 312 dL � 31.2 dL � 3.12 dL

f. 1701 kg � 50 kg � 43 kg

f. 1000 kg

g. 180. mm

h. 0.4936 L

i. 0.020 700 s

5. A rectangle measures 87.59 cm by 35.1 mm. Express its area with the proper

number of significant figures in the specified unit:

a. in cm 2

b. in mm 2

c. in m 2

e. 5.18 m � 0.77 m � 10.22 m

f. 34.95 g � 11.169 cm 3

Holt ChemFile: Problem-Solving Workbook 25 Significant Figures

6. A box measures 900. mm by 31.5 mm by 6.3 cm. State its volume with the

proper number of significant figures in the specified unit:

a. in cm 3

b. in m 3

c. in mm 3

7. A 125 mL sample of liquid has a mass of 0.16 kg. What is the density of the

liquid in the following measurements?

a. kg/m 3

b. g/mL

c. kg/dm 3

8. Perform the following calculations, and express the results in the correct

units and with the proper number of significant figures.

a. 13.75 mm � 10.1 mm � 0.91 mm

b. 89.4 cm 2 � 4.8 cm

c. 14.9 m 3 � 3.0 m 2

d. 6.975 m � 30 m � 21.5 m

9. What is the volume of a region of space that measures 752 m � 319 m � 110 m?

Give your answer in the correct unit and with the proper number of significant

10. Perform the following calculations, and express the results in the correct

a. 7.382 g � 1.21 g � 4.7923 g

b. 51.3 mg � 83 mg � 34.2 mg

c. 0.007 L � 0.0037 L � 0.012 L

d. 253.05 cm 2 � 33.9 cm 2 � 28 cm 2

e. 14.77 kg � 0.086 kg � 0.391 kg

f. 319 mL � 13.75 mL � 20. mL

11. A container measures 30.5 mm � 202 mm � 153 mm. When it is full of a

liquid, it has a mass of 1.33 kg. When it is empty, it has a mass of 0.30 kg.

What is the density of the liquid in kilograms per liter?

12. If 7.76 km of wire has a mass of 3.3 kg, what is the mass of the wire in g/m?

What length in meters would have a mass of 1.0 g?

13. A container of plant food recommends an application rate of 52 kg/ha. If the

container holds 10 kg of plant food, how many square meters will it cover

(1 ha � 10 000 m 2 )?

14. A chemical process produces 974 550 kJ of energy as heat in 37.0 min. What is

the rate in kilojoules per minute? What is the rate in kilojoules per second?

Holt ChemFile: Problem-Solving Workbook 26 Significant Figures

15. A water pipe fills a container that measures 189 cm � 307 cm � 272 cm in 97 s.

a. What is the volume of the container in cubic meters?

b. What is the rate of flow in the pipe in liters per minute?

c. What is the rate of flow in cubic meters per hour?

16. Perform the following calculations, and express the results in the correct

units and with the proper number of significant figures. Note, in problems

with multiple steps, it is better to perform the entire calculation and then

round to significant figures.

a. (0.054 kg � 1.33 kg) � 5.4 m 2

b. 67.35 cm 2 � (1.401 cm � 0.399 cm)

c. 4.198 kg � (1019 m 2 � 40 m 2 ) � (54.2 s � 31.3 s)

d. 3.14159 m � (4.17 m � 2.150 m)

e. 690 000 m � (5.022 h � 4.31 h)

f. (6.23 cm � 3.111 cm � 0.05 cm) � 14.99 cm

Holt ChemFile: Problem-Solving Workbook 27 Significant Figures

Scientific Notation

People who work in scientific fields often have to use very large and very small

numbers. Look at some examples in the following table:

Measurement Value

Density of air at 27°C and 1 atm pressure 0.001 61 g/cm 3

Radius of a calcium atom 0.000 000 000 197 m

One light-year 9 460 000 000 000 km

The mass of a neutron 0.000 000 000 000 000 000 000 001 675 g

You can see that measurements such as these would be awkward to write out

repeatedly. Also, calculating with very long numbers is likely to lead to errors

because it’s so easy to miscount zeros and decimal places. To make these numbers

easier to handle, scientists express them in a form known as scientific notation,

which uses powers of 10 to reduce the number of zeros to a minimum.

Look at a simple example of the way that scientific notation works. Following

are some powers of 10 and their decimal equivalents.

10 �2 � 000.01

10 �1 � 000.1

10 0� � 001

10 1� � 010

10 2� � 100

Suppose we rewrite the values in the table using scientific notation. The numbers

become much less cumbersome.

Density of air at 27°C and 1 atm pressure 1.61 � 10 �3 g/cm 3

Radius of a calcium atom 1.97 � 10 �10 m

One light-year 9.46 � 10 12 km

Mass of a neutron 1.675 � 10 �24 g

Holt ChemFile: Problem-Solving Workbook 28 Scientific Notation

CONVERTING QUANTITIES TO SCIENTIFIC NOTATION

General Plan for Converting Quantities to Scientific Notation

Use the resulting

number as the

coefficient M.

Quantity expressed in long form

Move the

decimal point

right or left until

there is only one

nonzero digit

to the left

of it.

Count the number of

places the decimal

point moved, and

call that number n.

Make the number n

negative if the

decimal moved to

the right.

M, the coefficient of 10 n n, the exponent of 10

Quantity expressed in the form

M � 10 n

(scientific notation)

Holt ChemFile: Problem-Solving Workbook 29 Scientific Notation

Express the following measurements in scientific notation.

a. 310 000 L

b. 0.000 49 kg

What is given in the problem? two measured quantities

What are you asked to find? the measured quantities expressed in scientific

notation

What steps are needed to rewrite the quantities in scientific notation?

Move the decimal point in each value until there is only one nonzero digit to the

left of it. This number becomes the coefficient, M.

Count the number of places the decimal was moved. If it moved to the left, the

count is a positive number. If it moved to the right, the count is a negative

number. Make this number, n, the exponent of 10.

Quantity written in long form � M � 10 n

a. Express 310 000 L in scientific notation

M � 3.1

310 000 L � 3.1 � 10 5 L

b. Express 0.000 49 kg in scientific notation.

M � 4.9

0.000 49 kg � 4.9 � 10 �4 kg

3 1 0 0 0 0

0. 0 0 0 4 9

decimal point moves

4 places to the right

n � �4

Holt ChemFile: Problem-Solving Workbook 30 Scientific Notation

a b

Measured quantity 310 000 L 0.000 49 kg

Quantity expressed in scientific notation ? L ? kg

5 places to the left

Are units correct?

Is the quantity correctly

expressed?

1. Express the following quantities in scientific notation:

a. 8 800 000 000 m ans: 8.8 � 10 9 m

b. 0.0015 kg ans: 1.5 � 10 �3 kg

c. 0.000 000 000 06 kg/m 3 ans: 6 � 10 �11 kg/m 3

d. 8 002 000 Hz ans: 8.002 � 10 6 Hz

e. 0.009 003 amp ans: 9.003 � 10 �3 amp

f. 70 000 000 000 000 000 km ans: 7 � 10 16 km

g. 6028 L ans: 6.028 � 10 3 L

h. 0.2105 g ans: 2.105 � 10 �1 g

i. 600 005 000 kJ/h ans: 6.000 05 � 10 8 kJ/h

j. 33.8 m 2 ans: 3.38 � 10 1 m 2

Yes; the original measurement

was in liters.

Yes; the decimal was

moved to the left five

places to give a coefficient

of 3.1 and an exponent

of �5.

was in kilograms.

moved to the right four

of 4.9 and an exponent

of �4.

Holt ChemFile: Problem-Solving Workbook 31 Scientific Notation

CALCULATING WITH QUANTITIES IN SCIENTIFIC NOTATION

What is the total of the measurements 3.61 � 10 4 mm, 5.88 � 10 3 mm, and

8.1 � 10 2 mm?

What is given in the problem? three measured quantities expressed in scientific

What are you asked to find? the sum of those quantities

Measured quantity 3.61 � 10 4 mm 5.88 � 10 3 mm 8.1 � 10 2 mm

What steps are needed to add the quantities?

Convert each quantity so that each exponent is the same as that on the quantity

with the largest exponent. The quantities can then be added together. Make sure

the result has the correct number of significant figures.

(P � 10 Q ) � (R � 10 S ) � (T � 10 V ) � ? (P � R� � T�) � 10 Q � ?

if the exponents are different,

convert the quantities so that

they have the same exponent

as the term with the largest

exponent

3.61 � 10 4 mm � 5.88 � 10 3 mm � 8.1 � 10 2 mm � ? mm

Convert the second and third quantities to multiples of 10 4 .

To convert 5.88 � 10 3 mm:

5.88 � 10 3 mm � M � 10 4 mm

Because one was added to the exponent, the decimal point must be moved

one place to the left.

M � 0.588

To convert 8.1 � 10 2 mm:

8.1 � 10 2 mm � M � 10 4 mm

Because two was added to the exponent, the decimal point must be moved

two places to the left.

add the quantities P,

R�, and T�, and

multiply them by the

factor 10 Q

(P � 10 Q ) � (R� � 10 Q ) � (T� � 10 Q ) � ?

Holt ChemFile: Problem-Solving Workbook 32 Scientific Notation

M � 0.081

Now the three quantities can be added, as follows:

� 3.610 � 10 4 mm

� 0.588 � 10 4 mm

� 0.081 � 10 4 mm

� 4.275 � 10 4 mm

To express the result in the correct number of significant figures, note that the

result should only contain two decimal places.

Yes; units of all quantities were millimeters.

Is the quantity correctly expressed in scientific notation?

Yes; there is only one number to the left of the decimal point.

Is the quantity expressed in the correct number of significant figures?

Yes; the result was rounded to give two decimal places to match the least accurate

measurement.

4.275 � 10 4 mm � 4.28 � 10 4 mm

this digit is 5, so

round up

1. Carry out the following calculations. Express the results in scientific notation

and with the correct number of significant figures.

a. 4.74 � 10 4 km � 7.71 � 10 3 km � 1.05 � 10 3 km ans: 5.62 � 10 4 km

b. 2.75 � 10 �4 m � 8.03 � 10 �5 m � 2.122 � 10 �3 m ans: 2.477 � 10 �3 m

Holt ChemFile: Problem-Solving Workbook 33 Scientific Notation

c. 4.0 � 10 �5 m 3 � 6.85 � 10 �6 m 3 � 1.05 � 10 �5 m 3 ans: 3.6 � 10 �5 m 3

d. 3.15 � 10 2 mg � 3.15 � 10 3 mg � 3.15 � 10 4 mg ans: 3.50 � 10 4 mg

e. 3.01 � 10 22 atoms � 1.19 � 10 23 atoms � 9.80 � 10 21 atoms

ans: 1.59 � 10 23 atoms

f. 6.85 � 10 7 nm � 4.0229 � 10 8 nm � 8.38 � 10 6 nm ans: 4.624 � 10 8 nm

Holt ChemFile: Problem-Solving Workbook 34 Scientific Notation

Perform the following calculation, and express the result in scientific

notation:

3.03 � 10 4 cm 2 � 6.29 � 10 2 cm

What is given in the problem? two quantities expressed in scientific notation

What are you asked to find? the product of the two quantities

Measured quantity 3.03 � 10 4 cm 2

6.29 � 10 2 cm

What steps are needed to multiply quantities expressed in scientific notation?

Multiply the coefficients, and add the exponents. Then transform to the correct

scientific notation form with the correct units and number of significant figures.

(P � 10 Q ) � (R � 10 S ) Result correctly written

in scientific notation,

with the correct units

and number of

multiply the

coefficients and add

the exponents if there is more or less than

one nonzero digit to the left

of the decimal, move the

decimal and change the

exponent to account for the

move, then round to the

correct number of

(P � 10 Q ) unit 1 � (R � 10 S ) unit 2

coefficients

exponents

(P � R) � 10 (Q � S)

product of

3.03 � 10 4 cm 2 � 6.29 � 10 2 cm �

(3.03 � 6.29) � 10 (4�2) (cm 2 � cm) � 19.0587 � 10 6 cm 3

To transform the result to the correct form for scientific notation, move the

decimal point left one place and increase the exponent by one.

sum of

(Q � S)

product of units

� (P � R) � 10 (unit 1 � unit 2)

Holt ChemFile: Problem-Solving Workbook 35 Scientific Notation

19.0587 � 10 6 cm 3 � 1.90587 � 10 (6�1) cm 3 � 1.90587 � 10 7 cm 3

To express the result to the correct number of significant figures, note that

both of the original quantities have three significant figures. Therefore, round off

the result to three significant figures.

Yes; the units cm 2 and cm are multiplied to give cm 3 .

Is the quantity expressed to the correct number of significant figures?

Yes; the number of significant figures is correct because the data were given to

Is the quantity expressed correctly in scientific notation?

Yes; moving the decimal point decreases the coefficient by a factor of 10, so the

exponent increases by one to compensate.

1.905 87 � 10 7 cm 3 � 1.91 � 10 7 cm 3

this digit is 5,

so round up

1. Carry out the following computations, and express the result in scientific

a. 7.20 � 10 3 cm � 8.08 � 10 3 cm ans: 5.82 � 10 7 cm 2

b. 3.7 � 10 4 mm � 6.6 � 10 4 mm � 9.89 � 10 3 mm ans: 2.4 � 10 13 mm 3

c. 8.27 � 10 2 m � 2.5 � 10 �3 m � 3.00 � 10 �4 m ans: 6.2 � 10 �4 m 3

d. 4.44 � 10 �35 m � 5.55 � 10 19 m � 7.69 � 10 �12 kg ans: 1.89 � 10 �26 kg�m 2

e. 6.55 � 10 4 dm � 7.89 � 10 9 dm � 4.01893 � 10 5 dm ans: 2.08 � 10 20 dm 3

Holt ChemFile: Problem-Solving Workbook 36 Scientific Notation

Sample Problem 4

3.803 � 10 3 g � 5.3 � 10 6 mL

What are you asked to find? the quotient of the two quantities

Measured quantity 3.803 � 10 3 g 5.3 � 10 6 mL

What steps are needed to divide the quantities expressed in scientific notation?

Divide the coefficients, and subtract the exponents. Then transform the result

to the correct form for scientific notation with the correct units and number of

(P � 10 Q ) � (R � 10 S ) Result written correctly

divide the coefficients

and subtract the exponents

in the order shown

P � 10 Q unit 1

R � 10 S unit 2

3.803 � 10 3 g � 5.3 � 10 6 mL � � 3. 803

� � 10

5.

(3�6) g

�� 0.717 547 � 10

mL

�3 g/mL

The measurement 5.3 � 10 6 mL has the fewest significant figures; round the

result accordingly.

0.717 547 � 10 �

�3 g/mL 0.72 � 10�3 g/mL

this digit is greater

than 5, so round up

quotient of

P

R

� 10

difference

of exponents

if there is more or less than

quotient

of units

unit 1

unit 2

Holt ChemFile: Problem-Solving Workbook 37 Scientific Notation

decimal point to the right one place and decrease the exponent by one.

0.72 � 10 �3 g/mL � 7.2 � 10 (�3�1) g/mL � 7.2 � 10 �4 g/mL

Yes; grams divided by milliliters gives g/mL, a unit of density.

Yes; the result was limited to two significant figures by the data given.

Yes; there is only one nonzero digit to the left of the decimal point.

a. 2.290 � 10 7 cm � 4.33 � 10 3 s ans: 5.29 � 10 3 cm/s

b. 1.788 � 10 �5 L � 7.111 � 10 �3 m 2 ans: 2.514 � 10 �3 L/m 2

c. 5.515 � 10 4 L � 6.04 � 10 3 km ans: 9.13 L/km

d. 3.29 � 10 �4 km � 1.48 � 10 �2 min ans: 2.22 � 10 �2 km/min

e. 4.73 � 10 �4 g � (2.08 � 10 �3 km � 5.60 � 10 �4 km) ans: 4.06 � 10 2 g/km 2

Holt ChemFile: Problem-Solving Workbook 38 Scientific Notation

a. 158 000 km

b. 0.000 009 782 L

c. 837 100 000 cm 3

d. 6 500 000 000 mm 2

e. 0.005 93 g

2. Perform the following calculations, and express the result in scientific notation

with the correct number of significant figures:

a. 2.48 � 10 2 kg � 9.17 � 10 3 kg � 7.2 � 10 1 kg

b. 4.07 � 10 �5 mg � 3.966 � 10 �4 mg � 7.1 � 10 �2 mg

c. 1.39 � 10 4 m 3 � 6.52 � 10 2 m 3 � 4.8 � 10 3 m 3

d. 7.70 � 10 �9 m � 3.95 � 10 �8 m � 1.88 � 10 �7 m

e. 1.111 � 10 5 J � 5.82 � 10 4 J � 3.01 � 10 6 J

f. 9.81 � 10 27 molecules � 3.18 � 10 25 molecules � 2.09 � 10 26 molecules

g. 1.36 � 10 7 cm � 3.456 � 10 6 cm � 1.01 � 10 7 cm � 5.122 � 10 5 cm

3. Perform the following computations, and express the result in scientific

notation with the correct number of significant figures:

a. 1.54 � 10 �1 L � 2.36 � 10 �4 s

b. 3.890 � 10 4 mm � 4.71 � 10 2 mm 2

c. 9.571 � 10 3 kg � 3.82 � 10 �1 m 2

d. 8.33 � 10 3 km � 1.97 � 10 2 s

e. 9.36 � 10 2 m � 3.82 � 10 3 m � 9.01 � 10 �1 m

f. 6.377 � 10 4 J � 7.35 � 10 �3 s

4. Your electric company charges you for the electric energy you use, measured

in kilowatt-hours (kWh). One kWh is equivalent to 3 600 000 J. Express this

quantity in scientific notation.

5. The pressure in the deepest part of the ocean is 11 200 000 Pa. Express this

pressure in scientific notation.

6. Convert 1.5 km to millimeters, and express the result in scientific notation.

7. Light travels at a speed of about 300 000 km/s.

a. Express this value in scientific notation.

b. Convert this value to meters per hour.

c. What distance in centimeters does light travel in 1 �s?

f. 0.000 000 006 13 m

g. 12 552 000 J

h. 0.000 008 004 g/L

i. 0.010 995 kg

j. 1 050 000 000 Hz

Holt ChemFile: Problem-Solving Workbook 39 Scientific Notation

8. There are 7.11 � 10 24 molecules in 100.0 cm 3 of a certain substance.

a. What is the number of molecules in 1.09 cm 3 of the substance?

b. What would be the number of molecules in 2.24 � 10 4 cm 3 of the substance?

c. What number of molecules are in 9.01 � 10 �6 cm 3 of the substance?

9. The number of transistors on a particular integrated circuit is 3 578 000, and

the integrated circuit measures 9.5 mm � 8.2 mm.

a. What is the area occupied by each transistor?

b. Using your answer from (a), how many transistors could be formed on a

silicon sheet that measures 353 mm � 265 mm?

10. A solution has 0.0501 g of a substance in 1.00 L. Express this concentration in

grams per microliter.

11. Cesium atoms are the largest of the naturally occurring elements. They have a

diameter of 5.30 � 10 �10 m. Calculate the number of cesium atoms that

would have to be lined up to give a row of cesium atoms 2.54 cm (1 in.) long.

12. The neutron has a volume of approximately 1.4 � 10 �44 m 3 and a mass of

1.675 � 10 �24 g. Calculate the density of the neutron in g/m 3 . What is the

mass of 1.0 cm 3 of neutrons in kilograms?

13. The pits in a compact disc are some of the smallest things ever mass-produced

mechanically by humans. These pits represent the 1s and 0s of digital

information on a compact disc. These pits are only 1.6 � 10 �8 m deep (1/4 the

wavelength of red laser light). How many of these pits would have to be

stacked on top of each other to make a hole 0.305 m deep?

14. 22 400 mL of oxygen gas contains 6.022 � 10 23 oxygen molecules at 0°C and

standard atmospheric pressure.

a. How many oxygen molecules are in 0.100 mL of gas?

b. How many oxygen molecules are in 1.00 L of gas?

c. What is the average space in milliters occupied by one oxygen molecule?

15. The mass of the atmosphere is calculated to be 5.136 � 10 18 kg, and there are

6 500 000 000 people living on Earth. Calculate the following values.

a. The mass of atmosphere in kilograms per person.

b. The mass of atmosphere in metric tons per person.

c. If the number of people increases to 9 500 000 000, what is the mass in

kilograms per person?

16. The mass of the sun is 1.989 � 10 30 kg, and the mass of Earth is 5.974 � 10 24

kilograms. How many Earths would be needed to equal the mass of the sun?

Holt ChemFile: Problem-Solving Workbook 40 Scientific Notation

17. A new landfill has dimensions of 2.3 km � 1.4 km � 0.15 km.

a. What is the volume in cubic kilometer?

b. What is the volume in cubic meters?

c. If 250 000 000 objects averaging 0.060 m 3 each are placed into the landfill

each year, how many years will it take to fill the landfill?

18. A dietary calorie (C) is exactly equal to 1000 cal. If your daily intake of food

gives you 2400 C, what is your intake in joules per day? (1 cal � 4.184 J)

Holt ChemFile: Problem-Solving Workbook 41 Scientific Notation

Four Steps for Solving Quantitative Problems

Maybe you have noticed that the sample problems in the first three chapters of

this book are solved in a four-step process. The steps in the process are as

follows.

1. ANALYZE

2. PLAN

3. COMPUTE

4. EVALUATE

Now it’s time to examine each step to learn how this process can help you

solve more-complex problems, like those you will encounter in your chemistry

course.

In this first step, you should read the problem carefully and then reread it. You

must determine what specific information and data you are given in the problem

and what you need to find.

Try to visualize the situation the problem describes. Look closely at the words

in the problem statement for clues to understanding the problem. Are you working

with elements, compounds, or mixtures? Are they solids, liquids, or gases?

What change is taking place? Is it a chemical change? What are the reactants?

What are the products?

It is always a good idea to collect and organize all of your information in a

table, where you can see it at a glance. Include the things you want to find in the

table, too. Be sure to include units with both the data and the quantities you must

find. Scanning the quantities and their units will often provide clues about how to

set up the problem. Remember, you may be given information not needed to

solve the problem. You must analyze the data to determine what is useful and

what is not.

In the planning step, you develop a method to solve the problem. Always keep in

mind what you want to find and its units. Chances are good that an approach that

gives an answer with the correct units is the correct one to use. In any case, a

setup that gives an answer with the wrong units is certain to be wrong. You may

find it helpful to diagram your solution method. This process helps you organize

your thoughts.

As you work out your problem-solving method, write down a trial calculation

without numbers but with units. When you complete your setup, see if the trial

calculation will give you a quantity with the correct units. As stated above, if the

setup gives the needed units, it is probably correct.

Holt ChemFile: Problem-Solving Workbook 42 Four Steps for Solving Quantitative Problems

During this planning process you may discover that you need more information,

such as atomic masses from the periodic table, the boiling point of alcohol,

or the density of tin. You will need to look up such information in the appropriate

tables.

In this step, you follow your plan, set up a calculation using the data you have

assembled, and compute the result.

It is a good strategy to write out and check your calculation setup before you

start working with your calculator. First reconfirm that the calculation will give a

result with the correct units. Go through your calculation and lightly strike

through the units that cancel. Be sure the remaining units are those that you want

in your answer. Whenever possible, use your calculator in a way that lets you

complete the entire problem without writing down numbers and then re-entering

them.

Everyone makes errors, but good problem solvers always develop strategies to

check their work. Confidence in your problem-solving ability will come from

knowing how to determine on your own whether your answers are correct.

One evaluation strategy is to estimate the numerical value of the answer. In

simple problems, you can probably do this in your head. With calculations having

several terms, round off each numerical value to the nearest simple value, and

then write and compute the estimation.

Suppose you had to make the following calculation.

� ?

The numerical calculation can be estimated as follows.

30 30

��

� ��

6 � 20 120

� � 0.25

Once you have done the actual computation, compare your result with the

estimate. In this case, the calculation gives 0.242 g/cm 2 , which is close to the estimated

result. Therefore, it is likely that you made no mistakes in the calculation.

Next, check that your answer is expressed to the correct number of significant

figures. Look at the data values you used in the calculation. Usually, significant

figures will be limited by the measurement that has the fewest significant figures.

Finally, ask yourself the simple question, does this answer make sense based

on what you know? If you are calculating the circumference of Earth, an answer

of 50 km is obviously much too small. If you are calculating the density of air, a

value of 340 g/cm 3 is much too large because air density is usually less than 1

g/cm 3 28.8 g

��

6.30 cm � 18.9 cm

. You will detect many errors by asking if the answer makes sense.

Holt ChemFile: Problem-Solving Workbook 43 Four Steps for Solving Quantitative Problems

A 10.0% sodium hydroxide solution has a density of 1.11 g/mL. What

volume in liters will 2280 g of the solution have?

What is given in the problem? the density of the sodium hydroxide solution in

g/mL, and the mass of the solution whose volume

is to be determined

What are you asked to find? the volume of the specified mass of solution

Next, bring together in a table everything you might need in the problem. Include

what you want to find in the table. The fact that the solution is 10.0% sodium

hydroxide is unimportant in the solution of this problem. This piece of data need

not appear in your table. Notice that the problem asks you for a volume in liters

and that density is given in grams per milliliter. You will need a factor to convert

between these units. Therefore, include in the table any relationships between

units that will be helpful.

What steps are needed to calculate the volume of 2280 g of the solution?

Apply the relationship D � � m

�. Rearrange to solve for V, substitute data, and

convert to liters.

Mass of solution

Solve the density equation for V.

Density of solution 1.11 g/mL

Mass of solution 2280 g

Volume of solution ? L

Relationship between mL and L 1000 mL � 1 L

rearrange

D �

to solve for V

Volume of solution

in mL

V � � m

mass of solution in g

density of solution in g/ml

in L

convert using

the factor

1 L

1000 mL

� volume in mL

Holt ChemFile: Problem-Solving Workbook 44 Four Steps for Solving Quantitative Problems

To change the result to liters, multiply by the conversion factor.

mass of solution in g 1 L

�� volume in L

density of solution g/ml

Yes; units canceled to give liters.

Is the number of significant figures correct?

Yes; the number of significant figures is correct because data were given to

Yes; the calculation can be approximated as 2000/1000 � 2. Also, considering

that 2000 g of water occupies 2 L, you would expect 2 L of a slightly more dense

material to have a mass slightly greater than 2000 g.

2280 g

1.11 g/mL

1. Gasoline has a density of 0.73 g/cm 3 . How many liters of gasoline would be

required to increase the mass of an automobile from 1271 kg to 1305 kg?

ans: 47 L

2. A swimming pool measures 9.0 m long by 3.5 m wide by 1.75 m deep. What

mass of water in metric tons (1 metric ton � 1000 kg) does the pool contain

when filled? The density of the water in the pool is 0.997 g/cm 3 .

ans: 55 metric tons

3. A tightly packed box of crackers contains 250 g of crackers and measures 7.0

cm � 17.0 cm � 19.0 cm. What is the average density in kilograms per liter of

the crackers in the package? Assume that the unused volume is negligible.

ans: 0.11 kg/L

� 2.05 L

Holt ChemFile: Problem-Solving Workbook 45 Four Steps for Solving Quantitative Problems

Solve these problems by using the Four Steps for Solving Quantitative Problems.

1. The aluminum foil on a certain roll has a total area of 18.5 m 2 and a mass of

1275 g. Using a density of 2.7 g per cubic centimeter for aluminum, determine

the thickness in millimeters of the aluminum foil.

2. If a liquid has a density of 1.17 g/cm 3 , how many liters of the liquid have a

mass of 3.75 kg?

3. A stack of 500 sheets of paper measuring 28 cm � 21 cm is 44.5 mm high and

has a mass of 2090 g. What is the density of the paper in grams per cubic

centimeter?

4. A triangular-shaped piece of a metal has a mass of 6.58 g. The triangle is 0.560

mm thick and measures 36.4 mm on the base and 30.1 mm in height. What is

the density of the metal in grams per cubic centimeter?

5. A packing crate measures 0.40 m � 0.40 m � 0.25 m. You must fill the crate

with boxes of cookies that each measure 22.0 cm � 12.0 cm � 5.0 cm. How

many boxes of cookies can fit into the crate?

6. Calculate the unknown quantities in the following table. Use the following

relationships for volumes of the various shapes.

Volume of a cube � l � l � l

Volume of a rectangle � l � w � h

Volume of a sphere � 4/3�r 3

Volume of a cylinder ��r 2 � h

D m V Shape Dimensions

a. 2.27 g/cm 3

b. 1.85 g/cm 3

7. When a sample of a metal alloy that has a mass of 9.65 g is placed into a

graduated cylinder containing water, the volume reading in the cylinder

increases from 16.0 mL to 19.5 mL. What is the density of the alloy sample in

grams per cubic centimeter?

8. Pure gold can be made into extremely thin sheets called gold leaf. Suppose

that 50.0 kg of gold is made into gold leaf having an area of 3620 m 2 . The

density of gold is 19.3 g/cm 3 .

a. How thick in micrometers is the gold leaf?

b. A gold atom has a radius of 1.44 � 10 �10 m. How many atoms thick is the

gold leaf?

3.93 kg ? L cube ? m � ? m � ? m

? g ? cm 3

c. 3.21 g/L ? kg ? dm 3

d. ? g/cm 3

e. 0.92 g/cm 3

497 g ? m 3

? kg ? cm 3

rectangle 33 mm � 21 mm � 7.2 mm

sphere 3.30 m diameter

cylinder 7.5 cm diameter � 12 cm

rectangle 3.5 m � 1.2 m � 0.65 m

Holt ChemFile: Problem-Solving Workbook 46 Four Steps for Solving Quantitative Problems

9. A chemical plant process requires that a cylindrical reaction tank be filled

with a certain liquid in 238 s. The tank is 1.2 m in diameter and 4.6 m high.

What flow rate in liters per minute is required to fill the reaction tank in the

specified time?

10. The radioactive decay of 2.8 g of plutonium-238 generates 1.0 joule of energy

as heat every second. Plutonium has a density of 19.86 g/cm 3 . How many

calories (1 cal � 4.184 J) of energy as heat will a rectangular piece of plutonium

that is 4.5 cm � 3.05 cm � 15 cm generate per hour?

11. The mass of Earth is 5.974 � 10 24 kg. Assume that Earth is a sphere of diameter

1.28 � 10 4 km and calculate the average density of Earth in grams per

cubic centimeter.

12. What volume of magnesium in cubic centimeters would have the same mass

as 1.82 dm 3 of platinum? The density of magnesium is 1.74 g/cm 3 , and the

density of platinum is 21.45 g/cm 3 .

13. A roll of transparent tape has 66 m of tape on it. If an average of 5.0 cm of

tape is needed each time the tape is used, how many uses can you get from a

case of tape containing 24 rolls?

14. An automobile can travel 38 km on 4.0 L of gasoline. If the automobile is

driven 75% of the days in a year and the average distance traveled each day is

86 km, how many liters of gasoline will be consumed in one year (assume the

year has 365 days)?

15. A hose delivers water to a swimming pool that measures 9.0 m long by 3.5 m

wide by 1.75 m deep. It requires 97 h to fill the pool. At what rate in liters per

minute will the hose fill the pool?

16. Automobile batteries are filled with a solution of sulfuric acid, which has a

density of 1.285 g/cm 3 . The solution used to fill the battery is 38% (by mass)

sulfuric acid. How many grams of sulfuric acid are present in 500 mL of

battery acid?

Holt ChemFile: Problem-Solving Workbook 47 Four Steps for Solving Quantitative Problems

Mole Concept

Suppose you want to carry out a reaction that requires combining one atom of

iron with one atom of sulfur. How much iron should you use? How much sulfur?

When you look around the lab, there is no device that can count numbers of

atoms. Besides, the merest speck (0.001 g) of iron contains over a billion billion

atoms. The same is true of sulfur.

Fortunately, you do have a way to relate mass and numbers of atoms. One iron

atom has a mass of 55.847 amu, and 55.847 g of iron contains 6.022 137 � 10 23

atoms of iron. Likewise, 32.066 g of sulfur contains 6.022 137 � 10 23 atoms of sulfur.

Knowing this, you can measure out 55.847 g of iron and 32.066 g of sulfur and

be pretty certain that you have the same number of atoms of each.

The number 6.022 137 � 10 23 is called Avogadro’s number. For most purposes

it is rounded off to 6.022 � 10 23 . Because this is an awkward number to write

over and over again, chemists refer to it as a mole (abbreviated mol). 6.022 �

10 23 objects is called a mole, just as you call 12 objects a dozen.

Look again at how these quantities are related.

55.847 g of iron � 6.022 � 10 23 iron atoms � 1 mol of iron

32.066 g of sulfur � 6.022 � 10 23 sulfur atoms � 1 mol of sulfur

Mass of

substance

Amount of

in moles

General Plan for Converting Mass, Amount,

and Numbers of Particles

Convert using

the molar mass of

the substance.

2 3

Use Avogadro's

number for conversion.

Number of atoms,

molecules, or formula

units of substance

Holt ChemFile: Problem-Solving Workbook 48 Mole Concept

PROBLEMS INVOLVING ATOMS AND ELEMENTS

A chemist has a jar containing 388.2 g of iron filings. How many moles of

iron does the jar contain?

What is given in the problem? mass of iron in grams

What are you asked to find? amount of iron in moles

What step is needed to convert from grams of Fe to number of moles of Fe?

The molar mass of iron can be used to convert mass of iron to amount of iron in

moles.

Mass of iron 388.2 g

Molar mass of iron* 55.85 g/mol

Amount of iron ? mol

* determined from the periodic table

Mass of Fe in g

1 mol Fe

388.2 g Fe � � 6.951 mol Fe

55.85 g Fe

Yes; the answer has the correct units of moles of Fe.

multiply by the inverse

molar mass of Fe

Amount of Fe in mol

molar mass Fe

given 1 mol Fe � mol Fe

g Fe � 55.85 g Fe

Holt ChemFile: Problem-Solving Workbook 49 Mole Concept

Yes; the number of significant figures is correct because there are four significant

figures in the given value of 388.2 g Fe.

Yes; 388.2 g Fe is about seven times the molar mass. Therefore, the sample

contains about 7 mol.

1. Calculate the number of moles in each of the following masses:

a. 64.1 g of aluminum ans: 2.38 mol Al

b. 28.1 g of silicon ans: 1.00 mol Si

c. 0.255 g of sulfur ans: 7.95 � 10 �3 mol S

d. 850.5 g of zinc ans: 13.01 mol Zn

Holt ChemFile: Problem-Solving Workbook 50 Mole Concept

A student needs 0.366 mol of zinc for a reaction. What mass of zinc in

grams should the student obtain?

What is given in the problem? amount of zinc needed in moles

What are you asked to find? mass of zinc in grams

What step is needed to convert from moles of Zn to grams of Zn?

The molar mass of zinc can be used to convert amount of zinc to mass of zinc.

Amount of zinc 0.366 mol

Molar mass of zinc 65.39 g/mol

Mass of zinc ? g

Amount of Zn in mol

0.366 mol Zn �

multiply by the

molar mass of Zn

molar mass Zn

given 65.39 g Zn � g Zn

mol Zn � 1 mol Zn

65.39 g Zn

1 mol Zn

Yes; the answer has the correct units of grams of Zn.

Yes; the number of significant figures is correct because there are three significant

figures in the given value of 0.366 mol Zn.

Yes; 0.366 mol is about 1/3 mol. 23.9 g is about 1/3 the molar mass of Zn.

Mass of Zn in mol

� 23.9 g Zn

Holt ChemFile: Problem-Solving Workbook 51 Mole Concept

1. Calculate the mass of each of the following amounts:

a. 1.22 mol sodium ans: 28.0 g Na

b. 14.5 mol copper ans: 921 g Cu

c. 0.275 mol mercury ans: 55.2 g Hg

d. 9.37 � 10 �3 mol magnesium ans: 0.228 Mg

Holt ChemFile: Problem-Solving Workbook 52 Mole Concept

How many moles of lithium are there in 1.204 � 10 24 lithium atoms?

What is given in the problem? number of lithium atoms

What are you asked to find? amount of lithium in moles

Number of lithium atoms 1.204 � 10 24 atoms

Avogadro’s number—the 6.022 � 10 23 number of atoms per mole

atoms/mol

Amount of lithium ? mol

What step is needed to convert from number of atoms of Li to moles of Li?

Avogadro’s number is the number of atoms per mole of lithium and can be used

to calculate the number of moles from the number of atoms.

Number of Li atoms

multiply by the inverse of

Avogadro's number

1 mol Li

6.022 � 10 23 � 1.999 mol Li

atoms Li 1.204 � 1024 atoms Li �

Yes; the answer has the correct units of moles of Li.

Yes; four significant figures is correct.

Yes; 1.204 � 10 24 is approximately twice Avogadro’s number. Therefore, it is

reasonable that this number of atoms would equal about 2 mol.

Amount of Li in mol

given 1 mol Li

atoms Li � � mol Li

6.022 � 1023 atoms Li

Holt ChemFile: Problem-Solving Workbook 53 Mole Concept

1. Calculate the amount in moles in each of the following quantities:

a. 3.01 � 10 23 atoms of rubidium ans: 0.500 mol Rb

b. 8.08 � 10 22 atoms of krypton ans: 0.134 mol Kr

c. 5 700 000 000 atoms of lead ans: 9.5 � 10 �15 mol Pb

d. 2.997 � 10 25 atoms of vanadium ans: 49.77 mol V

Holt ChemFile: Problem-Solving Workbook 54 Mole Concept

CONVERTING THE AMOUNT OF AN ELEMENT IN MOLES

TO THE NUMBER OF ATOMS

In Sample Problem 3, you were asked to determine the number of moles in

1.204 � 10 24 atoms of lithium. Had you been given the amount in moles and

asked to calculate the number of atoms, you would have simply multiplied by

Avogadro’s number. Steps 2 and 3 of the plan for solving Sample Problem 3 would

have been reversed.

1. Calculate the number of atoms in each of the following amounts:

a. 1.004 mol bismuth ans: 6.046 � 10 23 atoms Bi

b. 2.5 mol manganese ans: 1.5 � 10 24 atoms Mg

c. 0.000 000 2 mol helium ans: 1 � 10 17 atoms He

d. 32.6 mol strontium ans: 1.96 � 10 25 atoms Sr

Holt ChemFile: Problem-Solving Workbook 55 Mole Concept

How many boron atoms are there in 2.00 g of boron?

What is given in the problem? mass of boron in grams

What are you asked to find? number of boron atoms

Mass of boron 2.00 g

Molar mass of boron 10.81 g/mol

Avogadro’s number—the number 6.022 � 10 23 of boron atoms per mole of boron

Number of boron atoms ? atoms

What steps are needed to convert from grams of B to number of atoms of B?

First, you must convert the mass of boron to moles of boron by using the molar

mass of boron. Then you can use Avogadro’s number to convert amount in moles

to number of atoms of boron.

Mass of B in g

inverse of the

molar mass of

boron

molar mass B

Amount of B in mol

1 mol B

10.81 g B � � 1.11 � 1023 6.022 � 10

2.00 g B �

atoms B

23 atoms B

Yes; the answer has the correct units of atoms of boron.

Yes; the mass of boron was given to three significant figures.

Number of B atoms

multiply by

Avogadro's

number

given 6.022 � 1023 1 mol B

g B � �

� atoms B

10.81 g B 1 mol B

Holt ChemFile: Problem-Solving Workbook 56 Mole Concept

Yes; 2 g of boron is about 1/5 of the molar mass of boron. Therefore, 2.00 g

boron will contain about 1/5 of an Avogadro’s constant of atoms.

1. Calculate the number of atoms in each of the following masses:

a. 54.0 g of aluminum ans: 1.21 � 10 24 atoms Al

b. 69.45 g of lanthanum ans: 3.011 � 10 23 atoms La

c. 0.697 g of gallium ans: 6.02 � 10 21 atoms Ga

d. 0.000 000 020 g beryllium ans: 1.3 � 10 15 atoms Be

Holt ChemFile: Problem-Solving Workbook 57 Mole Concept

CONVERTING NUMBER OF ATOMS OF AN ELEMENT TO MASS

Sample Problem 4 uses the progression of steps 1 → 2 → 3 to convert from the

mass of an element to the number of atoms. In order to calculate the mass from a

given number of atoms, these steps will be reversed. The number of moles in the

sample will be calculated. Then this value will be converted to the mass in grams.

1. Calculate the mass of the following numbers of atoms:

a. 6.022 � 10 24 atoms of tantalum ans: 1810. g Ta

b. 3.01 � 10 21 atoms of cobalt ans: 0.295 g Co

c. 1.506 � 10 24 atoms of argon ans: 99.91 g Ar

d. 1.20 � 10 25 atoms of helium ans: 79.7 g He

Holt ChemFile: Problem-Solving Workbook 58 Mole Concept

PROBLEMS INVOLVING MOLECULES, FORMULA UNITS,

AND IONS

How many water molecules are there in 200.0 g of water? What is the mass of

15.7 mol of nitrogen gas? Both of these substances consist of molecules, not single

atoms. Look back at the diagram of the General Plan for Converting Mass,

Amount, and Numbers of Particles. You can see that the same conversion methods

can be used with molecular compounds and elements, such as CO 2, H 2O,

H 2SO 4, and O 2.

For example, 1 mol of water contains 6.022 � 10 23 H 2O molecules. The mass

of a molecule of water is the sum of the masses of two hydrogen atoms and one

oxygen atom, and is equal to 18.02 amu. Therefore, 1 mol of water has a mass of

18.02 g. In the same way, you can relate amount, mass, and number of formula

units for ionic compounds, such as NaCl, CaBr 2, and Al 2(SO 4) 3.

Sample Problem 5

How many moles of carbon dioxide are in 66.0 g of dry ice, which is solid

CO 2?

What is given in the problem? mass of carbon dioxide

What are you asked to find? amount of carbon dioxide

Mass of CO2 66.0 g

Molar mass of CO2 44.0 g/mol

Amount of CO2 ? mol

What step is needed to convert from grams of CO2 to moles of CO2? The molar mass of CO2 can be used to convert mass of CO2 to moles of CO2. COMPUTE

Mass of CO2 in g

multiply by the inverse of the

molar mass of CO 2

molar mass CO 2

given 1 mol CO 2 � mol CO2

g CO 2 � 44.01 g CO2

1 mol CO2 66.0 g CO2 �

Amount of CO2 in mol

Holt ChemFile: Problem-Solving Workbook 59 Mole Concept

Yes; the answer has the correct units of moles CO2. Is the number of significant figures correct?

Yes; the number of significant figures is correct because the mass of CO2 was

given to three significant figures.

Yes; 66 g is about 3/2 the value of the molar mass of CO2. It is reasonable that

the sample contains 3/2 (1.5) mol.

a. 3.00 g of boron tribromide, BBr3 ans: 0.0120 mol BBr3 b. 0.472 g of sodium fluoride, NaF ans: 0.0112 mol NaF

c. 7.50 � 10 2 g of methanol, CH 3OH ans: 23.4 mol CH 3OH

d. 50.0 g of calcium chlorate, Ca(ClO 3) 2 ans: 0.242 mol Ca(ClO 3) 2

Holt ChemFile: Problem-Solving Workbook 60 Mole Concept

CONVERTING MOLES OF A COMPOUND TO MASS

Perhaps you have noticed that Sample Problems 1 and 5 are very much alike. In

each case, you multiplied the mass by the inverse of the molar mass to calculate

the number of moles. The only difference in the two problems is that iron is an

element and CO2 is a compound containing a carbon atom and two oxygen

atoms.

In Sample Problem 2, you determined the mass of 1.366 mol of zinc. Suppose

that you are now asked to determine the mass of 1.366 mol of the molecular compound

ammonia, NH3. You can follow the same plan as you did in Sample

Problem 2, but this time use the molar mass of ammonia.

1. Determine the mass of each of the following amounts:

a. 1.366 mol of NH3 ans: 23.28 g NH3 b. 0.120 mol of glucose, C 6H 12O 6 ans: 21.6 g C 6H 12O 6

c. 6.94 mol barium chloride, BaCl 2 ans: 1.45 � 10 3 g or 1.45 kg BaCl 2

d. 0.005 mol of propane, C 3H 8 ans: 0.2 g C 3H 8

Holt ChemFile: Problem-Solving Workbook 61 Mole Concept

Sample Problem 6

Determine the number of molecules in 0.0500 mol of hexane, C 6H 14.

What is given in the problem? amount of hexane in moles

What are you asked to find? number of molecules of hexane

Amount of hexane 0.0500 mol

Avogadro’s number—the number 6.022 � 10 23 of molecules per mole of hexane

molecules/mol

Molecules of hexane ? molecules

What step is needed to convert from moles of C 6H 14 to number of molecules of

C 6H 14?

Avogadro’s number is the number of molecules per mole of hexane and can be

used to calculate the number of molecules from number of moles.

Amount of C6H14 in mol

0.0500 mol C 6H 14 �

6.022 � 10 23 molecules C 6H 14

1 mol C 6H 14

Yes; the answer has the correct units of molecules of C6H14. Is the number of significant figures correct?

Yes; multiplying Avogadro’s number by 0.05 would yield a product that is a factor

of 10 less with a value of 3 � 10 22 .

Number of C6H14 molecules

mol C6H14 � � molecules C6H14 1 mol C6H14 6.022 � 1023 molecules C6H14 � 3.01 � 10 22 molecules C 6H 14

Holt ChemFile: Problem-Solving Workbook 62 Mole Concept

1. Calculate the number of molecules in each of the following amounts:

a. 4.99 mol of methane, CH 4 ans: 3.00 � 10 24 molecules CH 4

b. 0.005 20 mol of nitrogen gas, N 2 ans: 3.13 � 10 21 molecules N 2

c. 1.05 mol of phosphorus trichloride, PCl 3 ans: 6.32 � 10 23 molecules PCl 3

d. 3.5 � 10 �5 mol of vitamin C, ascorbic acid, C 6H 8O 6 ans: 2.1 � 10 19 molecules

C 6H 8O 6

Holt ChemFile: Problem-Solving Workbook 63 Mole Concept

USING FORMULA UNITS OF IONIC COMPOUNDS

Ionic compounds do not exist as molecules. A crystal of sodium chloride, for

example, consists of Na � ions and Cl � ions in a 1:1 ratio. Chemists refer to a

combination of one Na � ion and one Cl � ion as one formula unit of NaCl. A mole

of an ionic compound consists of 6.022 � 10 23 formula units. The mass of one

formula unit is called the formula mass. This mass is used in the same way

atomic mass or molecular mass is used in calculations.

1. Calculate the number of formula units in the following amounts:

a. 1.25 mol of potassium bromide, KBr ans: 7.53 � 10 23 formula units KBr

b. 5.00 mol of magnesium chloride, MgCl 2 ans: 3.01 � 10 24 formula units

MgCl 2

c. 0.025 mol of sodium carbonate, Na 2CO 3 ans: 1.5 � 10 22 formula units

Na 2CO 3

d. 6.82 � 10 �6 mol of lead(II) nitrate, Pb(NO 3) 2 ans: 4.11 � 10 18 formula

units Pb(NO 3) 2

Holt ChemFile: Problem-Solving Workbook 64 Mole Concept

CONVERTING NUMBER OF MOLECULES OR FORMULA UNITS

TO AMOUNT IN MOLES

In Sample Problem 3, you determined the amount in moles of the element

lithium. Suppose that you are asked to determine the amount in moles of copper(II)

hydroxide in 3.34 � 10 34 formula units of Cu(OH) 2. You can follow the

same plan as you did in Sample Problem 3.

1. Calculate the amount in moles of the following numbers of molecules or

formula units:

a. 3.34 � 10 34 formula units of Cu(OH) 2 ans: 5.55 � 10 10 mol Cu(OH) 2

b. 1.17 � 10 16 molecules of H 2S ans: 1.94 � 10 �8 mol H 2S

c. 5.47 � 10 21 formula units of nickel(II) sulfate, NiSO 4 ans: 9.08 � 10 �3 mol

NiSO 4

d. 7.66 � 10 19 molecules of hydrogen peroxide, H 2O 2 ans: 1.27 � 10 �4 mol

H 2O 2

Holt ChemFile: Problem-Solving Workbook 65 Mole Concept

Sample Problem 7

What is the mass of a sample consisting of 1.00 � 10 22 formula units of

MgSO 4?

What is given in the problem? number of magnesium sulfate formula units

What are you asked to find? mass of magnesium sulfate in grams

Number of formula units of magnesium sulfate 1.00 � 10 22 formula units

Avogadro’s number—the number of 6.022 � 10 23 formula units of magnesium sulfate per mole

formula units/mol

Molar mass of magnesium sulfate 120.37 g/mol

Mass of magnesium sulfate ? g

What steps are needed to convert from formula units of MgSO 4 to grams of

First, you must convert the number of formula units of MgSO 4 to amount of

MgSO 4 by using Avogadro’s number. Then you can use the molar mass of MgSO 4

to convert amount in moles to mass of MgSO 4.

Number of MgSO4 formula units

inverse of

6.022 � 1023 1

1 mol MgSO4 formula units MgSO4 �

Amount of MgSO4 in mol

Mass of MgSO4 in g

molar mass MgSO 4

the molar

mass of

MgSO 4

120.37 g MgSO4 � � g MgSO4 1 mol MgSO4

Holt ChemFile: Problem-Solving Workbook 66 Mole Concept

1 mol MgSO4 6.022 � 10 23 1.00 � 10

formula units MgSO4 22 formula units MgSO4 �

Yes; the answer has the correct units of grams of MgSO4. Is the number of significant figures correct?

Yes; 2 g of MgSO4 is about 1/60 of the molar mass of MgSO4. Therefore, 2.00 g

MgSO4 will contain about 1/60 of an Avogadro’s number of formula units.

1. Calculate the mass of each of the following quantities:

a. 2.41 � 10 24 molecules of hydrogen, H2 ans: 8.08 g H2 b. 5.00 � 10 21 formula units of aluminum hydroxide, Al(OH) 3 ans: 0.648 g

Al(OH) 3

c. 8.25 � 10 22 molecules of bromine pentafluoride, BrF 5 ans: 24.0 g BrF 5

d. 1.20 � 10 23 formula units of sodium oxalate, Na 2C 2O 4 ans: 26.7 g Na 2C 2O 4

� 120.37 g MgSO 4

1 mol MgSO 4

� 2.00 g MgSO 4

Holt ChemFile: Problem-Solving Workbook 67 Mole Concept

CONVERTING MOLECULES OR FORMULA UNITS OF A COMPOUND TO MASS

In Sample Problem 4, you converted a given mass of boron to the number of

boron atoms present in the sample. You can now apply the same method to convert

mass of an ionic or molecular compound to numbers of molecules or formula

units.

1. Calculate the number of molecules or formula units in each of the following

masses:

a. 22.9 g of sodium sulfide, Na2S ans: 1.77 � 10 23 formula units Na2S b. 0.272 g of nickel(II) nitrate, Ni(NO 3) 2 ans: 8.96 � 10 20 formula units

Ni(NO 3) 2

c. 260 mg of acrylonitrile, CH 2CHCN ans: 3.0 � 10 21 molecules CH 2CHCN

Holt ChemFile: Problem-Solving Workbook 68 Mole Concept

a. 0.039 g of palladium

b. 8200 g of iron

c. 0.0073 kg of tantalum

d. 0.006 55 g of antimony

e. 5.64 kg of barium

f. 3.37 � 10 �6 g of molybdenum

2. Calculate the mass in grams of each of the following amounts:

a. 1.002 mol of chromium

b. 550 mol of aluminum

c. 4.08 � 10 �8 mol of neon

d. 7 mol of titanium

e. 0.0086 mol of xenon

f. 3.29 � 10 4 mol of lithium

3. Calculate the number of atoms in each of the following amounts:

a. 17.0 mol of germanium

b. 0.6144 mol of copper

c. 3.02 mol of tin

d. 2.0 � 10 6 mol of carbon

e. 0.0019 mol of zirconium

f. 3.227 � 10 �10 mol of potassium

4. Calculate the number of moles in each of the following quantities:

a. 6.022 � 10 24 atoms of cobalt

b. 1.06 � 10 23 atoms of tungsten

c. 3.008 � 10 19 atoms of silver

d. 950 000 000 atoms of plutonium

e. 4.61 � 10 17 atoms of radon

f. 8 trillion atoms of cerium

5. Calculate the number of atoms in each of the following masses:

a. 0.0082 g of gold

b. 812 g of molybdenum

c. 2.00 � 10 2 mg of americium

d. 10.09 kg of neon

e. 0.705 mg of bismuth

f. 37 �g of uranium

Holt ChemFile: Problem-Solving Workbook 69 Mole Concept

6. Calculate the mass of each of the following:

a. 8.22 � 10 23 atoms of rubidium

b. 4.05 Avogadro’s numbers of manganese atoms

c. 9.96 � 10 26 atoms of tellurium

d. 0.000 025 Avogadro’s numbers of rhodium atoms

e. 88 300 000 000 000 atoms of radium

f. 2.94 � 10 17 atoms of hafnium

7. Calculate the number of moles in each of the following masses:

a. 45.0 g of acetic acid, CH 3COOH

b. 7.04 g of lead(II) nitrate, Pb(NO 3) 2

c. 5000 kg of iron(III) oxide, Fe 2O 3

d. 12.0 mg of ethylamine, C 2H 5NH 2

e. 0.003 22 g of stearic acid, C 17H 35COOH

f. 50.0 kg of ammonium sulfate, (NH 4) 2SO 4

8. Calculate the mass of each of the following amounts:

a. 3.00 mol of selenium oxybromide, SeOBr 2

b. 488 mol of calcium carbonate, CaCO 3

c. 0.0091 mol of retinoic acid, C 20H 28O 2

d. 6.00 � 10 �8 mol of nicotine, C 10H 14N 2

e. 2.50 mol of strontium nitrate, Sr(NO 3) 2

f. 3.50 � 10 �6 mol of uranium hexafluoride, UF 6

9. Calculate the number of molecules or formula units in each of the following

amounts:

a. 4.27 mol of tungsten(VI) oxide, WO3 b. 0.003 00 mol of strontium nitrate, Sr(NO3) 2

c. 72.5 mol of toluene, C6H5CH3 d. 5.11 � 10 �7 mol of �-tocopherol (vitamin E), C29H50O2 e. 1500 mol of hydrazine, N 2H 4

f. 0.989 mol of nitrobenzene C 6H 5NO 2

10. Calculate the number of molecules or formula units in each of the following

a. 285 g of iron(III) phosphate, FePO4 b. 0.0084 g of C5H5N c. 85 mg of 2-methyl-1-propanol, (CH3) 2CHCH2OH d. 4.6 � 10 �4 g of mercury(II) acetate, Hg(C2H3O2) 2

e. 0.0067 g of lithium carbonate, Li 2CO 3

Holt ChemFile: Problem-Solving Workbook 70 Mole Concept

11. Calculate the mass of each of the following quantities:

a. 8.39 � 10 23 molecules of fluorine, F 2

b. 6.82 � 10 24 formula units of beryllium sulfate, BeSO 4

c. 7.004 � 10 26 molecules of chloroform, CHCl 3

d. 31 billion formula units of chromium(III) formate, Cr(CHO 2) 3

e. 6.3 � 10 18 molecules of nitric acid, HNO 3

f. 8.37 � 10 25 molecules of freon 114, C 2Cl 2F 4

12. Precious metals are commonly measured in troy ounces. A troy ounce is

equivalent to 31.1 g. How many moles are in a troy ounce of gold? How many

moles are in a troy ounce of platinum? of silver?

13. A chemist needs 22.0 g of phenol, C6H5OH, for an experiment. How many

moles of phenol is this?

14. A student needs 0.015 mol of iodine crystals, I2, for an experiment. What mass

of iodine crystals should the student obtain?

15. The weight of a diamond is given in carats. One carat is equivalent to 200. mg.

A pure diamond is made up entirely of carbon atoms. How many carbon

atoms make up a 1.00 carat diamond?

16. 8.00 g of calcium chloride, CaCl2, is dissolved in 1.000 kg of water.

a. How many moles of CaCl2 are in solution? How many moles of water are

present?

b. Assume that the ionic compound, CaCl2, separates completely into Ca 2�

and Cl � ions when it dissolves in water. How many moles of each ion are

present in the solution?

17. How many moles are in each of the following masses?

a. 453.6 g (1.000 pound) of sucrose (table sugar), C12H22O11 b. 1.000 pound of table salt, NaCl

18. When the ionic compound NH4Cl dissolves in water, it breaks into one ammo-

� �

nium ion, NH4 , and one chloride ion, Cl . If you dissolved 10.7 g of NH4Cl in

water, how many moles of ions would be in solution?

19. What is the total amount in moles of atoms in a jar that contains 2.41 � 10 24

atoms of chromium, 1.51 � 10 23 atoms of nickel, and 3.01 � 10 23 atoms of

copper?

20. The density of liquid water is 0.997 g/mL at 25°C.

a. Calculate the mass of 250.0 mL (about a cupful) of water.

b. How many moles of water are in 250.0 mL of water? Hint: Use the result of

(a).

c. Calculate the volume that would be occupied by 2.000 mol of water at 25°C.

d. What mass of water is 2.000 mol of water?

Holt ChemFile: Problem-Solving Workbook 71 Mole Concept

21. An Avogadro’s number (1 mol) of sugar molecules has a mass of 342 g, but an

Avogadro’s number (1 mol) of water molecules has a mass of only 18 g.

Explain why there is such a difference between the mass of 1 mol of sugar

and the mass of 1 mol of water.

22. Calculate the mass of aluminum that would have the same number of atoms

as 6.35 g of cadmium.

23. A chemist weighs a steel cylinder of compressed oxygen, O2, and finds that it

has a mass of 1027.8 g. After some of the oxygen is used in an experiment, the

cylinder has a mass of 1023.2 g. How many moles of oxygen gas are used in

the experiment?

24. Suppose that you could decompose 0.250 mol of Ag2S into its elements.

a. How many moles of silver would you have? How many moles of sulfur

would you have?

b. How many moles of Ag2S are there in 38.8 g of Ag2S? How many moles of

silver and sulfur would be produced from this amount of Ag2S? c. Calculate the masses of silver and sulfur produced in (b).

Holt ChemFile: Problem-Solving Workbook 72 Mole Concept

Percentage Composition

Suppose you are working in an industrial laboratory. Your supervisor gives you a

bottle containing a white crystalline compound and asks you to determine its

identity. Several unlabeled drums of this substance have been discovered in a

warehouse, and no one knows what it is. You take it into the laboratory and carry

out an analysis, which shows that the compound is composed of the elements

sodium, carbon, and oxygen. Immediately, you think of the compound sodium

carbonate, Na 2CO 3, a very common substance found in most laboratories and

used in many industrial processes.

Before you report your conclusion to your boss, you decide to check a reference

book to see if there are any other compounds that contain only the elements

sodium, carbon, and oxygen. You discover that there is another compound,

sodium oxalate, which has the formula Na 2C 2O 4. When you read about this compound,

you find that it is highly poisonous and can cause serious illness and even

death. Mistaking sodium carbonate for sodium oxalate could have very serious

consequences. What can you do to determine the identity of your sample? Is it

the common industrial substance or the dangerous poison?

Fortunately, you can determine not only which elements are in the compound,

but also how much of each element is present. As you have learned, every compound

has a definite composition. Every water molecule is made up of two

hydrogen atoms and one oxygen atom, no matter where the water came from. A

formula unit of sodium chloride is composed of one sodium atom and one chlorine

atom, no matter whether the salt came from a mine or was obtained by evaporating

sea water.

Likewise, sodium carbonate always has two sodium atoms, one carbon atom,

and three oxygen atoms per formula unit, giving it the formula Na 2CO 3; and a formula

unit of sodium oxalate always contains two sodium atoms, two carbon

atoms, and four oxygen atoms, giving it the formula Na 2C 2O 4. Because each atom

has a definite mass, each compound will have a distinct composition by mass.

This composition is usually expressed as the percentage composition of the compound—the

percentage by mass of each element in a compound. To identify a

compound, you can compare the percentage composition obtained by laboratory

analysis with a calculated percentage composition of each possible compound.

Holt ChemFile: Problem-Solving Workbook 73 Percentage Composition

Molar mass

of

element

5

element in a

sample of

compound

General Plan for Determining Percentage

Composition of a Compound

Convert using the

formula of the compound.

Convert by expressing

percentage as a fraction

and then multiplying by

the mass of the sample.

Convert by multiplying

by the inverse of the

molar mass of the

compound. Then convert

to a percentage by

multiplying by 100.

element per

mole of

Percentage

element in the

composition of

the compound

Repeat 1, 2,

and 3 for each

remaining

compound.

Holt ChemFile: Problem-Solving Workbook 74 Percentage Composition

Determine the percentage composition of sodium carbonate, Na 2CO 3.

What is given in the problem? the formula of sodium carbonate

What are you asked to find? the percentage of each element in sodium carbonate

(the percentage composition)

Formula of sodium carbonate Na2CO3 Molar mass of each element* Na � 22.99 g/mol

C � 12.01 g/mol

O � 16.00 g/mol

Molar mass of sodium carbonate 105.99 g/mol

Percentage composition of sodium carbonate ?%

What step is needed to determine the mass of each element per mole of compound?

Multiply the molar mass of each element by the ratio of the number of moles of

that element in a mole of the compound (the subscript of that element in the

compound’s formula).

What steps are needed to determine the portion of each element as a percentage

of the mass of the compound?

Multiply the mass of each element by the inverse of the molar mass of the

compound, and then multiply by 100 to convert to a percentage.

Step 1

Molar mass of Na

multiply by the subscript of Na

in Na 2 CO 3

ratio of mol Na per mol

molar mass Na Na2CO3 from formula

22.99 g Na 2 mol Na g Na

Mass Na per mole

Na2CO3 Holt ChemFile: Problem-Solving Workbook 75 Percentage Composition

Step 2

Now you can combine Step 1 and Step 2 into one calculation.

combining Steps 1 and 2

22.99 g Na 2 mol Na 1 mol Na

2CO3 �� 100 � percentage Na in Na2CO3 1 mol Na 1 mol Na2CO3 105.99 g Na2CO3 Finally, determine the percentage of carbon and oxygen in Na 2CO 3 by repeating

the calculation above with each of those elements.

percentage sodium

22.99 g Na

1 mol Na

percentage carbon

12.01 g C

1 mol C

percentage oxygen

Percentage of each

element in Na2CO3 16.00 g O

1 mol O

Mass Na per

mole Na2CO3 multiply by the inverse of the

molar mass of Na 2 CO 3 and

multiply by 100

molar mass Na 2 CO 3

from Step 1

g Na 1 mol Na2CO3 � � 100 � percentage Na in Na2CO3 1 mol Na2CO3 105.99 g Na2CO3 repeat Steps 1 and 2 for

each remaining element

2 mol Na 1 mol Na

2CO3 �

� 100 � 43.38% Na

1 mol Na2CO3 105.99 g Na2CO3 1 mol C 1 mol Na

� 100 � 11.33% C

1 mol Na2CO3 105.99 g Na2CO3 3 mol O 1 mol Na

� 100 � 45.29% O

1 mol Na2CO3 105.99 g Na2CO3 Element Percentage

sodium 43.38% Na

carbon 11.33% C

oxygen 45.29% O

Yes; the composition is given in percentages.

Percentage Na in

Na2CO3 4

Percentage composition

Holt ChemFile: Problem-Solving Workbook 76 Percentage Composition

Yes; four significant figures is correct because the molar masses have four

Yes; the percentages add up to 100 percent.

1. Determine the percentage composition of each of the following compounds:

a. sodium oxalate, Na2C2O4 ans: 34.31% Na, 17.93% C, 47.76% O

b. ethanol, C 2H 5OH ans: 52.13% C, 13.15% H, 34.72% O

c. aluminum oxide, Al 2O 3 ans: 52.92% Al, 47.08% O

d. potassium sulfate, K 2SO 4 ans: 44.87% K, 18.40% S, 36.72% O

2. Suppose that your laboratory analysis of the white powder discussed at the

beginning of this chapter showed 42.59% Na, 12.02% C, and 44.99% oxygen.

Would you report that the compound is sodium oxalate or sodium carbonate

(use the results of Practice Problem 1 and Sample Problem 1)? ans: sodium

carbonate

Holt ChemFile: Problem-Solving Workbook 77 Percentage Composition

Calculate the mass of zinc in a 30.00 g sample of zinc nitrate, Zn(NO 3) 2.

What is given in the problem? the mass in grams of zinc nitrate

What are you asked to find? the mass in grams of zinc in the sample

What steps are needed to determine the mass of Zn in a given mass of Zn(NO 3) 2?

The percentage of Zn in Zn(NO 3) 2 can be calculated and used to find the mass of

Zn in the sample.

multiply

by the mole

ratio of Zn to

Zn(NO 3 ) 2

Mass of zinc nitrate 30.00 g

Formula of zinc nitrate Zn(NO3) 2

Molar mass of zinc nitrate 189.41 g/mol

Mass of zinc in the sample ? g

Molar mass of Zn

Mass Zn per

mole Zn(NO3 ) 2

ratio of mol Zn per mol

Zn(NO 3 ) 2 from formula

65.39 g Zn 1 mol Zn

1 mol

� � � 100 � percentage Zn

1 mol Zn 1 mol Zn(NO3 ) 2 189.41 g Zn(NO3 ) 2

percentage Zn

expressed as a fraction

Zn(NO 3 ) 2 , then multiply

by 100

g Zn

� g Zn(NO3 ) 2 � g Zn in sample

100 g Zn(NO3 ) 2

65.39 g Zn 1 mol Zn 1 mol Zn(NO

3) 2

� 100 � 34.52% Zn

1 mol Zn 1 mol Zn(NO3) 2 189.41 g Zn(NO3) 2

molar mass Zn(NO3 ) 2

Mass Zn in g in sample

express percentage

as a fraction and

mass of the sample

Percentage Zn

in Zn(NO3 ) 2

Holt ChemFile: Problem-Solving Workbook 78 Percentage Composition

Note that mass percentage is the same as grams per 100 g, so 34.52% Zn in

Zn(NO 3) 2 is the same as 34.52 g Zn in 100 g Zn(NO 3) 2.

Yes; units cancel to give the correct units, grams of zinc.

Yes; four significant figures is correct because the data given have four significant

Yes; the molar mass of zinc is about one third of the molar mass of Zn(NO3) 2,

and 10.36 g Zn is about one third of 30.00 g of Zn(NO3) 2.

34.52 g Zn

� 30.00 g Zn(NO3) 2 � 10.36 g Zn

100 g Zn(NO3) 2

1. Calculate the mass of the given element in each of the following compounds:

a. bromine in 50.0 g potassium bromide, KBr ans: 33.6 g Br

b. chromium in 1.00 kg sodium dichromate, Na 2Cr 2O 7 ans: 397 g Cr

c. nitrogen in 85.0 mg of the amino acid lysine, C 6H 14N 2O 2 ans: 16.3 mg N

d. cobalt in 2.84 g cobalt(II) acetate, Co(C 2H 3O 2) 2 ans: 0.945 g Co

Holt ChemFile: Problem-Solving Workbook 79 Percentage Composition

HYDRATES

Many compounds, especially ionic compounds, are produced and purified by

crystallizing them from water solutions. When this happens, some compounds

incorporate water molecules into their crystal structure. These crystalline compounds

are called hydrates because they include water molecules. The number of

water molecules per formula unit is specific for each type of crystal. When you

have to measure a certain quantity of the compound, it is important to know how

much the water molecules contribute to the mass.

You may have seen blue crystals of copper(II) sulfate in the laboratory. When

this compound is crystallized from water solution, the crystals include five water

molecules for each formula unit of CuSO 4. The true name of the substance is

copper(II) sulfate pentahydrate, and its formula is written correctly as

CuSO 4�5H 2O. Notice that the five water molecules are written separately. They

are preceded by a dot, which means they are attached to the copper sulfate molecule.

On a molar basis, a mole of CuSO 4�5H 2O contains 5 mol of water per mole

of CuSO 4�5H 2O. The water molecules contribute to the total mass of

CuSO 4�5H 2O. When you determine the percentage water in a hydrate, the water

molecules are treated separately, as if they were another element.

Determine the percentage water in copper(II) sulfate pentahydrate,

CuSO 4�5H 2O.

What is given in the problem? the formula of copper(II) sulfate pentahydrate

What are you asked to find? the percentage water in the hydrate

Formula of copper(II) sulfate pentahydrate CuSO4�5H2O Molar mass of H 2O 18.02 g/mol

Molar mass of copper(II) sulfate pentahydrate* 249.72 g/mol

Percentage H2O in CuSO4�5H2O ?%

* molar mass of CuSO 4 � mass of 5 mol H 2O

Holt ChemFile: Problem-Solving Workbook 80 Percentage Composition

What steps are needed to determine the percentage of water in CuSO 4�5H 2O?

Find the mass of water per mole of hydrate, multiply by the inverse molar mass

of the hydrate, and multiply that by 100 to convert to a percentage.

1 3

Molar mass of H2O multiply by the

mole ratio of

H 2 O to

CuSO 4� 5H 2 O

molar mass H 2 O

ratio of moles H 2 O per mole

CuSO 4� 5H 2 O from formula

18.01 g H2O 5 mol H2O � �

1 mol H2O 1 mol CuSO4�5H2O COMPUTE

18.01 g H2O 5 mol H2O 1 mol CuSO

4�5H2O � 100 � 36.08% H2O 1 mol H2O 1 mol CuSO4�5H2O 249.72 g CuSO4�5H2O EVALUATE

Yes; the percentage of water in copper(II) sulfate pentahydrate was needed.

Yes; four significant figures is correct because molar masses were given to at

least four significant figures.

Yes; five water molecules have a mass of about 90 g, and 90 g is a little more

than 1/3 of 250 g; the calculated percentage is a little more than 1/3.

Mass H 2 O per mole

Percentage H 2 O in CuSO 4� 5H 2 O

CuSO 4� 5H 2 O; then

molar mass CuSO4�5H2O 1 mol CuSO4�5H2O 249.72 g CuSO4�5H2O � 100 � percentage H 2 O

Holt ChemFile: Problem-Solving Workbook 81 Percentage Composition

1. Calculate the percentage of water in each of the following hydrates:

a. sodium carbonate decahydrate, Na 2CO 3�10H 2O ans: 62.97% H 2O in

Na 2CO 3�10H 2O

b. nickel(II) iodide hexahydrate, NiI 2�6H 2O ans: 25.71% H 2O in NiI 2�6H 2O

c. ammonium hexacyanoferrate(III) trihydrate (commonly called ammonium

ferricyanide), (NH 4) 2Fe(CN) 6�3H 2O ans: 17.89 % H 2O in

(NH 4) 2Fe(CN) 6�3H 2O

d. aluminum bromide hexahydrate ans: 28.85% H 2O in AlBr 3�6H 2O

Holt ChemFile: Problem-Solving Workbook 82 Percentage Composition

1. Write formulas for the following compounds and determine the percentage

composition of each:

a. nitric acid

b. ammonia

c. mercury(II) sulfate

d. antimony(V) fluoride

2. Calculate the percentage composition of the following compounds:

a. lithium bromide, LiBr

b. anthracene, C14H10 c. ammonium nitrate, NH4NO3 d. nitrous acid, HNO2 e. silver sulfide, Ag2S f. iron(II) thiocyanate, Fe(SCN) 2

g. lithium acetate

h. nickel(II) formate

3. Calculate the percentage of the given element in each of the following compounds:

a. nitrogen in urea, NH2CONH2 b. sulfur in sulfuryl chloride, SO2Cl2 c. thallium in thallium(III) oxide, Tl2O3 d. oxygen in potassium chlorate, KClO3 e. bromine in calcium bromide, CaBr2 f. tin in tin(IV) oxide, SnO2 4. Calculate the mass of the given element in each of the following quantities:

a. oxygen in 4.00 g of manganese dioxide, MnO2 b. aluminum in 50.0 metric tons of aluminum oxide, Al2O3 c. silver in 325 g silver cyanide, AgCN

d. gold in 0.780 g of gold(III) selenide, Au2Se3 e. selenium in 683 g sodium selenite, Na2SeO3 f. chlorine in 5.0 � 10 4 g of 1,1-dichloropropane, CHCl2CH2CH3 5. Calculate the percentage of water in each of the following hydrates:

a. strontium chloride hexahydrate, SrCl 2�6H 2O

b. zinc sulfate heptahydrate, ZnSO 4�7H 2O

c. calcium fluorophosphate dihydrate, CaFPO 3�2H 2O

d. beryllium nitrate trihydrate, Be(NO 3) 2�3H 2O

Holt ChemFile: Problem-Solving Workbook 83 Percentage Composition

6. Calculate the percentage of the given element in each of the following

hydrates. You must first determine the formulas of the hydrates.

a. nickel in nickel(II) acetate tetrahydrate

b. chromium in sodium chromate tetrahydrate

c. cerium in cerium(IV) sulfate tetrahydrate

7. Cinnabar is a mineral that is mined in order to produce mercury. Cinnabar is

mercury(II) sulfide, HgS. What mass of mercury can be obtained from 50.0 kg

of cinnabar?

8. The minerals malachite, Cu2(OH) 2CO3, and chalcopyrite, CuFeS2, can be

mined to obtain copper metal. How much copper could be obtained from

1.00 � 10 3 kg of each? Which of the two has the greater copper content?

9. Calculate the percentage of the given element in each of the following

hydrates:

a. vanadium in vanadium oxysulfate dihydrate, VOSO4�2H2O b. tin in potassium stannate trihydrate, K 2SnO 3�3H 2O

c. chlorine in calcium chlorate dihydrate, CaClO 3�2H 2O

10. Heating copper sulfate pentahydrate will evaporate the water from the crystals,

leaving anhydrous copper sulfate, a white powder. Anhydrous means

“without water.” What mass of anhydrous CuSO 4 would be produced by

heating 500.0 g of CuSO 4�5H 2O?

11. Silver metal may be precipitated from a solution of silver nitrate by placing a

copper strip into the solution. What mass of AgNO3 would you dissolve in

water in order to get 1.00 g of silver?

12. A sample of Ag2S has a mass of 62.4 g. What mass of each element could be

obtained by decomposing this sample?

13. A quantity of epsom salts, magnesium sulfate heptahydrate, MgSO4�7H2O, is

heated until all the water is driven off. The sample loses 11.8 g in the process.

What was the mass of the original sample?

14. The process of manufacturing sulfuric acid begins with the burning of sulfur.

What mass of sulfur would have to be burned in order to produce 1.00 kg of

H2SO4? Assume that all of the sulfur ends up in the sulfuric acid.

Holt ChemFile: Problem-Solving Workbook 84 Percentage Composition

Empirical Formulas

Suppose you analyze an unknown compound that is a white powder and find that

it is composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. You can use

those percentages to determine the mole ratios among sodium, sulfur, and oxygen

and write a formula for the compound.

To begin, the mass percentages of each element can be interpreted as “grams

of element per 100 grams of compound.” To make things simpler, you can assume

you have a 100 g sample of the unknown compound. The unknown compound

contains 36.5% sodium by mass. Therefore 100.0 g of the compound would contain

36.5 g of sodium. You already know how to convert mass of a substance into

number of moles, so you can calculate the number of moles of sodium in 36.5 g.

After you find the number of moles of each element, you can look for a simple

ratio among the elements and use this ratio of elements to write a formula for the

The chemical formula obtained from the mass percentages is in the simplest

form for that compound. The mole ratios for each element, which you determined

from the analytical data given, are reduced to the smallest whole numbers.

This simplest formula is also called the empirical formula. The actual formula for

the compound could be a multiple of the empirical formula. For instance, suppose

you analyze a compound and find that it is composed of 40.0% carbon, 6.7%

hydrogen, and 53.3% oxygen. If you determine the formula for this compound

based only on the analytical data, you will determine the formula to be CH 2O.

There are, however, other possibilities for the formula. It could be C 2H 4O 2 and

still have the same percentage composition. In fact, it could be any multiple of

CH 2O.

It is possible to convert from the empirical formula to the actual chemical formula

for the compound as long as the molar mass of the compound is known.

Look again at the CH 2O example. If the true compound were CH 2O, it would have

a molar mass of 30.03 g/mol. If you do more tests on the unknown compound and

find that its molar mass is 60.06, you know that CH 2O cannot be its true identity.

The molar mass 60.06 is twice the molar mass of CH 2O. Therefore, you know that

the true chemical formula must be twice the empirical formula, (CH 2O) � 2, or

C 2H 4O 2. Any correct molecular formula can be determined from an empirical formula

and a molar mass in this same way.

Holt ChemFile: Problem-Solving Workbook 85 Empirical Formulas

General Plan for Determining Empirical Formulas

and Molecular Formulas

Percentage of

element expressed

as grams of

element per 100 g

unknown

Empirical formula

of the compound

the experimental

unknown and the

simplest formula.

Molecular formula

each element.

The calculated ratio

is the simplest

formula.

Amount of each

of unknown

Use the amount of

the least-abundant

element to calculate

the simplest wholenumber

ratio among

the elements.

Calculated

whole-number

ratio among the

elements

Holt ChemFile: Problem-Solving Workbook 86 Empirical Formulas

Determine the empirical formula for an unknown compound composed of

36.5% sodium, 38.1% oxygen, and 25.4% sulfur by mass.

What is given in the problem? the percentage composition of the compound

What are you asked to find? the empirical formula for the compound

The percentage composition of the unknown subtance 36.5% sodium

38.1% oxygen

25.4% sulfur

The molar mass of each element* 22.99 g Na/mol Na

16.00 g O/mol O

32.07 g S/mol S

Amount of each element per 100.0 g of the unknown ? mol

Simplest mole ratio of elements in the unknown ?

What steps are needed to calculate the amount in moles of each element per

100.0 g of unknown?

State the percentage of the element in grams and multiply by the inverse of the

molar mass of the element.

What steps are needed to determine the whole-number mole ratio of the elements

in the unknown (the simplest formula)?

Divide the amount of each element by the amount of the least-abundant element.

If necessary, multiply the ratio by a small integer that will produce a

whole-number ratio.

Mass of Na per

100.0 g unknown

the molar mass of Na

percent of Na stated as grams 1

Na per 100 g unknown molar mass Na

Amount Na in mol per

36.5 g Na 1 mol Na mol Na

100.0 g unknown 22.99 g Na 100.0 g unknown

Holt ChemFile: Problem-Solving Workbook 87 Empirical Formulas

Repeat this step for the remaining elements.

36.5 g Na

38.1 g O

25.4 g S

Divide the amount of each element by the amount of the least-abundant element,

which in this example is S. This can be accomplished by multiplying the

amount of each element by the inverse of the amount of the least abundant element.

1.59 mol Na

2.38 mol O

0.792 mol S

From the calculations, the simplest mole ratio is 2 mol Na:3 mol O:1 mol S.

The simplest formula is therefore Na 2O 3S. Seeing the ratio 3 mol O:1 mol S,

you can use your knowledge of chemistry to suggest that this possibly represents

a sulfite group, –SO 3 and propose the formula Na 2SO 3.

Yes; units canceled throughout the calculation, so it is reasonable to assume

that the resulting ratio is accurate.

Yes; ratios were calculated to three significant figures because percentages were

16.00 g O

1 mol S

32.07 g S

Amount of Na in mol per

divide by the

amount of the

least-abundant

Whole-number ratio

among the elements

2.01 mol Na

3.01 mol O

1.00 mol S

Holt ChemFile: Problem-Solving Workbook 88 Empirical Formulas

Yes; the formula, Na 2SO 3 is plausible, given the mole ratios and considering that

the sulfite ion has a 2� charge and the sodium ion has a 1� charge.

1. Determine the empirical formula for compounds that have the following

analyses:

a. 28.4% copper, 71.6% bromine ans: CuBr2 b. 39.0% potassium, 12.0% carbon, 1.01% hydrogen, and 47.9% oxygen

ans: KHCO 3

c. 77.3% silver, 7.4% phosphorus, 15.3% oxygen ans: Ag 3PO 4

d. 0.57% hydrogen, 72.1% iodine, 27.3% oxygen ans: HIO 3

Holt ChemFile: Problem-Solving Workbook 89 Empirical Formulas

38.4% oxygen, 23.7% carbon, and 1.66% hydrogen.

If necessary, multiply the ratio by a small integer to produce a wholenumber

ratio.

Mass of K in g per

38.4 g K

the molar mass of K

1 mol K

39.10 g K

Proceed to find the amount in moles per 100.0 g of unknown for the elements

carbon, oxygen, and hydrogen, as in Sample Problem 1.

When determining the formula of a compound having more than two elements,

it is usually advisable to put the data and results in a table.

Amount of K in mol per

divide by the amount of the

least-abundant element,

and multiply by an integer

that will produce a wholenumber

ratio

0.982 mol K

Holt ChemFile: Problem-Solving Workbook 90 Empirical Formulas

Mass per 100.0 g Amount in mol per

Element of unknown Molar mass 100.0 g of unknown

Potassium 38.4 g K 39.10 g/mol 0.982 mol K

Carbon 23.7 g C 12.01 g/mol 1.97 mol C

Oxygen 36.3 g O 16.00 g/mol 2.27 mol O

Hydrogen 1.66 g H 1.01 g/mol 1.64 mol H

Again, as in Sample Problem 1, divide each result by the amount in moles of

the least-abundant element, which in this example is K.

You should get the following results:

Amount in mol of element Amount in mol of element

Element per 100.0 g of unknown per mol of potassium

Potassium 0.982 mol K 1.00 mol K

Carbon 1.97 mol C 2.01 mol C

Oxygen 2.27 mol O 2.31 mol O

Hydrogen 1.64 mol H 1.67 mol H

In contrast to Sample Problem 1, this calculation does not give a simple

whole-number ratio among the elements. To solve this problem, multiply by a

small integer that will result in a whole-number ratio. You can pick an integer that

you think might work, or you can convert the number of moles to an equivalent

fractional number. At this point, you should keep in mind that analytical data is

never perfect, so change the number of moles to the fraction that is closest to the

decimal number. Then, choose the appropriate integer factor to use. In this case,

the fractions are in thirds so a factor of 3 will change the fractions into whole

numbers.

Amount in mol

of element per Fraction nearest the Integer Whole-number

mole of potassium decimal value factor mole ratio

1.00 mol K 1 mol K �3 3 mol K

2.01 mol C 2 mol C �3 6 mol C

2.31 mol O 2 1/3 mol O �3 7 mol O

1.67 mol H 1 2/3 mol H �3 5 mol H

Thus, the simplest formula for the compound is K 3C 6H 5O 7, which happens to

be the formula for potassium citrate.

Holt ChemFile: Problem-Solving Workbook 91 Empirical Formulas

Yes; the formula, K 3C 6H 5O 7 is plausible, considering that the potassium ion has

a 1� charge and the citrate polyatomic ion has a 3� charge.

1. Determine the simplest formula for compounds that have the following analyses.

The data may not be exact.

a. 36.2% aluminum and 63.8% sulfur ans: Al2S3 b. 93.5% niobium and 6.50% oxygen ans: Nb 5O 2

c. 57.6% strontium, 13.8% phosphorus, and 28.6% oxygen ans: Sr 3P 2O 8 or

Sr 3(PO 4) 2

d. 28.5% iron, 48.6% oxygen, and 22.9% sulfur ans: Fe 2S 3O 12 or Fe 2(SO 4) 3

Holt ChemFile: Problem-Solving Workbook 92 Empirical Formulas

A compound is analyzed and found to have the empirical formula CH 2O.

The molar mass of the compound is found to be 153 g/mol. What is the

compound’s molecular formula?

What is given in the problem? the empirical formula, and the experimental

molar mass

What are you asked to find? the molecular formula of the compound

Empirical formula of unknown CH 2O

Experimental molar mass of unknown 153 g/mol

Molar mass of empirical formula 30.03 g/mol

Molecular formula of the compound ?

What steps are needed to determine the molecular formula of the unknown compound?

Multiply the experimental molar mass by the inverse of the molar mass of the

empirical formula. The subscripts of the empirical formula are multiplied by the

whole-number factor obtained.

multiply the experimental molar mass

by the inverse of the molar mass of the

empirical formula, and multiply each

subscript in the empirical formula by

the resulting factor

empirical formula

153 g 1 mol CH2O � �

1 mol unknown 30.03 g

153 g� 1 mol CH 5.09 mol CH

�� 2O

2O

1 mol unknown 30.03 g� 1 mol unknown

Allowing for a little experimental error, the molecular formula must be five

times the empirical formula.

factor that shows the number

of times the empirical formula

must be multiplied to get the

molecular formula

mol CH 2 O

1 mol unknown

Holt ChemFile: Problem-Solving Workbook 93 Empirical Formulas

Yes; the calculated molar mass of C 5H 10O 5 is 150.15, which is close to the experimental

molar mass of the unknown. Reference books show that there are several

different compounds with the formula C 5H 10O 5.

1. Determine the molecular formula of each of the following unknown substances:

a. empirical formula CH2, experimental molar mass 28 g/mol ans: C2H4 b. empirical formula B 2H 5, experimental molar mass 54 g/mol ans: B 4H 10

c. empirical formula C 2HCl, experimental molar mass 179 g/mol ans: C 6H 3Cl 3

d. empirical formula C 6H 8O, experimental molar mass 290 g/mol ans:

C 18H 24O 3

e. empirical formula C 3H 2O, experimental molar mass 216 g/mol ans: C 12H 8O 4

Holt ChemFile: Problem-Solving Workbook 94 Empirical Formulas

a. 66.0% barium and 34.0% chlorine

b. 80.38% bismuth, 18.46% oxygen, and 1.16% hydrogen

c. 12.67% aluminum, 19.73% nitrogen, and 67.60% oxygen

d. 35.64% zinc, 26.18% carbon, 34.88% oxygen, and 3.30% hydrogen

e. 2.8% hydrogen, 9.8% nitrogen, 20.5% nickel, 44.5% oxygen, and 22.4% sulfur

f. 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine

2. Sometimes, instead of percentage composition, you will have the composition

of a sample by mass. Use the same method shown in Sample Problem 1, but

use the actual mass of the sample instead of assuming a 100 g sample.

Determine the empirical formula for compounds that have the following

a. a 0.858 g sample of an unknown substance is composed of 0.537 g of copper

and 0.321 g of fluorine

b. a 13.07 g sample of an unknown substance is composed of 9.48 g of barium,

1.66 g of carbon, and 1.93 g of nitrogen

c. a 0.025 g sample of an unknown substance is composed of 0.0091 g manganese,

0.0106 g oxygen, and 0.0053 g sulfur

3. Determine the empirical formula for compounds that have the following

a. a 0.0082 g sample contains 0.0015 g of nickel and 0.0067 g of iodine

b. a 0.470 g sample contains 0.144 g of manganese, 0.074 g of nitrogen, and

0.252 g of oxygen

c. a 3.880 g sample contains 0.691 g of magnesium, 1.824 g of sulfur, and 1.365

g of oxygen

d. a 46.25 g sample contains 14.77 g of potassium, 9.06 g of oxygen, and 22.42

g of tin

4. Determine the empirical formula for compounds that have the following

a. 60.9% As and 39.1% S

b. 76.89% Re and 23.12% O

c. 5.04% H, 35.00% N, and 59.96% O

d. 24.3% Fe, 33.9% Cr, and 41.8% O

e. 54.03% C, 37.81% N, and 8.16% H

f. 55.81% C, 3.90% H, 29.43% F, and 10.85% N

Holt ChemFile: Problem-Solving Workbook 95 Empirical Formulas

5. Determine the molecular formulas for compounds having the following empirical

formulas and molar masses:

a. C2H4S; experimental molar mass 179

b. C2H4O; experimental molar mass 176

c. C2H3O2; experimental molar mass 119

d. C2H2O, experimental molar mass 254

6. Use the experimental molar mass to determine the molecular formula for

compounds having the following analyses:

a. 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen; experimental molar

mass 116.07

b. 54.53% carbon, 9.15% hydrogen, and 36.32% oxygen; experimental molar

mass 88

c. 64.27% carbon, 7.19% hydrogen, and 28.54% oxygen; experimental molar

mass 168.19

7. A 0.400 g sample of a white powder contains 0.141 g of potassium, 0.115 g of

sulfur, and 0.144 g of oxygen. What is the empirical formula for the compound?

8. A 10.64 g sample of a lead compound is analyzed and found to be made up of

9.65 g of lead and 0.99 g of oxygen. Determine the empirical formula for this

9. A 2.65 g sample of a salmon-colored powder contains 0.70 g of chromium,

0.65 g of sulfur, and 1.30 g of oxygen. The molar mass is 392.2. What is the

formula of the compound?

10. Ninhydrin is a compound that reacts with amino acids and proteins to produce

a dark-colored complex. It is used by forensic chemists and detectives

to see fingerprints that might otherwise be invisible. Ninhydrin’s composition

is 60.68% carbon, 3.40% hydrogen, and 35.92% oxygen. What is the empirical

formula for ninhydrin?

11. Histamine is a substance that is released by cells in response to injury, infection,

stings, and materials that cause allergic responses, such as pollen.

Histamine causes dilation of blood vessels and swelling due to accumulation

of fluid in the tissues. People sometimes take antihistamine drugs to counteract

the effects of histamine. A sample of histamine having a mass of 385 mg is

composed of 208 mg of carbon, 31 mg of hydrogen, and 146 mg of nitrogen.

The molar mass of histamine is 111 g/mol. What is the molecular formula for

histamine?

12. You analyze two substances in the laboratory and discover that each has the

empirical formula CH2O. You can easily see that they are different substances

because one is a liquid with a sharp, biting odor and the other is an odorless,

crystalline solid. How can you account for the fact that both have the same

empirical formula?

Holt ChemFile: Problem-Solving Workbook 96 Empirical Formulas

Stoichiometry

So far in your chemistry course, you have learned that chemists count quantities

of elements and compounds in terms of moles and that they relate moles of a

substance to mass by using the molar mass. In addition, you have learned to

write chemical equations so that they represent the rearrangements of atoms that

take place during chemical reactions, and you have learned to balance these

equations. In this chapter you will be able to put these separate skills together to

accomplish one of the most important tasks of chemistry—using chemical equations

to make predictions about the quantities of substances that react or are

given off as products and relating those quantities to one another. This process of

relating quantities of reactants and products in a chemical reaction to one

another is called stoichiometry.

First, look at an analogy.

Suppose you need to make several sandwiches to take on a picnic with

friends. You decide to make turkey-and-cheese sandwiches using the following

“equation:”

2 bread slices � 2 turkey slices � 1 lettuce leaf � 1 cheese slice

→ 1 turkey-and-cheese sandwich

This equation shows that you need those ingredients in a ratio of 2:2:1:1,

respectively. You can use this equation to predict that you would need 30 turkey

slices to make 15 sandwiches or 6 cheese slices to go with 12 turkey slices.

Zinc reacts with oxygen according to the following balanced chemical equation:

2Zn � O2 → 2ZnO

Like the sandwich recipe, this equation can be viewed as a “recipe” for zinc

oxide. It tells you that reacting two zinc atoms with a molecule of oxygen will

produce two formula units of zinc oxide. Can you predict how many zinc oxide

units could be formed from 500 zinc atoms? Could you determine how many

moles of oxygen molecules it would take to react with 4 mol of zinc atoms? What

if you had 22 g of zinc and wanted to know how many grams of ZnO could be

made from it? Keep in mind that the chemical equation relates amounts, not

masses, of products and reactants. The problems in this chapter will show you

how to solve problems of this kind.

Holt ChemFile: Problem-Solving Workbook 97 Stoichiometry

General Plan for Solving Stoichiometry Problems

substance A

the molar mass

of A.

mole ratio A 2

Amount in

mol of

, given in the

B

balanced chemical equation.

substance B

Holt ChemFile: Problem-Solving Workbook 98 Stoichiometry

Ammonia is made industrially by reacting nitrogen and hydrogen under

pressure, at high temperature, and in the presence of a catalyst. The

equation is N 2(g) � 3H 2(g) → 2NH 3(g). If 4.0 mol of H 2 react, how many

moles of NH 3 will be produced?

What is given in the problem? the balanced equation, and the amount of H2 in

moles

What are you asked to find? the amount of NH3 produced in moles

Organization of data is extremely important in dealing with stoichiometry problems.

You will find that it is most helpful to make data tables such as the following

one.

Substance H 2 NH 3

Coefficient in balanced equation 3 2

Molar mass NA* NA

Amount 4.0 mol ? mol

Mass of substance NA NA

* NA means not applicable to the problem

What steps are needed to calculate the amount of NH 3 that can be produced from

4.0 mol H 2?

Multiply by the mole ratio of NH 3 to H 2 determined from the coefficients of the

balanced equation.

H2 in mol

multiply by mole ratio:

mole ratio

given 2 mol NH 3 � mol NH3

mol H 2 � 3 mol H2

NH3 in mol

Holt ChemFile: Problem-Solving Workbook 99 Stoichiometry

NH 3

Yes; the answer has the correct units of moles NH3. Is the number of significant figures correct?

Yes; two significant figures is correct because data were given to two significant

Yes; the answer is 2/3 of 4.0.

2 mol NH3 4.0 mol H2 � � 2.7 mol NH3 3 mol H2 1. How many moles of sodium will react with water to produce 4.0 mol of

hydrogen in the following reaction?

2Na(s) � 2H2O(l) → 2NaOH(aq) � H2(g) ans: 8.0 mol Na

2. How many moles of lithium chloride will be formed by the reaction of chlorine

with 0.046 mol of lithium bromide in the following reaction?

2LiBr(aq) � Cl2(g) → 2LiCl(aq) � Br2(l) ans: 0.046 mol LiCl

Holt ChemFile: Problem-Solving Workbook 100 Stoichiometry

3. Aluminum will react with sulfuric acid in the following reaction.

2Al(s) � 3H2SO4(l) → Al2(SO4) 3(aq) � 3H2(g) a. How many moles of H 2SO 4 will react with 18 mol Al? ans: 27 mol H 2SO 4

b. How many moles of each product will be produced? ans: 27 mol H 2, 9 mol

Al 2(SO 4) 3

4. Propane burns in excess oxygen according to the following reaction.

C3H8 � 5O2 → 3CO2 � 4H2O a. How many moles each of CO 2 and H 2O are formed from 3.85 mol of

propane? ans: 11.6 mol CO 2, 15.4 mol H 2O

b. If 0.647 mol of oxygen is used in the burning of propane, how many moles

each of CO 2 and H 2O are produced? How many moles of C 3H 8 are consumed?

ans: 0.388 mol CO 2, 0.518 mol H 2O, 0.129 mol C 3H 8

Holt ChemFile: Problem-Solving Workbook 101 Stoichiometry

Potassium chlorate is sometimes decomposed in the laboratory to generate

oxygen. The reaction is 2KClO 3(s) → 2KCl(s) � 3O 2(g). What mass

of KClO 3 do you need to produce 0.50 mol O 2?

What is given in the problem? the amount of oxygen in moles

What are you asked to find? the mass of potassium chlorate

What steps are needed to calculate the mass of KClO 3 needed to produce 0.50

mol O 2?

Use the mole ratio to convert amount of O 2 to amount of KClO 3. Then convert

amount of KClO 3 to mass of KClO 3.

Substance KClO 3 O 2

Coefficient in balanced equation 2 3

Molar mass* 122.55 g/mol NA

Amount ? mol 0.50 mol

Mass ? g NA

Amount of O2 in mol

multiply by mole ratio

2 mol KClO3 3 mol O2 mole ratio

2 mol KClO3 0.50 mol O2 � �

3 mol O2 122.55 g KClO3 � 41 g KClO3 1 mol KClO3 EVALUATE

Yes; units canceled to give grams of KClO 3.

molar mass KClO 3

Amount of KClO3 in mol

of KClO 3

Mass of KClO3 in g

given 2 mol KClO3 122.55 g KClO3 mol O2 � � � g KClO3

3 mol O2 1 mol KClO3

Holt ChemFile: Problem-Solving Workbook 102 Stoichiometry

Yes; two significant figures is correct.

Yes; 41 g is about 1/3 of the molar mass of KClO 3, and 0.5 � 2/3 � 1/3.

1. Phosphorus burns in air to produce a phosphorus oxide in the following

reaction:

4P(s) � 5O2(g) → P4O10(s) a. What mass of phosphorus will be needed to produce 3.25 mol of P 4O 10?

ans: 403 g P

b. If 0.489 mol of phosphorus burns, what mass of oxygen is used? What mass

of P 4O 10 is produced? ans: 19.6 g O 2, 15.4 g P 2O 4

2. Hydrogen peroxide breaks down, releasing oxygen, in the following reaction:

2H2O2(aq) → 2H2O(l) � O2(g) a. What mass of oxygen is produced when 1.840 mol of H 2O 2 decomposes?

ans: 29.44 g O 2

b. What mass of water is produced when 5.0 mol O 2 is produced by this

reaction? ans: 180 g H 2O

Holt ChemFile: Problem-Solving Workbook 103 Stoichiometry

How many moles of aluminum will be produced from 30.0 kg Al2O3 in the

following reaction?

2Al2O3 → 4Al � 3O2 Solution

What is given in the problem? the mass of aluminum oxide

What are you asked to find? the amount of aluminum produced

Substance Al 2O 3 Al

Coefficient in balanced equation 2 4

Molar mass 101.96 g/mol NA

Amount ? mol ? mol

Mass 30.0 kg NA

What steps are needed to calculate the amount of Al produced from 30.0 kg of

Al 2O 3?

The molar mass of Al 2O 3 can be used to convert to moles Al 2O 3. The mole ratio

of Al:Al 2O 3 from the coefficients in the equation will convert to moles Al from

moles Al 2O 3.

Mass of Al2O3 in g

Al 2 O 3

Amount of Al2O3 in mol

multiply by the mole ratio

4 mol Al

2 mol Al2O3 1

molar mass Al 2 O 3

given 1000 g 1 mol Al2O3 4 mol Al

kg Al2 O3 � �

� mol Al

kg 101.96 g Al2O3 2 mol Al2O2 COMPUTE

30.0 kg Al2O3 � �

kg

1 mol Al2O3 4 mol Al

� � 588 mol Al

Mass of Al 2 O 3 in kg

Amount of Al in mol

Holt ChemFile: Problem-Solving Workbook 104 Stoichiometry

Yes; units canceled to give moles of Al.

Yes; the molar mass of Al2O3 is about 100, so 30 kg of Al2O3 is about 300 mol.

The mole ratio of Al:Al2O3 is 2:1, so the answer should be about 600 mol Al.

1. Sodium carbonate reacts with nitric acid according to the following equation.

Na2CO3(s) � 2HNO3 → 2NaNO3 � CO2 � H2O a. How many moles of Na 2CO 3 are required to produce 100.0 g of NaNO 3?

ans: 0.5882 mol Na 2CO 3

b. If 7.50 g of Na 2CO 3 reacts, how many moles of CO 2 are produced? ans:

0.0708 mol CO 2

2. Hydrogen is generated by passing hot steam over iron, which oxidizes to form

Fe3O4, in the following equation.

3Fe(s) � 4H2O(g) → 4H2(g) � Fe3O4(s) a. If 625 g of Fe 3O 4 is produced in the reaction, how many moles of hydrogen

are produced at the same time? ans: 10.8 mol H 2

b. How many moles of iron would be needed to generate 27 g of hydrogen?

ans: 10. mol Fe

Holt ChemFile: Problem-Solving Workbook 105 Stoichiometry

Methane burns in air by the following reaction:

CH4(g) � 2O2(g) → CO2(g) � 2H2O(g) What mass of water is produced by burning 500. g of methane?

What is given in the problem? the mass of methane in grams

What are you asked to find? the mass of water produced

Substance CH 4 H 2O

Coefficient in balanced equation 1 2

Molar mass 16.05 g/mol 18.02 g/mol

Mass 500. g ? g

What steps are needed to calculate the mass of H 2O produced from the burning

of 500. g of CH 4?

Convert grams of CH 4 to moles CH 4 by using the molar mass of CH 4. Use the

mole ratio from the balanced equation to determine moles H 2O from moles CH 4.

Use the molar mass of H 2O to calculate grams H 2O.

Mass of CH4 in g

CH 4

Amount of CH4 in mol

molar mass CH 4

2 mol H 2 O

1 mol CH 4

mass of H 2 O

given 1 mol CH4 2 mol H2O 18.02 g H2O g CH4 � � �

16.05 g CH4 1 mol CH4 1 mol H2O 3

Amount of H2O in mol

Mass of H2O in g

� g H 2 O

Holt ChemFile: Problem-Solving Workbook 106 Stoichiometry

500. g CH4 � 1 mol CH4 �

16.05 g CH4 2 mol H2O �

1 mol CH4 18.02 g H2O 1 mol H2O � 1.12 � 103 g H2O EVALUATE

Yes; mass of H2O was required, and units canceled to give grams H2O. Is the number of significant figures correct?

Yes; three significant figures is correct because the mass of CH4 was given to

Yes; CH4 and H2O have similar molar masses, and twice as many moles of H2O are produced as moles CH4 burned. So, you would expect to get a little more

than 1000 g of H2O. Practice

1. Calculate the mass of silver bromide produced from 22.5 g of silver nitrate in

the following reaction:

2AgNO3(aq) � MgBr2(aq) → 2AgBr(s) � Mg(NO3) 2(aq) ans: 24.9 g AgBr

2. What mass of acetylene, C2H2, will be produced from the reaction of 90. g of

calcium carbide, CaC2, with water in the following reaction?

CaC2(s) � 2H2O(l) → C2H2(g) � Ca(OH) 2(s) ans: 37 g C2H2 3. Chlorine gas can be produced in the laboratory by adding concentrated

hydrochloric acid to manganese(IV) oxide in the following reaction:

MnO2(s) � 4HCl(aq) → MnCl2(aq) � 2H2O(l) � Cl2(g) a. Calculate the mass of MnO 2 needed to produce 25.0 g of Cl 2. ans: 30.7 g

MnO 2

b. What mass of MnCl 2 is produced when 0.091 g of Cl 2 is generated?

ans: 0.16 g MnCl 2

Holt ChemFile: Problem-Solving Workbook 107 Stoichiometry

1. How many moles of ammonium sulfate can be made from the reaction of 30.0

mol of NH3 with H2SO4 according to the following equation?

2NH3 � H2SO4 → (NH4) 2SO4 2. In a very violent reaction called a thermite reaction, aluminum metal reacts

with iron(III) oxide to form iron metal and aluminum oxide according to the

following equation:

Fe2O3 � 2Al → 2Fe � Al2O3 a. What mass of Al will react with 150 g of Fe 2O 3?

b. If 0.905 mol Al2O3 is produced in the reaction, what mass of Fe is produced?

c. How many moles of Fe2O3 will react with 99.0 g of Al?

3. As you saw in Sample Problem 1, the reaction N2(g) � 3H2(g) → 2NH3(g) is

used to produce ammonia commercially. If 1.40 g of N2 are used in the reaction,

how many grams of H2 will be needed?

4. What mass of sulfuric acid, H2SO4, is required to react with 1.27 g of potassium

hydroxide, KOH? The products of this reaction are potassium sulfate

and water.

5. Ammonium hydrogen phosphate, (NH4) 2HPO4, a common fertilizer, is made

from reacting phosphoric acid, H3PO4, with ammonia.

a. Write the equation for this reaction.

b. If 10.00 g of ammonia react, how many moles of fertilizer will be produced?

c. What mass of ammonia will react with 2800 kg of H3PO4? 6. The following reaction shows the synthesis of zinc citrate, a ingredient in

toothpaste, from zinc carbonate and citric acid.

3ZnCO3(s) � 2C6H8O7(aq) → Zn3(C6H5O7) 2(aq) � 3H2O(l) � 3CO2(g) a. How many moles of ZnCO3 and C6H8O7 are required to produce 30.0 mol of

Zn3(C6H5O7) 2?

b. What quantities, in kilograms, of H2O and CO2 are produced by the reaction

of 500. mol of citric acid?

7. Methyl butanoate, an oily substance with a strong fruity fragrance, can be

made by reacting butanoic acid with methanol according to the following

equation:

C3H7COOH � CH3OH → C3H7COOCH3 � H2O a. What mass of methyl butanoate is produced from the reaction of 52.5 g of

butanoic acid?

b. In order to purify methyl butanoate, water must be removed. What mass of

water is produced from the reaction of 5800. g of methanol?

Holt ChemFile: Problem-Solving Workbook 108 Stoichiometry

8. Ammonium nitrate decomposes to yield nitrogen gas, water, and oxygen gas

in the following reaction:

2NH4NO3 → 2N2 � O2 � 4H2O a. How many moles of nitrogen gas are produced when 36.0 g of NH 4NO 3

reacts?

b. If 7.35 mol of H 2O are produced in this reaction, what mass of NH 4NO 3

reacted?

9. Lead(II) nitrate reacts with potassium iodide to produce lead(II) iodide and

potassium nitrate. If 1.23 mg of lead nitrate are consumed, what is the mass of

the potassium nitrate produced?

10. A car battery produces electrical energy with the following chemical reaction:

Pb(s) � PbO2(s) � 2H2SO4(aq) → 2PbSO4(s) � 2H2O(l) If the battery loses 0.34 kg of lead in this reaction, how many moles of lead(II)

sulfate are produced?

11. In a space shuttle, the CO2 that the crew exhales is removed from the air by a

reaction within canisters of lithium hydroxide. On average, each astronaut

exhales about 20.0 mol of CO2 daily. What mass of water will be produced

when this amount reacts with LiOH? The other product of the reaction is

Li2CO3. 12. Water is sometimes removed from the products of a reaction by placing them

in a closed container with excess P4O10. Water is absorbed by the following

P4O10 � 6H2O → 4H3PO4 a. What mass of water can be absorbed by 1.00 � 10 2 g of P 4O 10?

b. If the P4O10 in the container absorbs 0.614 mol of water, what mass of

H3PO4 is produced?

c. If the mass of the container of P4O10 increases from 56.64 g to 63.70 g, how

many moles of water are absorbed?

13. Ethanol, C2H5OH, is considered a clean fuel because it burns in oxygen to

produce carbon dioxide and water with few trace pollutants. If 95.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol

were present at the beginning of the reaction?

14. Sulfur dioxide is one of the major contributors to acid rain. Sulfur dioxide can

react with oxygen and water in the atmosphere to form sulfuric acid, as

shown in the following equation:

2H2O(l) � O2(g) � 2SO2(g) → 2H2SO4(aq) If 50.0 g of sulfur dioxide from pollutants reacts with water and oxygen found

in the air, how many grams of sulfuric acid can be produced? How many

grams of oxygen are used in the process?

Holt ChemFile: Problem-Solving Workbook 109 Stoichiometry

15. When heated, sodium bicarbonate, NaHCO3, decomposes into sodium carbonate,

Na2CO3, water, and carbon dioxide. If 5.00 g of NaHCO3 decomposes,

what is the mass of the carbon dioxide produced?

16. A reaction between hydrazine, N2H4, and dinitrogen tetroxide, N2O4, has

been used to launch rockets into space. The reaction produces nitrogen gas

and water vapor.

a. Write a balanced chemical equation for this reaction.

b. What is the mole ratio of N2O4 to N2? c. How many moles of N2 will be produced if 20 000 mol of N2H4 are used by

a rocket?

d. How many grams of H2O are made when 450. kg of N2O4 are consumed?

17. Joseph Priestley is credited with the discovery of oxygen. He produced O2 by

heating mercury(II) oxide, HgO, to decompose it into its elements. How many

moles of oxygen could Priestley have produced if he had decomposed 517.84

g of mercury oxide?

18. Iron(III) chloride, FeCl3, can be made by the reaction of iron with chlorine

gas. How much iron, in grams, will be needed to completely react with 58.0 g

of Cl2? 19. Sodium sulfide and cadmium nitrate undergo a double-displacement reaction,

as shown by the following equation:

Na2S � Cd(NO3) 2 → 2NaNO3 � CdS

What is the mass, in milligrams, of cadmium sulfide that can be made from

5.00 mg of sodium sulfide?

20. Potassium permanganate and glycerin react explosively according to the

14KMnO4 � 4C3H5(OH) 3 → 7K2CO3 � 7Mn2O3 � 5CO2 � 16H2O a. How many moles of carbon dioxide can be produced from 4.44 mol of

KMnO4? b. If 5.21 g of H2O are produced, how many moles of glycerin, C3H5(OH) 3,

were used?

c. If 3.39 mol of potassium carbonate are made, how many grams of manganese(III)

oxide are also made?

d. How many grams of glycerin will be needed to react with 50.0 g of KMnO4? How many grams of CO2 will be produced in the same reaction?

21. Calcium carbonate found in limestone and marble reacts with hydrochloric

acid to form calcium chloride, carbon dioxide, and water according to the

CaCO3(s) � 2HCl(aq) → CaCl2(aq) � CO2(g) � H2O(l) a. What mass of HCl will be needed to produce 5.00 � 10 3 kg of CaCl 2?

b. What mass of CO 2 could be produced from the reaction of 750 g of CaCO 3?

Holt ChemFile: Problem-Solving Workbook 110 Stoichiometry

22. The fuel used to power the booster rockets on the space shuttle is a mixture of

aluminum metal and ammonium perchlorate. The following balanced equation

represents the reaction of these two ingredients:

3Al(s) � 3NH4ClO4(s) → Al2O3(s) � AlCl3(g) � 3NO(g) � 6H2O(g) a. If 1.50 � 10 5 g of Al react, what mass of NH 4ClO 4, in grams, is required?

b. If aluminum reacts with 620 kg of NH4ClO4, what mass of nitrogen monoxide

is produced?

23. Phosphoric acid is typically produced by the action of sulfuric acid on rock

that has a high content of calcium phosphate according to the following

3H2SO4 � Ca3(PO4) 2 � 6H2O → 3[CaSO4�2H2O] � 2H3PO4 a. If 2.50 � 10 5 kg of H 2SO 4 react, how many moles of H 3PO 4 can be made?

b. What mass of calcium sulfate dihydrate is produced by the reaction of 400.

kg of calcium phosphate?

c. If the rock being used contains 78.8% Ca3(PO4) 2, how many metric tons of

H3PO4 can be produced from 68 metric tons of rock?

24. Rusting of iron occurs in the presence of moisture according to the following

4Fe(s) � 3O2(g) → 2Fe2O3(s) Suppose that 3.19% of a heap of steel scrap with a mass of 1650 kg rusts in a

year. What mass will the heap have after one year of rusting?

Holt ChemFile: Problem-Solving Workbook 111 Stoichiometry

Limiting Reactants

At the beginning of Chapter 8, a comparison was made between solving stoichiometry

problems and making turkey sandwiches. Look at the sandwich recipe

once more:

2 bread slices � 2 turkey slices � 1 lettuce leaf � 1 cheese slice →

1 turkey-and-cheese sandwich

If you have 24 slices of turkey, you can make 12 sandwiches at 2 slices per sandwich

if you have enough of all the other ingredients. If, however, you have only

16 slices of bread, you can make only 8 sandwiches, even though you may an

ample supply of the other ingredients. The bread is the limiting ingredient that

prevents you from making more than 8 sandwiches.

The same idea applies to chemical reactions. Look at a reaction used to generate

hydrogen gas in the laboratory:

Zn(s) � H2SO4(aq) → ZnSO4(aq) � H2(g) The balanced equation tells you that 1 mol Zn reacts with 1 mol H2SO4 to produce

1 mol ZnSO2 and 1 mol H2. Suppose you have 1 mol Zn and 5 mol H2SO4. What will happen, and what will you get? Only 1 mol of H2SO4 will react and only

1 mol of each of the products will be produced because only 1 mol Zn is available

to react. In this situation, zinc is the limiting reactant. When it is used up the reaction

stops even though more H2SO4 is available.

It is difficult to directly observe molar amounts of reactants as they are used

up. It is much easier to determine when a certain mass of a reactant has been

completely used. Use molar masses to restate the equation in terms of mass, as

follows:

65.39 g Zn � 98.09 g H2SO4 → 161.46 g ZnSO4 � 2.02 g H2 This version of the equation tells you that zinc and sulfuric acid will always react

in a mass ratio of 65.39 g of Zn:98.09 g of H 2SO 4 or 0.667 g of Zn:1.000 g of

H 2SO 4. If you have 65.39 g of Zn but only 87.55 g of H 2SO 4, you will not be able

to make 2.02 g of hydrogen. Sulfuric acid will be the limiting reactant, preventing

the zinc from reacting completely. Suppose you place 20 g of zinc and 100 g of

sulfuric acid into a flask. Which would be used up first? In other words, is the

limiting reactant zinc or sulfuric acid? How much of each product will be produced?

The sample problems in this chapter will show you how to answer these

questions.

Holt ChemFile: Problem-Solving Workbook 112 Limiting Reactants

General Plan for Solving Limiting Reactant Problems

1A

reactant A

available

2A

Convert

using the

in mol

reactant B

needed to

react with A

6

product

using the mole

ratio, .

A

If there are more moles of B

available than needed,

A is the limiting reactant.

If there are fewer moles of B

than needed, B is the

limiting reactant.

of B.

mole ratio,

limiting reactant .

1B

Holt ChemFile: Problem-Solving Workbook 113 Limiting Reactants

2B

Limiting

Reactant

Convert using the molar

mass of the product. 5

product in mol

Calcium hydroxide, used to neutralize acid spills, reacts with hydrochloric

acid according to the following equation:

Ca(OH) 2 � 2HCl → CaCl2 � 2H2O If you have spilled 6.3 mol of HCl and put 2.8 mol of Ca(OH) 2 on it,

which substance is the limiting reactant?

What is given in the problem? the balanced equation, the amounts of Ca(OH) 2

and HCl in moles

What are you asked to find? the limiting reactant

Reactant Ca(OH) 2 HCl

Amount of reactant 2.8 mol 6.3 mol

Mass of reactant NA NA

Limiting reactant ? ?

* not applicable to the problem

What steps are needed to determine the limiting reactant?

Choose one of the reactants. Use the mole ratio between the two reactants to

compute the amount of the other reactant that would be needed to react with it.

Compare that amount with the amount available.

Amount of Ca(OH) 2

HCl

Ca(OH) 2

2A 2B

Amount of HCl needed to

react with Ca(OH) 2

compare moles of HCl

needed with moles of HCl

Choose one of the reactants, for instance, Ca(OH) 2

Amount of HCl

Limiting reactant

Holt ChemFile: Problem-Solving Workbook 114 Limiting Reactants

The computation shows that more HCl (6.3 mol) is available than is needed (5.6

mol) to react with the 2.8 mol Ca(OH) 2 available. Therefore, HCl is present in

excess, making Ca(OH) 2 the limiting reactant.

Yes; you can see that 6.3 mol HCl is more than is needed to react with 2.8 mol

Ca(OH) 2.

given 2 mol HCl

mol Ca(OH) 2 � � mol HCl needed

1 mol Ca(OH) 2

2 mol HCl

2.8 mol Ca(OH) 2 � � 5.6 mol HCl needed

1. Aluminum oxidizes according to the following equation:

4Al � 3O2 → 2Al2O3 Powdered Al (0.048 mol) is placed into a container containing 0.030 mol O 2.

What is the limiting reactant? ans: O 2

Holt ChemFile: Problem-Solving Workbook 115 Limiting Reactants

Chlorine can replace bromine in bromide compounds forming a chloride

compound and elemental bromine. The following equation is an example

of this reaction.

2KBr(aq) � Cl2(aq) → 2KCl(aq) � Br2(l) When 0.855 g of Cl2 and 3.205 g of KBr are mixed in solution, which is

the limiting reactant? How many grams of Br2 are formed?

What is given in the problem? the balanced equation, and the masses of Cl2 and KBr available

What are you asked to find? which reactant is limiting, and the mass of Br2 produced

Substance KBr Cl2 Br2 Coefficient in balanced equation 2 1 1

Molar mass* 119.00 g/mol 70.90 g/mol 159.80 g/mol

Amount of substance ? mol ? mol ? mol

Mass of substance 3.205 g 0.855 g ? g

Limiting reactant ? ? NA

Convert mass of each reactant to amount in moles. Choose one of the reactants.

Compute the amount of the other reactant needed. Compare that with the

amount available.

What steps are needed to determine the mass of Br2 produced in the reaction?

Use amount of the limiting reactant and the mole ratio given in the equation to

determine the amount of Br2. Convert the amount of Br2 to the mass of Br2 using the molar mass.

Holt ChemFile: Problem-Solving Workbook 116 Limiting Reactants

given 1 mol Cl2 g Cl2 � � mol Cl2 70.90 g Cl2 Choose one of the reactants, KBr for instance.

Determine the limiting reactant.

1a 1b

Mass of KBr in g

molar mass of KBr

2a

Amount of KBr in mol

Cl2 KBr

Amount of Cl2 needed

to react with KBr

Mass of Br2 in g

1 mol Cl2 mol KBr � � mol Cl2 needed

1 mol KBr

mol Br2 159.80 g Br2 mol limiting reactant � �

� g Br2 mol limiting reactant 1 mol Br2 3.205 g KBr �

compare moles of reactant

needed with moles of

reactant available

multiply by the molar mass of

Br 2

molar mass KBr

given 1 mol KBr

g KBr � � mol KBr

119.00 g KBr

molar mass Cl2 mole ratio

0.855 g Cl2 � 1 mol Cl2 � 0.0121 mol Cl2 70.90 g Cl2 Choose one of the reactants, KBr, for instance.

0.02693 mol KBr � 1 mol Cl 2

2 mol KBr � 0.01346 mol Cl 2 needed

Mass of Cl2 in g

molar mass of Cl2 multiply moles of limiting reactant

Br2 by mole ratio:

limiting reactant

2b

Amount of Cl2 in mol

Amount of Br2 in mol

molar mass Br 2

� 0.02693 mol KBr

Holt ChemFile: Problem-Solving Workbook 117 Limiting Reactants

Only 0.0121 mol Cl 2 is available. For all of the KBr to react, 0.0136 mol Cl 2 is

needed. Therefore, Cl 2 is the limiting reactant.

Is the determination of limiting reactant reasonable?

Yes; the mass of 2 mol KBr is 238 g and the mass of 1 mol Cl2 is about 71 g, so

they react in roughly a 3:1 ratio by mass. 3.2 g KBr would require about 1 g of

Cl2, but only 0.855 g is available.

Are the units and significant figures of the mass of Br2 correct?

The number of significant figures is correct because the mass of Cl2 was given to

three significant figures. Units cancel to give grams of Br2. Practice

0.0121 mol Cl2 � 1 mol Br2 �

1 mol Cl2 159.80 g Br2 � 1.93 g Br2 1 mol Br2 1. A process by which zirconium metal can be produced from the mineral zirconium(IV)

orthosilicate, ZrSiO4, starts by reacting it with chlorine gas to form

zirconium(IV) chloride.

ZrSiO4 � 2Cl2 → ZrCl4 � SiO2 � O2 What mass of ZrCl 4 can be produced if 862 g of ZrSiO 4 and 950. g of Cl 2 are

available? You must first determine the limiting reactant. ans: ZrSiO 4, 1.10 �

10 3 g ZrCl 4

Holt ChemFile: Problem-Solving Workbook 118 Limiting Reactants

1. Heating zinc sulfide in the presence of oxygen yields the following:

ZnS � O2 → ZnO � SO2 If 1.72 mol of ZnS is heated in the presence of 3.04 mol of O2, which reactant

will be used up? Balance the equation first.

2. Use the following equation for the oxidation of aluminum in the following

problems.

4Al � 3O2 → 2Al2O3 a. Which reactant is limiting if 0.32 mol Al and 0.26 mol O 2 are available?

b. How many moles of Al 2O 3 are formed from the reaction of 6.38 � 10 �3 mol

of O 2 and 9.15 � 10 �3 mol of Al?

c. If 3.17 g of Al and 2.55 g of O 2 are available, which reactant is limiting?

3. In the production of copper from ore containing copper(II) sulfide, the ore is

first roasted to change it to the oxide according to the following equation:

2CuS � 3O2 → 2CuO � 2SO2 a. If 100 g of CuS and 56 g of O 2 are available, which reactant is limiting?

b. What mass of CuO can be formed from the reaction of 18.7 g of CuS and

12.0 g of O2? 4. A reaction such as the one shown here is often used to demonstrate a single

replacement reaction.

3CuSO4(aq) � 2Fe(s) → 3Cu(s) � Fe2(SO4) 3(aq)

If you place 0.092 mol of iron filings in a solution containing 0.158 mol of

CuSO4, what is the limiting reactant? How many moles of Cu will be formed?

5. In the reaction BaCO3 � 2HNO3 → Ba(NO3) 2 � CO2 � H2O, what mass of

Ba(NO3) 2 can be formed by combining 55 g BaCO3 and 26 g HNO3? 6. Bromine displaces iodine in magnesium iodide by the following process:

MgI2 � Br2 → MgBr2 � I2 a. Which is the excess reactant when 560 g of MgI2 and 360 g of Br2 react, and

what mass remains?

b. What mass of I2 is formed in the same process?

7. Nickel displaces silver from silver nitrate in solution according to the following

2AgNO3 � Ni → 2Ag � Ni(NO3) 2

a. If you have 22.9 g of Ni and 112 g of AgNO3, which reactant is in excess?

b. What mass of nickel(II) nitrate would be produced given the quantities

above?

Holt ChemFile: Problem-Solving Workbook 119 Limiting Reactants

8. Carbon disulfide, CS2, is an important industrial substance. Its fumes can

burn explosively in air to form sulfur dioxide and carbon dioxide.

CS2(g) � O2(g) → SO2(g) � CO2(g) If 1.60 mol of CS2 burns with 5.60 mol of O2, how many moles of the excess

reactant will still be present when the reaction is over?

9. Although poisonous, mercury compounds were once used to kill bacteria in

wounds and on the skin. One was called “ammoniated mercury” and is made

from mercury(II) chloride according to the following equation:

HgCl2(aq) � 2NH3(aq) → Hg(NH2)Cl(s) � NH4Cl(aq) a. What mass of Hg(NH2)Cl could be produced from 0.91 g of HgCl2 assuming

plenty of ammonia is available?

b. What mass of Hg(NH2)Cl could be produced from 0.91 g of HgCl2 and 0.15 g

of NH3 in solution?

10. Aluminum chips are sometimes added to sodium hydroxide-based drain

cleaners because they react to generate hydrogen gas which bubbles and

helps loosen material in the drain. The equation follows.

Al(s) � NaOH(aq) � H2O(l) → NaAlO2(aq) � H2(g) a. Balance the equation.

b. How many moles of H2 can be generated from 0.57 mol Al and 0.37 mol

NaOH in excess water?

c. Which reactant should be limiting in order for the mixture to be most

effective as a drain cleaner? Explain your choice.

11. Copper is changed to copper(II) ions by nitric acid according to the following

4HNO3 � Cu → Cu(NO3) 2 � 2NO2 � 2H2O a. How many moles each of HNO3 and Cu must react in order to produce

0.0845 mol of NO2? b. If 5.94 g of Cu and 23.23 g of HNO3 are combined, which reactant is in

excess?

12. One industrial process for producing nitric acid begins with the following

4NH3 � 5O2 → 4NO � 6H2O a. If 2.90 mol NH3 and 3.75 mol O2 are available, how many moles of each

product are formed?

b. Which reactant is limiting if 4.20 � 10 4 g of NH3 and 1.31 � 10 5 g of O2 are

available?

c. What mass of NO is formed in the reaction of 869 kg of NH3 and 2480 kg

Holt ChemFile: Problem-Solving Workbook 120 Limiting Reactants

13. Acetaldehyde CH3CHO is manufactured by the reaction of ethanol with

copper(II) oxide according to the following equation:

CH3CH2OH � CuO → CH3CHO � H2O � Cu

What mass of acetaldehyde can be produced by the reaction between 620 g of

ethanol and 1020 g of CuO? What mass of which reactant will be left over?

14. Hydrogen bromide can be produced by a reaction among bromine, sulfur

dioxide, and water as follows.

SO2 � Br2 � H2O → 2HBr � H2SO4 If 250 g of SO2 and 650 g of Br2 react in the presence of excess water, what

mass of HBr will be formed?

15. Sulfur dioxide can be produced in the laboratory by the reaction of

hydrochloric acid and a sulfite salt such as sodium sulfite.

Na2SO3 � 2HCl → 2NaCl � SO2 � H2O What mass of SO2 can be made from 25.0 g of Na2SO3 and 22.0 g of HCl?

16. The rare-earth metal terbium is produced from terbium(III) fluoride and

calcium metal by the following displacement reaction:

2TbF3 � 3Ca → 3CaF2 � 2Tb

a. Given 27.5 g of TbF3 and 6.96 g of Ca, how many grams of terbium could be

produced?

b. How many grams of the excess reactant are left over?

Holt ChemFile: Problem-Solving Workbook 121 Limiting Reactants

Percentage Yield

Although we can write perfectly balanced equations to represent perfect reactions,

the reactions themselves are often not perfect. A reaction does not always

produce the quantity of products that the balanced equation seems to guarantee.

This happens not because the equation is wrong but because reactions in the real

world seldom produce perfect results.

As an example of an imperfect reaction, look again at the equation that shows

the industrial production of ammonia.

N2(g) � 3H2(g) → 2NH3(g) In the manufacture of ammonia, it is nearly impossible to produce 2 mol (34.08 g)

of NH3 from the simple reaction of 1 mol (28.02 g) of N2 and 3 mol (6.06 g) of H2 because some ammonia molecules begin breaking down into N2 and H2 molecules

as soon as they are formed.

There are several reasons that real-world reactions do not produce products at

a yield of 100%. Some are simple mechanical reasons, such as:

• Reactants or products leak out, especially when they are gases.

• The reactants are not 100% pure.

• Some product is lost when it is purified.

There are also many chemical reasons, including:

• The products decompose back into reactants (as with the ammonia process).

• The products react to form different substances.

• Some of the reactants react in ways other than the one shown in the equation.

These are called side reactions.

• The reaction occurs very slowly. This is especially true of reactions involving

organic substances.

Chemists are very concerned with the yields of reactions because they must

find ways to carry out reactions economically and on a large scale. If the yield of

a reaction is too small, the products may not be competitive in the marketplace.

If a reaction has only a 50% yield, it produces only 50% of the amount of product

that it theoretically should. In this chapter, you will learn how to solve problems

involving real-world reactions and percentage yield.

Holt ChemFile: Problem-Solving Workbook 122 Percentage Yield

General Plan for Solving Percentage-Yield Problems

reactants Convert using reactants Convert using

in mol the mole

mass of the

ratio of the

reactants.

limiting

reactant to

the product.

yield

actual yield

% yield � � 100

theoretical yield

Theoretical

amount of

product in mol

of the

product.

Holt ChemFile: Problem-Solving Workbook 123 Percentage Yield

Actual

Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating

agent in research. It can be produced by passing chlorine gas over heated

mercury(II) oxide according to the following equation:

HgO � Cl2 → HgCl2 � Cl2O What is the percentage yield, if the quantity of reactants is sufficient to

produce 0.86 g of Cl2O but only 0.71 g is obtained?

What is given in the problem? the balanced equation, the actual yield of Cl2O, and the theoretical yield of Cl2O What are you asked to find? the percentage yield of Cl2O Items Data

Substance Cl 2O

Mass available NA*

Molar mass NA

Amount of reactant NA

Coefficient in balanced equation NA

Actual yield 0.71 g

Theoretical yield (moles) NA

Theoretical yield (grams) 0.86 g

Percentage yield ?

* Although this table has many Not Applicable entries,

you will need much of this information in other kinds of

percentage-yield problems.

What steps are needed to calculate the percentage yield of Cl 2O?

Compute the ratio of the actual yield to the theoretical yield, and multiply by

100 to convert to a percentage.

Holt ChemFile: Problem-Solving Workbook 124 Percentage Yield

0.

71

g Cl2O

� 100 � 83% yield

86

g Cl

O

Yes; the ratio was converted to a percentage.

two significant figures.

Yes; 83% is about 5/6, which appears to be close to the ratio 0.71/0.86.

Theoretical mass of Cl2O in g

Actual mass of Cl2O in g

actual mass

theoretical mass

g Cl 2 O produced � 100 � percentage yield

theoretical g Cl 2 O

Percentage yield of Cl2O 1. Calculate the percentage yield in each of the following cases:

a. theoretical yield is 50.0 g of product; actual yield is 41.9 g ans: 83.8% yield

b. theoretical yield is 290 kg of product; actual yield is 270 kg ans: 93% yield

Holt ChemFile: Problem-Solving Workbook 125 Percentage Yield

c. theoretical yield is 6.05 � 10 4 kg of product; actual yield is 4.18 � 10 4 kg

ans: 69.1% yield

d. theoretical yield is 0.00192 g of product; actual yield is 0.00089 g ans: 46%

Holt ChemFile: Problem-Solving Workbook 126 Percentage Yield

Acetylene, C 2H 2, can be used as an industrial starting material for the

production of many organic compounds. Sometimes, it is first brominated

to form 1,1,2,2-tetrabromoethane, CHBr 2CHBr 2, which can then be

reacted in many different ways to make other substances. The equation

for the bromination of acetylene follows:

acetylene

1,1,2,2-tetrabromoethane

C2H2� 2Br2 → CHBr2CHBR2 If 72.0 g of C 2H 2 reacts with excess bromine and 729 g of the product

is recovered, what is the percentage yield of the reaction?

What is given in the problem? the balanced equation, the mass of acetylene

that reacts, and the mass of tetrabromoethane

produced

What are you asked to find? the percentage yield of tetrabromoethane

Substance C2H2 CHBr2CHBr2 Mass available 72.0 g available NA

Molar mass* 26.04 g/mol 345.64 g/mol

Amount of reactant ? NA

Coefficient in balanced equation 1 1

Actual yield NA 729 g

Theoretical yield (moles) NA ?

Theoretical yield (grams) NA ?

Percentage yield NA ?

What steps are needed to calculate the theoretical yield of tetrabromoethane?

Set up a stoichiometry calculation to find the amount of product that can be

formed from the given amount of reactant.

What steps are needed to calculate the percentage yield of tetrabromoethane?

Holt ChemFile: Problem-Solving Workbook 127 Percentage Yield

Mass of C2H2 in g

Percentage yield

CHBr 2 CHBr 2

molar mass C 2 H 2

of the molar mass of C 2 H 2

72.0 g C2H2 �

729 g CHBr2CHBr2 �� 100 � 76.3% yield

956 g CHBr2CHBr2 1 mol C2H2 �

26.04 g C2H2 1 mol CHBr2CHBr2 �

1 mol C2H2 345.64 g CHBr2CHBr2 1 mol CHBr2CHBr2 � 956 g CHBr2CHBr2 EVALUATE

Yes; units canceled to give grams of CHBr2CHBr2. Also, the ratio was converted

to a percentage.

Yes; about 3 mol of acetylene were used and the theoretical yield is the mass of

about 3 mol tetrabromoethane.

Amount of C2H2 in mol

CHBr2CHBr2 C2H2 3

Theoretical amount of

CHBr2CHBr2 in mol

multiply by the molar

mass of CHBr2CHBr2 4

Theoretical mass of

CHBr2CHBr2 5

Actual mass of

CHBr2CHBr2 molar mass CHBr 2 CHBr 2

given 1 mol C2H2 1 mol CHBr2CHBr2 345.64 g CHBr2CHBr2 g C2H2 � � �

26.04 g C2H2 1 mol C2H2 1 mol CHBr2CHBr2 � theoretical g CHBr2CHBr2 actual grams

theoretical grams

g CHBr2CHBr2 produced

� 100 � percentage yield CHBr2CHBr2 theoretical g CHBr2CHBr2 Holt ChemFile: Problem-Solving Workbook 128 Percentage Yield

1. In the commercial production of the element arsenic, arsenic(III) oxide is

heated with carbon, which reduces the oxide to the metal according to the

2As2O3 � 3C → 3CO2 � 4As

a. If 8.87 g of As 2O 3 is used in the reaction and 5.33 g of As is produced, what

is the percentage yield? ans: 79.3% yield

b. If 67 g of carbon is used up in a different reaction and 425 g of As is produced,

calculate the percentage yield of this reaction. ans: 76% yield

Holt ChemFile: Problem-Solving Workbook 129 Percentage Yield

1. Ethyl acetate is a sweet-smelling solvent used in varnishes and fingernailpolish

remover. It is produced industrially by heating acetic acid and ethanol

together in the presence of sulfuric acid, which is added to speed up the

reaction. The ethyl acetate is distilled off as it is formed. The equation for the

process is as follows.

acetic acid ethanol H2SO ethyl acetate

CH3COOH � CH3CH2OH CH3COOCH2CH3 � H2O Determine the percentage yield in the following cases:

a. 68.3 g of ethyl acetate should be produced but only 43.9 g is recovered.

b. 0.0419 mol of ethyl acetate is produced but 0.0722 mol is expected. (Hint:

Percentage yield can also be calculated by dividing the actual yield in moles

by the theoretical yield in moles.)

c. 4.29 mol of ethanol is reacted with excess acetic acid, but only 2.98 mol of

ethyl acetate is produced.

d. A mixture of 0.58 mol ethanol and 0.82 mol acetic acid is reacted and 0.46

mol ethyl acetate is produced. (Hint: What is the limiting reactant?)

2. Assume the following hypothetical reaction takes place.

2A � 7B → 4C � 3D

Calculate the percentage yield in each of the following cases:

a. The reaction of 0.0251 mol of A produces 0.0349 mol of C.

b. The reaction of 1.19 mol of A produces 1.41 mol of D.

c. The reaction of 189 mol of B produces 39 mol of D.

d. The reaction of 3500 mol of B produces 1700 mol of C.

3. Elemental phosphorus can be produced by heating calcium phosphate from

rocks with silica sand (SiO2) and carbon in the form of coke. The following

reaction takes place.

Ca3(PO4) 2 � 3SiO2 � 5C → 3CaSiO3 � 2P � 5CO

a. If 57 mol of Ca3(PO4) 2 is used and 101 mol of CaSiO3 is obtained, what is

the percentage yield?

b. Determine the percentage yield obtained if 1280 mol of carbon is consumed

and 622 mol of CaSiO3 is produced.

c. The engineer in charge of this process expects a yield of 81.5%. If 1.4 � 10 5

mol of Ca3(PO4) 2 is used, how many moles of phosphorus will be produced?

4. Tungsten (W) can be produced from its oxide by reacting the oxide with

hydrogen at a high temperature according to the following equation:

WO3 � 3H2 → W � 3H2O a. What is the percentage yield if 56.9 g of WO 3 yields 41.4 g of tungsten?

Holt ChemFile: Problem-Solving Workbook 130 Percentage Yield

b. How many moles of tungsten will be produced from 3.72 g of WO3 if the

yield is 92.0%?

c. A chemist carries out this reaction and obtains 11.4 g of tungsten. If the

percentage yield is 89.4%, what mass of WO3 was used?

5. Carbon tetrachloride, CCl4, is a solvent that was once used in large quantities

in dry cleaning. Because it is a dense liquid that does not burn, it was also

used in fire extinguishers. Unfortunately, its use was discontinued because it

was found to be a carcinogen. It was manufactured by the following reaction:

CS2 � 3Cl2 → CCl4 � S2Cl2 The reaction was economical because the byproduct disulfur dichloride,

S2Cl2, could be used by industry in the manufacture of rubber products and

other materials.

a. What is the percentage yield of CCl4 if 719 kg is produced from the reaction

of 410. kg of CS2. b. If 67.5 g of Cl2 are used in the reaction and 39.5 g of S2Cl2 is produced,

what is the percentage yield?

c. If the percentage yield of the industrial process is 83.3%, how many kilograms

of CS2 should be reacted to obtain 5.00 � 10 4 kg of CCl4? How many

kilograms of S2Cl2 will be produced, assuming the same yield for that

product?

6. Nitrogen dioxide, NO2, can be converted to dinitrogen pentoxide, N2O5, by

reacting it with ozone, O3. The reaction of NO2 takes place according to the

2NO2(g) � O3(g) → N2O5(s or g) � O2(g) a. Calculate the percentage yield for a reaction in which 0.38 g of NO2 reacts

and 0.36 g of N2O5 is recovered.

b. What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there

is a 61.1% yield in the reaction?

7. In the past, hydrogen chloride, HCl, was made using the salt-cake method as

2NaCl(s) � H2SO4(aq) → Na2SO4(s) � 2HCl(g)

If 30.0 g of NaCl and 0.250 mol of H2SO4 are available, and 14.6 g of HCl is

made, what is the percentage yield?

8. Cyanide compounds such as sodium cyanide, NaCN, are especially useful in

gold refining because they will react with gold to form a stable compound

that can then be separated and broken down to retrieve the gold. Ore containing

only small quantities of gold can be used in this form of “chemical mining.”

The equation for the reaction follows.

4Au � 8NaCN � 2H2O � O2 → 4NaAu(CN) 2 � 4NaOH

a. What percentage yield is obtained if 410 g of gold produces 540 g of

NaAu(CN) 2?

Holt ChemFile: Problem-Solving Workbook 131 Percentage Yield

b. Assuming a 79.6% yield in the conversion of gold to NaAu(CN) 2, what mass

of gold would produce 1.00 kg of NaAu(CN) 2?

c. Given the conditions in (b), what mass of gold ore that is 0.001% gold

would be needed to produce 1.00 kg of NaAu(CN) 2?

9. Diiodine pentoxide is useful in devices such as respirators because it reacts

with the dangerous gas carbon monoxide, CO, to produce relatively harmless

CO2 according to the following equation:

I2O5 � 5CO → I2 � 5CO2 a. In testing a respirator, 2.00 g of carbon monoxide gas is passed through

diiodine pentoxide. Upon analyzing the results, it is found that 3.17 g of I2 was produced. Calculate the percentage yield of the reaction.

b. Assuming that the yield in (a) resulted because some of the CO did not

react, calculate the mass of CO that passed through.

10. Sodium hypochlorite, NaClO, the main ingredient in household bleach, is

produced by bubbling chlorine gas through a strong lye (sodium hydroxide,

NaOH) solution. The following equation shows the reaction that occurs.

2NaOH(aq) � Cl2(g) → NaCl(aq) � NaClO(aq) � H2O(l) a. What is the percentage yield of the reaction if 1.2 kg of Cl2 reacts to form

0.90 kg of NaClO?

b. If a plant operator wants to make 25 metric tons of NaClO per day at a

yield of 91.8%, how many metric tons of chlorine gas must be on hand each

day?

c. What mass of NaCl is formed per mole of chlorine gas at a yield of 81.8%?

d. At what rate in kg per hour must NaOH be replenished if the reaction

produces 370 kg/h of NaClO at a yield of 79.5%? Assume that all of the

NaOH reacts to produce this yield.

11. Magnesium burns in oxygen to form magnesium oxide. However, when magnesium

burns in air, which is only about 1/5 oxygen, side reactions form other

products, such as magnesium nitride, Mg3N2. a. Write a balanced equation for the burning of magnesium in

oxygen.

b. If enough magnesium burns in air to produce 2.04 g of magnesium oxide

but only 1.79 g is obtained, what is the percentage yield?

c. Magnesium will react with pure nitrogen to form the nitride, Mg3N2. Write a

balanced equation for this reaction.

d. If 0.097 mol of Mg react with nitrogen and 0.027 mol of Mg3N2 is produced,

what is the percentage yield of the reaction?

Holt ChemFile: Problem-Solving Workbook 132 Percentage Yield

12. Some alcohols can be converted to organic acids by using sodium dichromate

and sulfuric acid. The following equation shows the reaction of 1-propanol to

propanoic acid.

3CH 3CH2CH2OH � 2Na2Cr2O7 � 8H2SO4 →

3CH 3CH2COOH � 2Cr2(SO4) 3 � 2Na2SO4 � 11H2O a. If 0.89 g of 1-propanol reacts and 0.88 g of propanoic acid is produced, what

is the percentage yield?

b. A chemist uses this reaction to obtain 1.50 mol of propanoic acid. The

reaction consumes 136 g of propanol. Calculate the percentage yield.

c. Some 1-propanol of uncertain purity is used in the reaction. If 116 g of

Na2Cr2O7 are consumed in the reaction and 28.1 g of propanoic acid are

produced, what is the percentage yield?

13. Acrylonitrile, C3H3N(g), is an important ingredient in the production of various

fibers and plastics. Acrylonitrile is produced from the following reaction:

C3H6(g) � NH3(g) � O2(g) → C3H3N(g) � H2O(g) If 850. g of C3H6 is mixed with 300. g of NH3 and unlimited O2, to produce 850. g

of acrylonitrile, what is the percentage yield? You must first balance the equation.

14. Methanol, CH3OH, is frequently used in race cars as fuel. It is produced as the

sole product of the combination of carbon monoxide gas and hydrogen gas.

a. If 430. kg of hydrogen react, what mass of methanol could be produced?

b. If 3.12 � 10 3 kg of methanol are actually produced, what is the percentage

yield?

15. The compound, C6H16N2, is one of the starting materials in the production of

nylon. It can be prepared from the following reaction involving adipic acid,

C6H10O4: C6H10O4(l) � 2NH3(g) � 4H2(g) → C6H16N2(l) � 4H2O(l) What is the percentage yield if 750. g of adipic acid results in the production

of 578 g of C6H16N2? 16. Plants convert carbon dioxide to oxygen during photosynthesis according to

the following equation:

CO2 � H2O → C6H12O6 � O2 Balance this equation, and calculate how much oxygen would be produced if

1.37 � 10 4 g of carbon dioxide reacts with a percentage yield of 63.4%.

17. Lime, CaO, is frequently added to streams and lakes which have been polluted

by acid rain. The calcium oxide reacts with the water to form a base that can

neutralize the acid as shown in the following reaction:

CaO(s) � H2O(l) → Ca(OH) 2(s)

If 2.67 � 10 2 mol of base are needed to neutralize the acid in a lake, and the

above reaction has a percentage yield of 54.3%, what is the mass, in kilograms,

of lime that must be added to the lake?

Holt ChemFile: Problem-Solving Workbook 133 Percentage Yield

Thermochemistry

Thermochemistry deals with the changes in energy that accompany a chemical

reaction. Energy is measured in a quantity called enthalpy, represented as H. The

change in energy that accompanies a chemical reaction is represented as �H.

Hess’s law provides a method for calculating the �H of a reaction from tabulated

data. This law states that if two or more chemical equations are added, the �H of

the individual equations may also be added to find the �H of the final equation.

As an example of how this law operates, look at the three reactions below.

(1) 2H2(g) � O2(g) → 2H2O(l) �H ��571.6 kJ/mol

(2) 2H2O2(l) → 2H2(g) � 2O2(g) �H ��375.6 kJ/mol

(3) 2H2O2(l) → 2H2O(l) � O2(g) �H � ? kJ/mol

When adding equations 1 and 2, the 2 mol of H 2(g) will cancel each other out,

while only 1 mol of O 2(g) will cancel.

2H2(g) � O2(g) 3 2H2O(l) 1

2H2O2(l) 3 2H2(g) � 2O2(g) Combining what is left yields the following equation.

2H2O2(l) → 2H2O(l) � O2(g) Notice that this is the same equation as the third equation shown above. Adding

the two �H values for the reactions 1 and 2 gives the �H value for reaction 3.

Using Hess’s law to calculate the enthalpy of this reaction, the following answer

is obtained.

�571.6 kJ/mol � 375.6 kJ/mol ��196.0 kJ/mol

Thus, the �H value for the reaction is �196.0 kJ/mol.

Equation 1 represents the formation of water from its elemental components.

If equation 2 were written in reverse, it would represent the formation of hydrogen

peroxide from its elemental components. Therefore, adding equations 1 and 2

is the equivalent of subtracting the equation for the formation of the reactants of

equation 3 from the equation for the formation of the products of equation 3.

2H2(g) � O2(g) → 2H2O(l) �[2H2(g) � 2O2(g) → 2H2O2(l)] 2H2O2(l) → 2H2O(l) � O2(g) The enthalpy of the final reaction can be rewritten using the following equation.

�Hreaction � sum of �H 0 f � sum of �H products 0 freactants The equation states that the enthalpy change of a reaction is equal to the sum

of the enthalpies of formation of the products minus the sum of the enthalpies of

formation of the reactants. This allows Hess’s law to be extended to state that the

Holt ChemFile: Problem-Solving Workbook 134 Thermochemistry

enthalpy change of any reaction can be calculated by looking up the standard

molar enthalpy of formation, �H 0 f, of each substance involved. Some common

enthalpies of formation may be found in Table 1.

The enthalpy change, however, does not account for all of the energy change

of a reaction. Changes in the disorder (entropy) of a system can add to or detract

from the energy involved in the enthalpy change. This amount of energy is given

by the expression T�S, where T is the Kelvin temperature and �S is the change in

entropy during the reaction. A large increase in entropy, such as when a gas is

produced from a reaction of liquids or solids, can contribute significantly to the

overall energy change. The total amount of energy available from a reaction is

called Gibbs energy and is denoted by �G. Gibbs energy is given by the following

equation.

�G reaction ��H reaction � T�S reaction

TABLE 1 STANDARD ENTHALPIES OF FORMATION

Substance �H 0 f (kJ/mol) Substance �H 0 f (kJ/mol)

NH 3(g) �45.9 HF(g) �273.3

NH 4Cl(s) �314.4 H 2O(g) �241.82

NH 4F(s) �125 H 2O(l) �285.8

NH 4NO 3(s) �365.56 H 2O 2(l) �187.8

Br 2(l) 0.00 H 2SO 4(l) �813.989

CaCO 3(s) �1207.6 FeO(s) �825.5

CaO(s) �634.9 Fe 2O 3(s) �1118.4

CH 4(g) �74.9 MnO 2(s) �520.0

C 3H 8(g) �104.7 N 2O(g) �82.1

CO 2(g) �393.5 O 2(g) 0.00

F 2(g) 0.00 Na 2O(s) �414.2

H 2(g) 0.00 Na 2SO 3(s) �1101

HBr(g) �36.29 SO2(g) �296.8

HCl(g) �92.3 SO3(g) �395.7

Holt ChemFile: Problem-Solving Workbook 135 Thermochemistry

General Plan for Solving Thermochemistry Problems

sum of �H0 �H reaction

fproducts

� sum of �H0

freactants � �H reaction

Look up �H0 f for each reactant and

product. Be sure the physical states

in the reaction match those given in

the reference table. Solve for

�Hreaction. T �S reaction

�H reaction � T�S reaction � �G reaction

Holt ChemFile: Problem-Solving Workbook 136 Thermochemistry

Given the following two reactions and enthalpy data, calculate the

enthalpy change for the reaction in which methane and oxygen combine

to form ketene, CH2CO, and water.

CH2CO(g) � 2O2(g) 3 2CO2(g) � H2O(g) �H ��981.1 kJ

CH4(g) � 2O2(g) 3 CO2(g) � 2H2O(g) �H ��802.3 kJ

What is given in the problem? the desired product, chemical equations that

can be added to obtain the desired product, and

enthalpy changes for these chemical equations

What are you asked to find? enthalpy change for the reaction in which

methane and oxygen combine to form ketene

and water

�H for reaction 1 �981.1 kJ

�H for reaction 2 �802.3 kJ

What steps are needed to calculate �H for the reaction between methane and

oxygen to form CH2CO? First, the two equations must be added to produce the final reaction. The first

equation must be reversed so that ketene is a product, as shown in the final

equation. The second equation must be multiplied by 2 so that carbon dioxide

cancels out of the final equation. Then the individual enthalpies for the reactions

must be added, adjusting for the fact that equation 1 is reversed and

equation 2 is doubled.

2CO2(g) � H2O(g) → CH2CO(g) � 2O2(g) 2 � [CH4(g) � 2O2(g) → CO2(g) � 2H2O(g)] 2CH4(g) � 2O2(g) → CH2CO(g) � 3H2O(g) ��Hreaction 1

�(2 ��Hreaction 2)

�H final reaction

Holt ChemFile: Problem-Solving Workbook 137 Thermochemistry

Yes; adding terms in kilojoules gives an answer in kilojoules.

Yes; the significant figures are correct. Rules for adding and rounding measurements

give a result to four significant figures.

Yes; the result can be approximated as �1600 kJ � 1000 kJ ��600 kJ.

2CO2(g) � H2O(g) 3 CH2CO(g) � 2O2(g) 2CH4(g) � 4O2(g) 3 2CO2(g) � 4H2O(g) 2

2CO2(g) � H2O(g) � 2CH4(g) � 4O2(g) 3

CH2CO(g) � 2O2(g) � 2CO2(g) � 4H2O(g) 2CH4(g) � 2O2(g) 3 CH2CO(g) � 3H2O(g) �(�981.1 kJ)

�(2 � �802.3 kJ)

�623.5 kJ

1. Calculate the reaction enthalpy for the following reaction.

5CO2(g) � Si3N4(s) → 3SiO(s) � 2N2O(g) � 5CO(g)

Use the following equations and data.

(1) CO(g) � SiO2(s) → SiO(g) � CO2(g) (2) 8CO2(g) � Si3N4(s) → 3SiO2(s) � 2N2O(g) � 8CO(g)

�Hreaction 1 ��520.9 kJ

�Hreaction 2 ��461.05 kJ ans: 2024 kJ

Holt ChemFile: Problem-Solving Workbook 138 Thermochemistry

Calculate the enthalpy of reaction for the decomposition of hydrogen

peroxide to water and oxygen gas according to the following equation.

H2O2(l) 3 H2O(l) � O2(g) Use Table 1 for the necessary enthalpies of formation.

What is given in the problem? the equation for the decomposition of H2O2, enthalpies of formation given in Table 1

What are you asked to find? enthalpy of reaction for the decomposition of

H2O2 Items Data

�H decomposition of H 2O 2(l) ? kJ/mol

�H 0 f H 2O 2(l) �187.8 kJ/mol*

�H 0 f O2(g) 0.00 kJ/mol**

�H 0 f H2O(l) �285.8 kJ/mol*

* from Table 1

** any pure elemental substance has a �H 0 f of zero

What steps are needed to calculate �H for the decomposition reaction?

First, the equation must be balanced. Add up the enthalpies of formation for the

products. From this quantity, subtract the enthalpy of formation for the reactant.

Balance the equation for the decomposition of hydrogen peroxide.

2H2O2(l) → 2H2O(l) � O2(g) look up ∆H

reactant and product,

and solve

0 sum of �H

f for each

0 fproducts � sum of �H0 freactants � �Hreaction ( 2�H0

f H2O

given in Table 1

�H reaction

[2(�285.8 kJ/mol) � 0.00 kJ/mol] � [2(�187.8 kJ/mol)] � �196.0 kJ/mol

��H 0 ) � ( 2�H0 f H2O2) � �H reaction

f O2

Holt ChemFile: Problem-Solving Workbook 139 Thermochemistry

Yes; adding terms in kJ/mol gives kJ/mol.

Yes; rules for adding and rounding measurements give a result to four significant

Yes; the result can be approximated as �600 kJ/mol � 400 kJ/mol ��200 kJ/mol.

Determine �H for each of the following reactions.

1. The following reaction is used to make CaO from limestone.

CaCO3(s) → CaO(s) � CO2(g) ans: 179.2 kJ/mol

2. The following reaction represents the oxidation of FeO to Fe2O3. 2FeO(s) � O2(g) → Fe2O3(s) ans: 533 kJ/mol

3. The following reaction of ammonia and hydrogen fluoride produces ammonium

fluoride.

NH3(g) � HF(g) → NH4F(s) ans: 194 kJ/mol

Holt ChemFile: Problem-Solving Workbook 140 Thermochemistry

Calculate the Gibbs energy change for the following reaction at 25°C.

Ca(s) � 2H2O(l) 3 Ca(OH) 2(s) � H2(g) Use the following data.

�Hreaction ��411.6 kJ/mol; �Sreaction � 31.8 J/mol�K

What is given in the problem? T, �H, and �S for the reaction

What are you asked to find? the Gibbs energy of the reaction, �G

What steps are needed to calculate �G for the given reaction?

Apply the relationship �G ��H � T�S.

�H for the reaction at 25°C �411.6 kJ/mol

�S for the reaction at 25°C 31.8 J/mol�K

Temperature 25°C � 298 K

�G for the reaction at 25°C ? kJ/mol

Kelvin

temperature, T

�Hreaction � T�Sreaction � �Greaction given

J

mol�K

1 kJ

calculated

given given above

1000 J

Holt ChemFile: Problem-Solving Workbook 141 Thermochemistry

�S

kJ

�S reaction in

J/mol� K

31.8 J� 1kJ

� �� 0.0318 kJ/mol�K

1000 J�

�411.1 kJ/mol � (298 K� � 0.0318 kJ/mol�K�) ��411.6 kJ/mol � 9.5 kJ/mol

�Greaction ��421.1 kJ/mol

Yes; the kelvin units canceled. Adding terms in kJ/mol gives kJ/mol.

Yes; the number of significant figures is correct. Rules for adding and rounding

measurements give a result with four significant figures.

Yes; values were given for T, �H, and �S and the computation was carried out

correctly.

1. Calculate the Gibbs energy change, �G, for the combustion of hydrogen

sulfide according to the following chemical equation. Assume reactants and

products are at 25°C.

H2S(g) � O2(g) → H2O(l) � SO2(g) �Hreaction ��562.1 kJ/mol

�Sreaction ��0.09278 kJ/mol�K ans: �534.5 kJ/mol

2. Calculate the Gibbs energy change for the decomposition of sodium chlorate.

Assume reactants and products are at 25°C.

NaClO3(s) → NaCl(s) � O2(g) �Hreaction ��19.1 kJ/mol

�Sreaction � 0.1768 kJ/mol�K ans: �71.8 kJ/mol

3. Calculate the Gibbs energy change for the combustion of 1 mol of ethane.

C2H6(g) � O2(g) → 2CO2(g) � 3H2O(l) �Hreaction ��1561 kJ/mol

�Sreaction � 0.4084 kJ/mol�K ans: �1683 kJ/mol

Holt ChemFile: Problem-Solving Workbook 142 Thermochemistry

1. Calculate �H for the violent reaction of fluorine with water.

F2(g) � H2O(l) → 2HF(g) � O2(g) 2. Calculate �H for the reaction of calcium oxide and sulfur trioxide.

CaO(s) � SO3(g) → CaSO4(s) Use the following equations and data.

H2O(l) � SO3(g) → H2SO4(l) �H ��132.5 kJ/mol

H2SO4(l) � Ca(s) → CaSO4(s) � H2(g) �H ��602.5 kJ/mol

Ca(s) � O2(g) → CaO(s) �H ��634.9 kJ/mol

H2(g) � O2(g) → H2O(l) �H ��285.8 kJ/mol

3. Calculate �H for the reaction of sodium oxide with sulfur dioxide.

Na2O(s) � SO2(g) → Na2SO3(s) 4. Use enthalpies of combustion to calculate �H for the oxidation of 1-butanol

to make butanoic acid.

C4H9OH(l) � O2(g) → C3H7COOH(l) � H2O(l) Combustion of butanol:

C4H9OH(l) � 6O2(g) → 4CO2(g) � 5H2O(l) �Hc ��2675.9 kJ/mol

Combustion of butanoic acid:

C3H7COOH(l) � 5O2(g) → 4CO2(g) � 4H2O(l) �Hc ��2183.6 kJ/mol

5. Determine the free energy change for the reduction of CuO with hydrogen.

Products and reactants are at 25°C.

CuO(s) � H2(g) → Cu(s) � H2O(l) �H ��128.5 kJ/mol

�S ��70.1 J/mol�K

6. Calculate the enthalpy change at 25°C for the reaction of sodium iodide and

chlorine. Use only the data given.

NaI(s) � Cl2(g) → NaCl(s) � I2(l) �S ��79.9 J/mol�K

�G ��98.0 kJ/mol

7. The element bromine can be produced by the reaction of hydrogen bromide

and manganese(IV) oxide.

4HBr(g) � MnO2(s) → MnBr2(s) � 2H2O(l) � Br2(l) �H for the reaction is �291.3 kJ/mol at 25°C. Use this value and values of

Holt ChemFile: Problem-Solving Workbooky 143 Thermochemistry

8. Calculate the change in entropy, �S, at 25°C for the reaction of calcium carbide

with water to produce acetylene gas.

CaC2(s) � 2H2O(l) → C2H2(g) � Ca(OH) 2(s)

�G ��147.7 kJ/mol

�H ��125.6 kJ/mol

9. Calculate the Gibbs energy change for the explosive decomposition of ammonium

nitrate at 25°C. Note that H2O is a gas in this reaction.

NH4NO3(s) → N2O(g) � 2H2O(g) �S � 446.4 J/mol�K

10. In locations where natural gas, which is mostly methane, is not available,

many people burn propane, which is delivered by truck and stored in a tank

under pressure.

a. Write the chemical equations for the complete combustion of 1 mol of

methane, CH4, and 1 mol of propane, C3H8. b. Calculate the enthalpy change for each reaction to determine the amount of

energy released off by burning 1 mol of each fuel.

c. Using the enthapies of combustion you calculated, determine the energy

output per kilogram of each fuel. Which fuel yields more energy per unit

mass?

11. The hydration of acetylene to form acetaldehyde is shown in the following

C2H2(g) � H2O(l) → CH3CHO(l) Use enthapies of combustion for acetylene and acetaldehyde to compute the

enthalpy of the above reaction.

C2H2(g) � 2O2(g) → 2CO2(g) � H2O(l) �Hc ��1299.6 kJ/mol

CH3CHO(l) � 2O2(g) → 2CO2(g) � 2H2O(l) �Hc ��1166.9 kJ/mol

12. Calculate the enthalpy for the combustion of decane. �H 0 f for liquid decane is

�300.9 kJ/mol.

C10H22(l) � 15O2(g) → 10CO2(g) � 11H2O(l) 13. Find the enthalpy of the reaction of magnesium oxide with hydrogen chloride.

MgO(s) � 2HCl(g) → MgCl2(s) � H2O(l) Use the following equations and data.

Mg(s) � 2HCl(g) → MgCl2(s) � H2(g) �H ��456.9 kJ/mol

Mg(s) � O2(g) → MgO(s) �H ��601.6 kJ/mol

H2O(l) → H2(g) � O2(g) �H ��285.8 kJ/mol

Holt ChemFile: Problem-Solving Workbook 144 Thermochemistry

14. What is the Gibbs energy change for the following reaction at 25°C?

2NaOH(s) � 2Na(s) ∆

O3 2Na2O(s) � H2(g) �S � 10.6 J/mol�K

�H 0 f ��425.9 kJ/mol

NaOH

15. The following equation represents the reaction between gaseous HCl and

gaseous ammonia to form solid ammonium chloride.

NH3(g) � HCl(g) → NH4Cl(s) Calculate the entropy change in J/mol�K for the reaction of hydrogen chloride

and ammonia at 25°C using the following data and the values fround in

Table 1.

�G ��91.2 kJ/mol

16. The production of steel from iron involves the removal of many impurities in

the iron ore. The following equations show some of the purifying reactions.

Calculate the enthalpy for each reaction. Use Table 1 and the data given.

a. 3C(s) � Fe 2O 3(s) → 3CO(g) � 2Fe(s)

�H 0 f CO(g) ��110.53 kJ/mol

b. 3Mn(s) � Fe 2O 3(s) → 3MnO(s) � 2Fe(s)

�H 0 f MnO(s) ��384.9 kJ/mol

c. 12P(s) � 10Fe 2O 3(s) → 3P 4O 10(s) � 20Fe(s)

�H 0 f P4O10(s) ��3009.9 kJ/mol

d. 3Si(s) � 2Fe 2O 3(s) → 3SiO 2(s) � 4Fe(s)

�H 0 f SiO2(s) ��910.9 kJ/mol

e. 3S(s) � 2Fe 2O 3(s) → 3SO 2(g) � 4Fe(s)

Holt ChemFile: Problem-Solving Workbook 145 Thermochemistry

Gas Laws

Chemists found that there were relationships among temperature, volume, pressure,

and quantity of a gas that could be described mathematically. This chapter

deals with Boyle’s law, Charles’s law, Gay-Lussac’s law, the combined gas law, and

Dalton’s law of partial pressures. These laws have one condition in common.

They all assume that the molar amount of gas does not change. In other words,

these laws work correctly only when no additional gas enters a system and when

no gas leaks out of it. Remember also that a law describes a fact of nature. Gases

do not “obey” laws. The law does not dictate the behavior of the gas. Rather, each

gas law describes a certain behavior of gas that occurs if conditions are right.

BOYLE’S LAW

Robert Boyle, a British chemist who lived from 1627 to 1691 formulated the first

gas law, now known as Boyle’s law. This law describes the relationship between

the pressure and volume of a sample of gas confined in a container. Boyle found

that gases compress, much like a spring, when the pressure on the gas is

increased. He also found that they “spring back” when the pressure is lowered.

By “springing back” he meant that the volume increases when pressure is lowered.

It’s important to note that Boyle’s law is true only if the temperature of the

gas does not change and no additional gas is added to the container or leaks out

of the container.

Boyle’s law states that the volume and pressure of a sample of gas are

inversely proportional to each other at constant temperature. This statement

can be expressed as follows.

Proportionality

symbol. It means “is

proportional to.”

V � 1

P and

According to Boyle’s law, when the pressure on a gas is increased, the volume

of the gas decreases. For example, if the pressure is doubled, the volume

decreases by half. If the volume quadruples, the pressure decreases to one-fourth

of its original value.

The expression PV � k means that the product of the pressure and volume of

any sample of gas is a constant, k. If this is true, then P � V under one set of conditions

is equal to P � V for the same sample of gas under a second set of conditions,

as long as the temperature remains constant.

constant

PV � k or V � k

Holt ChemFile: Problem-Solving Workbook 146 Gas Laws

Boyle’s law can be expressed by the following mathematical equation.

Pressure under

the first set of

conditions

General Plan for Solving Boyle's-Law Problems

Given three of the following

four quantities:

P 1 , V 1 , P 2 , V 2

An equation that can be used to

calculate the unknown quantity

It will be one of the following four:

Volume under

P 2

P 1 V 1 � P 2 V 2

V 2

Rearrange the equation

algebraically to solve for

the unknown quantity.

P1V1 P1V1 P2V2 V2 � , P2 � , V1 � , P1 �

Substitute each of the

known quantities,

and calculate.

Unknown

P or V

the second set

of conditions

Holt ChemFile: Problem-Solving Workbook 147 Gas Laws

P 1

� the second set

P 2 V 2

A sample of nitrogen collected in the laboratory occupies a volume of

725 mL at a pressure of 0.971 atm. What volume will the gas occupy at a

pressure of 1.40 atm, assuming the temperature remains constant?

What is given in the problem? the original volume and pressure of the nitrogen

sample, and the new pressure of the sample

What are you asked to find? the volume at the new pressure

What steps are needed to calculate the new volume of the gas?

Rearrange the Boyle’s law equation to solve for V 2, substitute known quantities,

P1V1 � P2V2 Items Data

Original pressure, P 1

Original pressure, V 1

New pressure, P2 New volume, V2 to solve for V2 ,

divide both sides of

the equation by P2 to

isolate V2 insert data and solve for V2 COMPUTE

Substitute data for the terms of the equation, and compute the result.

V2 �

1V1 0.971 �atm � 725 mL N2 �P2 �� 503 mL N2 1.40 atm �

Yes; units canceled to give mL N2. Is the number of significant figures correct?

Yes; pressure increased by about 1/3, volume must decrease by about 1/3.

0.971 atm

725 mL N 2

1.40 atm

? mL N 2

P1V1 V2 �

P2 Holt ChemFile: Problem-Solving Workbook 148 Gas Laws

In each of the following problems, assume that the temperature and molar quantity

of gas do not change.

1. Calculate the unknown quantity in each of the following measurements of

gases.

P1 V1 P2 V2 a. 3.0 atm 25 mL 6.0 atm ? mL

b. 99.97 kPa 550. mL ? kPa 275 mL

c. 0.89 atm ? L 3.56 atm 20.0 L

d. ? kPa 800. mL 500. kPa 160. mL

e. 0.040 atm ? L 250 atm 1.0 � 10 �2 L

ans: 13 mL

ans: 200. kPa

ans: 80. L

ans: 100. kPa

ans: 63 L

Holt ChemFile: Problem-Solving Workbook 149 Gas Laws

2. A sample of neon gas occupies a volume of 2.8 L at 1.8 atm. What will its

volume be at 1.2 atm? ans: 4.2 L

3. To what pressure would you have to compress 48.0 L of oxygen gas at 99.3

kPa in order to reduce its volume to 16.0 L? ans: 298 kPa

4. A chemist collects 59.0 mL of sulfur dioxide gas on a day when the atmospheric

pressure is 0.989 atm. On the next day, the pressure has changed to

0.967 atm. What will the volume of the SO 2 gas be on the second day?

ans: 60.3 mL

5. 2.2 L of hydrogen at 6.5 atm pressure is used to fill a balloon at a final pressure

of 1.15 atm. What is its final volume? ans: 12 L

CHARLES’S LAW

The French physicist Jacques Charles carried out experiments in 1786 and 1787

that showed a relationship between the temperature and volume of gases at constant

pressure. You know that most matter expands as its temperature rises.

Gases are no different. When Benjamin Thomson and Lord Kelvin proposed an

absolute temperature scale in 1848, it was possible to set up the mathematical

expression of Charles’s law.

Holt ChemFile: Problem-Solving Workbook 150 Gas Laws

Charles’s law states that the volume of a sample of gas is directly proportional

to the absolute temperature when pressure remains constant. Charles’s

law can be expressed as follows.

V � T and � V

� � k, or V � kT

T

According to Charles’s law, when the temperature of a sample of gas increases,

the volume of the gas increases by the same factor. Therefore, doubling the

Kelvin temperature of a gas will double its volume. Reducing the Kelvin temperature

by 25% will reduce the volume by 25%.

The expression V/T � k means that the result of volume divided by temperature

is a constant, k, for any sample of gas. If this is true, then V/T under one set

of conditions is equal to V/T for the same sample of gas under another set of conditions,

as long as the pressure remains constant.

Charles’s law can be expressed by the following mathematical equation.

V1

V2

General Plan for Solving Charles’s-Law Problems

Convert Celsius

temperatures to

Kelvin temperatures

if necessary.

T 1 , V 1 , T 2 , V 2

Holt ChemFile: Problem-Solving Workbook 151 Gas Laws

T or V

V1 T1 �

V2 T2 algebraically to solve for

T2V1 T1V2 T1V2 T2V1 V2 � , T2 � , V1 � , T1 �

T1 V1 T2 V2 Substitute each of the

and compute.

K to �C if

needed.

A container of oxygen has a volume of 349 mL at a temperature of 22°C.

What volume will the gas occupy at 50.°C?

What is given in the problem? the original volume and temperature of the

oxygen sample, and the new temperature of the

sample

What are you asked to find? the volume at the new temperature

Original volume, V 1

Original temperature, t 1

Original Kelvin temperature, T 1

New volume, V 2

New temperature, t2 New Kelvin temperature, T2 PLAN

Rearrange the Charles’s law equation to solve for V 2, substitute known quantities,

V 1

T 1

V2 � � V1T2

349 mL O2 � 323 K�

� �� 382 mL O2 T

295 K�

V2 T2 1

to solve for V2 ,

multiply both sides of

the equation by T2 to

isolate V2 insert data and solve for V2 349 mL O 2

22°C

(22 � 273) K � 295 K

? mL O 2

50.°C

? (50. � 273) K � 323 K

Yes; units canceled to give mL O2. Is the number of significant figures correct?

Yes; the temperature increased, so the volume must also increase.

Holt ChemFile: Problem-Solving Workbook 152 Gas Laws

V 1 T 2

In each of the following problems, assume that the pressure and molar quantity of

gas do not change.

V1 T1 V2 T2 a. 40.0 mL 280. K ? mL 350. K

b. 0.606 L 300. K 0.404 L ? K

c. ? mL 292 K 250. mL 365 K

d. 100. mL ? K 125 mL 305 K

e. 0.0024 L 22°C ? L �14°C

ans: 50.0 mL

ans: 200. K

ans: 200. mL

ans: 244 K

ans: 0.0021 L

Holt ChemFile: Problem-Solving Workbook 153 Gas Laws

2. A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is

the balloon’s volume at 45°C? ans: 3.01 L

3. A sample of argon has a volume of 0.43 mL at 24°C. At what temperature in

degrees Celsius will it have a volume of 0.57 mL? ans: 121°C

GAY-LUSSAC’S LAW

You may have noticed a warning on an aerosol spray can that says something

similar to Do not incinerate! Do not expose to temperatures greater than 140°F!

Warnings such as this appear because the pressure of a confined gas increases

with increasing temperature. If the temperature of the can increases enough, the

can will explode because of the pressure that builds up inside of it.

The relationship between the pressure and temperature of a gas is described

by Gay-Lussac’s law. Gay-Lussac’s law states that the pressure of a sample of gas

is directly proportional to the absolute temperature when volume remains constant.

Gay-Lussac’s law can be expressed as follows.

P � T and � P

� � k, or P � kT

According to Gay-Lussac’s law, when the temperature of a sample of gas

increases, the pressure of the gas increases by the same factor. Therefore, doubling

the temperature of a gas will double its pressure. Reducing the temperature

of a gas to 75% of its original value will also reduce the pressure to 75% of its

original value.

The expression P/T � k means that the result of pressure divided by temperature

is a constant, k, for any sample of gas. If this is true, then P/T under one set

of conditions is equal to P/T for the same sample of gas under another set of conditions,

as long as the volume remains constant.

Holt ChemFile: Problem-Solving Workbook 154 Gas Laws

Gay-Lussac’s law can be expressed by the following mathematical equation.

P1

P2

General Plan for Solving Gay-Lussac’s-Law Problems

T 1 , P 1 , T 2 , P 2

Holt ChemFile: Problem-Solving Workbook 155 Gas Laws

T or P

P1 T1 �

P2 T2 algebraically to solve for

T2P1 P2 � ,

T1 T1P2 T2 � ,

P1 T1P2 P1 � ,

T2 T2P1 T1 �

P2 Substitute each of the

A cylinder of gas has a pressure of 4.40 atm at 25°C. At what temperature

in °C will it reach a pressure of 6.50 atm?

What is given in the problem? the original pressure and temperature of the

gas in the cylinder, and the new pressure of the

What are you asked to find? the temperature at which the gas reaches the

specified pressure

New pressure, P 2

What steps are needed to calculate the new temperature of the gas?

Rearrange the Gay-Lussac’s law equation to solve for T 2, substitute known

quantities, and calculate.

P1 P2 �

T1 T2 to solve for T2 , divide

both sides of the equation

by P2 and invert the

equality to isolate T2 insert data and solve for T2 COMPUTE

T2 � � T1P2

298 K � 6.

50atm�

� �� 440. K

4.40�

atm

Convert back to °C.

440. K � (440. � 273)°C � 167°C

Yes; units canceled to give Kelvin temperature, which was converted to °C.

4.40 atm

25°C

(25 � 273) K � 298 K

6.50 atm

Holt ChemFile: Problem-Solving Workbook 156 Gas Laws

?°C

? K

T 1 P 2

Yes; the pressure increases, so the temperature must also increase.

In each of the following problems, assume that the volume and molar quantity of

gases:

P1 T1 P2 T2 a. 1.50 atm 273 K ? atm 410 K

b. 0.208 atm 300. K 0.156 atm ? K

c. ? kPa 52°C 99.7 kPa 77°C

d. 5.20 atm ?°C 4.16 atm �13°C

e. 8.33 � 10 �4 atm �84°C 3.92 � 10 �3 atm ?°C

ans: 2.25 atm

ans: 225 K

ans: 92.6 kPa

ans: 52°C

ans: 616°C

Holt ChemFile: Problem-Solving Workbook 157 Gas Laws

2. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The

next day, the same cylinder of gas has a pressure of 4.690 atm, and its temperature

is 8°C. What was the temperature on the previous day in °C? ans: 20.°C

3. A mylar balloon is filled with helium gas to a pressure of 107 kPa when the

temperature is 22°C. If the temperature changes to 45°C, what will be the

pressure of the helium in the balloon? ans: 115 kPa

THE COMBINED GAS LAW

Look at the relationships among temperature, volume, and pressure of a gas that

you have studied so far.

Boyle’s law

At constant

temperature:

PV � k

Charles’s law

At constant pressure:

� k

Notice in these proportions that while P and V are inversely proportional to

each other, they are each directly proportional to temperature. These three gas

laws can be combined in one combined gas law. This law can be expressed as

PV

If PV/T equals a constant, k, then PV/T for a sample of gas under one set of

conditions equals PV/T under another set of conditions, assuming the amount of

gas remains the same.

Therefore, the combined gas law can be expressed by the following mathematical

equation. This equation can be used to solve problems in which pressure,

Gay-Lussac’s law

At constant volume:

Holt ChemFile: Problem-Solving Workbook 158 Gas Laws

volume, and temperature of a gas vary. Only the molar quantity of the gas must

be constant.

1V1

P2V2

General Plan for Solving Combined-Gas-Law Problems

Given five of the following

six quantities:

V 1 ,T 1 , P 1 , V 2 ,T 2 , P 2

Holt ChemFile: Problem-Solving Workbook 159 Gas Laws

P1V1 P2V2 �

T1 T2 algebraically to solve for

V, T, or P

A sample of hydrogen gas has a volume of 65.0 mL at a pressure of 0.992

atm and a temperature of 16°C. What volume will the hydrogen occupy at

0.984 atm and 25°C?

What is given in the problem? the original volume, pressure, and temperature

of the hydrogen, and the new pressure and temperature

of the hydrogen

What are you asked to find? the volume at the new temperature and

pressure

Rearrange the combined gas law equation to solve for V 2, substitute known

P1V1 �

T1 COMPUTE

V2 � � T2P1V1

298 K� � 0.992 �atm � 65.0 mL H2 � �� 67.6 mL H2 P2T1

0.984 atm � � 289 K�

Yes; units canceled to give mL of H 2.

the equation by T2 /P2 to

T 2

65.0 mL

0.992 atm

16°C

(16 � 273) K � 289 K

?mL

0.984 atm

T2P1V1 � V2 P2T1 Holt ChemFile: Problem-Solving Workbook 160 Gas Laws

Yes; the temperature increases, and the pressure decreases, both of which have

the effect of making the volume larger.

In each of the following problems, it is assumed that the molar quantity of gas

does not change.

P1 V1 T1 P2 V2 T2 a. 99.3 kPa 225 mL 15°C 102.8 kPa ? mL 24°C

b. 0.959 atm 3.50 L 45°C ? atm 3.70 L 37°C

c. 0.0036 atm 62 mL 373 K 0.0029 atm 64 mL ? K

d. 100. kPa 43.2 mL 19°C 101.3 kPa ? mL 0°C

2. A student collects 450. mL of HCl(g) hydrogen chloride gas at a pressure of

100. kPa and a temperature of 17°C. What is the volume of the HCl at 0°C and

101.3 kPa? ans: 418 mL

ans: 224 mL

ans: 0.884 atm

ans: 310 K

ans: 39.9 mL

Holt ChemFile: Problem-Solving Workbook 161 Gas Laws

DALTON’S LAW OF PARTIAL PRESSURES

Air is a mixture of approximately 78% N2, 20% O2, 1% Ar, and 1% other gases by

volume, so at any barometric pressure 78% of that pressure is exerted by nitrogen,

20% by oxygen, and so on. This phenomenon is described by Dalton’s law of

partial pressures, which says that the total pressure of a mixture of gases is

equal to the sum of the partial pressures of the component gases. It can be

stated mathematically as follows.

PTotal � PGas 1 � PGas 2 � PGas 3 � PGas 4 � …

A common method of collecting gas samples in the laboratory is to bubble the

gas into a bottle filled with water and allow it to displace the water. When this

technique is used, however, the gas collected in the bottle contains a small but

significant amount of water vapor. As a result, the pressure of the gas that has

displaced the liquid water is the sum of the pressure of the gas plus the vapor

pressure of water at that temperature. The vapor pressures of water at various

temperatures are given in Table 1 below.

TABLE 1 VAPOR PRESSURE OF WATER

Temp Vapor pressure Temp Vapor pressure Temp Vapor pressure

in °C in kPa in °C in kPa in °C in kPa

10 1.23 17 1.94 24 2.98

11 1.31 18 2.06 25 3.17

12 1.40 19 2.19 26 3.36

13 1.50 20 2.34 27 3.57

14 1.60 21 2.49 28 3.78

15 1.71 22 2.64 29 4.01

16 1.82 23 2.81 30 4.25

P Total � P Gas � P H2O vapor

To find the true pressure of the gas alone, the pressure of the water vapor

must be subtracted from the total pressure.

PGas � PTotal � PH2O vapor

You can use this corrected pressure in gas-law calculations to determine what the

volume of the gas alone would be.

Holt ChemFile: Problem-Solving Workbook 162 Gas Laws

A student collects oxygen gas by water displacement at a temperature of

16°C. The total volume is 188 mL at a pressure of 92.3 kPa. What is the

pressure of oxygen collected?

What is given in the problem? the total pressure, the fact that the gas was collected

by water displacement, and the temperature

What are you asked to find? the pressure of oxygen collected

Total pressure, P total

What step is needed to calculate the pressure of the oxygen collected?

Use Dalton’s law of partial pressures to determine the pressure of the oxygen

alone in the container.

Use the following equation to determine the pressure of oxygen alone.

P O2 � P total �

PO2 � Ptotal � PH2O vapor � 92.3 kPa � 1.82 kPa � 90.5 kPa

Yes; the units should be kPa.

Yes; the number of significant figures is correct because the total pressure was

given to one decimal place.

Yes; the pressure of the collected oxygen and the water vapor add up to the

total pressure of the system.

from Table 1

P H2O vapor

92.3 kPa

Temperature 16°C

Vapor pressure of water at 16°C, PH2O* 1.82 kPa

Pressure of collected O2, PO2 ? kPa

* determined from Table 1

Holt ChemFile: Problem-Solving Workbook 163 Gas Laws

1. A chemist collects a sample of H 2S(g) over water at a temperature of 27°C.

The total pressure of the gas that has displaced a volume of 15 mL of water is

207.33 kPa. What is the pressure of the H 2S gas collected? ans: 203.76 kPa

Holt ChemFile: Problem-Solving Workbook 164 Gas Laws

Chlorine gas is collected by water displacement at a temperature of 19°C.

The total volume is 1.45 L at a pressure of 156.5 kPa. Assume that no

chlorine gas dissolves in the water. What is the volume of chlorine corrected

to STP?

What is given in the problem? the total initial volume, pressure, and temperature;

the fact that the gas was collected by

water displacement; and the final temperature

and pressure of the chlorine

What are you asked to find? the volume of chlorine at STP

What steps are needed to calculate the volume of the chlorine at STP?

Use Dalton’s law of partial pressures to determine the pressure of the chlorine

alone in the container. Use the combined gas law to solve for the new volume.

given from table

P1 � Ptotal � PH2O vapor

Calculate the volume of chlorine at STP using the following equation.

Solve for V 2.

Total original volume, V 1

Total original pressure, P total

Vapor pressure of water at 19°C, P H2O

Pressure of collected Cl 2, P 1

above given

P1 T2V1 P 2 T 1

Holt ChemFile: Problem-Solving Workbook 165 Gas Laws

1.45 L

156.5 kPa

19°C

(19 � 273) K � 292 K

2.19 kPa

? kPa

101.3 kPa

0°C

Determine the pressure of chlorine alone.

P1 � Ptotal � PH2O vapor � 156.5 kPa � 2.19 kPa � 154.3 kPa

Calculate the volume of Cl2 at STP.

154.3 kPa � � 273 K� � 1.45 L

V2 �� 2.06 L

101.3 kPa � � 292 K�

Yes; the units canceled leaving the correct units, L.

Yes; the number of significant figures is correct because the data were given to a

minimum of three significant figures.

Yes; the pressure decreased, so the volume had to increase.

In each of the following problems, assume that the molar quantity of gas does not

change.

1. Some hydrogen is collected over water at 10°C and 105.5 kPa pressure. The

total volume of the sample was 1.93 L. Calculate the volume of the hydrogen

corrected to STP. ans: 1.92 L

2. One student carries out a reaction that gives off methane gas and obtains a

total volume by water displacement of 338 mL at a temperature of 19°C and a

pressure of 0.9566 atm. Another student does the identical experiment on

another day at a temperature of 26°C and a pressure of 0.989 atm. Which

student collected more CH 4? ans: The second student collected more CH 4.

Holt ChemFile: Problem-Solving Workbook 166 Gas Laws

P1 V1 P2 V2 a. 127.3 kPa 796 cm 3

? kPa 965 cm 3

b. 7.1 � 10 2 atm ? mL 9.6 � 10 �1 atm 3.7 � 10 3 mL

c. ? kPa 1.77 L 30.79 kPa 2.44 L

d. 114 kPa 2.93 dm 3

2. A gas cylinder contains 0.722 m 3 of hydrogen gas at a pressure of 10.6 atm. If

the gas is used to fill a balloon at a pressure of 0.96 atm, what is the volume in

m 3 of the filled balloon?

3. A weather balloon has a maximum volume of 7.50 � 10 3 L. The balloon

contains 195 L of helium gas at a pressure of 0.993 atm. What will be the

pressure when the balloon is at maximum volume?

4. A rubber ball contains 5.70 � 10 �1 dm 3 of gas at a pressure of 1.05 atm. What

volume will the gas occupy at 7.47 atm?

5. Calculate the unknown quantity in each of the following measurements of

6. A bubble of carbon dioxide gas in some unbaked bread dough has a volume

of 1.15 cm 3 at a temperature of 22°C. What volume will the bubble have when

the bread is baked and the bubble reaches a temperature of 99°C?

7. A perfectly elastic balloon contains 6.75 dm 3 of air at a temperature of 40.°C.

What is the temperature if the balloon has a volume of 5.03 dm 3 ?

4.93 � 10 4 kPa ? dm 3

e. 1.00 atm 120. mL ? atm 97.0 mL

f. 0.77 atm 3.6 m 3

1.90 atm ? m 3

V 1 T 1 V 2 T 2

a. 26.5 mL ? K 32.9 mL 290. K

b. ? dm 3

c. 7.44 � 10 4 mm 3

100.°C 0.83 dm 3

870.°C 2.59 � 10 2 mm 3

�9°C

Holt ChemFile: Problem-Solving Workbook 167 Gas Laws

d. 5.63 � 10 �2 L 132 K ? L 190. K

e. ?cm 3

f. 679 m 3

243 K 819 cm 3

�3°C ? m 3

409 K

8. Calculate the unknown quantity in each of the following measurements of

P1 T1 P2 T2 a. 0.777 atm ?°C 5.6 atm 192°C

b. 152 kPa 302 K ? kPa 11 K

c. ? atm �76°C 3.97 atm 27°C

d. 395 atm 46°C 706 atm ?°C

e. ? atm �37°C 350. atm 2050°C

f. 0.39 atm 263 K 0.058 atm ? K

9. A 2 L bottle containing only air is sealed at a temperature of 22°C and a

pressure of 0.982 atm. The bottle is placed in a freezer and allowed to cool to

� 3°C. What is the pressure in the bottle?

10. The pressure in a car tire is 2.50 atm at a temperature of 33°C. What would

the pressure be if the tire were allowed to cool to 0°C? Assume that the tire

does not change volume.

11. A container filled with helium gas has a pressure of 127.5 kPa at a temperature

of 290. K. What is the temperature when the pressure is 3.51 kPa?

12. Calculate the unknown quantity in each of the following measurements of

P 1 V 1 T 1 P 2 V 2 T 2

a. 1.03 atm 1.65 L 19°C 0.920 atm ? L 46°C

b. 107.0 kPa 3.79 dm 3

13. A scientist has a sample of gas that was collected several days earlier. The

sample has a volume of 392 cm 3 at a pressure of 0.987 atm and a temperature

of 21°C. On the day the gas was collected, the temperature was 13°C and the

pressure was 0.992 atm. What volume did the gas have on the day it was

collected?

14. Hydrogen gas is collected by water displacement. Total volume collected is

0.461 L at a temperature of 17°C and a pressure of 0.989 atm. What is the

pressure of dry hydrogen gas collected?

73°C ? kPa 7.58 dm 3

217°C

c. 0.029 atm 249 mL ? K 0.098 atm 197 mL 293 K

d. 113 kPa ? mm 3

e. 1.15 atm 0.93 m 3

f. ? atm 156 cm 3

12°C 149 kPa 3.18 � 10 3 mm 3

�22°C 1.01 atm 0.85 m 3

195 K 2.25 atm 468 cm 3

�18°C

Holt ChemFile: Problem-Solving Workbook 168 Gas Laws

15. One container with a volume of 1.00 L contains argon at a pressure of 1.77

atm, and a second container of 1.50 L volume contains argon at a pressure of

0.487 atm. They are then connected to each other so that the pressure can

become equal in both containers. What is the equalized pressure? Hint: Each

sample of gas now occupies the total space. Dalton’s law of partial pressures

applies here.

16. Oxygen gas is collected over water at a temperature of 10.°C and a pressure

of 1.02 atm. The volume of gas plus water vapor collected is 293 mL. What

volume of oxygen at STP was collected?

17. A 500 mL bottle is partially filled with water so that the total volume of gases

(water vapor and air) remaining in the bottle is 325 cm 3 , measured at 20.°C

and 101.3 kPa. The bottle is sealed and taken to a mountaintop where the

pressure is 76.24 kPa and the temperature is 10°C. If the bottle is upside down

and the seal leaks, how much water will leak out? The key to this problem is

to determine the pressure in the 325 cm 3 space when the bottle is at the top

of the mountain.

18. An air thermometer can be constructed by using a glass bubble attached to a

piece of small-diameter glass tubing. The tubing contains a small amount of

colored water that rises when the temperature increases and the trapped air

expands. You want a 0.20 cm 3 change in volume to equal a 1°C change in

temperature. What total volume of air at 20.°C should be trapped in the apparatus

below the liquid?

19. A sample of nitrogen gas is collected over water, yielding a total volume of

62.25 mL at a temperature of 22°C and a total pressure of 97.7 kPa. At what

pressure will the nitrogen alone occupy a volume of 50.00 mL at the same

temperature?

20. The theoretical yield of a reaction that gives off nitrogen trifluoride gas is

844 mL at STP. What total volume of NF 3 plus water vapor will be collected

over water at 25°C and a total pressure of 1.017 atm?

21. A weather balloon is inflated with 2.94 kL of helium at a location where the

pressure is 1.06 atm and the temperature is 32°C. What will be the volume of

the balloon at an altitude where the pressure is 0.092 atm and the temperature

is �35°C?

22. The safety limit for a certain can of aerosol spray is 95°C. If the pressure of

the gas in the can is 2.96 atm when it is 17°C, what will the pressure be at the

safety limit?

23. A chemistry student collects a sample of ammonia gas at a temperature of

39°C. Later, the student measures the volume of the ammonia as 108 mL, but

its temperature is now 21°C. What was the volume of the ammonia when it

was collected?

24. A quantity of CO 2 gas occupies a volume of 624 L at a pressure of 1.40 atm. If

this CO 2 is pumped into a gas cylinder that has a volume of 80.0 L, what

pressure will the CO 2 exert on the cylinder?

Holt ChemFile: Problem-Solving Workbook 169 Gas Laws

The Ideal Gas Law

In 1811, the Italian chemist Amedeo Avogadro proposed the principle that equal

volumes of gases at the same temperature and pressure contain equal numbers

of molecules. He determined that at standard temperature and pressure, one mole

of gas occupies 22.414 10 L (usually rounded to 22.4 L).

At this point, if you know the number of moles of a gas, you can use the molar

volume of 22.4 L/mol to calculate the volume that amount of gas would occupy at

STP. Then you could use the combined gas law to determine the volume of the

gas under any other set of conditions. However, a much simpler way to accomplish

the same task is by using the ideal gas law.

The ideal gas law is a mathematical relationship that has the conditions of

standard temperature (273 K) and pressure (1 atm or 101.3 kPa) plus the molar

gas volume (22.4 L/mol) already combined into a single constant. The following

equation is the mathematical statement of the ideal gas law.

PV � nRT

in which

P � the pressure of a sample of gas

V � the volume of a sample of gas

n � the number of moles of gas present

T � the Kelvin temperature of the gas

R � the ideal gas constant, which combines standard conditions

and molar volume into a single constant

The value of the ideal gas constant, R, depends on the units of P and V being

used in the equation. Temperature is always in kelvins and amount of gas is

always in moles. The most common values used for R are shown below.

Units of P and V Value of R

L� atm

Atmospheres and liters 0.0821 �� mol�

K

L� kPa

Kilopascals and liters 8.314 �� mol�K

Holt Chemistry 170 The Ideal Gas Law

If you have volume units other than liters or pressure units other than atmospheres

or kilopascals, it is best to convert volume to liters and pressure to atmospheres

or kilopascals.

General Plan for Solving Ideal-Gas-Law Problems

The equation for

the ideal gas law

An equation that can

be used to calculate

the unknown quantity

P, V, n, or T

Determine from the data which

is the unknown quantity.

algebraically to solve for the

unknown quantity.

Choose the gas constant, R, that

best fits the units of the data.

Substitute each of the data values

in the equation and calculate.

Holt ChemFile: Problem-Solving Workbook 171 The Ideal Gas Law

An engineer pumps 5.00 mol of carbon monoxide gas into a cylinder that

has a capacity of 20.0 L. What is the pressure in kPa of CO inside the

cylinder at 25°C?

What is given in the problem? the amount in moles of gas pumped into the

cylinder, the volume of the cylinder, and the

temperature

What are you asked to find? the pressure of the gas in the cylinder

Amount of gas, n 5.00 mol

Volume of gas in cylinder, V 20.0 L

Temperature of gas, t 25°C

Kelvin temperature of gas, T (25 � 273) K � 298 K

Ideal gas constant, R 0.0821 L�atm/mol�K or

8.314 L�kPa/mol�K

Pressure in cylinder, P ? kPa

What steps are needed to calculate the new pressure of the gas?

Rearrange the ideal-gas-law equation to solve for P, substitute known quantities,

Ideal-gas-law

equation,

P � nRT

solve the idealgas-law

equation

for pressure

the problem asks for

answer in kPa, so choose

the appropriate R,

substitute known values,

Unknown pressure,

Holt ChemFile: Problem-Solving Workbook 172 The Ideal Gas Law

Solve the ideal-gas-law equation for P, the unknown quantity.

P � � nRT

The problem asks for pressure in kPa, so use R � 8.314 L�kPa/mol�K.

Yes; the ideal gas constant was selected so that the units canceled to give kPa.

Yes; the calculation can be approximated as (1/4) � (8 � 300), or 2400/4, which

equals 600. Thus, 619 kPa is in the right range.

P �

5.00 mol � 8.314 L�kPa/mol�K � 298 K

20.0 L

1. A student collects 425 mL of oxygen at a temperature of 24°C and a pressure

of 0.899 atm. How many moles of oxygen did the student collect?

ans: 1.57 � 10 �2 mol O 2

� 619 kPa

Holt ChemFile: Problem-Solving Workbook 173 The Ideal Gas Law

2. Use the ideal-gas-law equation to calculate the unknown quantity in each of

the following sets of measurements. You will need to convert Celsius temperatures

to Kelvin temperatures and volume units to liters.

P V n T

a. 1.09 atm ? L 0.0881 mol 302 K

b. 94.9 kPa 0.0350 L ? mol 55°C

c. ? kPa 15.7 L 0.815 mol �20.°C

d. 0.500 atm 629 mL 0.0337 mol ? K

e. 0.950 atm ? L 0.0818 mol 19°C

f. 107 kPa 39.0 mL ? mol 27°C

ans: 2.00 L

ans: 1.22 � 10 �3 mol

ans: 109 kPa

ans: 114 K

ans: 2.06 L

ans: 1.67 � 10 �3 mol

Holt ChemFile: Problem-Solving Workbook 174 The Ideal Gas Law

APPLICATIONS OF THE IDEAL GAS LAW

You have seen that you can use the ideal gas law to calculate the moles of gas, n,

in a sample when you know the pressure, volume, and temperature of the sample.

When you know the amount and identity of the substance, you can use its molar

mass to calculate its mass. You did this when you learned how to convert

between mass and moles. The relationship is expressed as follows.

Amount in moles

n � m

M

If you substitute the expression m/M for n in the ideal-gas-law equation, you

get the following equation.

PV � � m

M �RT

This version of the ideal gas law can be solved for any of the five variables P,

V, m, M, or T. It is especially useful in determining the molecular mass of a substance.

This equation can also be related to the density of a gas. Density is mass

per unit volume, as shown in the following equation.

D � � m

V �

Solve for m:

m � DV

Then, substitute DV for m in the gas law equation:

PV � � DV

� RT

The two V terms cancel and the equation is rearranged to give:

PM � DRT or D � � PM

RT

This equation can be used to compute the density of a gas under any conditions

of temperature and pressure. It can also be used to calculate the molar

mass of an unknown gas if its density is known.

Mass in grams

Molar mass in grams per mole

Holt ChemFile: Problem-Solving Workbook 175 The Ideal Gas Law

1a

General Plan for Solving Problems Involving

Applications of the Ideal Gas Law

PV � RT

Determine which

equation fits the

problem.

algebraically to

solve for the

P, V, m, M, D or T

D � PM

Holt ChemFile: Problem-Solving Workbook 176 The Ideal Gas Law

Determine the molar mass of an unknown gas that has a volume of 72.5

mL at a temperature of 68°C, a pressure of 0.980 atm, and a mass of

0.207 g.

What is given in the problem? the mass, pressure, volume, and temperature of

the gas

What are you asked to find? the molar mass of the gas

Volume of gas, V 72.5 mL

Temperature of gas, t 68°C

Kelvin temperature of gas, T (68 � 273) K � 341 K

Pressure of gas, P 0.980 atm

Mass of gas, m 0.207 g

Ideal gas constant, R 8.314 L�kPa/mol�K or

0.0821 L�atm/mol�K

Molar mass of gas, M ? g/mol

Select the equation that will give the desired result. Solve the equation for the

unknown quantity. Substitute data values into the solved equation, and calculate.

solve this

equation for

M � mRT

the problem gives

pressure in atm, so

choose the appropriate R,

Unknown molar

mass, M

Holt ChemFile: Problem-Solving Workbook 177 The Ideal Gas Law

Use the equation that includes m and M.

Solve the equation for M, the unknown quantity.

M � � mRT

Convert the volume in milliliters to liters

72.5 mL � � �� 0.0725 L

1000�

The data give pressure in atm, so use R � 0.0821 L�atm/mol�K.

M � mRT 0.207 g � 0.0821 L�atm/mol�K � 341 K

� � 81.6 g/mol

PV 0.980 atm � 0.0725 L

Yes; units canceled to give g/mol, the correct units for molar mass.

Yes; 81.6 g/mol is a reasonable molar mass. The calculation can be approximated

as 0.2 � 341 � (8/7), which is roughly 80.

1. A sample of an unknown gas has a mass of 0.116 g. It occupies a volume of

25.0 mL at a temperature of 127°C and has a pressure of 155.3 kPa. Calculate

the molar mass of the gas. ans: 99.4 g/mol

2. Determine the mass of CO 2 gas that has a volume of 7.10 L at a pressure of

1.11 atm and a temperature of 31°C. Hint: Solve the equation for m, and

calculate the molar mass using the chemical formula and the periodic table.

ans: 13.9 g

Holt ChemFile: Problem-Solving Workbook 178 The Ideal Gas Law

Determine the density of hydrogen bromide gas at 3.10 atm and �5°C.

What is given in the problem? the pressure and temperature of the HBr gas

Temperature of HBr, t �5°C

Kelvin temperature of HBr, T (�5 � 273) K � 268 K

Pressure of HBr, P 3.10 atm

Molar mass of HBr, M* 80.91 g/mol

Density of HBr, D ? g/L

What steps are needed to calculate the density of HBr under the conditions

given?

Select the equation that will give the desired result. Rearrange the equation to

solve for the unknown quantity. Substitute data values into the correct equation,

equation is already

written correctly to solve

for the unknown

Unknown density, D

Holt ChemFile: Problem-Solving Workbook 179 The Ideal Gas Law

Use the equation that includes density.

D � � PM

Yes; units canceled to give g/L, the correct units for gas density.

Yes; 11.4 g/L is a reasonable density for a heavy gas compressed to 3 atm. The

calculation can be approximated as 3 � 80/(0.08 � 270) � 3 � 1000/270 � 11.

RT �

3.10 atm � 80.91 g/mol

� 11.4 g/L

0.0821 L�atm/mol�K � 268 K

1. What is the density of silicon tetrafluoride gas at 72°C and a pressure of 144.5

kPa? ans: 5.24 g/L

2. At what temperature will nitrogen gas have a density of 1.13 g/L at a pressure

of 1.09 atm? ans: 329 K or 56°C

Holt ChemFile: Problem-Solving Workbook 180 The Ideal Gas Law

1. Use the ideal-gas-law equation to calculate the unknown quantity in each of

the following sets of measurements.

P V n t

a. 0.0477 atm 15 200 L ? mol �15°C

b. ? kPa 0.119 mL 0.000 350 mol 0°C

c. 500.0 kPa 250. mL 0.120 mol ?°C

d. 19.5 atm ? 4.7 � 10 4 mol 300.°C

P V m M t

a. 0.955 atm 3.77 L 8.23 g ? g/mol 25°C

b. 105.0 kPa 50.0 mL ? g 48.02 g/mol 0°C

c. 0.782 atm ? L 3.20 � 10 �3 g 2.02 g/mol �5°C

d. ? atm 2.00 L 7.19 g 159.8 g/mol 185°C

e. 107.2 kPa 26.1 mL 0.414 g ? g/mol 45°C

3. Determine the volume of one mole of an ideal gas at 25°C and 0.915 kPa.

4. Calculate the unknown quantity in each of the following sets of measurements.

P Molar mass Density t

a. 1.12 atm ? g/mol 2.40 g/L 2°C

b. 7.50 atm 30.07 g/mol ? g/L 20.°C

c. 97.4 kPa 104.09 g/mol 4.37 g/L ?°C

d. ? atm 77.95 g/mol 6.27 g/L 66°C

5. What pressure in atmospheres will 1.36 kg of N 2O gas exert when it is compressed

in a 25.0 L cylinder and is stored in an outdoor shed where the temperature

can reach 59°C during the summer?

6. Aluminum chloride sublimes at high temperatures. What density will the

vapor have at 225°C and 0.939 atm pressure?

7. An unknown gas has a density of 0.0262 g/mL at a pressure of 0.918 atm and a

temperature of 10.°C. What is the molar mass of the gas?

8. A large balloon contains 11.7 g of helium. What volume will the helium

occupy at an altitude of 10 000 m, where the atmospheric pressure is 0.262

atm and the temperature is �50.°C?

Holt ChemFile: Problem-Solving Workbook 181 The Ideal Gas Law

9. A student collects ethane by water displacement at a temperature of 15°C

(vapor pressure of water is 1.5988 kPa) and a total pressure of 100.0 kPa. The

volume of the collection bottle is 245 mL. How many moles of ethane are in

the bottle?

10. A reaction yields 3.75 L of nitrogen monoxide. The volume is measured at

19°C and at a pressure of 1.10 atm. What mass of NO was produced by the

reaction?

11. A reaction has a theoretical yield of 8.83 g of ammonia. The reaction gives off

10.24 L of ammonia measured at 52°C and 105.3 kPa. What was the percent

yield of the reaction?

12. An unknown gas has a density of 0.405 g/L at a pressure of 0.889 atm and a

temperature of 7°C. Calculate its molar mass.

13. A paper label has been lost from an old tank of compressed gas. To help

identify the unknown gas, you must calculate its molar mass. It is known that

the tank has a capacity of 90.0 L and weighs 39.2 kg when empty. You find its

current mass to be 50.5 kg. The gauge shows a pressure of 1780 kPa when the

temperature is 18°C. What is the molar mass of the gas in the cylinder?

14. What is the pressure inside a tank that has a volume of 1.20 � 10 3 L and

contains 12.0 kg of HCl gas at a temperature of 18°C?

15. What pressure in kPa is exerted at a temperature of 20.°C by compressed

neon gas that has a density of 2.70 g/L?

16. A tank with a volume of 658 mL contains 1.50 g of neon gas. The maximum

safe pressure that the tank can withstand is 4.50 � 10 2 kPa. At what temperature

will the tank have that pressure?

17. The atmospheric pressure on Mars is about 6.75 millibars (1 bar � 100 kPa �

0.9869 atm), and the nighttime temperature can be about �75°C on the same

day that the daytime temperature goes up to �8°C. What volume would a bag

containing 1.00 g of H2 gas have at both the daytime and nighttime temperatures?

18. What is the pressure in kPa of 3.95 mol of Cl2 gas if it is compressed in a

cylinder with a volume of 850. mL at a temperature of 15°C?

19. What volume in mL will 0.00660 mol of hydrogen gas occupy at a pressure of

0.907 atm and a temperature of 9°C?

20. What volume will 8.47 kg of sulfur dioxide gas occupy at a pressure of 89.4

kPa and a temperature of 40.°C?

21. A cylinder contains 908 g of compressed helium. It is to be used to inflate a

balloon to a final pressure of 128.3 kPa at a temperature of 2°C. What will the

volume of the balloon be under these conditions?

22. The density of dry air at 27°C and 100.0 kPa is 1.162 g/L. Use this information

to calculate the molar mass of air (calculate as if air were a pure substance).

Holt ChemFile: Problem-Solving Workbook 182 The Ideal Gas Law

Stoichiometry of Gases

Now that you have worked with relationships among moles, mass, and volumes

of gases, you can easily put these to work in stoichiometry calculations. Many

reactions have gaseous reactants, gaseous products, or both.

Reactants and products that are not gases are usually measured in grams or

kilograms. As you know, you must convert these masses to amounts in moles

before you can relate the quantities by using a balanced chemical equation.

Gaseous products and reactants can be related to solid or liquid products and

reactants by using the mole ratio, just as solids and liquids are related to each

other.

Reactants and products that are gases are usually measured in liters. If the gas

is measured at STP, you will need only Avogadro’s law to relate the volume and

amount of a gas. One mole of any gas at STP occupies 22.4 L. If the gas is not at

STP, you will need to use the ideal gas law to determine the number of moles.

Once volume has been converted to amount in moles you can use the mole ratios

of products and reactants to solve stoichiometry problems involving multiple

phases of products and reactants.

n � �� RT

If the problem which you are trying to solve involves only gases, there is a

simpler way of dealing with the stoichiometric amounts. Look again at the

expression for the ideal gas law above; the molar amount of a gas is directly

related to its volume. Therefore, the mole ratios of gases given by the coefficients

in the balanced equation can be used as volume ratios of those gases to solve stoichiometry

problems. No conversion from volume to amount is required to determine

the volume of one gas from the volume of another gas in a balanced

chemical equation.

There is one condition that must be observed. Gas volumes can be related by

mole ratios only when the volumes are measured under the same conditions of

temperature and pressure. If they are not, then the volume of one of the gases

must be converted to the conditions of the other gas. Usually you will need to use

the combined gas law for this conversion.

V2 � � V1P1T2

T1P2

Holt Chemistry 183 Stoichiometry of Gases

substance A —

solid, liquid, or gas

General Plan for Solving Gas Stoichiometry Problems

in moles —

If A is a gas,

gas laws.

Volume of

if it is a gas

the mole ratio

between

A and B.

The volume ratio is the

same as the mole ratio if the

two gases are under the same

set of conditions. If the two gases

exist under different conditions,

convert using the gas laws.

substance B —

Holt ChemFile: Problem-Solving Workbook 184 Stoichiometry of Gases

If B is a gas,

if it is a gas

Ammonia can react with oxygen to produce nitrogen and water according

to the following equation.

4NH3(g) � 3O2(g) 3 2N2(g) � 6H2O(l) If 1.78 L of O2 reacts, what volume of nitrogen will be produced? Assume

that temperature and pressure remain constant.

What is given in the problem? the balanced equation, the volume of oxygen,

and the fact that the two gases exist under the

same conditions

What are you asked to find? the volume of N2 produced

Substance O 2 N 2

Molar mass NA NA

Moles NA NA

Volume of substance 1.78 L ? L

Temperature conditions NA NA

Pressure conditions NA NA

What steps are needed to calculate the volume of N 2 formed from a given volume

of O 2?

The coefficients of the balanced equation indicate the mole ratio of O 2 to N 2.

The volume ratio is the same as the mole ratio when volumes are measured

under the same conditions.

3 4

Volume of O 2

multiply by the volume ratio,

N2 O2 L O2 �

volume ratio,

given 2 L N 2

3 L O 2

� L N2 Volume of N 2

Holt ChemFile: Problem-Solving Workbook 185 Stoichiometry of Gases

Yes; units canceled to give L N2. Is the number of significant figures correct?

Yes; the volume of N2 should be 2/3 the volume of O2. Practice

1.78 L O2 � 2 L N2 � 1.19 L N2 3 L O2 1. In one method of manufacturing nitric acid, ammonia is oxidized to nitrogen

monoxide and water.

4NH3(g) � 5O2(g) → 4NO(g) � 6H2O(l) What volume of oxygen will be used in a reaction of 2800 L of NH3? What

volume of NO will be produced? All volumes are measured under the same

conditions. ans: 3500 L O2, 2800 L NO

2. Fluorine gas reacts violently with water to produce hydrogen fluoride and

ozone according to the following equation.

3F2(g) � 3H2O(l) → 6HF(g) � O3(g) What volumes of O3 and HF gas would be produced by the complete reaction

of 3.60 � 10 4 mL of fluorine gas? All gases are measured under the same

conditions. ans: 1.20 � 10 4 mL O3, 7.20 � 10 4 mL HF

Holt ChemFile: Problem-Solving Workbook 186 Stoichiometry of Gases

Ethylene gas burns in air according to the following equation.

C2H4(g) � 3O2(g) 3 2CO2(g) � 2H2O(l) If 13.8 L of C2H4 measured at 21°C and 1.038 atm burns completely with

oxygen, calculate the volume of CO2 produced, assuming the CO2 is

measured at 44°C and 0.989 atm.

What is given in the problem? the balanced equation, the volume of ethylene,

the conditions under which the ethylene was

measured, and the conditions under which the

CO2 is measured

What are you asked to find? the volume of CO2 produced as measured at the

specified conditions

Substance C2H4 CO2 Coefficient in balanced equation 1 2

Volume of substance 13.8 L ? L

Temperature conditions 21°C � 294 K 44°C � 317 K

Pressure conditions 1.083 atm 0.989 atm

What steps are needed to calculate the volume of CO 2 formed from the complete

burning of a given volume of C 2H 4?

Use the volume ratio of C 2H 4 to CO 2 to calculate the volume of CO 2 at the same

conditions as C 2H 4. Convert to the volume of CO 2 for the given conditions using

the combined gas law.

Holt ChemFile: Problem-Solving Workbook 187 Stoichiometry of Gases

* at 294 K and 1.083 atm

Neither pressure nor temperature is constant; therefore, the combined gas law

must be used to calculate the volume of CO2 at the final temperature and pressure.

Volume of C 2 H 4

in L at

initial conditions

CO 2

C 2 H 4

P1 �

P2 T1 given given

13.8 L C2H4* � 2 L CO2 � 27.6 L CO2* 1 L C2H4 * at 294 K and 1.083 atm

Volume of CO 2

in L at the same

as initial C 2 H 4

C2H4 given 2 L CO2 L C2H4 * �

1 L C 2 H 4

Solve the combined-gas-law equation for V2. 317 K� � 1.083 �atm � 27.6 L CO2 V2 �� 32.6 L CO2 0.989 �atm � 294 K�

Yes; units canceled to give L CO2. Is the number of significant figures correct?

Yes; the number of significant figures is correct because the data had a minimum

of three significant figures.

Yes; the changes in both pressure and temperature increased the volume by

small factors.

Holt ChemFile: Problem-Solving Workbook 188 Stoichiometry of Gases

above

V1 use the combined gas law to

convert from the initial

temperature and pressure to

the final temperature and

� L CO2 *

final conditions

1. A sample of ethanol burns in O2 to form CO2 and H2O according to the following

C2H5OH � 3O2 → 2CO2 � 3H2O If the combustion uses 55.8 mL of oxygen measured at 2.26 atm and 40.°C,

what volume of CO2 is produced when measured at STP? ans: 73.3 mL CO2 2. Dinitrogen pentoxide decomposes into nitrogen dioxide and oxygen. If 5.00 L

of N 2O 5 reacts at STP, what volume of NO 2 is produced when measured at

64.5°C and 1.76 atm? ans: 7.02 L

Holt ChemFile: Problem-Solving Workbook 189 Stoichiometry of Gases

When arsenic(III) sulfide is roasted in air, it reacts with oxygen to produce

arsenic(III) oxide and sulfur dioxide according to the following

2As2S3(s) � 9O2(g) 3 2As2O3(s) � 6SO2(g) When 89.5 g of As2S3 is roasted with excess oxygen, what volume of SO2 is produced? The gaseous product is measured at 20°C and 98.0 kPa.

What is given in the problem? the balanced equation, the mass of As2S3, and

the pressure and temperature conditions under

which the SO2 is measured

What are you asked to find? the volume of SO2 produced as measured at the

given conditions

Substance As 2S 3(s) SO 2(g)

Coefficient in balanced equation 2 6

Molar mass* 246.05 g/mol NA

Mass of substance 89.5 g NA

Volume of substance NA ? L

Temperature conditions NA 20°C � 293 K

Pressure conditions NA 98.0 kPa

What steps are needed to calculate the volume of SO 2 formed from the reaction

of a given mass of As 2S 3?

Use the molar mass of As 2S 3 to determine the number of moles that react. Use

the mole ratio from the balanced chemical equation to determine the amount in

moles of SO 2 formed. Use the ideal-gas-law equation to determine the volume of

SO 2 formed from the amount in moles.

Holt ChemFile: Problem-Solving Workbook 190 Stoichiometry of Gases

Rearrange the ideal-gas-law equation to solve for the unknown quantity, V.

V � � nRT

As2S3 (s) in g

inverse molar

mass of As2S3 Amount of

As 2 S 3 (s) in mol

mol SO 2

multiply by the mole ratio,

molar mass As 2 S 3

ideal gas constant

�8.314 L�kPa�mol�K �

kPa

89.5 g As 2S 3 � 1 mol As 2S 3

246.05 g As 2S 3

1.09 mol SO 2 � 8.314 L�kPa/mol�K � 293 K

98.0 kPa

Yes; units canceled to give liters of SO2. Is the number of significant figures correct?

Yes; computation of the amount of SO2 can be approximated as (9/25) � 3 � 27/25,

so you would expect an answer a little greater than 1. At a temperature slightly

above standard temperature, you would expect a volume a little greater than

22.4 L.

SO 2

As 2 S 3

mole ratio, SO 2

6 mol SO2 �

� 1.09 mol SO2 2 mol As2S3 � 27.1 L SO 2

Holt ChemFile: Problem-Solving Workbook 191 Stoichiometry of Gases

SO 2 (g) in mol

SO 2 (g) in L

g As2S3 � � � mol SO2 6 mol SO 1 mol As2S3 2

246.05 g As2S3 2 mol As2S3 � L SO 2

use the ideal

gas law to

1. Complete the table below using the following equation, which represents a

reaction that produces aluminum chloride.

2Al(s) � 3Cl2(g) → 2AlCl3(s) Mass Al Volume Cl 2 Conditions Mass AlCl 3

a. excess ? L STP 7.15 g

b. 19.4 g ? L STP NA

c. 1.559 kg ? L 20.°C and NA

0.945 atm

d. excess 920. L STP ? g

e. ? g 1.049 mL 37°C and NA

5.00 atm

f. 500.00 kg ? m 3

15°C and NA

83.0 kPa

ans: 1.80 L Cl 2

ans: 24.2 L Cl 2

ans: 2.21 � 10 3 L Cl 2

ans: 3.65 � 10 3 g AlCl 3

ans: 3.71 � 10 �3 g Al

ans: 8.02 � 10 2 m 3 Cl 2

Holt ChemFile: Problem-Solving Workbook 192 Stoichiometry of Gases

1. The industrial production of ammonia proceeds according to the following

N2(g) � 3H2(g) → 2NH3(g) a. What volume of nitrogen at STP is needed to react with 57.0 mL of hydrogen

measured at STP?

b. What volume of NH3 at STP can be produced from the complete reaction of

6.39 � 10 4 L of hydrogen?

c. If 20.0 mol of nitrogen is available, what volume of NH3 at STP can be

d. What volume of H2 at STP will be needed to produce 800. L of ammonia,

measured at 55°C and 0.900 atm?

2. Propane burns according to the following equation.

C3H8(g) � 5O2(g) → 3CO2(g) � 4H2O(g) a. What volume of water vapor measured at 250.°C and 1.00 atm is produced

when 3.0 L of propane at STP is burned?

b. What volume of oxygen at 20.°C and 102.6 kPa is used if 640. L of CO2 is

produced? The CO2 is also measured at 20.°C and 102.6 kPa.

c. If 465 mL of oxygen at STP is used in the reaction, what volume of CO 2,

measured at 37°C and 0.973 atm, is produced?

d. When 2.50 L of C 3H 8 at STP burns, what total volume of gaseous products

is formed? The volume of the products is measured at 175°C and 1.14 atm.

3. Carbon monoxide will burn in air to produce CO2 according to the following

2CO(g) � O2(g) → 2CO2(g) What volume of oxygen at STP will be needed to react with 3500. L of CO

measured at 20.°C and a pressure of 0.953 atm?

4. Silicon tetrafluoride gas can be produced by the action of HF on silica according

SiO2(s) � 4HF(g) → SiF4(g) � 2H2O(l) 1.00 L of HF gas under pressure at 3.48 atm and a temperature of 25°C reacts

completely with SiO2 to form SiF4. What volume of SiF4, measured at 15°C

and 0.940 atm, is produced by this reaction?

5. One method used in the eighteenth century to generate hydrogen was to pass

steam through red-hot steel tubes. The following reaction takes place.

3Fe(s) � 4H2O(g) → Fe3O4(s) � 4H2(g) a. What volume of hydrogen at STP can be produced by the reaction of 6.28 g

of iron?

Holt ChemFile: Problem-Solving Workbook 193 Stoichiometry of Gases

b. What mass of iron will react with 500. L of steam at 250.°C and 1.00 atm

pressure?

c. If 285 g of Fe3O4 are formed, what volume of hydrogen, measured at 20.°C

and 1.06 atm, is produced?

6. Sodium reacts vigorously with water to produce hydrogen and sodium

hydroxide according to the following equation.

2Na(s) � 2H2O(l) → 2NaOH(aq) � H2(g) If 0.027 g of sodium reacts with excess water, what volume of hydrogen at

STP is formed?

7. Diethyl ether burns in air according to the following equation.

C4H10O(l) � 6O2(g) → 4CO2(g) � 5H2O(l) If 7.15 L of CO2 is produced at a temperature of 125°C and a pressure of 1.02

atm, what volume of oxygen, measured at STP, was consumed and what mass

of diethyl ether was burned?

8. When nitroglycerin detonates, it produces large volumes of hot gases almost

instantly according to the following equation.

4C3H5N3O9(l) → 6N2(g) � 12CO2(g) � 10H2O(g) � O2(g) a. When 0.100 mol of nitroglycerin explodes, what volume of each gas measured

at STP is produced?

b. What total volume of gases is produced at 300.°C and 1.00 atm when 10.0 g

of nitroglycerin explodes?

9. Dinitrogen monoxide can be prepared by heating ammonium nitrate, which

decomposes according to the following equation.

NH4NO3(s) → N2O(g) � 2H2O(l) What mass of ammonium nitrate should be decomposed in order to produce

250. mL of N2O, measured at STP?

10. Phosphine, PH3, is the phosphorus analogue to ammonia, NH3. It can be

produced by the reaction between calcium phosphide and water according to

the following equation.

Ca3P2(s) � 6H2O(l) → 3Ca(OH) 2(s and aq) � 2PH3(g) What volume of phosphine, measured at 18°C and 102.4 kPa, is produced by

the reaction of 8.46 g of Ca3P2? 11. In one method of producing aluminum chloride, HCl gas is passed over aluminum

and the following reaction takes place.

2Al(s) � 6HCl(g) → 2AlCl3(g) � 3H2(g) What mass of Al should be on hand in order to produce 6.0 � 10 3 kg of AlCl3? What volume of compressed HCl at 4.71 atm and a temperature of 43°C

should be on hand at the same time?

Holt ChemFile: Problem-Solving Workbook 194 Stoichiometry of Gases

12. Urea, (NH2) 2CO, is an important fertilizer that is manufactured by the following

reaction.

2NH3(g) � CO2(g) → (NH2) 2CO(s) � H2O(g) What volume of NH3 at STP will be needed to produce 8.50 � 10 4 kg of urea if

there is an 89.5% yield in the process?

13. An obsolete method of generating oxygen in the laboratory involves the

decomposition of barium peroxide by the following equation.

2BaO2(s) → 2BaO(s) � O2(g) What mass of BaO2 reacted if 265 mL of O2 is collected by water displacement

at 0.975 atm and 10.°C?

14. It is possible to generate chlorine gas by dripping concentrated HCl solution

onto solid potassium permanganate according to the following equation.

2KMnO4(s) � 16HCl(aq) → 2KCl(aq) � 2MnCl2(aq) � 8H2O(l) � 5Cl2(g) If excess HCl is dripped onto 15.0 g of KMnO4, what volume of Cl2 will be

produced? The Cl2 is measured at 15°C and 0.959 atm.

15. Ammonia can be oxidized in the presence of a platinum catalyst according to

4NH3(g) � 5O2(g) → 4NO(g) � 6H2O(l) The NO that is produced reacts almost immediately with additional oxygen

according to the following equation.

2NO(g) � O2(g) → 2NO2(g) If 35.0 kL of oxygen at STP react in the first reaction, what volume of NH3 at

STP reacts with it? What volume of NO2 at STP will be formed in the second

reaction, assuming there is excess oxygen that was not used up in the first

16. Oxygen can be generated in the laboratory by heating potassium chlorate. The

reaction is represented by the following equation.

2KClO3(s) → 2KCl(s) � 3O2(g) What mass of KClO3 must be used in order to generate 5.00 L of O2, measured

at STP?

17. One of the reactions in the Solvay process is used to make sodium hydrogen

carbonate. It occurs when carbon dioxide and ammonia are passed through

concentrated salt brine. The following equation represents the reaction.

NaCl(aq) � H2O(l) � CO2(g) � NH3(g) → NaHCO3(s) � NH4Cl(aq) a. What volume of NH3 at 25°C and 1.00 atm pressure will be required if

38 000 L of CO2, measured under the same conditions, react to form

NaHCO3? b. What mass of NaHCO3 can be formed when the gases in (a) react with

NaCl?

Holt ChemFile: Problem-Solving Workbook 195 Stoichiometry of Gases

c. If this reaction forms 46.0 kg of NaHCO3, what volume of NH3, measured

at STP, reacted?

d. What volume of CO2, compressed in a tank at 5.50 atm and a temperature

of 42°C, will be needed to produce 100.00 kg of NaHCO3? 18. The combustion of butane is represented in the following equation.

2C4H10(g) � 13O2(g) → 8CO2(g) � 10H2O(l) a. If 4.74 g of butane react with excess oxygen, what volume of CO 2, measured

at 150.°C and 1.14 atm, will be formed?

b. What volume of oxygen, measured at 0.980 atm and 75°C, will be consumed

by the complete combustion of 0.500 g of butane?

c. A butane-fueled torch has a mass of 876.2 g. After burning for some time,

the torch has a mass of 859.3 g. What volume of CO2, at STP, was formed

while the torch burned?

d. What mass of H2O is produced when butane burns and produces 3720 L of

CO2, measured at 35°C and 0.993 atm pressure?

Holt ChemFile: Problem-Solving Workbook 196 Stoichiometry of Gases

Concentration of Solutions

There are three principal ways to express solution concentration in chemistry—

percentage by mass, molarity, and molality.

The following table compares these three ways of stating solution concentration.

Examining the method of preparation of the three types may help you

understand the differences among them.

Symbol Meaning How to prepare

Percentage %

Grams solute per 5%: Dissolve 5 g of

100 g of solution solute in 95 g solvent.

Molarity M Moles solute per 5 M: Dissolve 5 mol of

liter of solution solute in solvent and

add solvent to make

1 L of solution.

Molality m

PERCENTAGE CONCENTRATION

You will find percentages of solutes stated on the labels of many commercial

products, such as household cleaners, liquid pesticide solutions, and shampoos. If

your sink becomes clogged, you might buy a bottle of drain opener whose label

states that it is a 2.4% sodium hydroxide solution. This means that the bottle contains

2.4 g of NaOH for every 100 g of solution.

Computing percentage concentration is very much like computing percentage

composition. Both involve finding the percentage of a single component of a multicomponent

system. In each type of percentage calculation, the mass of the

important component (in percentage concentration, the solute) is divided by the

total mass of the system and multiplied by 100 to yield a percentage. In percentage

concentration, the solute is the important component, and the total mass of

the system is the mass of the solute plus the mass of the solvent.

General Plan for Solving Percentage Concentration Problems

solvent

Moles solute per

kilogram of solvent

� solute

concentration �

mass of solute

� 100

mass of solution

concentration

by mass

5 m: Dissolve 5 mol of

solute in 1 kg of solvent.

solution

Holt ChemFile: Problem-Solving Workbook 197 Concentration of Solutions

What is the percentage by mass of a solution made by dissolving 0.49 g of

potassium sulfate in 12.70 g of water?

What is given in the problem? the mass of solvent, and the mass of solute,

K2SO4 What are you asked to find? the concentration of the solution expressed as

a percentage by mass

Mass of solvent 12.70 g H 2O

Mass of solute 0.49 g K 2SO 4

Concentration (% by mass) ? %

What step is needed to calculate the concentration of the solution as a percentage

by mass?

Divide the mass of solute by the mass of the solution and multiply by 100.

Mass of water in g � Mass of K2SO4 in g �

0.49 g K

percentage concentration �

2SO4 �� 100 � 3.7% K2SO4 0.49 g K2SO4 � 12.70 g H2O EVALUATE

Yes; percentage K2SO4 was required.

of two significant figures.

Mass of K2SO4 solution in g

percentage solute mass

solution mass

K2SO4 by mass

Holt ChemFile: Problem-Solving Workbook 198 Concentration of Solutions

g K2SO4 percentage concentration � � 100

g K2SO4 � g H2O given given

Yes; the computation can be approximated as 0.5/13 � 100 � 3.8%.

1. What is the percentage concentration of 75.0 g of ethanol dissolved in 500.0 g

of water? ans: 13.0% ethanol

2. A chemist dissolves 3.50 g of potassium iodate and 6.23 g of potassium

hydroxide in 805.05 g of water. What is the percentage concentration of each

solute in the solution? ans: 0.430% KIO 3, 0.765% KOH

3. A student wants to make a 5.00% solution of rubidium chloride using 0.377 g

of the substance. What mass of water will be needed to make the solution?

ans: 7.16 g H 2O

4. What mass of lithium nitrate would have to be dissolved in 30.0 g of water in

order to make an 18.0% solution? ans: 6.59 g LiNO 3

Holt ChemFile: Problem-Solving Workbook 199 Concentration of Solutions

MOLARITY

Molarity is the most common way to express concentration in chemistry. Molarity

is the number of moles of solute per liter of solution and is given as a number followed

by a capital M. A 2 M solution of nitric acid contains 2 mol of HNO3 per

liter of solution. As you know, substances react in mole ratios. Knowing the

molar concentration of a solution allows you to measure a number of moles of a

dissolved substance by measuring the volume of solution.

General Plan for Solving Molarity Problems

solute

Amount

of solute

of the solute.

M �

moles solute

liter solution

Molar

concentration,

Volume

of solution

Holt ChemFile: Problem-Solving Workbook 200 Concentration of Solutions

What is the molarity of a solution prepared by dissolving 37.94 g of

potassium hydroxide in some water and then diluting the solution to a

volume of 500.00 mL?

What is given in the problem? the mass of the solute, KOH, and the final volume

of the solution

What are you asked to find? the concentration of the solution expressed as

molarity

What steps are needed to calculate the concentration of the solution as molarity?

Determine the amount in moles of solute; calculate the moles per liter of solution.

Mass of KOH

inverted molar

mass of KOH

Amount of KOH

Mass of solute 37.94 g KOH

Moles of solute ? mol KOH

Molar mass of solute* 56.11 g/mol

Volume of solution 500.00 mL

Concentration (molarity) ? M

Molarity of KOH

molar mass of KOH

given 1 mol KOH � mol KOH

g KOH � 56.11 g KOH

Volume of KOH

solution in mL

solution in L

Holt ChemFile: Problem-Solving Workbook 201 Concentration of Solutions

500.00 mL � solution � �� 0.500 00 L solution

0.6762 mol KOH

�� 1.352 M

0.500 00 L solution

Yes; units canceled to give moles KOH per liter of solution.

of four significant figures.

Yes; note that 0.6762 mol is approximately 2/3 mol and 0.500 00 L is 1/2 L. Thus,

the calculation can be estimated as (2/3)/(1/2) � 4/3, which is very close to the

result.

given 1 L � L solution

mL solution � 1000 mL

37.94 g KOH �

mol KOH

� M solution

L solution

1. Determine the molarity of a solution prepared by dissolving 141.6 g of citric

acid, C 3H 5O(COOH) 3, in water and then diluting the resulting solution to

3500.0 mL. ans: 0.2106 M

1 mol KOH

� 0.6762 mol KOH

56.11 g KOH

Holt ChemFile: Problem-Solving Workbook 202 Concentration of Solutions

2. What is the molarity of a salt solution made by dissolving 280.0 mg of NaCl in

2.00 mL of water? Assume the final volume is the same as the volume of the

water. ans: 2.40 M

3. What is the molarity of a solution that contains 390.0 g of acetic acid,

CH 3COOH, dissolved in enough acetone to make 1000.0 mL of solution?

ans: 6.494 M

Holt ChemFile: Problem-Solving Workbook 203 Concentration of Solutions

An analytical chemist wants to make 750.0 mL of a 6.00 M solution of

sodium hydroxide. What mass of NaOH will the chemist need to make this

solution?

What is given in the problem? the identity of the solute, the total volume of

solution, and the molarity of the solution

What are you asked to find? the mass of solute to dissolve

What steps are needed to calculate the mass of solute needed?

Determine the amount in moles needed for the solution required, and convert to

grams by multiplying by the molar mass of the solute.

Molarity of

NaOH solution

Mass of solute ? g NaOH

Molar mass of solute 40.00 g/mol

Moles of solute ? mol NaOH

Volume of solution 750.0 mL

Concentration (molarity) 6.00 M

mol NaOH

Volume of NaOH

750.0 �mL solution � �� 0.7500 L solution

mL �

NaOH solution in mL

Amount of NaOH

Mass of NaOH

molar mass of NaOH

above 40.00 g NaOH

� L solution � � g NaOH

1 mol NaOH

Holt ChemFile: Problem-Solving Workbook 204 Concentration of Solutions

Yes; units canceled to give grams of NaOH.

Yes; the calculation can be estimated as (3/4) � (6)(40) � (3/4) � 240 � 180.

6.00 mol NaOH

� 0.7500 L solution �

1. What mass of glucose, C 6H 12O 6, would be required to prepare 5.000 � 10 3 L of

a 0.215 M solution? ans: 1.94 � 10 5 g

2. What mass of magnesium bromide would be required to prepare 720. mL of a

0.0939 M aqueous solution? ans: 12.4 g

3. What mass of ammonium chloride is dissolved in 300. mL of a 0.875 M solution?

ans: 14.0 g

40.00 g NaOH

� 180. g NaOH

Holt ChemFile: Problem-Solving Workbook 205 Concentration of Solutions

MOLALITY

Molality is the amount in moles of solute per kilogram of solvent and is given by

a number followed by an italic lowercase m. A 5 m aqueous solution of glucose

contains 5 mol of C6H12O6 per kilogram of water. Molal concentration is important

primarily in working with colligative properties of solutions.

General Plan for Solving Molality Problems

m �

mol solute

kg solvent

Molal

solvent in

Holt ChemFile: Problem-Solving Workbook 206 Concentration of Solutions

1 kg � 1000 g.

Determine the molal concentration of a solution containing 81.3 g of

ethylene glycol, HOCH 2CH 2OH, dissolved in 166 g of water.

What is given in the problem? the mass of ethylene glycol dissolved, and the

mass of the solvent, water

What are you asked to find? the molal concentration of the solution

Mass of solute 81.3 g ethylene glycol

Molar mass of solute 62.08 g/mol ethylene glycol

Moles of solute ? mol ethylene glycol

Mass of solvent 166 g H2O Concentration (molality) ? m

What steps are needed to calculate the molal concentration of the ethylene glycol

Determine the amount of solute in moles and the mass of solvent in kilograms;

calculate the moles of solute per kilogram of solvent.

Mass of C2H6O2 in g

mass of C 2 H 6 O 2

Amount of C2H6O2 in mol

Molality of C 2 H 6 O 2

mol C2H6O2 � m C2H6O2 solution

kg H2O calculated above

moles C2H6O2 m �

molar mass of C 2 H 6 O 2

given 1 mol C 2 H 6 O 2 � mol C2 H 6 O 2

g C 2 H 6 O 2 � 62.08 g C2 H 6 O 2

given 1 kg � kg H2 O

g H 2 O � 1000 g

Mass of H2O in kg

Holt ChemFile: Problem-Solving Workbook 207 Concentration of Solutionss

166 g� H2O � �� 0.166 kg H2O 1000

1.31

mol

C2H6O2 ��

� 7.89 m

0.166

H2O

Yes; units canceled to give moles C2H6O2 per kilogram of solvent.

Yes; because 1.31 mol is approximately 4/3 mol and 0.166 kg is approximately 1/6

kg, the calculation can be estimated as (4/3)/(1/6) � 24/3 � 8, which is very

close to the result.

81.3 g C2H6O2 � 1 mol C2H6O2 � 1.31 mol C2H6O2 62.08 g C2H6O2 1. Determine the molality of a solution of 560 g of acetone, CH 3COCH 3, in 620 g

of water. ans: 16 m

2. What is the molality of a solution of 12.9 g of fructose, C 6H 12O 6, in 31.0 g of

water? ans: 2.31 m

3. How many moles of 2-butanol, CH 3CHOHCH 2CH 3, must be dissolved in 125 g

of ethanol in order to produce a 12.0 m 2-butanol solution? What mass of 2butanol

is this? ans: 1.50 mol 2-butanol, 111 g 2-butanol

Holt ChemFile: Problem-Solving Workbook 208 Concentration of Solutions

1. Complete the table below by determining the missing quantity in each example.

All solutions are aqueous. Any quantity that is not applicable to a given

solution is marked NA.

Quantity Quantity

Mass of of solution of solvent

Solution made solute used made used

a. 12.0% KMnO4 ? g KMnO4 500.0 g ? g H2O b. 0.60 M BaCl2 ? g BaCl2 1.750 L NA

c. 6.20 m glycerol,

HOCH2CHOHCH2OH ? g glycerol NA 800.0 g H2O d. ? M K 2Cr 2O 7 12.27 g K 2Cr 2O 7 650. mL NA

e. ? m CaCl 2 288 g CaCl 2 NA 2.04 kg H 2O

f. 0.160 M NaCl ? g NaCl 25.0 mL NA

g. 2.00 m glucose,

C6H12O6 ? g glucose ? g solution 1.50 kg H2O 2. How many moles of H2SO4 are in 2.50 L of a 4.25 M aqueous solution?

3. Determine the molal concentration of 71.5 g of linoleic acid, C18H32O2, in

525 g of hexane, C6H14. 4. You have a solution that is 16.2% sodium thiosulfate, Na2S2O3, by mass.

a. What mass of sodium thiosulfate is in 80.0 g of solution?

b. How many moles of sodium thiosulfate are in 80.0 g of solution?

c. If 80.0 g of the sodium thiosulfate solution is diluted to 250.0 mL with

water, what is the molarity of the resulting solution?

5. What mass of anhydrous cobalt(II) chloride would be needed in order to

make 650.00 mL of a 4.00 M cobalt(II) chloride solution?

6. A student wants to make a 0.150 M aqueous solution of silver nitrate, AgNO3 and has a bottle containing 11.27 g of silver nitrate. What should be the final

volume of the solution?

7. What mass of urea, NH2CONH2, must be dissolved in 2250 g of water in order

to prepare a 1.50 m solution?

8. What mass of barium nitrate is dissolved in 21.29 mL of a 3.38 M solution?

9. Describe what you would do to prepare 100.0 g of a 3.5% solution of ammonium

sulfate in water.

10. What mass of anhydrous calcium chloride should be dissolved in 590.0 g of

water in order to produce a 0.82 m solution?

11. How many moles of ammonia are in 0.250 L of a 5.00 M aqueous ammonia

solution? If this solution were diluted to 1.000 L, what would be the molarity

of the resulting solution?

Holt ChemFile: Problem-Solving Workbook 209 Concentration of Solutions

12. What is the molar mass of a solute if 62.0 g of the solute in 125 g of water

produce a 5.3 m solution?

13. A saline solution is 0.9% NaCl. What masses of NaCl and water would be

required to prepare 50. L of this saline solution? Assume that the density of

water is 1.000 g/mL and that the NaCl does not add to the volume of the

solution.

14. A student weighs an empty beaker on a balance and finds its mass to be

68.60 g. The student weighs the beaker again after adding water and finds the

new mass to be 115.12 g. A mass of 4.08 g of glucose is then dissolved in the

water. What is the percentage concentration of glucose in the solution?

15. The density of ethyl acetate at 20°C is 0.902 g/mL. What volume of ethyl

acetate at 20°C would be required to prepare a 2.0% solution of cellulose

nitrate using 25 g of cellulose nitrate?

16. Aqueous cadmium chloride reacts with sodium sulfide to produce brightyellow

cadmium sulfide. Write the balanced equation for this reaction and

answer the following questions.

a. How many moles of CdCl2 are in 50.00 mL of a 3.91 M solution?

b. If the solution in (a) reacted with excess sodium sulfide, how many moles

of CdS would be formed?

c. What mass of CdS would be formed?

17. What mass of H2SO4 is contained in 60.00 mL of a 5.85 M solution of sulfuric

acid?

18. A truck carrying 22.5 kL of 6.83 M aqueous hydrochloric acid used to clean

brick and masonry has overturned. The authorities plan to neutralize the acid

with sodium carbonate. How many moles of HCl will have to be neutralized?

19. A chemist wants to produce 12.00 g of barium sulfate by reacting a 0.600 M

BaCl2 solution with excess H2SO4, as shown in the reaction below. What

volume of the BaCl2 solution should be used?

BaCl2 � H2SO4 → BaSO4 � 2HCl

20. Many substances are hydrates. Whenever you make a solution, it is important

to know whether or not the solute you are using is a hydrate and, if it is a

hydrate, how many molecules of water are present per formula unit of the

substance. This water must be taken into account when weighing out the

solute. Something else to remember when making aqueous solutions from

hydrates is that once the hydrate is dissolved, the water of hydration is considered

to be part of the solvent. A common hydrate used in the chemistry

laboratory is copper sulfate pentahydrate, CuSO4�5H2O. Describe how you

would make each of the following solutions using CuSO4�5H2O. Specify

masses and volumes as needed.

a. 100. g of a 6.00% solution of CuSO4 b. 1.00 L of a 0.800 M solution of CuSO4 c. a 3.5 m solution of CuSO4 in 1.0 kg of water

Holt ChemFile: Problem-Solving Workbook 210 Concentration of Solutions

21. What mass of calcium chloride hexahydrate is required in order to make 700.0

mL of a 2.50 M solution?

22. What mass of the amino acid arginine, C6H14N4O2, would be required to make

1.250 L of a 0.00205 M solution?

23. How much water would you have to add to 2.402 kg of nickel(II) sulfate

hexahydrate in order to prepare a 25.00% solution?

24. What mass of potassium aluminum sulfate dodecahydrate, KAl(SO4) 2�12H2O, would be needed to prepare 35.00 g of a 15.00% KAl(SO4) 2 solution? What

mass of water would be added to make this solution?

Holt ChemFile: Problem-Solving Workbook 211 Concentration of Solutions

Dilutions

Suppose you work in the laboratory of a paint company where you use 100 mL of

a 0.1 M solution of zinc chloride in a quality-control test that you carry out 10

times a day. It would be tedious and time-consuming to continually measure out

small amounts of ZnCl2 to make 100 mL of this solution. Of course, you could

make many liters of the solution at one time, but that would require several large

containers to store the solution. The answer to the problem is to make a much

more concentrated solution and then dilute it with water to make the less concentrated

solution that you need. The more-concentrated solution is called a

stock solution. You could make a 1 M ZnCl2 solution by measuring out 1 mol of

zinc chloride, 136.3 g, and dissolving it in enough water to make a liter of solution.

This solution is 10 times as concentrated as the solution you need. Every

time you need the test solution, you can measure out 10 mL of the 1 M solution

and dilute it to 100 mL to yield 100 mL of 0.1 M ZnCl2 solution.

To make a solution by dilution, you must determine the volume of stock solution

to use and the amount of solvent needed to dilute to the concentration you

need. As you have learned, the molarity of a solution is its concentration in moles

of solute per liter of solution. Molarity is found by dividing the moles of solute by

the number of liters of solution.

M �� liter

So, for a measured volume of any solution:

amount of solute in mol � molarity � volume of solution

If this measured volume of solution is diluted to a new volume by adding solvent,

the new, larger volume still contains the same number of moles of solute.

Therefore, where 1 and 2 represent the concentrated and diluted solutions:

Therefore:

molarity 1 � volume 1 � moles solute � molarity 2 � volume 2

molarity 1 � volume 1 � molarity 2 � volume 2

This relationship applies whenever solution 2 is made from solution 1 by dilution.

Holt ChemFile: Problem-Solving Workbook 212 Dilutions

General Plan for Solving Dilution Problems

M 1 V 1 � M 2 V 2

The equation used to

will be one of the following four:

M1V1 M1V1 V2 � M2 �

M2 V2 Rearrange the equation

M2V2 M1 , , V1 � , M1 �

known quantities for its

symbol, and calculate.

molarity or volume

M 2 V 2

Holt ChemFile: Problem-Solving Workbook 213 Dilutions

What is the molarity of a solution that is made by diluting 50.00 mL of a

4.74 M solution of HCl to 250.00 mL?

What is given in the problem? the molarity of the stock solution, the volume

used to dilute, and the volume of the diluted

What are you asked to find? the molarity of the diluted solution

Concentration of the stock solution (M 1) 4.74 M HCl

Volume of stock solution used (V 1) 50.00 mL

Volume of diluted solution (V2) 250.00 mL

Concentration of the diluted solution (M2) ? M

What step is needed to calculate the concentration of the diluted solution?

Apply the principle that volume 1 � molarity 1 � volume 2 � molarity 2.

M1V1 � M2V2 rearrange the equation

algebraically to solve for M 2

substitute each of the known

quantities for its symbol and

calculate

M1 � V1 M2 �

V2 COMPUTE

Note: Even though molarity is moles per liter, you can use volumes in milliliters

along with molarity whenever the units cancel.

4.74 M � 50.00 �mL

M2 �� 0.948 M

250.00 �mL

Yes; molarity (mol/L) was required.

M 2 �

M 1 V 1

Holt ChemFile: Problem-Solving Workbook 214 Dilutions

Yes; the computation is the same as 4.74/5, which is a little less than 1.

1. Complete the table below by calculating the missing value in each row.

Molarity Volume Molarity Volume

of stock of stock of dilute of dilute

solution solution solution solution

a. 0.500 M KBr 20.00 mL ? M KBr 100.00 mL

b. 1.00 M LiOH ? mL 0.075 M LiOH 500.00 mL

c. ? M HI 5.00 mL 0.0493 M HI 100.00 mL

d. 12.0 M HCl 0.250 L 1.8 M HCl ? L

e. 7.44 M NH3 ? mL 0.093 M NH3 4.00 L

ans: 0.100 M

ans: 38 mL

ans: 0.986 M

ans: 1.7 L

ans: 50. mL

Holt ChemFile: Problem-Solving Workbook 215 Dilutions

What volume of water would you add to 15.00 mL of a 6.77 M solution of

nitric acid in order to get a 1.50 M solution?

of stock solution, and the molarity of the

diluted solution

What are you asked to find? the volume of water to add to make the dilute

Concentration of the stock solution (M 1) 6.77 M HNO 3

Volume of stock solution used (V 1) 15.00 mL

Molarity of the diluted solution (M 2) 1.50 M HNO 3

Volume of diluted solution (V2) ? mL

Volume of water to add ? mL

What steps are needed to calculate the amount of water to add to dilute a solution

to the given molarity?

Apply the principle that volume 1 � molarity 1 � volume 2 � molarity 2. Subtract

the stock solution volume from the final volume to determine the amount of

water to add.

V 2 �

solve the equation

algebraically for the V 2

subtract the original volume

from final diluted volume

V 2 � V 1 � V water added to dilute

substitute each of the known quantities

for its symbol and calculate

Volume of water

added to dilute

volume

Holt ChemFile: Problem-Solving Workbook 216 Dilutions

M 1 � V 1

M 2

� V water added to dilute

6.77 M� � 15.00 mL

V2 �� 67.7 mL

1.50 M�

67.7 mL � 15.00 mL � 52.7 mL H2O EVALUATE

Yes; volume of water was required.

Yes; the dilution was to a concentration of less than 1/4 of the original concentration.

Thus, the volume of the diluted solution should be more than four times

the original volume, 4 � 15 mL � 60 mL.

1. What volume of water would be added to 16.5 mL of a 0.0813 M solution of

sodium borate in order to get a 0.0200 M solution? ans: 50.6 mL H 2O

Holt ChemFile: Problem-Solving Workbook 217 Dilutions

1. What is the molarity of a solution of ammonium chloride prepared by diluting

50.00 mL of a 3.79 M NH4Cl solution to 2.00 L?

2. A student takes a sample of KOH solution and dilutes it with 100.00 mL of

water. The student determines that the diluted solution is 0.046 M KOH, but

has forgotten to record the volume of the original sample. The concentration

of the original solution is 2.09 M. What was the volume of the original sample?

3. A chemist wants to prepare a stock solution of H2SO4 so that samples of 20.00

mL will produce a solution with a concentration of 0.50 M when added to

100.0 mL of water.

a. What should the molarity of the stock solution be?

b. If the chemist wants to prepare 5.00 L of the stock solution from concentrated

H2SO4, which is 18.0 M, what volume of concentrated acid should be

used?

c. The density of 18.0 M H2SO4 is 1.84 g/mL. What mass of concentrated H2SO4 should be used to make the stock solution in (b)?

4. To what volume should 1.19 mL of an 8.00 M acetic acid solution be diluted in

order to obtain a final solution that is 1.50 M?

5. What volume of a 5.75 M formic acid solution should be used to prepare

2.00 L of a 1.00 M formic acid solution?

6. A 25.00 mL sample of ammonium nitrate solution produces a 0.186 M solution

when diluted with 50.00 mL of water. What is the molarity of the stock solution?

7. Given a solution of known percentage concentration by mass, a laboratory

worker can often measure out a calculated mass of the solution in order to

obtain a certain mass of solute. Sometimes, though, it is impractical to use the

mass of a solution, especially with fuming solutions, such as concentrated

HCl and concentrated HNO3. Measuring these solutions by volume is much

more practical. In order to determine the volume that should be measured, a

worker would need to know the density of the solution. This information

usually appears on the label of the solution bottle.

a. Concentrated hydrochloric acid is 36% HCl by mass and has a density of

1.18 g/mL. What is the volume of 1.0 kg of this HCl solution? What volume

contains 1.0 g of HCl? What volume contains 1.0 mol of HCl?

b. The density of concentrated nitric acid is 1.42 g/mL, and its concentration is

71% HNO3 by mass. What volume of concentrated HNO3 would be needed

to prepare 10.0 L of a 2.00 M solution of HNO3? c. What volume of concentrated HCl solution would be needed to prepare

4.50 L of 3.0 M HCl? See (a) for data.

8. A 3.8 M solution of FeSO4 solution is diluted to eight times its original volume.

What is the molarity of the diluted solution?

Holt ChemFile: Problem-Solving Workbook 218 Dilutions

9. A chemist prepares 480. mL of a 2.50 M solution of K2Cr2O7 in water. A week

later, the chemist wants to use the solution, but the stopper has been left off

the flask and 39 mL of water has evaporated. What is the new molarity of the

10. You must write out procedures for a group of lab technicians. One test they

will perform requires 25.00 mL of a 1.22 M solution of acetic acid. You decide

to use a 6.45 M acetic acid solution that you have on hand. What procedure

should the technicians use in order to get the solution they need?

11. A chemical test has determined the concentration of a solution of an

unknown substance to be 2.41 M. A 100.0 mL volume of the solution is evaporated

to dryness, leaving 9.56 g of crystals of the unknown solute. Calculate

the molar mass of the unknown substance.

12. Tincture of iodine can be prepared by dissolving 34 g of I2 and 25 g of KI in 25

mL of distilled water and diluting the solution to 500. mL with ethanol. What

is the molarity of I2 in the solution?

13. Phosphoric acid is commonly supplied as an 85% solution. What mass of this

solution would be required to prepare 600.0 mL of a 2.80 M phosphoric acid

14. Commercially available concentrated sulfuric acid is 18.0 M H2SO4. What

volume of concentrated H2SO4 would you use in order to make 3.00 L of a

4.0 M stock solution?

15. Describe how to prepare 1.00 L of a 0.495 M solution of urea, NH2CONH2, starting with a 3.07 M stock solution.

16. Honey is a solution consisting almost entirely of a mixture of the hexose

sugars fructose and glucose; both sugars have the formula C6H12O6, but they

differ in molecular structure.

a. A sample of honey is found to be 76.2% C6H12O6 by mass. What is the

molality of the hexose sugars in honey? Consider the sugars to be equivalent.

b. The density of the honey sample is 1.42 g/mL. What mass of hexose sugars

are in 1.00 L of honey? What is the molarity of the mixed hexose sugars in

honey?

17. Industrial chemicals used in manufacturing are almost never pure, and the

content of the material may vary from one batch to the next. For these reasons,

a sample is taken from each shipment and sent to a laboratory, where

its makeup is determined. This procedure is called assaying. Once the content

of a material is known, engineers adjust the manufacturing process to

account for the degree of purity of the starting chemicals.

Suppose you have just received a shipment of sodium carbonate, Na2CO3. You weigh out 50.00 g of the material, dissolve it in water, and dilute the

solution to 1.000 L. You remove 10.00 mL from the solution and dilute it to

50.00 mL. By measuring the amount of a second substance that reacts with

Holt ChemFile: Problem-Solving Workbook 219 Dilutions

Na2CO3, you determine that the concentration of sodium carbonate in the

diluted solution is 0.0890 M. Calculate the percentage of Na2CO3 in the original

batch of material. The molar mass of Na2CO3 is 105.99 g. (Hint: Determine

the number of moles in the original solution and convert to mass of Na2CO3.) 18. A student wants to prepare 0.600 L of a stock solution of copper(II) chloride

so that 20.0 mL of the stock solution diluted by adding 130.0 mL of water will

yield a 0.250 M solution. What mass of CuCl2 should be used to make the

stock solution?

19. You have a bottle containing a 2.15 M BaCl2 solution. You must tell other

students how to dilute this solution to get various volumes of a 0.65 M BaCl2 solution. By what factor will you tell them to dilute the stock solution? In

other words, when a student removes any volume, V, of the stock solution,

how many times V of water should be added to dilute to 0.65 M?

20. You have a bottle containing an 18.2% solution of strontium nitrate (density �

1.02 g/mL).

a. What mass of strontium nitrate is dissolved in 80.0 mL of this solution?

b. How many moles of strontium nitrate are dissolved in 80.0 mL of the solution?

c. If 80.0 mL of this solution is diluted with 420.0 mL of water, what is the

molarity of the solution?

Holt ChemFile: Problem-Solving Workbook 220 Dilutions

Colligative Properties

Colligative properties of solutions are properties that depend solely on the number

of particles of solute in solution. In other words, these properties depend

only on the concentration of solute particles, not on the identity of those particles.

Colligative properties result from the interference of solute particles with

the motion of solvent molecules.

Solute molecules can be either electrolytes or nonelectrolytes. When a nonelectrolyte

dissolves, the molecule remains whole in the solution. Glucose and

glycerol are examples of nonelectrolyte solutes.

Ionic solutes are electrolytes. When they dissolve, they dissociate into multiple

particles, or ions. When magnesium chloride dissolves in water, it dissociates as

MgCl2(s) Mg 2� (aq) � 2Cl � (aq)

As you can see, when a mole of MgCl2 completely dissociates in solution, it

produces 1 mol of Mg 2� ions and 2 mol of Cl � H2O O3

ions for a total of 3 mol of solute

particles. Because colligative properties depend on the number of particles in

solution, 1 mol of MgCl2 in solution should have three times the effect of 1 mol of

a nonelectrolyte solute, such as glucose.

Two important colligative properties are freezing-point depression and boilingpoint

elevation. A dissolved solute lowers the freezing point of the solution. The

freezing point of a solution differs from that of the pure solvent according to the

following equation, in which �tf is the change in freezing point.

�tf � Kfm Kf is a constant that differs for each solvent. Because the freezing point of the

solution is lower than that of the solvent alone, Kf is a negative number. The symbol

m represents the molality (moles of solute per kilogram of solvent) of the

Boiling-point elevation works in the same way. The equation to determine the

change in boiling point is as follows.

�tb � Kbm Like Kf, Kb is a constant that differs for each solvent. But unlike Kf, Kb is a

positive number because the boiling point of the solution is higher than that of

the solvent alone.

Holt ChemFile: Problem-Solving Workbook 221 Colligative Properties

General Plan for Solving Problems Involving Freezing-Point

Depression and Boiling-Point Elevation

Determine if

solute is an

electrolyte or

nonelectrolyte.

Mass of solute

solute.

Amount of solute

particles

in solution

mol particles

5a

6a

Molal concentration

of particles in

solution, m

Holt ChemFile: Problem-Solving Workbook 222 Colligative Properties

Multiply by the

molal freezingpoint

constant,

K f .

Add �t f to the

normal freezing

point of the

solvent.

Freezing-point

depression, �t f

Freezing point

5b

Boiling-point

elevation, �tb Add �t b to the

normal boiling

6b

molal boilingpoint

K b .

Boiling point

of solution

Table 1 lists freezing-point depression and boiling-point elevation constants

for common solvents.

TABLE 1

Solvent Normal f.p. K f Normal b.p. K b

Acetic acid 16.6°C �3.90°C/m 117.9°C 3.07°C/m

Camphor 178.8°C �39.7°C/m 207.4°C 5.61°C/m

Ether �116.3°C �1.79°C/m 34.6°C 2.02°C/m

Naphthalene 80.2°C �6.94°C/m 217.7°C 5.80°C/m

Phenol 40.9°C �7.40°C/m 181.8°C 3.60°C/m

Water 0.00°C �1.86°C/m 100.0°C 0.51°C/m

Holt ChemFile: Problem-Solving Workbook 223 Colligative Properties

What is the freezing point of a solution of 210.0 g of glycerol,

HOCH 2CHOHCH 2OH, dissolved in 350. g of water?

What is given in the problem? the formula and mass of solute, and the mass of

water used

What are you asked to find? the freezing point of the solution

Identity of solute glycerol, HOCH 2CHOHCH 2OH

Particles per mole of solute 1 mol

Identity of solvent water

Freezing point of solvent 0.00°C

Mass of solvent 350. g

Mass of solute 210.0 g

Molar mass of solute* 92.11 g/mol

Molal concentration of solute particles ? m

Molal freezing-point constant for water �1.86°C/m

Freezing-point depression ?°C

Freezing point of solution ?°C

What steps are needed to calculate the freezing point of the solution?

Use the molar mass of the solute to determine the amount of solute. Then apply

the mass of solvent to calculate the molality of the solution. From the molality,

use the molal freezing-point constant for water to calculate the number of

degrees the freezing point is lowered. Add this negative value to the normal

freezing point.

Holt ChemFile: Problem-Solving Workbook 224 Colligative Properties

Mass of glycerol in g

inverse of the molar

mass of glycerol

Amount of glycerol in mol

the solute is a

nonelectrolyte, so the

amount of solute equals

the amount of particles

210.0 g glycerol �

molar mass of glycerol above

given 1 mol glycerol 1

g glycerol �

92.11 g glycerol kg H2O freezing point

of H2O 0.00�C

350. g� H2O � �� 0.350 kg H2O 1000

1 mol glycerol

92.11 g glycerol

0.00°C � (�12.1°C) � �12.1°C

divide the amount of the

particles in solution by the

mass of the solvent in

kilograms

of glycerol

in water, m

molal freezing-point

constant, Kf , for

water

0.350 kg H2O � �t f

�1.86°C

� � �12.1°C

mol/kg

Holt ChemFile: Problem-Solving Workbook 225 Colligative Properties

�t f

Mass of water in g

Mass of water in kg

� tf �

Freezing-point depression, �tf add �tf to the

point of water

Freezing point of the

glycerol solution

freezingpoint

depression

�1.86�C

Yes; units canceled to give Celsius degrees.

Yes; three significant figures is correct because the data had a minimum of three

Yes; the calculation can be approximated as 200 � [90 � 3(350 � 1000)] ��2 �

�400/30 ��13, which is close to the calculated value.

1. Determine the freezing point of a solution of 60.0 g of glucose, C 6H 12O 6,

dissolved in 80.0 g of water. ans: �7.74°C

2. What is the freezing point of a solution of 645 g of urea, H 2NCONH 2, dissolved

in 980. g of water? ans: �20.4°C

Holt ChemFile: Problem-Solving Workbook 226 Colligative Properties

What is the boiling point of a solution containing 34.3 g of the ionic

compound magnesium nitrate dissolved in 0.107 kg of water?

What are you asked to find? the boiling point of the solution

Identity of solute magnesium nitrate

Equation for the dissociation of the solute Mg(NO3) 2 → Mg 2� �

� 2NO3 Amount of ions per mole of solute 3 mol

Boiling point of solvent 100.0°C

Mass of solvent 0.107 kg H 2O

Mass of solute 34.3 g

Molar mass of solute 148.32 g/mol

Molal boiling-point constant for solvent 0.51°C/m

Boiling-point depression ?°C

Boiling point of solution ?°C

What steps are needed to calculate the boiling point of the solution?

Use the molar mass to calculate the amount of solute in moles. Multiply the

amount of solute by the number of moles of ions produced per mole of solute.

Use the amount of ions with the mass of solvent to compute the molality of

particles in solution. Use this effective molality to determine the boiling-point

elevation and the boiling point of the solution.

Holt ChemFile: Problem-Solving Workbook 227 Colligative Properties

Mass of Mg(NO3 ) 2 in g

mass of Mg(NO3 ) 2

Amount of Mg(NO3 ) 2 in mol

the solute is an

electrolyte, so multiply

the amount of solute by

the number of particles

per mole of solute

particles in solution

molar mass of Mg(NO3 ) 2

given 1 mol Mg(NO3 ) 2 3 mol particles 1

g Mg(NO3 ) 2 �

148.32 g Mg(NO3 ) 2 1 mol Mg(NO3 ) 2 kg H2O boiling point

of H2O 100.0�C

34.3 g Mg(NO 3) 2 � 1 mol Mg(NO 3) 2

148.32 g Mg(NO 3) 2

100.0°C � 3.31°C � 103.3°C

3 mol particles

1 mol Mg(NO3) 2

� �t b

0.107 kg H 2O

0.51°C

� � 3.31°C

Holt ChemFile: Problem-Solving Workbook 228 Colligative Properties

�t b

Boiling-point elevation, �tb add �tb to the

� tb Mass of water in kg

Molal concentration of

particles in water, m

molal boiling-point

constant, Kb , for

Boiling point of the

Mg(NO3 ) 2 solution

constant for water

0.51�C

Yes; the number of significant figures is correct because the boiling point of

water was given to one decimal place.

Yes; the calculation can be approximated as [(35 � 3)/150] � 5 � (7/10) � 5 �

3.5, which is close to the calculated value for the boiling-point elevation.

1. What is the expected boiling point of a brine solution containing 30.00 g of

KBr dissolved in 100.00 g of water? ans: 102.6°C

2. What is the expected boiling point of a CaCl 2 solution containing 385 g of

CaCl 2 dissolved in 1.230 � 10 3 g of water? ans: 104.3°C

Holt ChemFile: Problem-Solving Workbook 229 Colligative Properties

A solution of 3.39 g of an unknown compound in 10.00 g of water has a

freezing point of �7.31°C. The solution does not conduct electricity.

What is the molar mass of the compound?

What is given in the problem? the freezing point of the solution, the mass of

the dissolved compound, the mass of solvent,

and the fact that the solution does not conduct

electricity

What are you asked to find? the molar mass of the unknown compound

Mass of solute 3.39 g

Molar mass of solute ? g/mol

Mass of solvent 10.00 g

Molal freezing-point constant for solvent �1.86°C/m

Freezing point of solution �7.31°C

What steps are needed to calculate the molar mass of the unknown solute?

Determine the molality of the solution from the freezing-point depression. Use

the molality and the solute and solvent masses to calculate the solute molar

mass.

Holt ChemFile: Problem-Solving Workbook 230 Colligative Properties

Freezing point of the solution

subtract the

Freezing-point depression, �tf multiply by the

constant, Kf ,

for water

Mass of water

multiply the molal

concentration by the mass of

the water

multiply by the conversion

tf given

g H 2 O

g solute

� molar mass of solute

�7.31°C � 0.00°C ��7.31°C 1 kg

10.00 g� H2O � �� 0.010 00 kg H2O 1000

�7.31°C � mol/kg

�1.86°C � 0.010 00 kg H 2O � 0.039 30 mol solute

3.39 g solute

0.039 30 mol solute

freezing point

of water

0.00�C

� �t f

� � kg H2O 1000 g

Molar mass of solute

� kg H2O � mol solute

� 86.3 g/mol

divide the mass of

the solute by the

amount of solute in

Amount of solute in mol

Amount of particles

Holt ChemFile: Problem-Solving Workbook 231 Colligative Propertiess

Yes; molar mass has units of g/mol.

Yes; the calculation can be approximated as (4/1) � (1/100) � 0.04, which is

close to the value of 0.0393 for the amount of solute.

1. A solution of 0.827 g of an unknown non-electrolyte compound in 2.500 g of

water has a freezing point of �10.18°C. Calculate the molar mass of the

compound. ans: 60.4 g/mol

2. A 0.171 g sample of an unknown organic compound is dissolved in ether. The

solution has a total mass of 2.470 g. The boiling point of the solution is found

to be 36.43°C. What is the molar mass of the organic compound?

ans: 82.1 g/mol

Holt ChemFile: Problem-Solving Workbook 232 Colligative Properties

In each of the following problems, assume that the solute is a nonelectrolyte

unless otherwise stated.

1. Calculate the freezing point and boiling point of a solution of 383 g of glucose

dissolved in 400. g of water.

2. Determine the boiling point of a solution of 72.4 g of glycerol dissolved in

122.5 g of water.

3. What is the boiling point of a solution of 30.20 g of ethylene glycol,

HOCH2CH2OH, in 88.40 g of phenol?

4. What mass of ethanol, CH3CH2OH, should be dissolved in 450. g of water to

obtain a freezing point of �4.5°C?

5. Calculate the molar mass of a nonelectrolyte that lowers the freezing point of

25.00 g of water to �3.9°C when 4.27 g of the substance is dissolved in the

water.

6. What is the freezing point of a solution of 1.17 g of 1-naphthol, C10H8O, dissolved

in 2.00 mL of benzene at 20°C? The density of benzene at 20°C is 0.876

g/mL. Kf for benzene is �5.12°C/m, and benzene’s normal freezing point is

5.53°C.

7. The boiling point of a solution containing 10.44 g of an unknown nonelectrolyte

in 50.00 g of acetic acid is 159.2°C. What is the molar mass of the

solute?

8. A 0.0355 g sample of an unknown molecular compound is dissolved in 1.000 g

of liquid camphor at 200.0°C. Upon cooling, the camphor freezes at 157.7°C.

Calculate the molar mass of the unknown compound.

9. Determine the boiling point of a solution of 22.5 g of fructose, C6H12O6, in

294 g of phenol.

10. Ethylene glycol, HOCH2CH2OH, is effective as an antifreeze, but it also raises

the boiling temperature of automobile coolant, which helps prevent loss of

coolant when the weather is hot.

a. What is the freezing point of a 50.0% solution of ethylene glycol in water?

b. What is the boiling point of the same 50.0% solution?

11. The value of Kf for cyclohexane is �20.0°C/m, and its normal freezing point is

6.6°C. A mass of 1.604 g of a waxy solid dissolved in 10.000 g of cyclohexane

results in a freezing point of �4.4°C. Calculate the molar mass of the solid.

12. What is the expected freezing point of an aqueous solution of 2.62 kg of nitric

acid, HNO3, in a solution with a total mass of 5.91 kg? Assume that the nitric

acid is completely ionized.

Holt Chemistry 233 Colligative Properties

13. An unknown organic compound is mixed with 0.5190 g of naphthalene crystals

to give a mixture having a total mass of 0.5959 g. The mixture is heated

until the naphthalene melts and the unknown substance dissolves. Upon

cooling, the solution freezes at a temperature of 74.8°C. What is the molar

mass of the unknown compound?

14. What is the boiling point of a solution of 8.69 g of the electrolyte sodium

acetate, NaCH3COO, dissolved in 15.00 g of water?

15. What is the expected freezing point of a solution of 110.5 g of H2SO4 in 225 g

of water? Assume sulfuric acid completely dissociates in water.

16. A compound called pyrene has the empirical formula C8H5. When 4.04 g of

pyrene is dissolved in 10.00 g of benzene, the boiling point of the solution is

85.1°C. Calculate the molar mass of pyrene and determine its molecular

formula. The molal boiling-point constant for benzene is 2.53°C/m. Its normal

boiling point is 80.1°C.

17. What mass of CaCl2, when dissolved in 100.00 g of water, gives an expected

freezing point of �5.0°C; CaCl2 is ionic? What mass of glucose would give the

same result?

18. A compound has the empirical formula CH2O. When 0.0866 g is dissolved in

1.000 g of ether, the solution’s boiling point is 36.5°C. Determine the molecular

formula of this substance.

19. What is the freezing point of a 28.6% (by mass) aqueous solution of HCl?

Assume the HCl is 100% ionized.

20. What mass of ethylene glycol, HOCH2CH2OH, must be dissolved in 4.510 kg of

water to result in a freezing point of �18.0°C? What is the boiling point of the

same solution?

21. A water solution containing 2.00 g of an unknown molecular substance dissolved

in 10.00 g of water has a freezing point of �4.0°C.

a. Calculate the molality of the solution.

b. When 2.00 g of the substance is dissolved in acetone instead of in water, the

boiling point of the solution is 58.9°C. The normal boiling point of acetone

is 56.00°C, and its Kb is 1.71°C/m. Calculate the molality of the solution

from this data.

22. A chemist wants to prepare a solution with a freezing point of �22.0°C and

has 100.00 g of glycerol on hand. What mass of water should the chemist mix

with the glycerol?

23. An unknown carbohydrate compound has the empirical formula CH2O. A

solution consisting of 0.515 g of the carbohydrate dissolved in 1.717 g of

acetic acid freezes at 8.8°C. What is the molar mass of the carbohydrate?

What is its molecular formula?

24. An unknown organic compound has the empirical formula C2H2O. A solution

of 3.775 g of the unknown compound dissolved in 12.00 g of water is cooled

until it freezes at a temperature of �4.72°C. Determine the molar mass and

the molecular formula of the compound.

Holt ChemFile: Problem-Solving Workbook 234 Colligative Properties

Equilibrium

Not all processes in nature proceed to completion. In fact, most changes hover

somewhere between the initial state and what would be the final state. Compare

a light switch and a dimmer. If the mechanical switch is working properly, it can

be stable in only two positions: on or off. Either current flows or it doesn’t. With

a dimmer you can regulate the flow of current so that it stays somewhere

between fully on and fully off. If you’ve used mechanical balances, you know that

to weigh an object accurately you must adjust the masses so that the pointer hovers

in the middle of its range. The balance is in a state of equilibrium.

Most chemical reactions also reach a state of equilibrium between no reaction

at all and the complete reaction to form products. Equilibrium states occur when

reactions are reversible, that is, when products react to re-form the original reactants.

When the products re-form reactants at the same rate as the reactants form

the products, then the equilibrium point of the reaction has been reached.

The progress of a reaction is gauged by measuring the concentrations in moles

per liter of reactants and products. At the equilibrium point, these concentrations

stop changing.

A � 2B ^ 2C

The equation above represents a reaction in which 1 mol of A reacts with

2 mol of B to produce 2 mol of C. As C is formed, it breaks down to re-form reactants.

The extent of the reaction at equilibrium is indicated by the equilibrium

constant, K eq.

K e

q � [C

[ A][B]

As you can see, the concentration of each reaction component is raised to the

power of its coefficient in the balanced equation. These concentration terms are

arranged in a fraction with products in the numerator and reactants in the

denominator. Pure substances (substances that appear in the chemical equation

as solids or pure liquids) are not included in the equilibrium expressions because

their concentrations are meaningless.

Problems involving chemical equilibrium will ask you to solve for either the

equilibrium constant, K eq, given the concentrations of all of the reaction components,

or the concentration of one of the reaction components, given K eq.

Holt ChemFile: Problem-Solving Workbook 235 Equilibrium

General Plan for Solving Equilibrium Problems

1 Balanced chemical

xA � yB 3 zC

Rearrange to solve for

the unknown quantity,

substitute known

values, and solve.

Expression for the

equilibrium constant, Keq [C] z

[A] x [B] y

Keq �

[C] � �Keq [A] x [B] y

z

Holt ChemFile: Problem-Solving Workbook 236 Equilibrium

The following equation represents the reversible decomposition of PCl5. PCl5(g) ^ PCl3(g) � Cl2(g) At 250°C, the equilibrium concentrations of the substances are as follows:

[PCl5] � 1.271 M

[PCl3] � 0.229 M

[Cl2] � 0.229 M

What is the value of the equilibrium constant, Keq, for this reaction?

What is given in the problem? the equilibrium concentrations of the products

and reactant at 250°C

What are you asked to find? the value of the equilibrium constant for the

reaction

Molar concentration of PCl 5 at equilibrium 1.271

Molar concentration of PCl 3 at equilibrium 0.229 M

Molar concentration of Cl2 at equilibrium 0.229 M

Equilibrium constant Keq ?

What steps are needed to calculate the equilibrium constant for the given reaction?

Set up the equilibrium expression for the reaction using the coefficients as

exponents. Substitute the concentration values, and calculate Keq. 1

PCl5 (g) ª PCl3 (g) � Cl2 (g) Keq �

Note that since all coefficients have the value 1, there is no need to write in the

exponent.

Keq � [PCl given given

3] [Cl

2]

[PCl5]

[PCl3 ][Cl2 ]

[PCl5 ]

Holt ChemFile: Problem-Solving Workbook 237 Equilibrium

Keq �� [0.229][

0.229]

� � 0.0413

[ 1.271]

Yes; the equilibrium constant has no units.

Yes; the number of significant figures is correct because data values were given

to a minimum of three significant figures.

Yes; the calculation may be approximated as (0.2 � 0.2)/1. This approximation

gives a result of 0.04, which is close to the calculated value.

1. Calculate the equilibrium constants for the following hypothetical reactions.

Assume that all components of the reactions are gaseous.

a. A ^ C � D

At equilibrium, the concentration of A is 2.24 � 10 �2 M and the concentrations

of both C and D are 6.41 � 10 �3 M. ans: Keq � 1.83 � 10 �3

b. A � B ^ C � D

At equilibrium, the concentrations of both A and B are 3.23 � 10 �5 M and the

concentrations of both C and D are 1.27 � 10 �2 M. ans: Keq � 1.55 � 10 5

Holt ChemFile: Problem-Solving Workbook 238 Equilibrium

c. A � B ^ 2C

At equilibrium, the concentrations of both A and B are 7.02 � 10 �3 M and the

concentration of C is 2.16 � 10 �2 M. ans: Keq � 9.47

d. 2A ^ 2C � D

At equilibrium, the concentration of A is 6.59 � 10 �4 M. The concentration of

C is 4.06 � 10 �3 M, and the concentration of D is 2.03 � 10 �3 M.

ans: Keq � 7.71 � 10 �2

e. A � B ^ C � D � E

At equilibrium, the concentrations of both A and B are 3.73 � 10 �4 M and the

concentrations of C, D, and E are 9.35 � 10 �4 M. ans: Keq � 5.88 � 10 �3

f. 2A � B ^ 2C

At equilibrium, the concentration of A is 5.50 � 10 �3 M, the concentration of

B is 2.25 � 10 �3 M, and the concentration of C is 1.02 � 10 �2 M.

ans: K eq � 1.53 � 10 3

Holt ChemFile: Problem-Solving Workbook 239 Equilibrium

The following equilibrium reaction is used in the manufacture of

methanol. The equilibrium constant at 400 K for the reaction is 1.609.

CO(g) � 2H2(g) ^ CH3OH(g) At equilibrium, the mixture in the reaction vessel has a concentration of

0.818 M of CH3OH and 1.402 M of CO. Calculate the concentration of H2 in the vessel.

What is given in the problem? the equilibrium concentrations of CO and

CH3OH, and the equilibrium constant at 400 K

What are you asked to find? the equilibrium concentration of H2 in the vessel

Molar concentration of CO at equilibrium 1.402 M

Molar concentration of H 2 at equilibrium ? M

Molar concentration of CH3OH at equilibrium 0.818 M

Equilibrium constant Keq 0.609 M

What steps are needed to calculate the concentration of H 2?

Set up the equilibrium expression for the reaction using coefficients as exponents.

Rearrange the expression to solve for [H 2]. Substitute known values for

[CO 2], [CH 3OH], and K eq, and solve for [H 2].

CO(g) � 2H2 (g) ª CH3OH(g) [H2] �

K eq �

[CH3OH] [CO][H 2] 2

[CH3OH] Keq � [CO]

Holt ChemFile: Problem-Solving Workbook 240 Equilibrium

rearrange to solve for [H 2 ],

substitute known values, and

[H2] �� [0.818]

�1.609 ��

� 0.602 M

[ 1.402]

Yes; concentrations are in moles per liter.

Yes; the calculation may be approximated as [1 � (1.5 � 1.5)] 1/2 � 0.67, which is

close to the calculated value.

1. Calculate the concentration of product D in the following hypothetical reaction:

2A(g) ^ 2C(g) � D(g)

At equilibrium, the concentration of A is 1.88 � 10 �1 M, the concentration of

C is 6.56 M, and the equilibrium constant is 2.403 � 10 2 . ans: 0.197 M

2. At a temperature of 700 K, the equilibrium constant is 3.164 � 10 3 for the

following reaction system for the hydrogenation of ethene, C 2H 4, to ethane,

C 2H 6.

C 2H 4(g) � H 2(g) ^ C 2H 6(g)

What will be the equilibrium concentration of ethene if the concentration of

H 2 is 0.0619 M and the concentration of C 2H 6 is 1.055 M? ans: 5.39 � 10 �3 M

Holt ChemFile: Problem-Solving Workbook 241 Equilibrium

Additional Problems—Equilibrium

1. Using the reaction A � 2B ^ C � 2D, determine Keq if the following equilibrium

concentrations are found. All components are gases.

[A] � 0.0567 M

[B] � 0.1171 M

[C] � 0.000 3378 M

[D] � 0.000 6756 M

2. In the reaction 2A ^ 2C � 2D, determine Keq when the following equilibrium

[A] � 0.1077 M

[C] � 0.000 4104 M

[D] � 0.000 4104 M

3. Calculate the equilibrium constant for the following reaction. Note the phases

of the components.

2A(g) � B(s) ^ C(g) � D(g)

The equilibrium concentrations of the components are

[A] � 0.0922 M

[C] � 4.11 � 10 �4 M

[D] � 8.22 � 10 �4 M

4. The equilibrium constant of the following reaction for the decomposition of

phosgene at 25°C is 4.282 � 10 �2 .

COCl2(g) ^ CO(g) � Cl2(g) a. What is the concentration of COCl 2 when the concentrations of both CO

and Cl 2 are 5.90 � 10 �3 M?

b. When the equilibrium concentration of COCl 2 is 0.003 70 M, what are the

concentrations of CO and Cl 2? Assume the concentrations are equal.

5. Consider the following hypothetical reaction.

A(g) � B(s) ^ C(g) � D(s)

a. If Keq � 1 for this reaction at 500 K, what can you say about the concentrations

of A and C at equilibrium?

b. If raising the temperature of the reaction results in an equilibrium with a

higher concentration of C than A, how will the value of Keq change?

Holt ChemFile: Problem-Solving Workbook 242 Equilibrium

6. The following reaction occurs when steam is passed over hot carbon. The

mixture of gases it generates is called water gas and is useful as an industrial

fuel and as a source of hydrogen for the production of ammonia

C(s) � H2O(g) ^ CO(g) � H2(g) The equilibrium constant for this reaction is 4.251 � 10 �2 at 800 K. If the

equilibrium concentration of H 2O(g) is 0.1990 M, what concentrations of CO

and H 2 would you expect to find?

7. When nitrogen monoxide gas comes in contact with air, it oxidizes to the

brown gas nitrogen dioxide according to the following equation:

2NO(g) � O2(g) ^ 2NO2(g) a. The equilibrium constant for this reaction at 500 K is 1.671 � 10 4 . What

concentration of NO 2 is present at equilibrium if [NO] � 6.200 � 10 �2 M

and [O 2] � 8.305 � 10 �3 M?

b. At 1000 K, the equilibrium constant, K eq, for the same reaction is

1.315 � 10 �2 . What will be the concentration of NO 2 at 1000 K given the

same concentrations of NO and O 2 as were in (a)?

8. Consider the following hypothetical reaction, for which Keq � 1 at 300 K:

A(g) � B(g) ^ 2C(g)

a. If the reaction begins with equal concentrations of A and B and a zero

concentration of C, what can you say about the relative concentrations of

the components at equilibrium?

b. Additional C is introduced at equilibrium, and the temperature remains

constant. When equilibrium is restored, how will the concentrations of all

components have changed? How will Keq have changed?

9. The equilibrium constant for the following reaction of hydrogen gas and

bromine gas at 25°C is 5.628 � 10 18 .

H 2(g) � Br 2(g) ^ 2HBr(g)

a. Write the equilibrium expression for this reaction.

b. Assume that equimolar amounts of H 2 and Br 2 were present at the beginning.

Calculate the equilibrium concentration of H 2 if the concentration of

HBr is 0.500 M.

c. If equal amounts of H 2 and Br 2 react, which reaction component will be

present in the greatest concentration at equilibrium? Explain your

reasoning.

Holt ChemFile: Problem-Solving Workbook 243 Equilibrium

10. The following reaction reaches an equilibrium state:

N2F4(g) ^ 2NF2(g) At equilibrium at 25°C the concentration of N 2F 4 is found to be 0.9989 M and

the concentration of NF 2 is 1.131 � 10 �3 M. Calculate the equilibrium constant

of the reaction.

11. The equilibrium between dinitrogen tetroxide and nitrogen dioxide is represented

by the following equation:

N2O4(g) ^ NO2(g) A student places a mixture of the two gases into a closed gas tube and allows

the reaction to reach equilibrium at 25°C. At equilibrium, the concentration of

N2O4 is found to be 5.95 � 10 �1 M and the concentration of NO2 is found to

be 5.24 � 10 �2 M. What is the equilibrium constant of the reaction?

12. Consider the following equilibrium system:

NaCN(s) � HCl(g) ^ HCN(g) � NaCl(s)

a. Write a complete expression for the equilibrium constant of this system.

b. The K eq for this reaction is 2.405 � 10 6 . What is the concentration of HCl

remaining when the concentration of HCN is 0.8959 M?

13. The following reaction is used in the industrial production of hydrogen gas:

CH4(g) � H2O(g) ^ CO(g) � 3H2(g) The equilibrium constant of this reaction at 298 K (25°C) is 3.896 � 10 �27 , but

at 1100 K the constant is 3.112 � 10 2 .

a. What do these equilibrium constants tell you about the progress of the

reaction at the two temperatures?

b. Suppose the reaction mixture is sampled at 1100 K and found to contain

1.56 M of hydrogen, 3.70 � 10 �2 M of methane, and 8.27 � 10 �1 M of

gaseous H2O. What concentration of carbon monoxide would you expect to

find?

14. Dinitrogen tetroxide, N 2O 4, is soluble in cyclohexane, a common nonpolar

solvent. While in solution, N 2O 4 can break down into NO 2 according to the

N2O4(cyclohexane) ^ NO2(cyclohexane) At 20°C, the following concentrations were observed for this equilibrium

[N2O4] � 2.55 � 10 �3 M

[NO2] � 10.4 � 10 �3 M

What is the value of the equilibrium constant for this reaction? Note, the

chemical equation must be balanced first.

Holt ChemFile: Problem-Solving Workbook 244 Equilibrium

15. The reaction given in item 14 also occurs when the dinitrogen tetroxide and

nitrogen dioxide are dissolved in carbon tetrachloride, CCl4, another nonpolar

N2O4(CCl4) ^ NO2(CCl4) The following experimental data were obtained at 20°C:

[N2O4] � 2.67 � 10 �3 M

[NO2] � 10.2 � 10 �3 M

Calculate the value of the equilibrium constant for this reaction occurring in

carbon tetrachloride.

Holt ChemFile: Problem-Solving Workbook 245 Equilibrium

Equilibrium of Salts, K sp

When you try to dissolve a solid substance in water, you expect the solid form to

disappear, forming ions or molecules in solution. Most substances, however, are

only slightly soluble in water. For example, when you stir silver chloride in water,

you may think none of the solid dissolves. Does this mean that some of the solid

dissolves, forms a saturated solution, and, after that, experiences no further

change between the solid and solution phases? It is true to say that there is no

further change in the amount of substance in either the solution or the solid

phase, but to say, that no further change occurs is inaccurate. An equilibrium

exists between the silver chloride and its dissolved ions, and a state of equilibrium

is a dynamic state. The following chemical equation shows this equilibrium.

AgCl(s) ^ Ag � (aq) � Cl � (aq)

Like the other examples of equilibria that you have studied, the extent to

which this solubility equilibrium proceeds toward the products (the ions in solution)

is indicated by an equilibrium constant. When an equilibrium constant is

written for a solubility equilibrium, it is called a solubility product constant and is

symbolized as K sp.

In solubility equilibrium problems, the reactants are pure substances, and pure

substances are never included in an equilibrium expression. That means that you

will not have anything in the denominator in the expression for the solubility

product constant. This K sp expression for the silver chloride example can be written

as follows.

K sp � [Ag � ][Cl � ]

Note that the coefficients in the balanced equation are understood to be 1 for

both silver and chlorine. Therefore, no exponents appear in the K sp expression.

As with any equilibrium expression, the concentration of each component is

raised to the power of its coefficient from the balanced chemical equation. The

value for this solubility product constant is 1.77 � 10 �10 at 25°C. The very small

value of K sp indicates that silver chloride is only very slightly soluble in aqueous

solution at this temperature. The value of K sp supports the observation that little

seems to occur when silver chloride is stirred into a water solution.

In this worksheet, you will learn to apply the solubility equilibrium relationship

to determine K sp for substances and to calculate concentrations of ions in

saturated solutions using their K sp values.

Holt ChemFile: Problem-Solving Workbook 246 Equilibrium of Salts, K sp

General Plan for Solving Solubility Equilibrium Problems

If the solution is

saturated, use the

balanced chemical

equation to write the

K sp expression.

4a

Ksp expression

If K sp is greater than the ion

product, the solution is not

saturated and no precipitation

occurs. If K sp is less than the ion

product, precipitation occurs.

Molarity

Write the balanced equation

for the dissociation of the

Balanced chemical

Use the mole ratios of

the original solute to

each ion to calculate each

concentration.

Concentrations of

each ion in solution

Use the balanced

chemical equation to

write the ion product

expression.

4b

Ion

Holt ChemFile: Problem-Solving Workbook 247 Equilibrium of Salts, K sp

A saturated solution of magnesium fluoride, MgF 2, contains 0.00741 g of

dissolved MgF 2 per 1.00 � 10 2 mL at 25°C. What is the K sp for magnesium

fluoride?

What is given in the problem? the mass of MgF2 dissolved in 1.00 � 10 2 mL of

a saturated solution

What are you asked to find? the solubility product constant, Ksp Items Data

Mass of dissolved MgF 2

*determined from the periodic table

What steps are needed to calculate the solubility product constant, K sp, of MgF 2?

Determine the molar concentration of the saturated MgF 2 solution. Write the

balanced chemical equation for the dissociation of MgF 2, and use this equation

to determine the concentrations of each ion in solution. Compute K sp.

0.00741 g

Volume of solution 1.00 � 10 2 mL

Molar mass of MgF 2* 62.30 g/mol

Molar concentration of MgF 2

Molar concentration of Mg 2�

Molar concentration of F �

K sp of MgF 2

Holt ChemFile: Problem-Solving Workbook 248 Equilibrium of Salts, K sp

? M

?M

Calculate the molarity of the saturated MgF 2 solution.

Write the balanced chemical equation for the dissociation to determine the mole

ratios of solute and ions.

Write the K sp expression.

Molarity of solution

use the balanced

equation to write

the K sp expression,

substitute, and

solve

molar mass MgF 2

given 1 mol MgF2 g MgF2 � � mol MgF2 62.30 g MgF2 given 1 L

mL solution � � L solution

mol MgF2 L solution

MgF 2(s) a Mg 2� (aq) � 2F � (aq)

[MgF 2] � [Mg 2� ]

2[MgF 2] � [F � ]

K sp � [Mg 2� ][F � ] 2

write a balanced chemical

equation for the dissociation

MgF2 (s) ^ Mg2� (aq) � 2F� 2

(aq)

use

from the

balanced

chemical

Molarity of Mg2� 3

Molarity of F� 3

Ksp � [Mg2� ][F� ] 2

� [MgF2 ]

Holt ChemFile: Problem-Solving Workbook 249 Equilibrium of Salts, K sp

Yes; K sp has no units.

100. mL � solution � �� 0.100 L solution

� [MgF2] � 1.19 � 10 �3 M

[Mg 2� ] � [MgF2] � 1.19 � 10 �3 M

[F � ] � 2[MgF2] � 2.38 � 10 �3 1.19 � 10

�4 0.007 41 g MgF2 �

mol MgF2 ��

0.100 L solution

1 mol MgF2 � 1.19 � 10

62.30 g MgF2 �4 mol MgF2 K sp � [1.19 � 10 �3 ][2.38 � 10 �3 ] 2 � 6.74 � 10 �9

Yes; the number of significant figures is correct because all data were given to

Yes; the calculation can be approximated as (1 � 10 �3 )(2.5 � 10 �3 ) 2 � 6 � 10 �9 ,

which is of the same order of magnitude as the calculated answer.

1. Silver bromate, AgBrO 3, is slightly soluble in water. A saturated solution is

found to contain 0.276 g AgBrO 3 dissolved in 150.0 mL of water. Calculate K sp

for silver bromate. ans: K sp � 6.09 � 10 �5

2. 2.50 L of a saturated solution of calcium fluoride leaves a residue of 0.0427 g

of CaF 2 when evaporated to dryness. Calculate the K sp of CaF 2.

ans: K sp � 4.20 � 10 �11

Holt ChemFile: Problem-Solving Workbook 250 Equilibrium of Salts, K sp

The K sp for lead(II) iodide is 7.08 � 10 �9 at 25°C. What is the molar

concentration of PbI 2 in a saturated solution?

What is given in the problem the solubility product constant, K sp of PbI 2

What are you asked to find? the concentration of PbI 2 in a saturated solution

What steps are needed to calculate the concentration of dissolved PbI2 in a saturated

Write the equation for the dissociation of PbI2. Set up the equation for Ksp, and

compute the concentrations of the ions. Determine the concentration of dissolved

Write the balanced chemical equation for the dissociation of lead(II) iodide, PbI2 in aqueous solution.

PbI2(s) ^ Pb 2� (aq) � 2I � (aq)

Write the Ksp expression.

Ksp � [Pb 2� ][I � ] 2

Substitute x for [Pb 2� ]. The balanced equation gives the following relationship:

2[Pb 2� ] � [I � ].

Therefore, [I � ] � 2x.

Ksp � [x][2x] 2

Rearrange, and solve for x.

Ksp of PbI 2

Concentration of Pb 2�

Concentration of I �

?

Concentration of PbI2 in solution ?

Ksp � [x][4x 2 ]

Ksp � 4x 3

Ksp x � � �3

Relate the substituted value to the unknown solution concentration using the

mole ratio from the original balanced chemical equation. The mole ratio shows

that [Pb 2� ] � [PbI 2].

[Pb 2� ]

7.08 � 10 �9

Holt ChemFile: Problem-Solving Workbook 251 Equilibrium of Salts, K sp

�� 4

� [Pb2� ] � 1.21 � 10 �3 M

[Pb 2� ] � [PbI2] � 1.21 � 10 �3 M

3 7.08 �

x ��

10�9

Yes; concentrations are in molarity (mol/L).

Yes; the best check is to use the result to calculate K sp and see if it gives (or is

very near) the K sp you started with. In this case, the calculated K sp is

7.08 � 10 �9 , the same value as was given.

1. The K sp of calcium sulfate, CaSO 4, is 9.1 � 10 �6 . What is the molar concentration

of CaSO 4 in a saturated solution? ans: 3.0 � 10 �3 M

2.A salt has the formula X 2Y, and its K sp is 4.25 � 10 �7 .

a. What is the molarity of a saturated solution of the salt?

ans: [X 2Y] � 4.74 � 10 �3 M

b. What is the molarity of a solution of AZ if its K sp is the same value?

ans: [AZ] � 6.52 � 10 �4 M

Holt ChemFile: Problem-Solving Workbook 252 Equilibrium of Salts, K sp

Will precipitation of strontium sulfate occur when 50.0 mL of 0.025 M

strontium nitrate solution is mixed with 50.0 mL of a 0.014 M copper(II)

sulfate solution? The K sp of strontium nitrate is 3.2 � 10 �7 .

What is given in the problem? the molar concentrations of the solutions to be

mixed, the identities of the solutes, and the

volumes of the solutions to be mixed

What are you asked to find? whether a precipitate of strontium sulfate

forms when the two solutions are mixed

Concentration of solution 1 0.025 M SrNO 3

Volume of solution 1 50.0 mL

Concentration of solution 2 0.014 M CuSO 4

Volume of solution 2 50.0 mL

Volume of combined solution 100.0 mL

Concentration of combined solution ? M SrSO 4

Potential precipitate SrSO 4

K sp of SrSO 4

Precipitate/no precipitate forms ?

What steps are needed to determine whether a precipitate will form?

Calculate the molar concentrations of the ions that can form a precipitate in the

new volume of solution. Use these concentrations to calculate the ion product.

3.2 � 10 �7

Holt ChemFile: Problem-Solving Workbook 253 Equilibrium of Salts, K sp

use the molarity of

Sr(NO 3 ) 2 , and

calculate the

molarity of Sr 2� in

the new diluted

Ksp SrSO4 (aq) ^ Sr2� 2

(aq) � SO 2�

4 (aq)

Molarity of Sr2� 3

if K sp � ion product,

precipitate forms,

no precipitate

forms

Precipitate or

Write the balanced equation for the dissociation of SrSO4. SrSO4(aq) ^ Sr 2� 2�

(aq) � SO4 (aq)

Calculate the molarities of Sr 2� 2�

and SO4 . This is a simple dilution calculation.

The subscript 1 in each case represents that individual solution; the subscript 2

represents the combined solution.

Sr(NO3) 2(aq) ^ Sr 2� �

(aq) � 2NO3 (aq)

[Sr 2� ] 1 � [Sr(NO3) 2]

[Sr 2� ] 1V1 � [Sr 2� ] 2V2 [Sr 2

2� ] �

CuSO 4(aq) ^ Cu 2� (aq) � SO 4 2� (aq)

[SO 4 2� ]1 � [CuSO 4]

[SO 4 2� ]1V 1 � [SO 4 2� ]2V 2

[SO 2� ] �

use coefficients

from the balanced

chemical equation

[Sr V1 V2 2� given

] 1

sum of volumes of

solutions mixed

[SO V1 V2 2� given

4 ] 1

CuSO 4 , and

molarity of SO 4 2� in

Molarity of SO 2�

ion product � [Sr2� 4b

][SO 2�

4 ]

Holt ChemFile: Problem-Solving Workbook 254 Equilibrium of Salts, K sp

Calculate the ion product for SrSO 4.

Compare the ion product to the K sp value to determine if precipitation occurs.

[Sr 2� ] 2 � � 1.2 � 10 �2 M

[SO 2 4 � ] 2 � � 7.0 � 10 �3 M

ion product � [1.2 � 10 �2 ][7.0 � 10 �3 ] � 8.4 � 10 �5

Ksp � 3.2 � 10 �7

0.025 M � 50.0 �mL

100.0 mL �

0.014 M � 50.0 �mL

Ksp � ion product

Precipitation will occur.

Yes; the ion product has no units.

Yes; the number of significant figures is correct because data were given to a

minimum of two significant figures.

Yes; the calculation can be approximated as 0.01 � 0.007 � 0.000 07 � 7 � 10 �5 ,

which is of the same order of magnitude as the calculated result.

ion product �

In each of the following problems, include the calculated ion product with your

answer.

1. Will a precipitate of Ca(OH) 2 form when 320. mL of a 0.046 M solution of

NaOH mixes with 400. mL of a 0.085 M CaCl2 solution? Ksp of Ca(OH) 2 is

5.5 � 10 �6 . ans: ion product � 1.9 � 10 �5 , precipitation occurs.

[SO2� [Sr 2 4 ] 2

2� ]

Holt ChemFile: Problem-Solving Workbook 255 Equilibrium of Salts, K sp

2. 20.00 mL of a 0.077 M solution of silver nitrate, AgNO 3, is mixed with 30.00

mL of a 0.043 M solution of sodium acetate, NaC 2H 3O 2. Does a precipitate

form? The K sp of AgC 2H 3O 2 is 2.5 � 10 �3 .

ans: ion product � 8.1 � 10 �4 , no precipitate

3. If you mix 100. mL of 0.036 M Pb(C 2H 3O 2) 2 with 50. mL of 0.074 M NaCl, will

a precipitate of PbCl 2 form? K sp of PbCl 2 is 1.9 � 10 �4 .

ans: ion product � 1.5 � 10 �5 , no precipitate

4. If 20.00 mL of a 0.0090 M solution of (NH 4) 2S is mixed with 120.00 mL of a

0.0082 M solution of Al(NO 3) 3, does a precipitate form? The K sp of Al 2S 3 is

2.00 � 10 �7 . ans: ion product � 1.1 � 10 �13 , no precipitate

Holt ChemFile: Problem-Solving Workbook 256 Equilibrium of Salts, K sp

Additional Problems—Equilibrium of Salts, K sp

1. The molar concentration of a saturated calcium chromate, CaCrO 4, solution is

0.010 M at 25°C. What is the K sp of calcium chromate?

2. A 10.00 mL sample of a saturated lead selenate solution is found to contain

0.00136 g of dissolved PbSeO 4 at 25°C. Determine the K sp of lead selenate.

3. A 22.50 mL sample of a saturated copper(I) thiocyanate, CuSCN, solution at

25°C is found to have a 4.0 � 10 �6 M concentration.

a. Determine the K sp of CuSCN.

b. What mass of CuSCN would be dissolved in 1.0 � 10 3 L of solution?

4. A saturated solution of silver dichromate, Ag 2Cr 2O 7, has a concentration of

3.684 � 10 �3 M. Calculate the K sp of silver dichromate.

5. The K sp of barium sulfite, BaSO 3, at 25°C is 8.0 � 10 �7 .

a. What is the molar concentration of a saturated solution of BaSO 3?

b. What mass of BaSO 3 would dissolve in 500. mL of water?

6. The K sp of lead(II) chloride at 25°C is 1.9 � 10 �4 . What is the molar concentration

of a saturated solution at 25°C?

7. The K sp of barium carbonate at 25°C is 1.2 � 10 �8 .

a. What is the molar concentration of a saturated solution of BaCO 3 at 25°C?

b. What volume of water would be needed to dissolve 0.10 g of barium carbonate?

8. The Ksp of SrSO4 is 3.2 � 10 �7 at 25°C.

a. What is the molar concentration of a saturated SrSO 4 solution?

b. If 20.0 L of a saturated solution of SrSO4 were evaporated to dryness, what

mass of SrSO4 would remain?

9. The Ksp of strontium sulfite, SrSO3, is 4.0 � 10 �8 at 25°C. If 1.0000 g of SrSO3 is stirred in 5.0 L of water until the solution is saturated and then filtered,

what mass of SrSO3 would remain?

10. The Ksp of manganese(II) arsenate is 1.9 � 10 �11 at 25°C. What is the molar

concentration of Mn3(AsO4) 2 in a saturated solution? Note that five ions are

produced from the dissociation of Mn3(AsO4) 2.

11. Suppose that 30.0 mL of a 0.0050 M solution of Sr(NO3) 2 is mixed with 20.0

mL of a 0.010 M solution of K2SO4 at 25°C. The Ksp of SrSO4 is 3.2 � 10 �7 .

a. What is the ion product of the ions that can potentially form a precipitate?

b. Does a precipitate form?

12. Lead(II) bromide, PbBr 2, is slightly soluble in water. Its K sp is 6.3 � 10 �6 at

25°C. Suppose that 120. mL of a 0.0035 M solution of MgBr 2 is mixed with

180. mL of a 0.0024 M Pb(C 2H 3O 2) 2 solution at 25°C.

a. What is the ion product of Br � and Pb 2� in the mixed solution?

Holt ChemFile: Problem-Solving Workbook 257 Equilibrium of Salts, K sp

13. The K sp of Mg(OH) 2 at 25°C is 1.5 � 10 �11 .

a. Write the equilibrium equation for the dissociation of Mg(OH) 2.

b. What volume of water would be required to dissolve 0.10 g of Mg(OH) 2?

c. Considering that magnesium hydroxide is essentially insoluble, why is it

possible to titrate a suspension of Mg(OH) 2 to an equivalence point with a

strong acid such as HCl?

14. Lithium carbonate is somewhat soluble in water; its Ksp at 25°C is 2.51 � 10 �2 .

a. What is the molar concentration of a saturated Li 2CO 3 solution?

b. What mass of Li2CO3 would you dissolve in order to make 3440 mL of

saturated solution?

15. A 50.00 mL sample of a saturated solution of barium hydroxide, Ba(OH) 2, is

titrated to the equivalence point by 31.61 mL of a 0.3417 M solution of HCl.

Determine the Ksp of Ba(OH) 2.

16. Calculate the Ksp for salts represented by QR that dissociate into two ions,

Q � and R � , in each of the following solutions:

a. saturated solution of QR is 1.0 M

b. saturated solution of QR is 0.50 M

c. saturated solution of QR is 0.1 M

d. saturated solution of QR is 0.001 M

17. Suppose that salts QR, X2Y, KL2, A3Z, and D2E3 form saturated solutions that

are 0.02 M in concentration. Calculate Ksp for each of these salts.

18. The K sp at 25°C of silver bromide is 5.0 � 10 �13 . What is the molar concentration

of a saturated AgBr solution? What mass of silver bromide would dissolve

in 10.0 L of saturated solution at 25°C?

19. The K sp at 25°C for calcium hydroxide is 5.5 � 10 �6 .

a. Calculate the molarity of a saturated Ca(OH) 2 solution.

b. What is the OH � concentration of this solution?

20. The K sp of magnesium carbonate is 3.5 � 10 �8 at 25°C. What mass of MgCO 3

would dissolve in 4.00 L of water at 25°C?

Holt ChemFile: Problem-Solving Workbook 258 Equilibrium of Salts, K sp

pH

In 1909, Danish biochemist S. P. L Sørensen introduced a system in which acidity

was expressed as the negative logarithm of the H � concentration. In this way, the

acidity of a solution having H � concentration of 10 �5 M would have a value of 5.

Because the power of 10 was now a part of the number, the system was called

pH, meaning power of hydrogen.

Taking the negative logarithm of the hydronium ion concentration will give

you a solution’s pH as given by the following equation.

pH ��log[H3O � ]

Likewise, the value pOH equals the negative logarithm of the hydroxide ion

pOH ��log[OH � ]

Water molecules interact with each other and ionize. At the same time, the

ions in solution reform molecules of water. This process is represented by the following

reversible equation.

H2O(l) � H2O(l) ^ H3O(aq) � OH � (aq)

In pure water the concentrations of hydroxide ions and hydronium ions will

always be equal. These two quantities are related by a term called the ion

product constant for water, Kw. Kw � [H3O � ][OH � ]

The ion product constant for water can be used to convert from pOH to pH.

The following equations derive a simple formula for this conversion.

Pure water has a pH of 7. Rearranging the equation for pH, you can solve for

the hydronium ion concentration of pure water, which is equal to the hydroxide

ion concentration. These values can be used to obtain a numerical value for Kw. Kw � [H3O � ][OH � ] � [1 � 10 �7 ][1 � 10 �7 ] = 1 � 10 �14

Rearrange the Kw expression to solve for [OH � ].

[OH � Kw

] � � ��

[H3O

]

Substitute this value into the equation for pOH.

Kw

pOH ��log� ��

The logarithm of a quotient is the difference of the logarithms of the numerator

and the denominator.

pOH ��logKw � log[H3O � ]

Substitute the value for Kw and pH for �log[H3O � ].

pOH ��log(1.0 � 10 �14 ) � pH

Holt ChemFile: Problem-Solving Workbook 259 pH

The negative logarithm of 10 �14 is 14. Substitute this value, and rearrange.

pOH � pH � 14

Multiply the

concentration of

the solution by

the moles of OH �

per mole of base.

General Plan for Solving pH Problems

Mass

of solute Convert using

of OH� 4b

pOH

This step

works only

for strong acids

or bases.

pOH � �log[OH � ]

the

relationship

Kw � [H3O� ][OH� ]

Kw � 1 � 10�14 pH � pOH � 14

of H3O� 4a

Holt ChemFile: Problem-Solving Workbook 260 pH

Divide the

solute by

the volume of

the solution.

concentration of the

solution by the moles

of H 3 O � per mole

of acid.

pH � �log[H 3 O � ]

A HCl solution has a concentration of 0.0050 M. Calculate [OH � ] and

[H 3O � ] for this solution. HCl is a strong acid, so assume it is 100% ionized.

What is given in the problem? the molarity of the HCl solution, and the fact

that HCl is a strong acid

What are you asked to find? [H 3O � ] and [OH � ]

What steps are needed to calculate the concentration of H 3O � and OH � ?

Determine [H 3O � ] from molarity and the fact that the acid is strong. Use K w to

calculate [OH � ].

rearrange the K w

equation to solve for

[OH � ], substitute known

values and solve

Identity of solute HCl

Concentration of solute 0.0050 M

Acid or base acid

1.0 � 10 �14

K w

[H3O � ] ? M

[OH � ] ? M

Molar concentration of OH� 4b

Molar concentration of

the HCl solution

[HCl] � [H3O � given

1 � 10 �14 � [OH � ] � [H 3O � ]

[OH � Kw 1 � 10

] �

�14

[H3O � ]

each HCl molecule

dissociates to

produce one H 3 O �

ion, so

[HCl] � [H 3 O � ]

Molar concentration of H3O� 4a

Holt ChemFile: Problem-Solving Workbook 261 pH

[H3O � ] � 0.0050 M

[OH � ] � � 2.0 � 10 �12 1.0 � 10

0.0050

Yes; molarity, or mol/L, was required.

Yes; the number of significant figures is correct because molarity of HCl was

given to two significant figures.

Yes; the two concentrations multiply to give 10 � 10 �15 , which is equal to

1.0 � 10 �14 .

1. The hydroxide ion concentration of an aqueous solution is 6.4 � 10 �5 M.

What is the hydronium ion concentration? ans: 1.6 � 10 �10 M

2. Calculate the H 3O � and OH � concentrations in a 7.50 � 10 �4 M solution of

HNO 3, a strong acid. ans: [H 3O � ] � 7.50 � 10 �4 M; [OH � ] � 1.33 � 10 �11 M

Holt ChemFile: Problem-Solving Workbook 262 pH

Calculate the pH of a 0.000 287 M solution of H 2SO 4. Assume 100% ionization.

What is given in the problem? the molarity of the H2SO4 solution, and the fact

that H2SO4 is completely ionized

What are you asked to find? pH

Identity of solute H 2SO 4

Concentration of solute 0.000 287 M

pH ?

What steps are needed to calculate the concentration of H 3O � and the pH?

Determine [H 3O � ] from molarity and the fact that the acid is 100% ionized.

Determine the pH, negative logarithm of the concentration.

the H2SO4 solution

each H 2 SO 4 molecule

dissociates to produce

two H 3 O � ions, so

2 � [H 2 SO 4 ] � [H 3 O � ]

pH � �log[H 3 O � ]

�log[H3O � pH

� ]

Holt ChemFile: Problem-Solving Workbook 263 pH

[H3O � ] � 2 � 0.000 287 M � 5.74 � 10 �4 M

�log[5.74 � 10 �4 ] � 3.24

Yes; there are no units on a pH value.

Yes; the number of significant figures is correct because molarity of H2SO4 was

Yes. You would expect the pH of a dilute H2SO4 solution to be below 7.

1. Determine the pH of a 0.001 18 M solution of HBr. ans: 2.93

2. What is the pH of a solution that has a hydronium ion concentration of 1.0 M?

ans: 0.00

Holt ChemFile: Problem-Solving Workbook 264 pH

3. What is the pH of a 2.0 M solution of HCl, assuming the acid remains 100%

ionized? ans: �0.30

4. What is the theoretical pH of a 10.0 M solution of HCl? ans: �1.00

Holt ChemFile: Problem-Solving Workbook 265 pH

A solution of acetic acid has a pH of 5.86. What are the pOH and [OH � ] of

the solution?

What is given in the problem? the pH of the acetic acid solution

What are you asked to find? pOH and [OH � ]

Identity of solute acetic acid

pH 5.86

pOH ?

What steps are needed to calculate the pOH?

The sum of the pH and pOH of any solution is 14.00. Use this relationship to find

the pOH of the acetic acid solution.

What steps are needed to calculate [OH � ]?

The pOH of a solution is the negative logarithm of the hydroxide ion concentration.

Therefore, calculate [OH � ] using the inverse logarithm of the negative

pOH.

Molar concentration of [OH� 4b

convert using the

pH � pOH � 14.00

14.00 � pH � pOH

10 �pOH

� �log[OH � ]

� [OH � ]

Holt ChemFile: Problem-Solving Workbook 266 pH

14.00 � 5.86 � 8.14

10 �8.14 � 7.2 � 10 �9 M

Yes; there are no units on a pOH value, and [OH � ] has the correct units of

molarity.

Yes; you would expect the pOH of an acid to be above 7, and the hydroxide ion

concentration to be small.

1. What is the pH of a solution with the following hydroxide ion concentrations?

a. 1 � 10 �5 M ans: 9.0

b. 5 � 10 �8 M ans: 6.7

c. 2.90 � 10 �11 M ans: 3.46

2. What are the pOH and hydroxide ion concentration of a solution with a pH of

8.92? ans: pOH � 5.08, [OH � ] � 8.3 � 10 �6 M

3. What are the pOH values of solutions with the following hydronium ion

concentrations?

a. 2.51 � 10 �13 M ans: 1.40

b. 4.3 � 10 �3 M ans: 11.6

c. 9.1 � 10 �6 M ans: 8.96

d. 0.070 M ans: 12.8

Holt ChemFile: Problem-Solving Workbook 267 pH

Determine the pH of a solution made by dissolving 4.50 g NaOH in a

0.400 L aqueous solution. NaOH is a strong base.

What is given in the problem? the mass of NaOH, and the solution volume

What steps are needed to calculate the pH?

First determine the concentration of the solution. Then find the concentration

of hydroxide ions. Calculate the pOH of the solution, and use this to find the pH.

Mass of NaOH in g

Molar concentration

Identity of solute NaOH

Mass of solute 4.50 g

Volume of solution 0.400 L

Concentration of solute ? M

Acid or base base

multiply the mass by the

inverse of the molar mass

of NaOH

NaOH dissociates to produce

one OH � per NaOH molecule,

so [NaOH] is equal to [OH � ]

convert using the relationship

pH � pOH � 14

Amount of NaOH in mol

divide the amount of

NaOH by the

volume of the

of NaOH(aq)

Holt ChemFile: Problem-Solving Workbook 268 pH

1 mol NaOH 1 1 mol OH�

4.50 g NaOH � �

40.00 g NaOH 0.400 L solution 1 mol NaOH

�log[0.281] � 0.551

14.00 � 0.551 � 13.45

Yes; pH has no units.

Yes; the number of significant figures is correct because the value 14.00 has two

decimal places.

Yes; NaOH is a strong base, so you would expect it to have a pH around 14.

given 1 mol NaOH

g NaOH � � � [OH� 1 1 mol OH

40.00 g NaOH L solution

14.00 � pOH � pH

1. A solution is prepared by dissolving 3.50 g of sodium hydroxide in water and

adding water until the total volume of the solution is 2.50 L. What are the

OH � and H 3O � concentrations? ans: [OH � ] � 0.0350 M, [H 3O � ] � 2.86 �

10 �13 M

2. If 1.00 L of a potassium hydroxide solution with a pH of 12.90 is diluted to

2.00 L, what is the pH of the resulting solution? ans: 13.20

� 0.281 M

Holt ChemFile: Problem-Solving Workbook 269 pH

Additional Problems—pH

1. Calculate the H3O � and OH � concentrations in the following solutions. Each

is either a strong acid or a stong base.

a. 0.05 M sodium hydroxide

b. 0.0025 M sulfuric acid

c. 0.013 M lithium hydroxide

d. 0.150 M nitric acid

e. 0.0200 M calcium hydroxide

f. 0.390 M perchloric acid

g. What is the pH of each solution in items 1a. to 1f.?

2. Calculate [H3O � ] and [OH � ] in a 0.160 M solution of potassium hydroxide.

Assume that the solute is 100% dissociated at this concentration.

3. The pH of an aqueous solution of sodium hydroxide is 12.9. What is the molarity

of the solution?

4. What is the pH of a 0.001 25 M HBr solution? If 175 mL of this solution is

diluted to a total volume of 3.00 L, what is the pH of the diluted solution?

5. What is the pH of a 0.0001 M solution of NaOH? What is the pH of a 0.0005 M

solution of NaOH?

6. A solution is prepared using 15.0 mL of 1.0 M HCl and 20.0 mL of

0.50 M HNO3. The final volume of the solution is 1.25 L. Answer the following

a. What are the [H3O � ] and [OH � ] in the final solution?

b. What is the pH of the final solution?

7. A container is labeled 500.0 mL of 0.001 57 M nitric acid solution. A chemist

finds that the container was not sealed and that some evaporation has taken

place. The volume of solution is now 447.0 mL.

a. What was the original pH of the solution?

b. What is the pH of the solution now?

8. Calculate the hydroxide ion concentration in an aqueous solution that has a

0.000 35 M hydronium ion concentration.

9. A solution of sodium hydroxide has a pH of 12.14. If 50.00 mL of the solution

is diluted to 2.000 L with water, what is the pH of the diluted solution?

10. An acetic acid solution has a pH of 4.0. What are the [H3O � ] and [OH � ] in this

11. What is the pH of a 0.000 460 M solution of Ca(OH) 2?

12. A solution of strontium hydroxide with a pH of 11.4 is to be prepared. What

mass of Sr(OH) 2 would be required to make 1.00 L of this solution?

13. A solution of NH3 has a pH of 11.00. What are the concentrations of hydronium

and hydroxide ions in this solution?

Holt ChemFile: Problem-Solving Workbook 270 pH

14. Acetic acid does not completely ionize in solution. Percent ionization of a

substance dissolved in water is equal to the moles of ions produced as a

percentage of the moles of ions that would be produced if the substance were

completely ionized. Calculate the percent ionization of acetic acid the following

solutions.

a. 1.0 M acetic acid solution with a pH of 2.40

b. 0.10 M acetic acid solution with a pH of 2.90

c. 0.010 M acetic acid solution, with a pH of 3.40

15. Calculate the pH of an aqueous solution that contains 5.00 g of HNO3 in

2.00 L of solution.

16. A solution of HCl has a pH of 1.50. Determine the pH of the solutions made in

each of the following ways.

a. 1.00 mL of the solution is diluted to 1000. mL with water.

b. 25.00 mL is diluted to 200 mL with distilled water.

c. 18.83 mL of the solution is diluted to 4.000 L with distilled water.

d. 1.50 L is diluted to 20.0 kL with distilled water.

17. An aqueous solution contains 10 000 times more hydronium ions than hydroxide

ions. What is the concentration of each ion?

18. A potassium hydroxide solution has a pH of 12.90. Enough acid is added to

react with half of the OH � ions present. What is the pH of the resulting solution?

Assume that the products of the neutralization have no effect on pH and

that the amount of additional water produced is negligible.

19. A hydrochloric acid solution has a pH of 1.70. What is the [H3O � ] in this

solution? Considering that HCl is a strong acid, what is the HCl concentration

20. What is the molarity of a solution of the strong base Ca(OH) 2 in a solution

that has a pH of 10.80?

21. You have a 1.00 M solution of the strong acid, HCl. What is the pH of this

solution? You need a solution of pH 4.00. To what volume would you dilute

1.00 L of the HCl solution to get this pH? To what volume would you dilute

1.00 L of the pH 4.00 solution to get a solution of pH 6.00? To what volume

would you dilute 1.00 L of the pH 4.00 solution to get a solution of pH 8.00?

22. A solution of perchloric acid, HClO3, a strong acid, has a pH of 1.28. How

many moles of NaOH would be required to react completely with the HClO3 in 1.00 L of the solution? What mass of NaOH is required?

23. A solution of the weak base NH3 has a pH of 11.90. How many moles of HCl

would have to be added to 1.00 L of the ammonia to react with all of the OH �

ions present at pH 11.90?

24. The pH of a citric acid solution is 3.15. What are the [H3O � ] and [OH � ] in this

Holt ChemFile: Problem-Solving Workbook 271 pH

Titrations

Chemists have many methods for determining the quantity of a substance present

in a solution or other mixture. One common method is titration, in which a solution

of known concentration reacts with a sample containing the substance of

unknown quantity. There are two main requirements for making titration possible.

Both substances must react quickly and completely with each other, and

there must be a way of knowing when the substances have reacted in precise

stoichiometric quantities.

The most common titrations are acid-base titrations. These reactions are easily

monitored by keeping track of pH changes with a pH meter or by choosing an

indicator that changes color when the acid and base have reacted in stoichiometric

quantities. This point is referred to as the equivalence point. Look at the following

equation for the neutralization of KOH with HCl.

KOH(aq) � HCl(aq) → KCl(aq) � H 2O(l)

Suppose you have a solution that contains 1.000 mol of KOH. All of the KOH

will have reacted when 1.000 mol of HCl has been added. This is the equivalence

point of this reaction.

Titration calculations rely on the relationship between volume, concentration,

and amount.

volume of solution � molarity of solution � amount of solute in moles

If a titration were carried out between KOH and HCl, according the reaction

above, the amount in moles of KOH and HCl would be equal at the equivalence

point. The following relationship applies to this system:

molarity KOH � volume KOH � amount of KOH in moles

amount of KOH in moles � amount of HCl in moles

amount of HCl in moles � molarity HCl � volume HCl

molarity KOH � volume KOH � molarity HCl � volume HCl

The following plan for solving titration problems may be applied to any acidbase

titration, regardless of whether the equivalence point occurs at equivalent

volumes.

Holt ChemFile: Problem-Solving Workbook 272 Titrations

of known

acid

3a

General Plan for Solving Titration Problems

The product of

molarity and

volume in liters

is the amount

in moles.

of acid

acid used in

titration

of base

Holt ChemFile: Problem-Solving Workbook 273 Titrations

1b

of acid to base.

Divide the amount

in moles by

the volume in liters

to compute molarity.

base

3b

base used in

A titration of a 25.00 mL sample of a hydrochloric acid solution of

unknown molarity reaches the equivalence point when 38.28 mL of

0.4370 M NaOH solution has been added. What is the molarity of the HCl

HCl(aq) � NaOH(aq) 3 NaCl(aq) � H2O(l) Solution

What is given in the problem? the volume of the HCl solution titrated, and the

molarity and volume of NaOH solution used in

the titration figures.

What are you asked to find? the molarity of the HCl solution

Volume of acid solution 25.00 mL

Molarity of acid solution ? M

Mole ratio of base to acid in titration reaction 1 mol base:

1 mol acid

Volume of base solution 38.28 mL

Molarity of base solution 0.4370 M

What steps are needed to calculate the molarity of the HCl solution?

Use the volume and molarity of the NaOH to calculate the number of moles of

NaOH that reacted. Use the mole ratio between base and acid to determine the

moles of HCl that reacted. Use the volume of the acid to calculate molarity.

Holt ChemFile: Problem-Solving Workbook 274 Titrations

1L

Volume of HCl

divide amount

of HCl by

volume to yield

Molarity of HCl

given 1L

L NaOH �

L NaOH

mL NaOH � � L NaOH

mL HCl ��

given in balanced

38.28 �mL NaOH �� 0.03828 L NaOH

1000 mL �

25.00 �mL HCl �� 0.02500 L HCl

0.4370 mol NaOH 1 mol HCl

0.03828 L NaOH � �

L NaOH 1 mol NaOH

0.02500 L HCl

Molarity of NaOH

the product of

molarity and volume

is the amount of NaOH

mol HCl

� L HCl

� � M HCl

L HCl

� 0.6691 M HCl

Holt ChemFile: Problem-Solving Workbook 275 Titrations

four significant figures.

Yes; a larger volume of base was required than the volume of acid used.

Therefore, the HCl must be more concentrated than the NaOH.

In each of the following problems, the acids and bases react in a mole ratio of

1mol base : 1 mol acid.

1. A student titrates a 20.00 mL sample of a solution of HBr with unknown

molarity. The titration requires 20.05 mL of a 0.1819 M solution of NaOH.

What is the molarity of the HBr solution? ans: 0.1824 M HBr

2. Vinegar can be assayed to determine its acetic acid content. Determine the

molarity of acetic acid in a 15.00 mL sample of vinegar that requires 22.70 mL

of a 0.550 M solution of NaOH to reach the equivalence point. ans: 0.832 M

Holt ChemFile: Problem-Solving Workbook 276 Titrations

A 50.00 mL sample of a sodium hydroxide solution is titrated with a

1.605 M solution of sulfuric acid. The titration requires 24.09 mL of the

acid solution to reach the equivalence point. What is the molarity of the

base solution?

H2SO4(aq) � 2NaOH(aq) 3 Na2SO4(aq) � 2H2O(l) Solution

What is given in the problem? the balanced chemical equation for the acidbase

reaction, the volume of the base solution,

and the molarity and volume of the acid used in

the titration

What are you asked to find? the molarity of the sodium hydroxide solution

Volume of acid solution 24.09 mL

Molarity of acid solution 1.605 M

Mole ratio of base to acid in titration reaction 2 mol base:

Volume of base solution 50.00 mL

Molarity of base solution ? M

What steps are needed to calculate the molarity of the NaOH solution?

Use the volume and molarity of the acid to calculate the number of moles of acid

that reacted. Use the mole ratio between base and acid to determine the moles

of base that reacted. Use the volume of the base to calculate molarity.

Holt ChemFile: Problem-Solving Workbook 277 Titrations

L H 2SO 4 �

Volume of H2SO4 in mL

Molarity of H2SO4 �

Volume of H2SO4 in L

is the amount of H2SO4 in moles

Amount of H2SO4 in mol

mL NaOH �

mol H 2SO 4

L H 2SO 4

50.00 �mL NaOH �� 0.05000 L NaOH

24.09 �mL H2SO4 �� 0.02409 L H2SO4 1000 mL �

1.605 mol H

0.02409 L H2SO4 �

2SO4 �

L H2SO4 given 1L

mL H2SO4 � � L H2SO4 1000 mL

mol H 2 SO 4

2 mol NaOH

1 mol H 2SO 4

� L NaOH

� 1.547 M NaOH

Holt ChemFile: Problem-Solving Workbook 278 Titrations

� M NaOH

0.05000 L NaOH

divide the amount

of NaOH by volume

to yield molarity

Yes; the volume of acid required was approximately half the volume of base

used. Because of the 1:2 mole ratio, the acid must be about the same as the

concentration of the base, which agrees with the result obtained.

1. A 20.00 mL sample of a solution of Sr(OH) 2 is titrated to the equivalence point

with 43.03 mL of 0.1159 M HCl. What is the molarity of the Sr(OH) 2 solution?

ans: 0.1247 M Sr(OH) 2

2. A 35.00 mL sample of ammonia solution is titrated to the equivalence point

with 54.95 mL of a 0.400 M sulfuric acid solution. What is the molarity of the

ammonia solution? ans: 1.26 M NH 3

Holt ChemFile: Problem-Solving Workbook 279 Titrations

A supply of NaOH is known to contain the contaminants NaCl and MgCl 2.

A 4.955 g sample of this material is dissolved and diluted to 500.00 mL

with water. A 20.00 mL sample of this solution is titrated with 22.26 mL

of a 0.1989 M solution of HCl. What percentage of the original sample is

NaOH? Assume that none of the contaminants react with HCl.

What is given in the problem? the mass of the original solute sample, the

volume of the solution of the sample, the

volume of the sample taken for titration, the

molarity of the acid solution, and the volume

of the acid solution used in the titration

What are you asked to find? the percentage by mass of NaOH in the original

Volume of acid solution 22.26 mL

Molarity of acid solution 0.1989 M

Mole ratio of base to acid in titration reaction ?

Volume of base solution titrated 20.00 mL

Moles of base in solution titrated ? mol NaOH

Volume of original sample solution 500.00 mL

Moles of base in original sample ? mol NaOH

Mass of original sample 4.955 g impure NaOH

Mass of base in original sample ? g NaOH

Percentage of NaOH in original sample ?% NaOH

What steps are needed to calculate the concentration of NaOH in the sample?

Determine the balanced chemical equation for the titration reaction. Use the

volume and molarity of the HCl to calculate the moles of HCl that reacted. Use

the mole ratio between base and acid to determine the amount of NaOH that

reacted. Divide by the volume titrated to obtain the concentration of NaOH.

Holt ChemFile: Problem-Solving Workbook 280 Titrations

What steps are needed to calculate the percentage of NaOH in the sample?

Convert the concentration of NaOH to the amount of NaOH in the original

sample by multiplying the concentration by the total volume. Convert amount of

NaOH to mass NaOH by using the molar mass of NaOH. Use the mass of NaOH

and the mass of the sample to calculate the percentage of NaOH.

You must first determine the equation for titration reaction.

Holt ChemFile: Problem-Solving Workbook 281 Titrations

Volume of HCl in mL

Volume of HCl in L

is the amount of HCl

Amount of NaOH in

the original solution

3b 4b

Amount of Volume of

NaOH NaOH used in

in mol titration in L

of NaOH by

Volume of �

original solution

Mass of NaOH in

divide by the total

mass of the solute

and multiply by 100

Percentage of NaOH

in original solution

Holt ChemFile: Problem-Solving Workbook 282 Titrations

L HCl �

mL NaOHtitrated � � L NaOHtitrated 1000 mL

mL HCl � � L HCl

mL NaOHoriginal �

� L NaOHoriginal �

� g NaOHoriginal mol NaOH

20.00 mL � NaOHtitrated �� 0.02000 L NaOHtitrated 1000 mL �

22.26 mL � HCl �� 0.02226 L HCl

0.1989 mol HCl 1 mol NaOH

0.02226 L HCl � �

L HCl 1 mol HCl

� 0.2214 M NaOH

0.02000 L NaOH

500.00 mL � NaOHoriginal �� 0.500 00 L NaOHoriginal 1000 �mL

0.2214 mol NaOH

� 0.500 00 L NaOH � � 4.428 g NaOH

428

g NaOH

� 100 � 89.35% NaOH

955

Yes; units canceled to give percentage of NaOH in sample.

g NaOHoriginal � 100 � percentage of NaOH in solute

1 mol NaOH 1

1 mol HCl L NaOHtitrated � L NaOH original

Holt ChemFile: Problem-Solving Workbook 283 Titrations

Yes; the calculation can be approximated as (0.02 � 0.2 � 25 � 40 � 100)/5 �

400/5 � 80, which is close to the calculated result.

In the problems below, assume that impurities are not acidic or basic and that

they do not react in an acid-base titration.

1. A supply of glacial acetic acid has absorbed water from the air. It must be

assayed to determine the actual percentage of acetic acid. 2.000 g of the acid

is diluted to 100.00 mL, and 20.00 mL is titrated with a solution of sodium

hydroxide. The base solution has a concentration of 0.218 M, and 28.25 mL is

used in the titration. Calculate the percentage of acetic acid in the original

sample. Write the titration equation to get the mole ratio.

ans: 92.5% acetic acid

2. A shipment of crude sodium carbonate must be assayed for its Na 2CO 3 content.

You receive a small jar containing a sample from the shipment and weigh

out 9.709 g into a flask, where it is dissolved in water and diluted to 1.0000 L

with distilled water. A 10.00 mL sample is taken from the flask and titrated to

the equivalence point with 16.90 mL of a 0.1022 M HCl solution. Determine

the percentage of Na 2CO 3 in the sample. Write the titration equation to get the

mole ratio. ans: 94.28% Na 2CO 3

Holt ChemFile: Problem-Solving Workbook 284 Titrations

Additional Problems—Titrations

1. A 50.00 mL sample of a potassium hydroxide is titrated with a 0.8186 M HCl

solution. The titration requires 27.87 mL of the HCl solution to reach the

equivalence point. What is the molarity of the KOH solution?

2. A 15.00 mL sample of acetic acid is titrated with 34.13 mL of 0.9940 M NaOH.

Determine the molarity of the acetic acid.

3. A 12.00 mL sample of an ammonia solution is titrated with 1.499 M HNO3 solution. A total of 19.48 mL of acid is required to reach the equivalence point.

What is the molarity of the ammonia solution?

4. A certain acid and base react in a 1:1 ratio.

a. If the acid and base solutions are of equal concentration, what volume of

acid will titrate a 20.00 mL sample of the base?

b. If the acid is twice as concentrated as the base, what volume of acid will be

required to titrate 20.00 mL of the base?

c. How much acid will be required if the base is four times as concentrated as

the acid, and 20.00 mL of base is used?

5. A 10.00 mL sample of a solution of hydrofluoric acid, HF, is diluted to

500.00 mL. A 20.00 mL sample of the diluted solution requires 13.51 mL of a

0.1500 M NaOH solution to be titrated to the equivalence point. What is the

molarity of the original HF solution?

6. A solution of oxalic acid, a diprotic acid, is used to titrate a 16.22 mL sample

of a 0.5030 M KOH solution. If the titration requires 18.41 mL of the oxalic

acid solution, what is its molarity?

7. A H2SO4 solution of unknown molarity is titrated with a 1.209 M NaOH solution.

The titration requires 42.27 mL of the NaOH solution to reach the equivalent

point with 25.00 mL of the H2SO4 solution. What is the molarity of the

acid solution?

8. Potassium hydrogen phthalate, KHC8H4O4, is a solid acidic substance that

reacts in a 1:1 mole ratio with bases that have one hydroxide ion. Suppose

that 0.7025 g of KHC8H4O4 is titrated to the equivalence point by 20.18 mL of a

KOH solution. What is the molarity of the KOH solution?

9. A solution of citric acid, a triprotic acid, is titrated with a sodium hydroxide

solution. A 20.00 mL sample of the citric acid solution requires 17.03 mL of a

2.025 M solution of NaOH to reach the equivalence point. What is the molarity

of the acid solution?

10. A flask contains 41.04 mL of a solution of potassium hydroxide. The solution

is titrated and reaches an equivalence point when 21.65 mL of a 0.6515 M

solution of HNO3 is added. Calculate the molarity of the base solution.

11. A bottle is labeled 2.00 M H2SO4. You decide to titrate a 20.00 mL sample with

a 1.85 M NaOH solution. What volume of NaOH solution would you expect to

use if the label is correct?

Holt ChemFile: Problem-Solving Workbook 285 Titrations

12. What volume of a 0.5200 M solution of H2SO4 would be needed to titrate

100.00 mL of a 0.1225 M solution of Sr(OH) 2?

13. A sample of a crude grade of KOH is sent to the lab to be tested for KOH

content. A 4.005 g sample is dissolved and diluted to 200.00 mL with water. A

25.00 mL sample of the solution is titrated with a 0.4388 M HCl solution and

requires 19.93 mL to reach the equivalence point. How many moles of KOH

were in the 4.005 g sample? What mass of KOH is this? What is the percent

KOH in the crude material?

14. What mass of magnesium hydroxide would be required for the magnesium

hydroxide to react to the equivalence point with 558 mL of 3.18 M hydrochloric

15. An ammonia solution of unknown concentration is titrated with a solution of

hydrochloric acid. The HCl solution is 1.25 M, and 5.19 mL are required to

titrate 12.61 mL of the ammonia solution. What is the molarity of the ammonia

16. What volume of 2.811 M oxalic acid solution is needed to react to the equivalence

point with a 5.090 g sample of material that is 92.10% NaOH? Oxalic acid

is a diprotic acid.

17. Standard solutions of accurately known concentration are available in most

laboratories. These solutions are used to titrate other solutions to determine

their concentrations. Once the concentration of the other solutions are accurately

known, they may be used to titrate solutions of unknowns.

The molarity of a solution of HCl is determined by titrating the solution

with an accurately known solution of Ba(OH) 2, which has a molar concentration

of 0.1529 M. A volume of 43.09 mL of the Ba(OH) 2 solution titrates 26.06

mL of the acid solution. The acid solution is in turn used to titrate 15.00 mL of

a solution of rubidium hydroxide. The titration requires 27.05 mL of the acid.

a. What is the molarity of the HCl solution?

b. What is the molarity of the RbOH solution?

18. A truck containing 2800 kg of a 6.0 M hydrochloric acid has been in an accident

and is in danger of spilling its load. What mass of Ca(OH) 2 should be

sent to the scene in order to neutralize all of the acid in case the tank bursts?

The density of the 6.0 M HCl solution is 1.10 g/mL.

19. A 1.00 mL sample of a fairly concentrated nitric acid solution is diluted to

200.00 mL. A 10.00 mL sample of the diluted solution requires 23.94 mL of a

0.0177 M solution of Ba(OH) 2 to be titrated to the equivalence point.

Determine the molarity of the original nitric acid solution.

20. What volume of 4.494 M H2SO4 solution would be required to react to the

equivalence point with 7.2280 g of LiOH(s)?

Holt ChemFile: Problem-Solving Workbook 286 Titrations

Equilibrium of Acids and Bases, K a and K b

The ionization of a strong acid, such as HCl, will proceed to completion in reasonably

dilute solutions. This process is written as follows.

HCl(g) � H 2O(l) → H 3O � (aq) � Cl � (aq)

goes 100% to the right; NOT an equilibrium process.

However, when weak acids and weak bases dissolve, they only partially ionize,

resulting in an equilibrium between ionic and molecular forms. The following

equation shows the equilibrium process that occurs when hydrogen fluoride, a

weak acid, dissolves in water.

HF(g) � H 2O(l) ^ H 3O � (aq) � F � (aq)

does not go 100% to the right; IS an equilibrium process

Most weak acids react in this way, that is, by donating a proton to a water molecule

to form a H 3O � ion.

The weak base ammonia establishes the following equilibrium in water.

NH 3(g) � H 2O(l) ^ NH 4 � (aq) � OH � (aq)

Most weak bases react in this way, that is, by accepting a proton from a water

molecule to leave an OH � ion.

You learned to write an equilibrium expression to solve for K, the equilibrium

constant of a reaction. As with equilibrium reactions, equilibrium constants can

be calculated for the ionization and dissociation processes shown above. The

equilibrium constants indicate how far the equilibrium goes toward the ionic

“products.” Recall that the concentration of each reaction component is raised to

the power of its coefficient in the balanced equation. These concentration terms

are arranged in a fraction with products in the numerator and reactants in the

denominator. The equilibrium constants calculated for acid ionization reactions

are called acid ionization constants and have the symbol K a. The equilibrium

constants calculated for base dissociation reactions are called base dissociation

constants and have the symbol K b. From the diagram on the next page and the

problems in this chapter, you will see how these constants are calculated and

how they relate to concentrations of the reactants and products and to pH.

Holt ChemFile: Problem-Solving Workbook 287 Equilibrium of Acids and Bases, K a and K b

General Plan for Solving K a and K b Problems

of H 3 O �

Calculate the

concentrations.

Balanced

HA � H2O º H3O� � A� 3a

Set up the equilibrium

expression and

substitute values.

[H 3 O � ][A � ]

[HA]

K w � [H 3 O � ][OH � ]

K w � 1 � 10 �14

4a 4b

Ka �

Holt ChemFile: Problem-Solving Workbook 288 Equilibrium of Acids and Bases, K a and K b

of OH� 2b

H2O � B º OH� � BH� 3b

[OH� ][BH� ]

Kb �

The hydronium ion concentration of a 0.500 M solution of HF at 25°C is

found to be 0.0185 M. Calculate the ionization constant for HF at this

temperature.

What is given in the problem? the molarity of the acid solution, the

equilibrium concentration of hydronium ions,

and the temperature

What are you asked to find? the acid ionization constant, Ka Items Data

Molar concentration of H 3O � at equilibrium 0.0185 M

Molar concentration of F � at equilibrium ? M

Initial molar concentration of HF 0.500 M

Molar concentration of HF at equilibrium ? M

Acid ionization constant, Ka, of HF ?

What steps are needed to calculate the acid ionization constant for HF?

Write the equation for the ionization reaction. Set up the equilibrium expression

for the ionization of HF in water. Determine the concentrations of all

components at equilbrium, and calculate K a.

[H 3 O � ]

concentrations

HF � H2O º H3O� � F� 3a

use the balanced equation to

relate the unknown quantities to

known quantities, and substitute

these values into the K a

expression

K a �

[HF] initial�[H3O� [F

� [H3O ] � ]

Holt ChemFile: Problem-Solving Workbook 289 Equilibrium of Acids and Bases, K a and K b

[F� [H3O ] � ]

Write the balanced chemical equation for the ionization of HF in aqueous

HF(aq) � H2O(l) → H3O � (aq) � F � (aq)

Write the acid ionization expression for Ka. Remember that pure substances are

not included in equilibrium expressions. For this reason, [H2O] does not appear

in the expression for Ka. Ka �� [H �

3O ][ F

[HF]

From the balanced chemical equation, you can see that one HF molecule ionizes

to give one fluoride ion and one hydronium ion. Therefore, at equilibrium, the

concentration of F � must equal the concentration of H3O � .

[F � ] � [H3O � ] � 0.0185 M

When the HF ionizes in solution, the HF concentration decreases from its initial

value. The amount by which it decreases is equal to the concentration of either

the fluoride ion or the hydronium ion.

[HF] equilibrium � [HF] initial � [H3O � ] equilibrium

Set up the equilibrium expression.

[0.0185][0.0185]

Ka �� 7.11 � 10

[0.500]

� [0.0185]

�4

Yes; the acid ionization constant has no units.

Yes; the number of significant figures is correct because data values have as few

as three significant figures.

Yes; the calculation can be approximated as (0.02 � 0.02)/0.5 � 0.0008, which is

of the same magnitude as the calculated result.

PRACTICE

1. At 25°C, a 0.025 M solution of formic acid, HCOOH, is found to have a hydronium

ion concentration of 2.03 � 10 �3 M. Calculate the ionization constant of

formic acid. ans: Ka � 1.8 � 10 �4

[H 3O � ] [F � ]

[HF] initial � [H 3O � ]

given given

Holt ChemFile: Problem-Solving Workbook 290 Equilibrium of Acids and Bases, K a and K b

At 25°C, the pH of a 0.315 M solution of nitrous acid, HNO 2, is 1.93.

Calculate the K a of nitrous acid at this temperature.

What is given in the problem? the pH of the acid solution and the original

concentration of HNO2 What are you asked to find? the acid ionization constant, Ka Items Data

pH of solution 1.93

Molar concentration of H 3O � at equilibrium ? M

Molar concentration of NO 2 � at equilibrium ? M

Initial molar concentration of HNO 2

What steps are needed to calculate the acid ionization constant for HNO 2?

Determine the H 3O � concentration from the pH. Write the equation for the

ionization reaction. Set up the equilibrium expression. Determine all equilibrium

concentrations and calculate K a.

pH � �log[H3O� 1a

[H3O� ] � 10�pH 2a

rearrange to solve

for [H 3 O � ]

HNO2 � H2O º H3O� 3a

� NO� 2

[H3O� ]

[HNO2] use the balanced equation to

0.315 M

Molar concentration of HNO2 at equilibrium ? M

Acid ionization constant, Ka, of HNO2 ?

[NO2 ] �

[H3O� ][NO2 ] �

[HNO 2] initial �[H 3 O � ]

Holt ChemFile: Problem-Solving Workbook 291 Equilibrium of Acids and Bases, K a and K b

Calculate the H3O � concentration from the pH.

pH ��log [H3O � ]

[H3O � ] � 10 �pH

Write the balanced chemical equation for the ionization of HNO2 in aqueous

HNO2(aq) � H2O(l) ^ H3O � �

(aq) � NO2 (aq)

Write the mathematical equation to compute Ka. Ka �� [NO

2 ][H3O

[HNO2]

Because each HNO2 molecule dissociates into one hydronium ion and one nitrite

ion,

[NO2 ] � [H3O � ]

the HNO2 concentration at equilibrium will be its initial concentration minus any

HNO2 that has ionized. The amount ionized will equal the concentration of H3O �

or NO2 .

[HNO2] equilibrium � [HNO2] initial � [H3O � ]

Substitute known values into the Ka expression.

[H 3O � ] � 10 �1.93 � 1.2 � 10 �2

Ka � = 4.4 � 10 �4

[1.2 � 10 �2 ][1.2 � 10 �2 ]

[0.315] � [1.2 � 10 �2 ]

Yes; the number of significant figures is correct because the pH was given to two

Yes; the calculation can be approximated as (0.01 � 0.01)/0.3 � 3 � 10 �4 , which

is of the same order of magnitude as the calculated result.

[H 3O � ] [NO 2 � ]

[HNO 2] initial � [H 3O � ]

given calculated

Holt ChemFile: Problem-Solving Workbook 292 Equilibrium of Acids and Bases, K a and K b

1. The pH of a 0.400 M solution of iodic acid, HIO 3, is 0.726 at 25°C. What is the

K a at this temperature? ans: K a � 0.167

2. The pH of a 0.150 M solution of hypochlorous acid, HClO, is found to be 4.55

at 25°C. Calculate the K a for HClO at this temperature. ans: K a � 5.2 � 10 �9

Holt ChemFile: Problem-Solving Workbook 293 Equilibrium of Acids and Bases, K a and K b

A 0.450 M ammonia solution has a pH of 11.45 at 25°C. Calculate the

[H 3O � ] and [OH � ] of the solution, and determine the base dissociation

constant, K b, of ammonia.

What is given in the problem? the pH of the base solution, the original

concentration of NH3, and the temperature

What are you asked to find? [H3O � ], [OH � ], and the base dissociation

constant, Kb Items Data

pH of solution 11.45

Molar concentration of OH � at equilibrium ? M

Molar concentration of NH 4 � at equilibrium ? M

Initial molar concentration of NH 3

What steps are needed to calculate the base dissociation constant for NH 3?

Determine the H 3O � concentration from the pH. Calculate the OH � concentration.

Write the balanced chemical equation for the dissociation reaction. Write

the mathematical equation for K b, substitute values, and calculate.

0.450 M

Molar concentration of NH3 at equilibrium ? M

Base dissociation constant, Kb, of NH3 ?

Holt ChemFile: Problem-Solving Workbook 294 Equilibrium of Acids and Bases, K a and K b

Determine [H 3O � ].

Determine [OH � ].

K w � [OH � ][H 3 O � ]

K w � [H 3O � ][OH � ]

[OH � ] �

Write the equilibrium expression for the ionization reaction.

NH3(aq) � H2O(l) ^ NH4 (aq) � OH (aq)

Kb �� [OH

] [NH4

[ NH3]

Because 1 mol of NH3 reacts with 1 mol of H2O to produce 1 mol of NH4 and

1 mol of OH � :

[OH � �

] � [NH4 ]

these values into the K b

Holt ChemFile: Problem-Solving Workbook 295 Equilibrium of Acids and Bases, K a and K b

NH3 � H2O º NH� 4 � OH� K b �

K b �

Kw [H3O� [OH

� ] �

[NH4][OH� � ]

[NH 3]

[NH4][OH� �

[NH 3] initial �[OH � ]

The equilibrium concentration of NH3 will be the initial concentration minus any

NH3 that has reacted. The amount reacted will equal the concentration of NH4 or OH � .

[NH3] equilibrium � [NH3] initial � [OH � ]

Substitute known quantities into the Kb expression.

[H 3O � ] � 10 �11.45 � 3.5 � 10 �12

0 �14

[OH � 1

] �� 2.9 � 10 �12

3.5 � 10

�3

Kb � � 1.9 � 10 �5

[2.9 � 10 �3 ][2.9 � 10 �3 ]

[0.450] � [2.9 � 10 �3 ]

Yes. The calculation can be approximated as (0.003 � 0.003)/0.5 � 0.00009/5 �

1.8 � 10 �5 .

[OH � ][NH 4 � ]

[NH 3] initial � [OH � ]

1. The compound propylamine, CH 3CH 2CH 2NH 2, is a weak base. At equilibrium,

a 0.039 M solution of propylamine has an OH � concentration of 3.74 � 10 �3

M. Calculate the pH of this solution and K b for propylamine.

ans: pH � 11.573; K b � 4.0 � 10 �4

Holt ChemFile: Problem-Solving Workbook 296 Equilibrium of Acids and Bases, K a and K b

A 1.00 M iodic acid, HIO 3, solution has an acid ionization constant of

0.169 at 25°C. Calculate the hydronium ion concentration of the solution

at this temperature.

What is given in the problem? the original concentration of HIO3, the

temperature, and Ka What are you asked to find? [H 3O � ]

Initial molar concentration of HIO 3

What steps are needed to calculate the acid ionization constant for HIO 3?

Write the balanced chemical equation for the ionization reaction. Write the

equation for K a. Substitute x for the unknown values. Rearrange the K a expression

so that a quadratic equation remains. Substitute known values, and solve

for x.

Write the equilibrium equation for the ionization reaction.

HIO3(aq) � H2O(l) ^ H3O � �

(aq) � IO3 (aq)

Use the chemical equation to write an expression for Ka.

Ka �� [H � �

3O

][ IO3

[ HIO

Since 1 mol of HIO3 reacts with 1 mol of H2O to produce 1 mol of H3O � and

1 mol of IO3 :

[H3O � �

] � [IO3 ]

Holt ChemFile: Problem-Solving Workbook 297 Equilibrium of Acids and Bases, K a and K b

1.00 M

Molar concentration of HIO3 at equilibrium ? M

Acid ionization constant, Ka, of HIO3 0.169

The equilibrium concentration of HIO3 will be the initial concentration minus any

HIO3 that has reacted. The amount reacted will equal the concentration of H3O �

or IO3 .

[HIO3] equilibrium � [HIO3] initial � [H3O � ]

Because [H3O � ] is an unknown quantity, substitute the variable x for it to solve.

[HIO3] equilibrium � [HIO3] initial � x

Ka x

��

[HIO3] nitial

� x

given i

Holt ChemFile: Problem-Solving Workbook 298 Equilibrium of Acids and Bases, K a and K b

x

0.169 � �

1.00

0.169(1.00 � x) � x 2

0.169 � 0.169x � x 2

Rearrange the above equation.

x 2 � 0.169x � 0.169 � 0

Notice that this equation fits the form for a general quadratic equation.

ax 2 � bx � c � 0

where a � 1, b � 0.169, c ��0.169

Use the formula for solving quadratic equations.

�b � � 2

b�ac

� 4�

x �� 2a

Substitute the values given above into the quadratic equation, and solve for x.

x �

�0.169 � �(0.169 �)

x � 0.335 M

2 �(1)(�0 � 4�.169)

��

2(1)

Yes; the value has units of mol/L, or M.

Yes; the number of significant figures is correct because the data values had

three significant figures.

Yes; substituting the calculated [H 3O � ] back into the equation for K a yields the

given value.

1. The K a of nitrous acid is 4.6 � 10 �4 at 25°C. Calculate the [H 3O � ] of a

0.0450 M nitrous acid solution. ans: 4.4 � 10 �3 M

Holt ChemFile: Problem-Solving Workbook 299 Equilibrium of Acids and Bases, K a and K b

Additional Problems—Equilibrium of Acids and Bases

1. Hydrazoic acid, HN 3, is a weak acid. The [H 3O � ] of a 0.102 M solution of

hydrazoic acid is 1.39 � 10 �3 M. Determine the pH of this solution, and

calculate K a at 25°C for HN 3.

2. Bromoacetic acid, BrCH 2COOH, is a moderately weak acid. A 0.200 M solution

of bromoacetic acid has a H 3O � concentration of 0.0192 M. Determine

the pH of this solution and the K a of bromoacetic acid at 25°C.

3. A base, B, dissociates in water according to the following equation:

B � H2O ^ BH � � OH �

Complete the following table for base solutions with the characteristics given.

Initial [B] [B] at equilibrium [OH � ] K b [H 3O � ] pH

a. 0.400 M NA 2.70 � 10 �4 M ? ? M ?

b. 0.005 50 M ? M 8.45 � 10 �4 M ? NA ?

c. 0.0350 M ? M ? M ? ? M 11.29

d. ? M 0.006 28 M 0.000 92 M ? NA ?

4. The solubility of benzoic acid, C6H5COOH, in water at 25°C is 2.9 g/L. The pH

of this saturated solution is 2.92. Determine Ka at 25°C for benzoic acid. (Hint:

first calculate the initial concentration of benzoic acid.)

5. A 0.006 50 M solution of ethanolamine, H2NCH2CH2OH, has a pH of 10.64 at

25°C. Calculate the Kb of ethanolamine. What concentration of undissociated

ethanolamine remains at equilibrium?

6. The weak acid hydrogen selenide, H2Se, has two hydrogen atoms that can

form hydronium ions. The second ionization is so small that the concentration

of the resulting H3O � is insignificant. If the [H3O � ] of a 0.060 M solution of

H2Se is 2.72 � 10 �3 M at 25°C, what is the Ka of the first ionization?

7. Pyridine, C5H5N, is a very weak base. Its Kb at 25°C is 1.78 x 10 –9 . Calculate

the [OH � ] and pH of a 0.140 M solution. Assume that the concentration of

pyridine at equilibrium is equal to its initial concentration because so little

pyridine is dissociated.

8. A solution of a monoprotic acid, HA, at equilibrium is found to have a

0.0208 M concentration of nonionized acid. The pH of the acid solution is 2.17.

Calculate the initial acid concentration and Ka for this acid.

9. Pyruvic acid, CH 3COCOOH, is an important intermediate in the metabolism of

carbohydrates in the cells of the body. A solution made by dissolving 438 mg

of pyruvic acid in 10.00 mL of water is found to have a pH of 1.34 at 25°C.

Calculate K a for pyruvic acid.

Holt ChemFile: Problem-Solving Workbook 300 Equilibrium of Acids and Bases, K a and K b

10. The [H 3O � ] of a solution of acetoacetic acid, CH 3COCH 2COOH, is

4.38 � 10 �3 M at 25°C. The concentration of nonionized acid is 0.0731 M at

equilibrium. Calculate K a for acetoacetic acid at 25°C.

11. The Ka of 2-chloropropanoic acid, CH3CHClCOOH, is 1.48 � 10 �3 . Calculate

the [H3O � ] and the pH of a 0.116 M solution of 2-chloropropionic acid. Let

x � [H3O � ]. The degree of ionization of the acid is too large to ignore. If your

set up is correct, you will have a quadratic equation to solve.

12. Sulfuric acid ionizes in two steps in water solution. For the first ionization

shown in the following equation, the Ka is so large that in moderately dilute

solution the ionization can be considered 100%.

H2SO4 � H2O → H3O � �

� HSO4 The second ionization is fairly strong, and Ka � 1.3 � 10 �2 .

HSO4 � H2O ^ H3O � 2�

� SO4 Calculate the total [H 3O � ] and pH of a 0.0788 M H 2SO 4 solution. Hint: If the

first ionization is 100%, what will [HSO 4 � ] and [H3O � ] be? Remember to

account for the already existing concentration of H 3O � in the second ionization.

Let x � [SO 4 2� ].

13. The hydronium ion concentration of a 0.100 M solution of cyanic acid, HOCN,

is found to be 5.74 � 10 �3 M at 25°C. Calculate the ionization constant of

cyanic acid. What is the pH of this solution?

14. A solution of hydrogen cyanide, HCN, has a 0.025 M concentration. The

cyanide ion concentration is found to be 3.16 � 10 �6 M.

a. What is the hydronium ion concentration of this solution?

b. What is the pH of this solution?

c. What is the concentration of nonionized HCN in the solution? Be sure to

use the correct number of significant figures.

d. Calculate the ionization constant of HCN.

e. How would you characterize the strength of HCN as an acid?

f. Determine the [H3O � ] for a 0.085 M solution of HCN.

15. A 1.20 M solution of dichloroacetic acid, CCl2HCOOH, at 25°C has a hydronium

ion concentration of 0.182 M.

a. What is the pH of this solution?

b. What is the Ka of dichloroacetic acid at 25°C?

c. What is the concentration of nonionized dichloroacetic acid in this solution?

d. What can you say about the strength of dichloroacetic acid?

Holt ChemFile: Problem-Solving Workbook 301 Equilibrium of Acids and Bases, K a and K b

16. Phenol, C 6H 5OH, is a very weak acid. The pH of a 0.215 M solution of phenol

at 25°C is found to be 5.61. Calculate the K a for phenol.

17. A solution of the simplest amino acid, glycine (NH2CH2COOH), is prepared by

dissolving 3.75 g in 250.0 mL of water at 25°C. The pH of this solution is found

to be 0.890.

a. Calculate the molarity of the glycine solution.

b. Calculate the Ka for glycine.

18. Trimethylamine, (CH 3) 3N, dissociates in water the same way that NH 3 does—

by accepting a proton from a water molecule. The [OH � ] of a 0.0750 M solution

of trimethylamine at 25°C is 2.32 � 10 �3 M. Calculate the pH of this

solution and the K b of trimethylamine.

19. Dimethylamine, (CH3) 2NH, is a weak base similar to the trimethylamine in

item 18. A 5.00 � 10 �3 M solution of dimethylamine has a pH of 11.20 at 25°C.

Calculate the Kb of dimethylamine. Compare this Kb with the Kb for trimethylamine

that you calculated in item 18. Which substance is the stronger base?

20. Hydrazine dissociates in water solution according to the following equations:

H2NNH2 � H2O(l) ^ NH2NNH3 (aq) � OH (aq)

� 2� �

H2NNH3 (aq) � H2O(l) ^ NH3NNH3 (aq) � OH (aq)

The Kb of this second dissociation is 8.9 � 10 �16 , so it contributes almost no

hydroxide ions in solution and can be ignored here.

a. The pH of a 0.120 M solution of hydrazine at 25°C is 10.50. Calculate Kb for

the first ionization of hydrazine. Assume that the original concentration of

H2NNH2 does not change.

b. Make the same assumption as you did in (a) and calculate the [OH � ] of a

0.020 M solution.

c. Calculate the pH of the solution in (b).

Holt ChemFile: Problem-Solving Workbook 302 Equilibrium of Acids and Bases, K a and K b

Redox Equations

The feature that distinguishes redox reactions from other types of reactions is

that elements change oxidation state by gaining or losing electrons. Compare the

equations for the following two reactions:

(1) KBr(aq) � AgNO3(aq) → AgBr(s) � KNO3(aq) (2) 2KBr(aq) � Cl2(g) → 2KCl(aq) � Br2(l) Equation 1 represents the combining of the salts potassium bromide solution

and a silver nitrate solution to form a precipitate of insoluble silver bromide,

leaving potassium nitrate in solution. The only change that occurs is that ions

trade places, forming an insoluble compound. This reaction is a typical doubledisplacement

reaction. It is driven by the removal of Ag � and Br � from solution

in the form of a precipitate.

You will recognize that Equation 2 is a single-displacement reaction in which

chlorine atoms replace bromine atoms in the salt KBr. Although it is not complex,

Equation 2 differs from Equation 1 in a fundamental way. In order for chlorine to

replace bromine, the uncharged atoms of elemental chlorine must change into

chloride ions, each having a 1� charge. Also, bromide ions with a 1� charge

must change into uncharged bromine atoms. The K � is a spectator ion that

doesn’t participate in the process. In fact, it could be Na � , Ca 2� , Fe 3� , H � , or any

other stable cation.

The loss of electrons by bromine is oxidation, and the gain of electrons by

chlorine is reduction.

Formation of the chloride ions and the bromine molecule involves the complete

transfer of two electrons. The two chlorine atoms gain two electrons, and

two bromide ions lose two electrons. Oxidation and reduction can involve the

partial transfer of electrons as well as the complete transfer seen in the preceding

example. The oxidation number of an atom is not synonymous with the charge on

that atom, so a change in oxidation number does not require a change in actual

charge. Take the example of the following half-reaction:

H 2C 2O 4(aq) → 2CO 2(g) � 2H � (aq) � 2e �

Carbon changes oxidation state from �3 to �4. Carbon is oxidized even though it

is not ionized.

The potassium bromide–chlorine reaction is simple, but many redox reactions

are not. In this worksheet, you will practice the art of balancing redox equations

and try your hand at more-complex ones.

Holt ChemFile: Problem-Solving Workbook 303 Redox Equations

There are seven simple steps to balancing redox equations. You will find these

steps in the General Plan for Balancing Redox Equations.

7

General Plan for Balancing Redox Equations

Write the unbalanced formula equation if it is not given. List

formulas for any ionic substances as their individual ions, and

write a total ionic equation.

Assign oxidation numbers to each element. Then rewrite the

equation, leaving out any ions or molecules whose elements do

not change oxidation state during the reaction.

Write the half-reaction for reduction. You must decide which

element of the ions and molecules left after item 2 is reduced.

Once you have written the half-reaction, you must balance it for

charge and mass.

Write the half-reaction for oxidation. You must decide which

element of the ions and molecules left after item 2 is oxidized.

Adjust the coefficients of the two half-reactions so that the same

number of electrons are gained in reduction as are lost in

oxidation.

Add the two half-reactions together. Cancel out anything common

to both sides of the new equation. Note that the electrons

should always cancel out of the total equation.

Combine ions to form the compounds shown in the original

formula equation. Check to ensure that all other ions and atoms

balance.

Holt ChemFile: Problem-Solving Workbook 304 Redox Equations

REACTIONS IN ACIDIC SOLUTION

Write a balanced redox equation for the reaction of hydrochloric acid

with nitric acid to produce aqueous hypochlorous acid and nitrogen

monoxide.

What is given in the problem? the reactants and products of a redox reaction

What are you asked to find? the balanced redox reaction

Reactants HCl, HNO 3

Products HClO, NO

Solution type acidic

Oxidized species ?

Reduced species ?

Balanced equation ?

What steps are needed to balance the redox equation?

1. Write the unbalanced formula equation followed by the ionic equation.

2. Assign oxidation numbers to each element. Delete any ion or molecule in

which there is no change in oxidation state.

3. Write the half-reaction for reduction, and balance the mass and charge. H �

and H 2O may be added to either side of the equation to balance mass.

4. Repeat step 3 for the oxidation half-reaction.

5. Adjust the coefficients so that the number of electrons lost equals the number

of electrons gained.

6. Combine the half-reactions, and cancel anything common to both sides of the

7. Combine ions to change the equation back to its original form, and check the

balance of everything.

1. Write the formula equation.

HCl(aq) � HNO3(aq) → HClO(aq) � NO(g)

Write the total ionic equation.

H � � Cl � � H � � � �

� NO3 → H � ClO � NO

Holt ChemFile: Problem-Solving Workbook 305 Redox Equations

2. Assign oxidation numbers to each element.

H �

�1

� Cl �

� H �

�5�2

� NO3 → H �

� OCl �

�2�1 �2�2

� NO

Delete any ion or molecule in which there is no change in oxidation state.

Cl �

3. Write the half-reaction for reduction.

� N �5 �1

O3 → OCl � N �2

�5 �2

NO3

→ NO Balance the mass by adding H � and H2O. 4H � � N �5 �2

O3 → O � 2H2O Balance the charge by adding electrons to the side with the higher positive

charge.

4H � � N �5

�2

O3 � 3e → O � 2H2O 4. Write the half-reaction for oxidation.

→ OCl �

Balance the mass by adding H � and H 2O.

� H2O → OCl �

� 2H �

Balance charge by adding electrons to the side with the higher positive charge.

� 2H � � 2e �

5. Multiply by factors so that the number of electrons lost equals the number of

electrons gained.

2e � are lost in oxidation; 3e � are gained in reduction. Therefore, to get 6e � on

both sides, calculate as follows:

2 � [4H � � �

� NO3 � 3e → NO � 2H2O]

3 � [Cl � � H2O → OCl � � 2H � � 2e � ]

6. Combine the half-reactions.

� 3 � [Cl � � H2O → OCl � � 2H � � e � ]

3Cl � � � � � � �

� 3H2O � 2NO3 � 8H � 6e → 3OCl � 6H � 6e � 2NO � 4H2O

Cancel out anything common to both sides of the equation.

3Cl � 2

� � � � � �

7. Combine ions to change the equation back to its original form. There must be

five H � ions on the reactant side of the equation to bind with the three chloride

ions and two nitrate ions, so three H � ions must be added to each side.

3Cl � � 2NO 3 � � 5H � → 3OCl � � 2NO � H2O � 3H �

3HCl � 2HNO3 → 3HOCl � 2NO � H2O Check the balance.

Holt ChemFile: Problem-Solving Workbook 306 Redox Equations

NA

Yes; the reaction has the reactants and products required and is balanced.

Balance the following redox equations. Assume that all reactions take place in an

acid environment where H � and H2O are readily available.

1. Fe � SnCl4 → FeCl3 � SnCl2 ans: 2Fe � 3SnCl4 3 2FeCl3 � 3SnCl2 2. H 2O 2 � FeSO 4 � H 2SO 4 → Fe 2(SO 4) 3 � H 2O

ans: H 2O 2 � 2FeSO 4 � H 2SO 4 3 Fe 2(SO 4) 3 � 2H 2O

3. CuS � HNO 3 → Cu(NO 3) 2 � NO � S � H 2O ans: 3CuS � 8HNO 3 3

3Cu(NO 3) 2 � 2NO � 3S � 4H 2O

4. K 2Cr 2O 7 � HI → CrI 3 � KI � I 2 � H 2O ans: K 2Cr 2O 7 � 14HI 3

2CrI 3 � 2KI � 3I 2 � 7H 2O

Holt ChemFile: Problem-Solving Workbook 307 Redox Equations

REACTIONS IN BASIC SOLUTION

Write a balanced equation for the reaction in a basic solution of NiO 2 and

Fe to produce Ni(OH) 2 and Fe(OH) 2.

Reactants NiO 2, Fe

Products Ni(OH) 2, Fe(OH) 2

Solution type basic

1. Write the formula equation followed by the ionic equation.

3. Write the half-reaction for reduction, and balance the mass and charge. OH �

and H 2O may be added to either side.

NiO2 � Fe → Ni(OH) 2 � Fe(OH) 2

NiO2 � Fe → Ni 2� � 2OH � � Fe 2� � 2OH �

Holt ChemFile: Problem-Solving Workbook 308 Redox Equations

�4�2

NiO2

� Fe

0

→ Ni

�2 �2�1 �2 �2�1

2� � 2� �

� 2OH � Fe � 2OH

Delete any ions or molecules in which there is no change in oxidation state.

�2 �2

2� 2�

�2 2�

Balance the mass by adding OH � and H2O. �4

� 2H2O → Ni

�2 2� �

� 4OH

Balance the charge by adding electrons to the side with the higher positive

� 2H2O � 2e � → Ni

4. Write the half-reaction for oxidation.

Fe

→ Fe

The mass is already balanced.

� 2e

5. The numbers of electrons lost and gained are already the same.

NiO 2 � 2H 2O � Fe → Ni 2� � Fe 2� � 4OH �

7. Combine ions to change the equation back to its original form. The four OH �

ions combine with the nickel and iron to make nickel(II) hydroxide and

iron(II) hydroxide.

NiO2 � 2H2O � Fe → Ni(OH) 2 � Fe(OH) 2

Check the balance.

Holt ChemFile: Problem-Solving Workbook 309 Redox Equations

Balance the following redox equations. Assume that all reactions take place in a

basic environment where OH � and H2O are readily available.

1. CO2 � NH2OH → CO � N2 � H2O ans: CO2 � 2NH2OH 3 CO � N2 � 3H2O 2. Bi(OH) 3 � K 2SnO 2 → Bi � K 2SnO 3 (Both of the potassium-tin-oxygen compounds

dissociate into potassium ions and tin-oxygen ions.)

ans: 2Bi(OH) 3 � 3K 2SnO 2 3 2Bi � 3K 2SnO 3 � 3H 2O

Holt ChemFile: Problem-Solving Workbook 310 Redox Equations

Balance each of the following redox equations. Unless stated otherwise, assume

that the reaction occurs in acidic solution.

1. Mg � N2 → Mg3N2 2. SO 2 � Br 2 � H 2O → HBr � H 2SO 4

3. H 2S � Cl 2 → S � HCl

4. PbO 2 � HBr → PbBr 2 � Br 2 � H 2O

5. S � HNO 3 → NO 2 � H 2SO 4 � H 2O

6. NaIO3 � N2H4 � HCl → N2 � NaICl2 � H2O (N2H4 is hydrazine; do not

separate it into ions.)

7. MnO2 � H2O2 � HCl → MnCl2 � O2 � H2O 8. AsH3 � NaClO3 → H3AsO4 � NaCl (AsH3 is arsine, the arsenic analogue of

ammonia, NH3.) 9. K2Cr2O7 � H2C2O4 � HCl → CrCl3 � CO2 � KCl � H2O (H2C2O4 is oxalic

acid; it can be treated as 2H � 2�

� C2O4 .)

10. Hg(NO 3) 2

O3

HgO � NO 2 � O 2 (The reaction is not in solution.)

11. HAuCl 4 � N 2H 4 → Au � N 2 � HCl (HAuCl 4 can be considered as H � � AuCl 4 � .)

12. Sb 2(SO 4) 3 � KMnO 4 � H 2O → H 3SbO 4 � K 2SO 4 � MnSO 4 � H 2SO 4

13. Mn(NO 3) 2 � NaBiO 3 � HNO 3 → Bi(NO 3) 2 � HMnO 4 � NaNO 3 � H 2O

14. H 3AsO 4 � Zn � HCl → AsH 3 � ZnCl 2 � H 2O

15. KClO 3 � HCl → Cl 2 � H 2O � KCl

16. The same reactants as in Item 15 can combine in the following way when

more KClO3 is present. Balance the equation.

KClO3 � HCl → Cl2 � ClO2 � H2O � KCl

17. MnCl 3 � H 2O → MnCl 2 � MnO 2 � HCl

18. NaOH � H 2O � Al → NaAl(OH) 4 � H 2 in basic solution

19. Br 2 � Ca(OH) 2 → CaBr 2 � Ca(BrO 3) 2 � H 2O in basic solution

20. N 2O � NaClO � NaOH → NaCl � NaNO 2 � H 2O in basic solution

21. Balance the following reaction, which can be used to prepare bromine in the

laboratory:

HBr � MnO2 → MnBr2 � H2O � Br2 22. The following reaction occurs when gold is dissolved in aqua regia. Balance

the equation.

Holt ChemFile: Problem-Solving Workbook 311 Redox Equations

Electrochemistry

The potential in volts has been measured for many different reduction halfreactions.

The potential value is measured against the standard hydrogen electrode,

which is assigned a value of zero. For consistency, these half-reactions are

always written in the direction of the reduction. A half-reaction that has a positive

reduction potential proceeds in the direction of the reduction when it is coupled

with a hydrogen electrode. A reaction that has a negative reduction potential

proceeds in the oxidation direction when it is coupled with a hydrogen electrode.

Table 1 gives some common standard reduction potentials.

Standard Standard

electrode electrode

Reduction potential, E 0

half-reaction (in volts) half-reaction (in volts)

MnO4 � 8H � 5e ^

Mn 2� � 4H2O �1.50 Fe 3� � 3e � ^ Fe �0.04

Au 3� � 3e � ^ Au �1.50 Pb 2� � 2e � ^ Pb �0.13

Cl2 � 2e � ^ 2Cl �

�1.36 Sn 2� � 2e � ^ Sn �0.14

2� � �

Cr2O7 � 14H � 6e ^

2Cr 3� � 7H2O �1.23 Ni 2� � 2e � ^ Ni �0.26

MnO2 � 4H � � 2e � ^

Mn 2� � 2H2O �1.22 Cd 2� � 2e � ^ Cd �0.40

Br 2 � 2e � ^ 2Br �

�1.07 Fe 2� � 2e � ^ Fe �0.45

Hg 2� � 2e � ^ Hg �0.85 S � 2e � ^ S 2�

�0.48

Ag � � e � ^ Ag �0.80 Zn 2� � 2e � ^ Zn �0.76

Hg 2 2� � 2e � ^ 2Hg �0.80 Al 3� � 3e � ^ Al �1.66

Fe 3� � e � ^ Fe 2�

MnO 4 � � e � ^ MnO4 2�

I 2 � 2e � ^ 2I �

�0.77 Mg 2� � 2e � ^ Mg �2.37

�0.56 Na � � e � ^ Na �2.71

�0.54 Ca 2� � 2e � ^ Ca �2.87

Cu 2� � 2e � ^ Cu �0.34 Ba 2� � 2e � S � 2H

^ Ba �2.91

� (aq) � 2e � ^

H2S(aq) �0.14 K � � e � ^ K �2.93

2H � (aq) � 2e � ^ H2 �0.00 Li � � e � ^ Li �3.04

Holt ChemFile: Problem-Solving Workbook 312 Electrochemistry

You can use reduction potentials to predict the direction in which any redox

reaction will be spontaneous. A spontaneous reaction occurs by itself, without

outside influence. The redox reaction will proceed in the direction for which the

difference between the two half-reaction potentials is positive. This direction is

the same as the direction of the more positive half-reaction.

General Plan for Solving Electrochemical Problems

reduction

Cathode

Reduction

occurs at the

cathode.

Look up the reduction

potential for the halfreaction

in Table 1.

E 0 cathode

Balanced ionic

equation for the redox

Write the

individual halfreactions

for both

oxidation and

reduction.

Substitute and

solve.

E 0 cell � E0 cathode � E0 anode

Oxidation

anode.

oxidation

Anode

Holt ChemFile: Problem-Solving Workbook 313 Electrochemistry

Calculate the cell potential to determine whether the following reaction

is spontaneous in the direction indicated.

Cd 2� (aq) � 2I � (aq) → Cd(s) � I2(s) Solution

What is given in the problem? reactants and products

What are you asked to find? whether the reaction is spontaneous

What steps are needed to determine whether the reaction is spontaneous?

Separate into oxidation and reduction half-reactions. Find reduction potentials

for each. Solve the equation for the cell potential to determine if the reaction is

spontaneous.

Cd2� (aq) � 2e� 2a

3 Cd(s)

E0 4a

cathode

Reactants Cd 2� (aq) � 2I � (aq)

Products Cd(s) � I 2(s)

E 0 anode

E 0 cell

reduction occurs

at the cathode

Cd2� (aq) � 2I� 1

(aq) 3 Cd(s) � I2 (s)

look up the

reduction potential

for the half-reaction

in Table 1

?V

? V

write the individual

half-reactions

substitute and solve

E0 cell � E0 cathode � E0 5

anode

2I� (aq) 3 I2 (s) � 2e� 2b

oxidation occurs

at the anode

E0 4b

Holt ChemFile: Problem-Solving Workbook 314 Electrochemistry

Write the given equation.

Cd 2� (aq) � 2I � (aq) → Cd(s) � I2(s) The oxidation number of cadmium decreases; it is reduced.

Cd 2� (aq) � 2e � → Cd(s)

The oxidation number of iodine increases; it is oxidized.

2I � (aq) → I2(s) � 2e �

Cadmium is the cathode, and iodine is the anode.

� �0.40 V

� �0.54 V

E0 � cathode � E0 E anode

0 given above given above

cell

Determine spontaneity. If the cell potential is positive, the reaction is spontaneous

as written. If the cell potential is negative, the reaction is not spontaneous

as written, but the reverse reaction is spontaneous.

E 0 cell ��0.40 V � 0.54 V ��0.94 V

The reaction potential is negative. Therefore, the reaction is not spontaneous.

The reverse reaction would have a positive potential and would, therefore, be

Cd 2� (aq) � 2I � (aq) → Cd(s) � I2(s) not spontaneous

Cd(s) � I2(s) → Cd 2� (aq) � 2I � (aq) spontaneous

Yes; cell potentials are in volts.

Yes; the number of significant figures is correct because the half-cell potentials

have two significant figures.

Yes; the reduction potential for the half-reaction involving iodine was more

positive than the potential for the reaction involving cadmium, which means that

I2 has a greater attraction for electrons than Cd 2� . Therefore, I2 is more likely

to be reduced than Cd 2� . The reverse reaction is favored.

Holt ChemFile: Problem-Solving Workbook 315 Electrochemistry

Use the reduction potentials in Table 1 to determine whether the following reactions

are spontaneous as written. Report the E 0 cell for the reactions.

1. Cu 2� � Fe → Fe 2� � Cu ans: �0.79 V; spontaneous

2. Pb 2� � Fe 2� → Fe 3� � Pb ans: �0.90 V; nonspontaneous

3. Mn 2� � 4H 2O � Sn 2� → MnO 4 � � 8H � � Sn ans: �1.64 V; nonspontaneous

4. MnO 4 2� � Cl2 → MnO 4 � � 2Cl � ans: �0.80 V; spontaneous

Holt ChemFile: Problem-Solving Workbook 316 Electrochemistry

5. Hg 2 2� � 2MnO4 2� → 2Hg � 2MnO4 � ans: �0.24 V; spontaneous

6. 2Li � � Pb → 2Li � Pb 2� ans: �2.91 V; nonspontaneous

7. Br 2 � 2Cl � → 2Br � � Cl 2 ans: �0.29 V; nonspontaneous

8. S � 2I � → S 2� � I 2 ans: �1.02 V; nonspontaneous

Holt ChemFile: Problem-Solving Workbook 317 Electrochemistry

A cell is constructed in which the following two half-reactions can occur

in either direction.

Zn 2� � 2e � ^ Zn

Br2 � 2e � ^ 2Br �

Write the full ionic equation for the cell in the spontaneous direction,

identify the reactions occurring at the anode and cathode, and determine

the cell’s voltage.

What is given in the problem? the reversible half-reactions of the cell

What are you asked to find? the equation in the spontaneous direction; the

voltage of the cell

Half-reaction 1 Zn 2� � 2e � ^ Zn

Half-reaction 2 Br 2 � 2e � ^ 2Br

Reduction potential of 1 �0.76 V

Reduction potential of 2 �1.07 V

Full ionic reaction ?

Cell voltage ?

What steps are needed to determine the spontaneous reaction of the cell and the

cell voltage?

Determine which half-reaction has the more positive reduction potential. This

will be the reduction half-reaction; it occurs at the cathode. Reverse the other

half-reaction so that it becomes an oxidation half-reaction; it occurs at the

anode. Adjust the half-reactions so that the same number of electrons are lost

as are gained. Add the reactions together. Compute the cell voltage by the

formula E 0 cell � E 0 cathode � E 0 anode, using the reduction potentials for the

reaction at each electrode.

Holt ChemFile: Problem-Solving Workbook 318 Electrochemistry

E 0 half-reaction

E

First, look up the reduction potentials for the two half-reactions in Table 1.

0 cell � E0 cathode � E0 anode

Br2 has the larger reduction potential; therefore, it is the cathode. Zn has the

smaller reduction potential; therefore, it is the anode.

The cathode half-reaction is

Br2 � 2e � → 2Br

The anode half-reaction is

Zn → Zn 2� � 2e �

Zn � E0 from Table 1

�0.76 V

The full-cell equation is

the half-cell with

the larger reduction

potential is the

Br 2 � 2e � → 2Br �

� Zn → Zn 2� � 2e �

Br 2 � Zn → 2Br � � Zn 2�

Substitute the reduction potentials for the anode and cathode into the cell potential

equation, and solve the equation.

combine to write the full

ionic equation

EBr �

2 0 from Table 1

�1.07 V

EBr2 0

E0 cathode

E0 EZn anode

Br 2 � 2e � 3 2Br � Zn 3 Zn 2� � 2e �

Br 2 � Zn 3 2Br � � Zn 2�

the smaller

is the anode

Cathode Anode

Holt ChemFile: Problem-Solving Workbook 319 Electrochemistry

E 0 cell � 1.07 V � (�0.76 V) � 1.83 V

Yes; the cell potential is in volts.

Yes; the half-cell potentials were given to two decimal places.

Yes; you would expect the reaction to have a positive cell potential because it

should be spontaneous.

If a cell is constructed in which the following pairs of reactions are possible, what

would be the cathode reaction, the anode reaction, and the overall cell voltage?

1. Ca 2� � 2e � ^ Ca

Fe 3� � 3e � ^ Fe

ans: cathode: Fe 3� � 3e � 3 Fe, anode: Ca 3 Ca 2� � 2e � , E 0 cell ��2.83 V

2. Ag � � e � ^ Ag

S � 2H � � 2e � ^ H 2S

ans: cathode: Ag � � e � 3 Ag, anode: H 2S 3 S � 2H � � 2e � , E 0 cell ��0.66 V

3. Fe 3� � e � ^ Fe 2�

Sn 2� � 2e � ^ Sn

ans: cathode: Fe 3� � e � 3 Fe 2� , anode: Sn 3 Sn 2� � 2e � , E 0 cell ��0.91 V

4. Cu 2� � 2e � ^ Cu

Au 3� � 3e � ^ Au

ans: cathode: Au 3� � 3e � 3 Au, anode: Cu 3 Cu 2� � 2e � , E 0 cell ��1.16 V

Holt ChemFile: Problem-Solving Workbook 320 Electrochemistry

Use reduction potentials to determine whether the reactions in the following 10

problems are spontaneous.

1. Ba � Sn 2� → Ba 2� � Sn

2. Ni � Hg 2� → Ni 2� � Hg

3. 2Cr 3� � 7H 2O � 6Fe 3� → Cr 2O 7 2� � 14H � � 6Fe 2�

4. Cl 2 � Sn → 2Cl � � Sn 2�

5. Al � 3Ag � → Al 3� � 3Ag

6. Hg 2 2� � S 2� → 2Hg � S

7. Ba � 2Ag � → Ba 2� � 2Ag

8. 2I � � Ca 2� → I 2 � Ca

9. Zn � 2MnO 4 � → Zn 2� � 2MnO4 2�

10. 2Cr 3� � 3Mg 2� � 7H 2O → Cr 2O 7 2� � 14H � � 3Mg

In the following problems, you are given a pair of reduction half-reactions. If a cell

were constructed in which the pairs of half-reactions were possible, what would

be the balanced equation for the overall cell reaction that would occur? Write the

half-reactions that occur at the cathode and anode, and calculate the cell voltage.

11. Cl2 � 2e � ^ 2Cl �

Ni 2� � 2e � ^ Ni

12. Fe 3� � 3e � ^ Fe

Hg 2� � 2e � ^ Hg

13. MnO 4 � � e � ^ MnO4 2�

Al 3� � 3e � ^ Al

14. MnO 4 � � 8H � � 5e � ^ Mn 2� � 4H2O

15. Ca 2� � 2e � ^ Ca

Li � � e � ^ Li

16. Br 2 � 2e � ^ 2Br �

MnO 4 � � 8H � � 5e � ^ Mn 2� � 4H2O

17. Sn 2� � 2e � ^ Sn

18. Zn 2� � 2e � ^ Zn

Cr 2O 7 2� � 14H � � 6e � ^ 2Cr 3� � 7H2O

19. Ba 2� � 2e � ^ Ba

Ca 2� � 2e � ^ Ca

2� �

20. Hg2 � 2e ^ 2Hg

Cd 2� � 2e � ^ Cd

Holt ChemFile: Problem-Solving Workbook 321 Electrochemistry

Answer Key

The Science of

Chemistry

CONVERSIONS

1. a. 12 750 km

b. 2.77 m

c. 3.056 hectares

d. 0.008 19 m 2

e. 300 Mm

2. a. 620 m

b. 3 875 000 mg

c. 3.6 �L

d. 342 kg

e. 68 710 L

3. a. 0.000 856 kg

b. 0.001 21 kg

c. 6.598 cm 3

d. 0.0806 mm

e. 0.010 74 L

4. a. 7930 cm 3

b. 590 cm

c. 4.19 dm 3

d. 74 800 cm 2

e. 197 L

5. 1 L

6. 370 �L

7. 6 L

8. 0.876 L of water per year; 876 kg of

9. a. 1674 km/h

b. 40 176 km/day

10. 7.8 kg of sodium hydroxide

11. 45 m of plastic tubing

12. a. 13.2 mL/day

b. 150 kg/min

c. 62 cm 3 /min

d. 1.7 m/s

13. a. 2.97 g/cm 3

b. 0.041 28 kg/cm 2

c. 5.27 kg/dm 3

d. 0.006 91 mg/mm 3

14. a. 750 mL

b. 5.56 kg

15. 1250 kg

16. a. 0.0028m 3

b. 1.05 g/cm 3

c. 0.056 m 2

17. a. 0.04 mL per drop

b. 1.48 mL

c. 17 000 drops

18. a. 0.5047 kg; 504.7 g

b. 0.0092 kg; 9.2 g

c. 0.000 122 kg; 0.122 g

d. 0.071 95 kg; 71.95 g

19. a. 0.582 L; 582 mL

b. 2.5 L; 2500 mL

c. 1.18 L; 1180 mL

d. 0.0329 L; 32.9 mL

20. a. 1370 g/L; 1370 kg/m 3

b. 692 g/L; 692 kg/m 3

c. 5200 g/L; 5200 kg/m 3

d. 38 g/L; 38 kg/m 3

e. 5790 g/L; 5790 kg/m 3

f. 0.0011 g/L; 0.0011 kg/m 3

21. a. 360 g/min

b. 518.4 kg/day

c. 6 mg/ms

22. 27.8 m/s

23. 4732 kcal/h

24. 620 kg

25. 3.9 mL 1 L 24 h

��

h

365 days

��

1 year

26. 40 doses

1000 mL 1 day

� 34.164 L/year

Matter and Energy

SIGNIFICANT FIGURES

1. a. 3

b. 4

c. 3

d. 2

e. 2

f. 1

g. 3

h. 4

i. 5

2. a. 5 490 000 m

b. 0.013 479 3 mL

c. 31 950 cm 2

d. 192.67 m 2

e. 790 cm

f. 389 278 000 J

g. 225 834.8 cm 3

3. a. 49 000 cm 2

b. 3.1 kg/L

c. 12.3 L/sec

d. 170 000 cm 3

Holt Chemistry 322 Answer Key

e. 41 m 3

f. 3.129 g/cm 3

4. a. 90.2 J

b. 0.0006 m

c. 900 g

d. 31.1 kPa

e. 278 dL

f. 1790 kg

5. a. 307 cm 2

b. 30 700 mm 2

c. 0.0307 m 2

6. a. 1800 cm 3

b. 0.0018 m 3

c. 1 800 000 mm 3

7. a. 1300 kg/m 3

b. 1.3 g/mL

c. 1.3 kg/dm 3

8. a. 130 mm 3

b. 430 cm 3

c. 5.0 m

d. 4000 m 3

9. 26 000 000 m 3

10. a. 13.38 g

b. 100. mg

c. 0.015 L

d. 315 cm 2

e. 14.47 kg

f. 353 mL

11. 1.09 kg/L

12. 0.43 g/m; 2.3 m

13. 2000 m 2

14. 26 300 kJ/min; 439 kJ/s

15. a. 15.8 m 3

b. 9800 L/min

c. 590 m 3 /h

16. a. 7.5 kg�m 2

b. 67.22 cm

c. 2.4 kg�m 2 /s 2

d. 19.9 m 2

e. 970 000 m/h

f. 139 cm 2

SCIENTIFIC NOTATION

1. a. 1.58 � 10 5 km

b. 9.782 � 10 �6 L

c. 8.371 � 10 8 cm 3

d. 6.5 � 10 9 mm 2

e. 5.93 � 10 �3 g

f. 6.13 � 10 �9 m

g. 1.2552 � 10 7 J

h. 8.004 � 10 �6 g/L

i. 1.0995 � 10 �2 kg

j. 1.05 � 10 9 Hz

2. a. 9.49 � 10 3 kg

b. 7.1 � 10 �2 mg

c. 9.8 � 10 3 m 3

d. 1.56 � 10 �7 m

e. 3.18 � 10 6 J

f. 9.63 � 10 27 molecules

g. 7.47 � 10 6 cm

3. a. 6.53 � 10 2 L/s

b. 1.83 � 10 7 mm 3

c. 2.51 � 10 4 kg/m 2

d. 4.23 � 10 1 km/s

e. 3.22 � 10 6 m 3

f. 8.68 � 10 6 J/s

4. 3.6 � 10 6 J

5. 1.12 � 10 7 Pa

6. 1.5 � 10 6 mm

7. a. 3 � 10 5 km/s

b. 1 � 10 12 m/h

c. 3 � 10 4 cm

8. a. 7.75 � 10 22 molecules

b. 1.59 � 10 27 molecules

c. 6.41 � 10 17 molecules

9. a. 2.2 � 10 �5 mm 2 /transistor

b. 4.3 � 10 9 transistors

10. 5.01 � 10 �8 g/�L

11. 4.79 � 10 7 cesium atoms

12. 1.2 � 10 20 g/m 3 ; 1.2 � 10 14 kg

13. 1.9 � 10 7 pits

14. a. 2.69 � 10 18 molecules of oxygen

b. 2.69 � 10 22 molecules of oxygen

c. 3.72 � 10 �20 mL/molecule

15. a. 7.9 � 10 8 kg/person

b. 7.9 � 10 5 metric ton/person

c. 5.4 � 10 8 kg/person

16. 3.329 � 10 5 Earths

17. a. 4.8 � 10 �1 km 3

b. 4.8 � 10 8 m 3

c. 32 years

18. 1.0 � 10 7 J/day

The Mole and Chemical

Composition

FOUR STEPS FOR SOLVING

QUANTITATIVE PROBLEMS

1. 0.026 mm

2. 3.21 L

3. 0.80 g/cm 3

4. 21.4 g/cm 3

5. 30 boxes

6. a. 1.73 L

0.120 m � 0.120 m � 0.120 m

b. 9.2 g; 5.0 cm 3

Holt Chemistry 323 Answer Key

c. 60.4 kg; 1.88 � 10 4 dm 3

d. 0.94 g/cm 3 ; 5.3 � 10 �4 m 3

e. 2.5 � 10 3 kg; 2.7 � 10 6 cm 3

7. 2.8 g/cm 3

8. a. 0.72 �m

b. 2.5 � 10 3 atoms

9. 1300 L/min

10. 1.3 � 10 6 cal/h

11. 5.44 g/ cm 3

12. 2.24 � 10 4 cm 3

13. 32 000 uses

14. 2500 L

15. 9.5 L/min

MOLE CONCEPT

1. a. 3.7 � 10 �4 mol Pd

b. 150 mol Fe

c. 0.040 mol Ta

d. 5.38 � 10 �5 mol Sb

e. 41.1 mol Ba

f. 3.51 � 10 �8 mol Mo

2. a. 52.10 g Cr

b. 1.5 � 10 4 g or 15 kg Al

c. 8.23 � 10 �7 g Ne

d. 3 � 10 2 g or 0.3 kg Ti

e. 1.1 g Xe

f. 2.28 � 10 5 g or 228 kg Li

3. a. 1.02 � 10 25 atoms Ge

b. 3.700 � 10 23 atoms Cu

c. 1.82 � 10 24 atoms Sn

d. 1.2 � 10 30 atoms C

e. 1.1 � 10 21 atoms Zr

f. 1.943 � 10 14 atoms K

4. a. 10.00 mol Co

b. 0.176 mol W

c. 4.995 � 10 �5 mol Ag

d. 1.6 � 10 �15 mol Pu

e. 7.66 � 10 �7 mol Rn

f. 1 � 10 �11 mol Ce

5. a. 2.5 � 10 19 atoms Au

b. 5.10 � 10 24 atoms Mo

c. 4.96 � 10 20 atoms Am

d. 3.011 � 10 26 atoms Ne

e. 2.03 � 10 18 atoms Bi

f. 9.4 � 10 16 atoms U

6. a. 117 g Rb

b. 223 g Mn

c. 2.11 � 10 5 g Te

d. 2.6 � 10 �3 g Rh

e. 3.31 � 10 �8 g Ra

f. 8.71 � 10 �5 g Hf

7. a. 0.749 mol CH 3COOH

b. 0.0213 mol Pb(NO 3) 2

c. 3 � 10 4 mol Fe 2O 3

d. 2.66 � 10 �4 mol C 2H 5NH 2

e. 1.13 � 10 �5 mol C 17H 35COOH

f. 378 mol (NH 4) 2SO 4

8. a. 764 g SeOBr 2

b. 4.88 � 10 4 g CaCO 3

c. 2.7 g C 20H 28O 2

d. 9.74 � 10 �6 g C 10H 14N 2

e. 529 g Sr(NO 3) 2

f. 1.23 � 10 �3 g UF 6

9. a. 2.57 � 10 24 formula units WO 3

b. 1.81 � 10 21 formula units Sr(NO 3) 2

c. 4.37 � 10 25 molecules C 6H 5CH 3

d. 3.08 � 10 17 molecules C 29H 50O 2

e. 9.0 � 10 26 molecules N 2H 4

f. 5.96 � 10 23 molecules C 6H 5NO 2

10. a. 1.14 � 10 24 formula units FePO 4

b. 6.4 � 10 19 molecules C 5H 5N

c. 6.9 � 10 20 molecules

(CH 3) 2CHCH 2OH

d. 8.7 � 10 17 formula units

Hg(C 2H 3O 2) 2

e. 5.5 � 10 19 formula units Li 2CO 3

11. a. 52.9 g F 2

b. 1.19 � 10 3 g or 1.19 kg BeSO 4

c. 1.388 � 10 5 g or 138.8 kg CHCl 3

d. 9.6 � 10 �12 g Cr(CHO 2) 3

e. 6.6 � 10 �4 g HNO 3

f. 2.38 � 10 4 g or 23.8 kg C 2Cl 2F 4

12. 0.158 mol Au

0.159 mol Pt

0.288 mol Ag

13.0.234 mol C 6H 5OH

14. 3.8 g I 2

15. 1.00 � 10 22 atoms C

16. a. 0.0721 mol CaCl 2

55.49 mol H 2O

b. 0.0721 mol Ca 2�

0.144 mol Cl �

17. a. 1.325 mol C 12H 22O 11

b. 7.762 mol NaCl

18. 0.400 mol ions

19. 4.75 mol atoms

20. a. 249 g H 2O

b. 13.8 mol H 2O

c. 36.1 mL H 2O

d. 36.0 g H 2O

21. The mass of a sugar molecule is much

greater than the mass of a water molecule.

Therefore, the mass of 1 mol of

sugar molecules is much greater than

the mass of 1 mol of water molecules.

22. 1.52 g Al

Holt Chemistry 324 Answer Key

23. 0.14 mol O 2

24. a. 0.500 mol Ag

0.250 mol S

b. 0.157 mol Ag 2S

0.313 mol Ag

0.157 mol S

c. 33.8 g Ag

5.03 g S

PERCENTAGE COMPOSITION

1. a. HNO 3

1.60% H

22.23% N

76.17% O

b. NH 3

82.22% N

17.78% H

c. HgSO 4

67.616% Hg

10.81% S

21.57% O

d. SbF 5

56.173% Sb

43.83% F

2. a. 7.99% Li

92.01% Br

b. 94.33% C

5.67% H

c. 35.00% N

5.05% H

59.96% O

d. 2.15% H

29.80% N

68.06% O

e. 87.059% Ag

12.94% S

f. 32.47% Fe

13.96% C

16.29% N

37.28% S

g. LiC 2H 3O 2

10.52% Li

36.40% C

4.59% H

48.49% O

h. Ni(CHO 2) 2

39.46% Ni

16.15% C

1.36% H

43.03% O

3. a. 46.65% N

b. 23.76% S

c. 89.491% Tl

d. 39.17% O

e. 79.95% Br in CaBr 2

f. 78.767% Sn in SnO 2

4. a. 1.47 g O

b. 26.5 metric tons Al

c. 262 g Ag

d. 0.487 g Au

e. 312 g Se

f. 3.1 � 10 4 g Cl

5. a. 40.55% H 2O

b. 43.86% H 2O

c. 20.70% H 2O

d. 28.90% H 2O

6. a. Ni(C 2H 3O 2) 2�4H 2O

23.58% Ni

b. Na 2CrO 4�4H 2O

22.22% Cr

c. Ce(SO 4) 2�4H 2O

34.65% Ce

7. 43.1 kg Hg

8. malachite: 5.75 � 10 2 kg Cu

chalcopyrite: 3.46 � 10 2 kg Cu

malachite has a greater Cu content

9. a. 25.59% V

b. 39.71% Sn

c. 22.22% Cl

10. 319.6 g anhydrous CuSO 4

11. 1.57 g AgNO 3

12. 54.3 g Ag

8.08 g S

13. 23.1 g MgSO 4�7H 2O

14. 3.27 � 10 2 g S

EMPIRICAL FORMULAS

1. a. BaCl 2

b. BiO 3H 3 or Bi(OH) 3

c. AlN 3O 9 or Al(NO 3) 3

d. ZnC 4H 6O 4 or Zn(CH 3COO) 2

e. NiN 2S 2H 8O 8 or Ni(NH 4) 2SO 4

f. C 2HBr 3O 2 or CBr 3COOH

2. a. CuF 2

b. Ba(CN) 2

c. MnSO 4

3. a. NiI 2

b. MgN 2O 6 or Mg(NO 3) 2

c. MgS 2O 3, magnesium thiosulfate

d. K 2SnO 3, potassium stannate

4. a. As 2S 3

b. Re 2O 7

c. N 2H 4O 3 or NH 4NO 3

d. Fe 2Cr 3O 12 or Fe 2(CrO 4) 3

e. C 5H 9N 3

f. C 6H 5F 2N or C 6H 3F 2NH 2

Holt Chemistry 325 Answer Key

5. a. C 6H 12S 3

b. C 8H 16O 4

c. C 4H 6O 4

d. C 12H 12O 6

6. a. C 4H 4O 4

b. C 4H 8O 2

c. C 9H 12O 3

7. K 2S 2O 5, potassium metabisulfite

8. Pb 3O 4

9. Cr 2S 3O 12 or Cr 2(SO 4) 3, chromium(III)

sulfate

10. C 9H 6O 4

11. C 5H 9N 3, the empirical formula and the

molecular formula are the same

12. The molecular formulas of the

compounds are different multiples of

the same empirical formula. (FYI: The

first could be acetic acid, C 2H 4O 2, and

the second could be glucose, C 6H 12O 6,

or some other simple sugar.)

STOICHIOMETRY

1. 15.0 mol (NH 4) 2SO 4

2. a. 51 g Al

b. 101 g Fe

c. 1.83 mol Fe 2O 3

3. 0.303 g H 2

4. H 2SO 4 � 2KOH 3 K 2SO 4 � 2H 2O; 1.11

g H 2SO 4

5. a. H 3PO 4 � 2NH 3 3 (NH 4) 2HPO 4

b. 0.293 mol (NH 4) 2HPO 4

c. 970 kg NH 3

6. a. 90.0 mol ZnCO 3; 60.0 mol C 6H 8O 7

b. 13.5 kg H 2O; 33.0 kg CO 2

7. a. 60.9 g methyl butanoate

b. 3261 g H 2O

8. a. 0.450 mol N 2

b. 294 g NH 4NO 3

9. Pb(NO 3) 2 � 2KI 3 PbI 2 � 2KNO 3;

0.751 mg KNO 3

10. 3.3 mol PbSO 4

11. 2LiOH � CO 2 3 H 2O � Li 2CO 3; 360 g

H 2O

12. a. 38.1 g H 2O

b. 40.1 g H 3PO 4

c. 0.392 mol H 2O

13. C 2H 5OH � 3O 2 3 2CO 2 � 3H 2O; 81.0

g C 2H 5OH

14. 76.5 g H 2SO 4; 12.5 g O 2

15. 2NaHCO 3 3 Na 2CO 3 � H 2O � CO 2 ;

1.31 g CO 2

16. a. 2N 2H 4 � N 2O 4 3 3N 2 � 4H 2O

b. 1 mol N 2O 4 to 3 mol N 2

c. 30 000 mol N 2

d. 3.52 � 10 5 g H 2O

17. 2HgO(s) 3 2Hg(l) � O 2(g); 1.1954 mol

O 2

18. 2Fe � 3Cl 2 3 2FeCl 3; 30.5 g Fe

19. 9.26 mg CdS

20. a. 1.59 mol CO 2

b. 0.0723 mol C 3H 5(OH) 3

c. 535 g Mn 2O 3

d. 8.33 g C 3H 5(OH) 3; 4.97 g CO 2

21. a. 3.29 � 10 3 kg of HCl

b. 330 g CO 2 (s)

22. a. 6.53 � 10 5 g NH 4ClO 4

b. 160 kg NO(g)

23. a. 1.70 � 10 6 mol H 3PO 4

b. 666 kg of CaSO 4�2H 2O

c. 34 metric tons of H 3PO 4

24. 1670 kg

LIMITING REACTANTS

1. 2ZnS � 3O 2 3 2ZnO � 2SO 2; ZnS is

2. a. Al is limiting

b. 4.25 � 10 �3 mol Al 2O 3

c. O 2 is limiting

3. a. CuS is limiting

b. 15.6 g CuO

4. Fe is limiting; 0.158 mol Cu

5. 54 g Ba(NO 3) 2

6. a. 38 g Br 2

b. 510 g I 2

7. a. Ni is in excess

b. 60.2 g Ni(NO 3) 2

8. CS 2(g) � 3O 2(g) 3 2SO 4(g) � CO 2(g)

0.80 mol O 2 remain

9. a. 0.84 g Hg(NH 2)Cl

b. 0.84 g

10. a. 2Al(s) � 2NaOH(aq) � 2H 2O(l)

3 2NaAlO 2(aq) � 3H 2(g)

b. NaOH is limiting; 0.56 mol H 2

c. Al should be limiting because you

would not want aluminum metal

remaining in the drain.

11. a. 0.0422 mol Cu; 0.169 mol HNO 3

b. Cu is in excess

c. 3.32 g H 2O

12. a. 2.90 mol NO;

4.35 mol H 2O

b. NH 3 is limiting

c. NH 3 is limiting; 1.53 � 10 3 kg NO

Holt Chemistry 326 Answer Key

13. 565 g CH 3CHO;

29 g CH 3CH 2OH remains

14. 630 g HBr

15. 12.7 g SO 2

16. a. 18.4 g Tb

b. 2.4 g TbF 3

PERCENTAGE YIELD

1. a. 64.3% yield

b. 58.0% yield

c. 69.5% yield

d. CH 3CH 2OH is limiting; 79% yield

2. a. 69.5% yield

b. 79.0% yield

c. 48% yield

d. 85% yield

3. a. 59% yield

b. 81.0% yield

c. 2.3 � 10 5 mol P

4. a. 91.8% yield

b. 0.0148 mol W

c. 16.1 g WO 3

5. a. 86.8% yield

b. 92.2% yield

c. 2.97 � 10 4 kg CS 2;

4.39 � 10 4 kg S 2Cl 2

6. a. 81% yield

b. 2.0 � 10 2 g N 2O 5

7. 80.1% yield

8. a. 95% yield

b. 9.10 � 10 2 g Au

c. 9 � 10 4 kg ore

9. a. 87.5% yield

b. 0.25 g CO

10. a. 71% yield

b. 26 metric tons

c. 47.8 g NaCl

d. 500 kg per hour NaOH

11. a. 2Mg � O 2 3 2MgO

b. 87.7% yield

c. 3Mg � N 2 3 Mg 3N 2

d. 56% yield

12. a. 80.% yield

b. 66.2% yield

c. 57.1% yield

13. 2C 3H 6(g) � 2NH 3(g) � 3O 2(g) 3

2C 3H 3N(g) � 6H 2O(g)

91.0% yield

14. a. CO � 2H 2 3 CH 3OH

3.41 � 10 3 kg

b. 91.5% yield

15. 96.9% yield

16. 6CO 2 � 6H 2O 3 C 6H 12O 6 � 6O 2

6.32 � 10 3 g O 2

17. 27.6 kg CaO

Causes of Change

THERMOCHEMISTRY

1. �260.8 kJ/mol

2. �385.9 kJ/mol

3. �390. kJ/mol

4. �492.3 kJ/mol

5. �107.6 kJ/mol

6. �121.8 kJ/mol

7. �384.9 kJ/mol

8. 74.2 kJ/mol

9. �169.0 kJ/mol

10. a. CH 4(g) � 2O 2(g) 3

CO 2(g) � 2H 2O(g)

C 3H 8(g) � 5O 2(g) 3

3CO 2(g) � 4H 2O(g)

b. for methane: �H ��802.2 kJ/mol

for propane: �H ��2043 kJ/mol

c. output methane � �4.998 � 10 4 kJ/kg

output propane � �4.632 � 10 4 kJ/kg

Methane yields more energy per mass.

11. �132.7 kJ/mol

12. �7171.4 kJ/mol

13. �141.1 kJ/mol

14. 20.2 kJ/mol

15. �285 kJ/mol�K

16. a. 786.8 kJ/mol

b. �36 kJ/mol

c. 2154 kJ/mol

d. �496 kJ/mol

e. 1346.4 kJ/mol

Gases

GAS LAWS

1. a. 105 kPa

b. 5.0 mL

c. 42.4 kPa

d. 6.78 � 10 �3 dm 3

e. 1.24 atm

f. 1.5 m 3

2. 8.0 m 3

3. 0.0258 atm

4. 8.01 � 10 �2 dm 3

5. a. 234 K

b. 1.2 dm 3

c. �269.17°C

d. 8.10 � 10 �2 L

e. 487 cm 3

f. 67.9 m 3

Holt Chemistry 327 Answer Key

6. 1.45 cm 3

7. �40.°C

8. a. �208.6°C

b. 5.5 kPa

c. 2.61 atm

d. 297°C

e. 35.6 atm

f. 39 K

9. 0.899 atm

10. 2.23 atm

11. 7.98 K

12. a. 2.02 L

b. 75.8 kPa

c. 110 K

d. 4.69 � 10 3 mm 3

e. �72°C

f. 2.25 atm

13. 379 cm 3

14. 98 kPa

15. 1.00 atm; use Boyle’s law to find the

pressure of each gas in the whole

space; add the partial pressures of

both gases when they occupy the

whole space.

16. 285 mL

17. 89 cm 3 The pressure in the bottle on

top of the mountain is the sum of

P O2 dry at the temperature of the

mountaintop and P H2 O vapor at the

temperature on top of the mountain.

18. 59 cm 3

Solve the equation:

V total/293 K � (V total � 0.20 cm 3 )/294 K

19. 118 kPa

20. 935 mL

21. 26.4 kL

22. 3.76 atm

23. 115 mL

24. 10.9 atm

IDEAL GAS LAW

1. a. 34.2 mol

b. 6.68 � 10 3 kPa

c. �148°C

d. 1.1 � 10 5 L

2. a. 55.9 g/mol

b. 0.111 g

c. 4.46 � 10 �2 L

d. 0.846 atm

e. 391 g/mol

3. 2.71 � 10 3 L

4. a. 48.4 g/mol

b. 9.38 g/L

c. 6°C

d. 2.24 atm or 227 kPa

5. 33.7 atm

6. 3.06 g/L

7. 663 g/mol

8. 204 L

9. 0.0101 mol ethane

10. 5.16 g NO

11. 77.0% yield

12. 10.5 g/mol

13. 171 g/mol

14. 6.55 atm

15. 326 kPa

16. 479 K or 206°C

17. 1210 L at �75°C; 1620 L at �8°C

18. 1.11 � 10 4 kPa

19. 168 mL

20. 3.85 � 10 3 L

21. 4.05 � 10 3 L

22. 29.0 g/mol

STOICHIOMETRY OF GASES

1. a. 19.0 mL N 2

b. 4.26 � 10 4 L NH 3

c. 896 L NH 3

d. 899 L H 2

2. a. 23 L H 2O

b. 1070 L O 2

c. 326 mL CO 2

d. 25.2 L total products

3. 1550 L O 2 at STP

4. 0.894 L SiF 4

5. a. 3.36 L H 2

b. 488 g Fe

c. 112 L H 2

6. 0.013 L H 2 or 13 mL H 2

7. 7.50 L O 2 at STP; 4.14 g diethyl ether

8. a. 3.36 L N 2; 6.72 L CO 2; 5.60 L H 2O;

0.560 L O 2

b. 15.0 L total volume all gases

9. 0.894 g NH 4NO 3

10. 2.19 L PH 3

11. 1.2 � 10 3 kg Al; 7.4 � 10 5 L HCl

12. 7.08 � 10 7 L NH 3

13. 3.77 g BaO 2

14. 5.85 L Cl 2

15. 28.0 kL NH 3; 28.0 kL NO 2 overall reaction

is: 4NH 3 � 7O 2 3 4NO 2 � 6H 2O

16. 18.2 g KClO 3

17. a. 38 000 L NH 3

b. 1.30 � 10 5 g NaHCO 3

c. 12.3 L NH 3

d. 5.60 � 10 3 L

Holt Chemistry 328 Answer Key

18. a. 9.93 L CO 2

b. 1.63 L O 2

c. 26.0 L CO 2

d. 3.29 � 10 3 g H 2O

Solutions

CONCENTRATION OF SOLUTIONS

1. a. 60.0 g KMnO 4; 440.0 g H 2O

b. 220 g BaCl 2

c. 457 g glycerol

d. 0.0642 M K 2Cr 2O 7

e. 1.27 m CaCl 2

f. 0.234 g NaCl

g. 541 g glucose; 2040 g total

2. 10.6 mol H 2SO 4

3. 0.486 m linoleic acid

4. a. 13.0 g Na 2S 2O 3

b. 0.0820 mol

c. 0. 328 M

5. 338 g CoCl 2

6. 0.442 L

7. 203 g urea

8. 18.8 g Ba(NO 3) 2

9. add 3.5 g (NH 4) 2SO 4 to 96.5 g H 2O

10. 54 g CaCl 2

11. 1.25 mol; 1.25 M

12. 93.6 g/mol

13. 49.6 kg water; 0.5 kg NaCl

14. 8.06%

15. 1.4 L ethyl acetate

16. CdCl 2(aq) � Na 2S(aq) 3 CdS(s) � 2

NaCl(aq)

a. 0.196 mol CdCl 2

b. 0.196 mol CdS

c. 28.3 g CdS

17. 34.4 g H 2SO 4

18. 1.54 � 10 5 mol HCl

19. 85.7 mL BaCl 2 solution

20. a. Measure out 9.39 g CuSO 4�5H 2O and

add 90.61 g H 2O to make 100. g of

solution. The 9.39 g of CuSO 4�5H 2O

contributes the 6.00 g of CuSO 4

b. Measure out 200. g CuSO 4�5H 2O,

dissolve in water, then add water to

make 1.00 L. Water of hydration

does not have to be considered here

as long as the molar mass of the

hydrate is used in determining the

mass to weigh out.

c. Measure out 870 g CuSO 4�5H 2O

and add 685 g H 2O. The hydrate

contributes 315 g of H 2O, so only

685 g H 2O must be added.

21. 383 g CaCl 2�6H 2O

22. 0.446 g arginine

23. 3254 g H 2O; the hydrate contributes

987.9 g H 2O.

24. 9.646 g KAl(SO 4) 2�12H 2O; 25.35 g H 2O

DILUTIONS

1. 0.0948 M

2. 0.44 mL

3. a. 3.0 M

b. 0.83 L

c. 1.5 � 10 3 g

4. 6.35 mL

5. 348 mL

6. 0.558 M

7. a. 850 mL; 2.4 mL; 86 mL

b. 1.3 L concentrated HNO 3

c. 1.16 L concentrated HCl

8. 0.48 M

9. 2.72 M

10. Dilute 4.73 mL of the 6.45 M acetic

acid to 25 mL. This means adding

20.27 mL of water.

11. 39.7 g/mol

12. 0.27 M

13. 1.9 � 10 2 g

14. 0.667 L

15. For 1.00 L of 0.495 M urea solution,

take 161 mL of 3.07 M stock solution

of urea and dilute with water (839 mL

of water) to make 1.00 L.

16. a. 17.8 m

b. 1080 g, 6.01 M

17. 47.17 g Na 2CO 3 per 50.00 g sample �

94.3% Na 2CO 3

18. 151 g CuCl 2

19. Add 2.3 volumes of H 2O per volume of

stock solution.

20. a. 14.9 g

b. 7.02 � 10 �2 mol

c. 0.167 M

COLLIGATIVE PROPERTIES

1. �9.88°C ; 102.7°C

2. 103.3°C

3. 201.6°C

4. 50. g ethanol

5. 82 g/mol

6. �18.2°C

7. 15.5 g/mol

8. 66.8 g/mol

9. 183.3°C

Holt Chemistry 329 Answer Key

10. a. �15.0°C

b. 104.1°C

11. 292 g/mol

12. �47.0°C

13. 190 g/mol

14. 107.2°C

15. �27.9°C

16. 204.4 g/mol; C 16H 10

17. 9.9 g CaCl 2; 49 g glucose

18. C 3H 6O 3

19. �29.2°C

20. 2.71 kg; 104.9°C

21. a. 2.2 m

b. 1.7 m

22. 92.0 g H 2O

23. 150 g/mol; C 5H 10O 5

24. 124 g/mol; C 6H 6O 3

Chemical Equilibrium

EQUILIBRIUM

1. 1.98 � 10 �7

2. 2.446 � 10 �12

3. 3.97 � 10 �5

4. a. 8.13 � 10 �4 M

b. 0.0126 M

5. a. The concentrations are equal.

b. K will increase.

6. 0.09198 M

7. a. 0.7304 M

b. 6.479 � 10 �4 M

8. a. [A] � [B] � [C] � 1/2[A] initial

b. [A], [B], and [C] will increase

equally. K remains the same.

9. a. K eq � [HBr] 2 /[H 2][Br 2]

b. 2.11 � 10 �10 M

c. Br 2 and H 2 will still have the same

concentration. HBr will have a

much higher concentration than the

two reactants; at equilibrium, essentially

only HBr will be present.

10. 1.281 � 10 �6

11. 4.61 � 10 �3

12. a. K eq � [HCN]/[HCl]

b. 3.725 � 10 �7 M

13. a. The reaction yields essentially no

products at 25°C; as a result, the

equilibrium constant is very small.

At 110 K, the reaction proceeds to

some extent.

b. 2.51 M

14. 0.0424

15. 0.0390

EQUILIBRIUM OF SALTS, K sp

1. 1.0 � 10 �4

2. 1.51 � 10 �7

3. a. 1.6 � 10 �11

b. 0.49 g

4. 2.000 � 10 �7

5. a. 8.9 � 10 �4 M

b. 0.097 g

6. 0.036 M

7. a. 1.1 � 10 �4 M

b. 4.6 L

8. a. 5.7 � 10 �4 M

b. 2.1 g

9. 0.83 g remains

10. 2.8 � 10 �3 M

11. a. 1.2 � 10 �5

b. Yes

12. a. 1.1 � 10 �8

b. No

13. a. Mg(OH) 2 3 Mg 2� �2OH �

b. 11 L

c. A suspension contains undissolved

Mg(OH) 2 suspended in a saturated

solution of Mg(OH) 2. As hydroxide

ions are depleted by titration, the

dissociation equilibrium continues

to replenish them until all of the

Mg(OH) 2 is used up.

14. a. 0.184 M

b. 46.8 g

15. 5.040 � 10 �3

16. a. 1.0

b. 0.25

c. 0.01

d. 1 � 10 �6

17. 4 � 10 �4 ; 3 � 10 �5 ; 3 � 10 �5 ;

1 � 10 �6 ; 3 � 10 �7

18. 7.1 � 10 �7 M; 1.3 � 10 �3 g

19. a. 0.011 M

b. 0.022 M

c. 12.35

20. 0.063 g

Acids and Bases

1. a. [OH � ] � 0.05 M,

[H 3O � ] � 2 � 10 �13 M

b. [OH � ] � 2.0 � 10 �12 M,

[H 3O � ] � 5.0 � 10 �3 M

c. [OH � ] � 0.013 M,

[H 3O � ] � 7.7 � 10 �13 M

Holt Chemistry 330 Answer Key

d. [OH � ] � 6.67 � 10 �14 M,

[H 3O � ] � 0.150 M

e. [OH � ] � 0.0400 M,

[H 3O � ] � 2.50 � 10 �13 M

f. [OH � ] � 2.56 � 10 �14 M,

[H 3O � ] � 0.390 M

g. 10, 2.3, 12.11, 0.824, 12.602, 0.409

2. [OH � ] � 0.160 M

[H 3O � ] � 6.25 � 10 �14 M

3. 0.08 M

4. 2.903; 4.137

5. 10.0; 10.7

6. a. [H 3O � ] � 0.020 M,

[OH � ] � 5.0 � 10 �13 M

b. 1.7

7. a. 2.804

b. 2.755

8. 2.9 � 10 �11 M

9. 10.54

10. [H 3O � ] � 1 � 10 �4 M

[OH � ] � 1 � 10 �10 M

11. 10.96

12. 0.2 g

13. [H 3O � ] � 1.0 � 10 �3 M

[OH � ] � 1.0 � 10 �4 M

14. a. 0.40%

b. 1.3%

c. 4.0%

15. 1.402

16. a. 4.50

b. 2.40

c. 3.83

d. 5.63

17. [H 3O � ] � 1 � 10 �9 M

[OH � ] � 1 � 10 �5 M

18. 11.70

19. [H 3O � ] � 0.020 M

[HCl] � 0.020 M

20. 3.2 � 10 �4 M

21.pH � 0.000

10. kL

1.0 � 10 2 L

22. 0.052 mol NaOH

2.1 g NaOH

23. 7.1 � 10 –3

24. [H 3O � ] � 7.1 � 10 –4 M

[OH � ] � 1.4 � 10 �11 M

TITRATIONS

1. 0.4563 M KOH

2. 2.262 M CH 3COOH

3. 2.433 M NH 3

4. a. 20.00 mL base

b. 10.00 mL acid

c. 80.00 mL acid

5. 5.066 M HF

6. 0.2216 M oxalic acid

7. 1.022 M H 2SO 4

8. 0.1705 M KOH

9. 0.5748 M citric acid

10. 0.3437 M KOH

11. 43.2 mL NaOH

12. 23.56 mL H 2SO 4

13. 0.06996 mol KOH; 3.926 g KOH;

98.02% KOH

14. 51.7 g Mg(OH) 2

15. 0.514 M NH 3

16. 20.85 mL oxalic acid

17. a. 0.5056 M HCl

b. 0.9118 M RbOH

18. 570 kg Ca(OH) 2

19. 16.9 M HNO 3

20. 33.58 mL

EQUILIBRIUM OF ACIDS AND

BASES, K a AND K b

1. pH � 2.857

K a � 1.92 � 10 �5

2. pH � 1.717

K a � 2.04 � 10 �3

3. a. [H 3O � ] � 3.70 � 10 �11 M

pH � 10.431

K b � 1.82 � 10 �7

b. [B] � 4.66 � 10 �3 M

K b � 1.53 � 10 �4

pH � 10.93

c. [OH � ] � 1.9 � 10 �3 M

[H 3O � ] � 5.3 � 10 �12 M

[B] � 0.0331 M

K b � 1.1 � 10 �4

d. [B] initial � 7.20 � 10 �3 M

K b � 1.35 � 10 �4

pH � 10.96

4. 6.4 � 10 �5

5. K b � 3.1 � 10 �5

[H 2NCH 2CH 2OH] � 6.06 � 10 �3 M

6. 1.3 � 10 �4

7. [OH � ] � 1.58 � 10 �5 M

pH � 9.20

8. K a � 2.2 � 10 �3

[HB] initial � 0.0276 M

9. 4.63 � 10 �3

10. 2.62 � 10 �4

11. [H 3O � ] � 0.0124 M

pH � 1.907

Holt Chemistry 331 Answer Key

12. [H 3O � ] � 0.0888 M

pH � 1.05

13. K a � 3.50 � 10 �4

pH � 2.241

14. a. 3.16 � 10 �6 M

b. 5.500

c. 0.025 M

d. 4.0 � 10 �10

e. HCN is a fairly weak acid.

f. 5.8 � 10 �6 M

15. a. 0.740

b. 0.0325

c. 1.02 M

d. It is moderately weak.

16. 2.80 � 10 �11

17. a. 0.200 M

b. 0.233

18. pH � 11.37

K b � 7.41 � 10 �5

19. 7.36 � 10 �4

dimethylamine is the stronger base

20. a. 8.3 � 10 �7

b. 1.3 � 10 �4

c. 10.11

Oxidation, Reduction, and

Electronegativity

REDOX EQUATIONS

1. 3Mg � N 2 3 Mg 3N 2

2. SO 2 � Br 2 � 2H 2O 3 2HBr � H 2SO 4

3. H 2S � Cl 2 3 S � 2HCl

4. PbO 2 � 4HBr 3 PbBr 2 � Br 2 � 2H 2O

5. S � 6HNO 3 3 6NO 2 � H 2SO 4 � 2H 2O

6. NaIO 3 � N 2H 4 � 2HCl 3

N 2 � NaICl 2 � 3H 2O

7. MnO 2 � H 2O 2 � 2HCl 3

MnCl 2 � O 2 � 2H 2O

8. 3AsH 3 � 4NaClO 3 3

3H 3AsO 4 � 4NaCl

9. K 2Cr 2O 7 � 3H 2C 2O 4 � 8HCl 3

2CrCl 3 � 6CO 2 � 2KCl � 7H 2O

10. 2Hg(NO 3) 2 3 2HgO � 4NO 2 � O 2

11. 4HAuCl 4 � 3N 2H 4 3

4Au � 3N 2 � 16HCl

12. 5Sb 2(SO 4) 3 � 4KMnO 4 � 24H 2O 3

10H 3SbO 4 � 2K 2SO 4 � 4MnSO 4 �

9H 2SO 4

13. 3Mn(NO 3) 2 � 5NaBiO 3 � 9HNO 3 3

5Bi(NO 3) 2 � 3HMnO 4 � 5NaNO 3 �

3H 2O

14. H 3AsO 4 � 4Zn � 8HCl 3

AsH 3 � 4ZnCl 2 � 4H 2O

15. KClO 3 � 6HCl 3 3Cl 2 � 3H 2O � KCl

16. 2KClO 3 � 4HCl 3

Cl 2 � 2ClO 2 � 2H 2O � 2KCl

17. 2MnCl 3 � 2H 2O 3

MnCl 2 � MnO 2 � 4HCl

18. 2NaOH � 6H 2O � 2Al 3

2NaAl(OH) 4 � 3H 2

19. 6Br 2 � 6Ca(OH) 2 3

5CaBr 2 � Ca(BrO 3) 2 � 6H 2O

20. N 2O � 2NaClO � 2NaOH 3

2NaCl � 2NaNO 2 � H 2O

21. 4HBr � MnO 2 3

MnBr 2 � 2H 2O � Br 2

22. Au � 4HCl � HNO 3 3

HAuCl 4 � NO � 2H 2O

ELECTROCHEMISTRY

1. E 0 ��2.77 V; spontaneous

2. E 0 ��1.11 V; spontaneous

3. E 0 ��0.46 V; not spontaneous

4. E 0 ��1.50 V; spontaneous

5. E 0 ��2.46 V; spontaneous

6. E 0 ��1.28 V; spontaneous

7. E 0 ��3.71 V; spontaneous

8. E 0 ��3.41 V; not spontaneous

9. E 0 ��1.32 V; spontaneous

10. E 0 ��3.60 V; not spontaneous

11. Overall reaction:

Cl 2 � Ni 3 Ni 2� � 2Cl �

Cathode reaction: Cl 2 �2e � 3 2Cl �

Anode reaction: Ni 3 Ni 2� � 2e �

Cell voltage: �1.62 V

12. Overall reaction:

3Hg 2� � 2Fe 3 3Hg � 2Fe 3�

Cathode reaction: Hg 2� � 2e � 3 Hg

Anode reaction: Fe 3 Fe 3� � 3e �

Cell voltage: �0.89 V

13. Overall reaction:

3MnO 4 � � Al 3 3MnO 2 4 � � Al 3�

Cathode reaction:

MnO 4 � � e � 3 MnO 2 4 �

Anode reaction: Al 3 Al 3� � 3e �

Cell voltage: �2.22 V

14. Overall reaction:

2MnO 4 � � 6H � � 5H2S 3

2Mn 2� � 8H 2O � 5S

MnO 4 � � 8H � � 5e � 3 Mn 2� � 4H2O

Anode reaction: H 2S 3 S � 2H � � 2e �

Cell voltage: �1.36 V

15. Overall reaction:

Ca 2� � 2Li 3 Ca � 2Li �

Cathode reaction: Ca 2� � 2e � 3 Ca

Holt Chemistry 332 Answer Key

Anode reaction: Li 3 Li � � e �

Cell voltage: �0.17 V

16. Overall reaction:

2MnO 4 � � 16H � � 10Br � 3

2Mn 2� � 8H 2O � 5Br 2

MnO 4 � � 8H � � 5e � 3

Mn 2� � 4H 2O

Anode reaction: 2Br � 3 Br 2 � 2e �

Cell voltage: �0.43 V

17. Overall reaction:

2Fe 3� � Sn 3 2Fe 2� � Sn 2�

Cathode reaction: Fe 3� � e � 3 Fe 2�

Anode reaction: Sn 3 Sn 2� � 2e �

Cell voltage: �0.91 V

18. Overall reaction:

Cr 2O 2 7 � � 14H � � 3Zn 3

2Cr 3� � 7H 2O � 3Zn 2�

Cr 2O 2 7 � � 14H � � 6e � 3

2Cr 3� � 7H 2O

Anode reaction: Zn 3 Zn 2� � 2e �

Cell voltage: �1.99 V

19. Overall reaction:

Ba � Ca 2� 3 Ba 2� � Ca

Anode reaction: Ba 3 Ba 2� � 2e �

Cell voltage: �0.04 V

20. Overall reaction:

Cd � Hg 2 2 � 3 Cd 2� � 2Hg

Cathode reaction: Hg 2 2 � � 2e � 3 2Hg

Anode reaction: Cd 3 Cd 2� � 2e �

Cell voltage: �1.20 V

Holt Chemistry 333 Answer Key

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e. 41 m 3 f. 3.129 g/cm 3 4. a. 90.

23. 0.14 mol O 2 24. a. 0.500 mol A

13. 565 g CH 3CHO; 29 g CH 3CH 2OH

18. a. 9.93 L CO 2 b. 1.63 L O 2 c.

d. [OH � ] � 6.67 � 10 �14

Anode reaction: Li 3 Li � � e

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