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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Holt ChemFile: Problem-Solving Workbook 48 Mole Concept
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 sul-
fur. 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
General Plan for Converting Mass, Amount,
and Numbers of Particles
Use Avogadro's
number for conversion.
Mass of
substance
Amount of
substance
in moles
Number of atoms,
molecules, or formula
units of substance
Convert using
the molar mass of
the substance.
1
2 3
Name Class Date
Problem Solving
Skills Worksheet
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Holt ChemFile: Problem-Solving Workbook 49 Mole Concept
Name Class Date
Problem Solving continued
PROBLEMS INVOLVING ATOMS AND ELEMENTS
Sample Problem 1
A chemist has a jar containing 388.2 g of iron filings. How many moles of
iron does the jar contain?
Solution
ANALYZE
What is given in the problem?
mass of iron in grams
What are you asked to find? amount of iron in moles
PLAN
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.
COMPUTE
EVALUATE
Are the units correct?
Yes; the answer has the correct units of moles of Fe.
1 mol Fe
55.85 g Fe
6.951 mol Fe388.2 g Fe
multiply by the inverse
molar mass of Fe
Mass of Fe in g
1
Amount of Fe in mol
2
1
molar mass Fe
given
1 mol Fe
mol Feg Fe
55.85
g
Fe
Items Data
Mass of iron 388.2 g
Molar mass of iron* 55.85 g/mol
Amount of iron ? mol
* determined from the periodic table
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Holt ChemFile: Problem-Solving Workbook 50 Mole Concept
Name Class Date
Problem Solving continued
Is the number of significant figures correct?
Yes; the number of significant figures is correct because there are four signifi-
cant figures in the given value of 388.2 g Fe.
Is the answer reasonable?
Yes; 388.2 g Fe is about seven times the molar mass. Therefore, the sample
contains about 7 mol.
Practice
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
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Holt ChemFile: Problem-Solving Workbook 51 Mole Concept
Name Class Date
Problem Solving continued
Sample Problem 2
A student needs 0.366 mol of zinc for a reaction. What mass of zinc in
grams should the student obtain?
Solution
ANALYZE
What is given in the problem?
amount of zinc needed in moles
What are you asked to find? mass of zinc in grams
PLAN
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.
COMPUTE
EVALUATE
Are the units correct?
Yes; the answer has the correct units of grams of Zn.
Is the number of significant figures correct?
Yes; the number of significant figures is correct because there are three signifi-
cant figures in the given value of 0.366 mol Zn.
Is the answer reasonable?
Yes; 0.366 mol is about 1/3 mol. 23.9 g is about 1/3 the molar mass of Zn.
65.39 g Zn
1 mol Zn
23.9 g Zn0.366 mol Zn
multiply by the
molar mass of Zn
Amount of Zn in mol
2
Mass of Zn in mol
1
molar mass Zn
given
65.39 g Zn
g Znmol Zn
1 mol Zn
Items Data
Amount of zinc 0.366 mol
Molar mass of zinc 65.39 g/mol
Mass of zinc ? g
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Holt ChemFile: Problem-Solving Workbook 52 Mole Concept
Name Class Date
Problem Solving continued
Practice
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
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Holt ChemFile: Problem-Solving Workbook 53 Mole Concept
Name Class Date
Problem Solving continued
Sample Problem 3
How many moles of lithium are there in 1.204 10
24
lithium atoms?
Solution
ANALYZE
What is given in the problem?
number of lithium atoms
What are you asked to find? amount of lithium in moles
PLAN
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.
COMPUTE
EVALUATE
Are the units correct?
Yes; the answer has the correct units of moles of Li.
Is the number of significant figures correct?
Yes; four significant figures is correct.
Is the answer reasonable?
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.
1 mol Li
6.022 10
23
atoms Li
1.999 mol Li1.204 10
24
atoms Li
multiply by the inverse of
Avogadro's number
Number of Li atoms
3
Amount of Li in mol
2
1
Avogadro's number
given
1 mol Li
mol Liatoms Li
6.022 10
23
atoms Li
Items Data
Number of lithium atoms 1.204 10
24
atoms
Avogadro’s number—the 6.022 10
23
atoms/mol
number of atoms per mole
Amount of lithium ? mol
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Holt ChemFile: Problem-Solving Workbook 54 Mole Concept
Name Class Date
Problem Solving continued
Practice
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
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Holt ChemFile: Problem-Solving Workbook 55 Mole Concept
Name Class Date
Problem Solving continued
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.
Practice
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
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Holt ChemFile: Problem-Solving Workbook 56 Mole Concept
Name Class Date
Problem Solving continued
Sample Problem 4
How many boron atoms are there in 2.00 g of boron?
Solution
ANALYZE
What is given in the problem?
mass of boron in grams
What are you asked to find? number of boron atoms
PLAN
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.
COMPUTE
EVALUATE
Are the units correct?
Yes; the answer has the correct units of atoms of boron.
Is the number of significant figures correct?
Yes; the mass of boron was given to three significant figures.
1 mol B
10.81 g B
 1.11 10
23
atoms B2.00 g B
6.022 10
23
atoms B
1 mol B
multiply by
Avogadro's
number
multiply by the
inverse of the
molar mass of
boron
Mass of B in g
1
Number of B atoms
3
Avogadro's number
given
6.022 10
23
atoms B
atoms Bg B 
1 mol B
1
molar mass B
1 mol B
10.81 g B
Amount of B in mol
2
Items Data
Mass of boron 2.00 g
Molar mass of boron 10.81 g/mol
Avogadro’s number—the number 6.022 10
23
atoms/mol
of boron atoms per mole of boron
Number of boron atoms ? atoms
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Holt ChemFile: Problem-Solving Workbook 57 Mole Concept
Name Class Date
Problem Solving continued
Is the answer reasonable?
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.
Practice
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
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Holt ChemFile: Problem-Solving Workbook 58 Mole Concept
Name Class Date
Problem Solving continued
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.
Practice
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
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Holt ChemFile: Problem-Solving Workbook 59 Mole Concept
Name Class Date
Problem Solving continued
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 sin-
gle 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 meth-
ods can be used with molecular compounds and elements, such as CO
2
, H
2
O,
H
2
SO
4
, and O
2
.
For example, 1 mol of water contains 6.022 10
23
H
2
O 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
?
Solution
ANALYZE
What is given in the problem?
mass of carbon dioxide
What are you asked to find? amount of carbon dioxide
PLAN
What step is needed to convert from grams of CO
2
to moles of CO
2
?
The molar mass of CO
2
can be used to convert mass of CO
2
to moles of CO
2
.
COMPUTE
1
molar mass CO
2
given
1 mol CO
2
mol CO
2
g CO
2
44.01 g CO
2
multiply by the inverse of the
molar mass of CO
2
Mass of CO
2
in g
1
Amount of CO
2
in mol
2
Items Data
Mass of CO
2
66.0 g
Molar mass of CO
2
44.0 g/mol
Amount of CO
2
? mol
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Holt ChemFile: Problem-Solving Workbook 60 Mole Concept
Name Class Date
Problem Solving continued
EVALUATE
Are the units correct?
Yes; the answer has the correct units of moles CO
2
.
Is the number of significant figures correct?
Yes; the number of significant figures is correct because the mass of CO
2
was
given to three significant figures.
Is the answer reasonable?
Yes; 66 g is about 3/2 the value of the molar mass of CO
2
. It is reasonable that
the sample contains 3/2 (1.5) mol.
Practice
1. Calculate the number of moles in each of the following masses:
a. 3.00 g of boron tribromide, BBr
3
ans: 0.0120 mol BBr
3
b. 0.472 g of sodium fluoride, NaF ans: 0.0112 mol NaF
c. 7.50 10
2
g of methanol, CH
3
OH ans: 23.4 mol CH
3
OH
d. 50.0 g of calcium chlorate, Ca(ClO
3
)
2
ans: 0.242 mol Ca(ClO
3
)
2
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Holt ChemFile: Problem-Solving Workbook 61 Mole Concept
Name Class Date
Problem Solving continued
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 CO
2
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 com-
pound ammonia, NH
3
. You can follow the same plan as you did in Sample
Problem 2, but this time use the molar mass of ammonia.
Practice
1. Determine the mass of each of the following amounts:
a. 1.366 mol of NH
3
ans: 23.28 g NH
3
b. 0.120 mol of glucose, C
6
H
12
O
6
ans: 21.6 g C
6
H
12
O
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
3
H
8
ans: 0.2 g C
3
H
8
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Holt ChemFile: Problem-Solving Workbook 62 Mole Concept
Name Class Date
Problem Solving continued
Sample Problem 6
Determine the number of molecules in 0.0500 mol of hexane, C
6
H
14
.
Solution
ANALYZE
What is given in the problem?
amount of hexane in moles
What are you asked to find? number of molecules of hexane
PLAN
What step is needed to convert from moles of C
6
H
14
to number of molecules of
C
6
H
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.
COMPUTE
EVALUATE
Are the units correct?
Yes; the answer has the correct units of molecules of C
6
H
14
.
Is the number of significant figures correct?
Yes; three significant figures is correct.
Is the answer reasonable?
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
.
6.022 10
23
molecules C
6
H
14
1 mol C
6
H
14
3.01 10
22
molecules C
6
H
14
0.0500 mol C
6
H
14
multiply by
Avogadro's number
Amount of C
6
H
14
in mol
2
Number of C
6
H
14
molecules
3
Avogadro's number
given
1 mol C
6
H
14
molecules C
6
H
14
mol C
6
H
14
6.022 10
23
molecules C
6
H
14
Items Data
Amount of hexane 0.0500 mol
Avogadro’s number—the number 6.022 10
23
molecules/mol
of molecules per mole of hexane
Molecules of hexane ? molecules
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Holt ChemFile: Problem-Solving Workbook 63 Mole Concept
Name Class Date
Problem Solving continued
Practice
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
6
H
8
O
6
ans: 2.1 10
19
mole-
cules C
6
H
8
O
6
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Holt ChemFile: Problem-Solving Workbook 64 Mole Concept
Name Class Date
Problem Solving continued
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.
Practice
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
2
CO
3
ans: 1.5 10
22
formula units
Na
2
CO
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
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Holt ChemFile: Problem-Solving Workbook 65 Mole Concept
Name Class Date
Problem Solving continued
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 cop-
per(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.
Practice
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
2
S ans: 1.94 10
8
mol H
2
S
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
2
O
2
ans: 1.27 10
4
mol
H
2
O
2
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Holt ChemFile: Problem-Solving Workbook 66 Mole Concept
Name Class Date
Problem Solving continued
Sample Problem 7
What is the mass of a sample consisting of 1.00 10
22
formula units of
MgSO
4
?
Solution
ANALYZE
What is given in the problem?
number of magnesium sulfate formula units
What are you asked to find? mass of magnesium sulfate in grams
PLAN
What steps are needed to convert from formula units of MgSO
4
to grams of
MgSO
4
?
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
.
multiply by
the molar
mass of
MgSO
4
multiply by the
inverse of
Avogadro's
number
Number of MgSO
4
formula units
3
Mass of MgSO
4
in g
1
molar mass MgSO
4
given
120.37 g MgSO
4
6.022 10
23
formula units MgSO
4
g MgSO
4
formula units MgSO
4
1 mol MgSO
4
1
Avogadro's number
1 mol MgSO
4
Amount of MgSO
4
in mol
2
Items Data
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/mol
formula units of magnesium sulfate per mole
Molar mass of magnesium sulfate 120.37 g/mol
Mass of magnesium sulfate ? g
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Holt ChemFile: Problem-Solving Workbook 67 Mole Concept
Name Class Date
Problem Solving continued
COMPUTE
EVALUATE
Are the units correct?
Yes; the answer has the correct units of grams of MgSO
4
.
Is the number of significant figures correct?
Yes; the number of significant figures is correct because data were given to
three significant figures.
Is the answer reasonable?
Yes; 2 g of MgSO
4
is about 1/60 of the molar mass of MgSO
4
. Therefore, 2.00 g
MgSO
4
will contain about 1/60 of an Avogadro’s number of formula units.
Practice
1. Calculate the mass of each of the following quantities:
a. 2.41 10
24
molecules of hydrogen, H
2
ans: 8.08 g H
2
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
2
C
2
O
4
ans: 26.7 g Na
2
C
2
O
4
1 mol MgSO
4
6.022 10
23
formula units MgSO
4
1.00 10
22
formula units MgSO
4
2.00 g MgSO
4
120.37 g MgSO
4
1 mol MgSO
4
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Holt ChemFile: Problem-Solving Workbook 68 Mole Concept
Name Class Date
Problem Solving continued
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 con-
vert mass of an ionic or molecular compound to numbers of molecules or for-
mula units.
Practice
1. Calculate the number of molecules or formula units in each of the following
masses:
a. 22.9 g of sodium sulfide, Na
2
S ans: 1.77 10
23
formula units Na
2
S
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
2
CHCN ans: 3.0 10
21
molecules CH
2
CHCN
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Holt ChemFile: Problem-Solving Workbook 69 Mole Concept
Name Class Date
Problem Solving continued
Additional Problems
1. Calculate the number of moles in each of the following masses:
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
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Holt ChemFile: Problem-Solving Workbook 70 Mole Concept
Name Class Date
Problem Solving continued
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
3
COOH
b. 7.04 g of lead(II) nitrate, Pb(NO
3
)
2
c. 5000 kg of iron(III) oxide, Fe
2
O
3
d. 12.0 mg of ethylamine, C
2
H
5
NH
2
e. 0.003 22 g of stearic acid, C
17
H
35
COOH
f. 50.0 kg of ammonium sulfate, (NH
4
)
2
SO
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
20
H
28
O
2
d. 6.00 10
8
mol of nicotine, C
10
H
14
N
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, WO
3
b. 0.003 00 mol of strontium nitrate, Sr(NO
3
)
2
c. 72.5 mol of toluene, C
6
H
5
CH
3
d. 5.11 10
7
mol of -tocopherol (vitamin E), C
29
H
50
O
2
e. 1500 mol of hydrazine, N
2
H
4
f. 0.989 mol of nitrobenzene C
6
H
5
NO
2
10. Calculate the number of molecules or formula units in each of the following
masses:
a. 285 g of iron(III) phosphate, FePO
4
b. 0.0084 g of C
5
H
5
N
c. 85 mg of 2-methyl-1-propanol, (CH
3
)
2
CHCH
2
OH
d. 4.6 10
4
g of mercury(II) acetate, Hg(C
2
H
3
O
2
)
2
e. 0.0067 g of lithium carbonate, Li
2
CO
3
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Holt ChemFile: Problem-Solving Workbook 71 Mole Concept
Name Class Date
Problem Solving continued
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
2
Cl
2
F
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, C
6
H
5
OH, for an experiment. How many
moles of phenol is this?
14. A student needs 0.015 mol of iodine crystals, I
2
, 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, CaCl
2
, is dissolved in 1.000 kg of water.
a. How many moles of CaCl
2
are in solution? How many moles of water are
present?
b. Assume that the ionic compound, CaCl
2
, 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), C
12
H
22
O
11
b. 1.000 pound of table salt, NaCl
18. When the ionic compound NH
4
Cl dissolves in water, it breaks into one ammo-
nium ion, NH
4
, and one chloride ion, Cl
. If you dissolved 10.7 g of NH
4
Cl 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?
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Holt ChemFile: Problem-Solving Workbook 72 Mole Concept
Name Class Date
Problem Solving continued
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, O
2
, 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 Ag
2
S 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 Ag
2
S are there in 38.8 g of Ag
2
S? How many moles of
silver and sulfur would be produced from this amount of Ag
2
S?
c. Calculate the masses of silver and sulfur produced in (b).
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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
3
COOH
b. 0.0213 mol Pb(NO
3
)
2
c. 3 10
4
mol Fe
2
O
3
d. 2.66 10
4
mol C
2
H
5
NH
2
e. 1.13 10
5
mol C
17
H
35
COOH
f. 378 mol (NH
4
)
2
SO
4
8. a. 764 g SeOBr
2
b. 4.88 10
4
g CaCO
3
c. 2.7 g C
20
H
28
O
2
d. 9.74 10
6
g C
10
H
14
N
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
6
H
5
CH
3
d. 3.08 10
17
molecules C
29
H
50
O
2
e. 9.0 10
26
molecules N
2
H
4
f. 5.96 10
23
molecules C
6
H
5
NO
2
10. a. 1.14 10
24
formula units FePO
4
b. 6.4 10
19
molecules C
5
H
5
N
c. 6.9 10
20
molecules
(CH
3
)
2
CHCH
2
OH
d. 8.7 10
17
formula units
Hg(C
2
H
3
O
2
)
2
e. 5.5 10
19
formula units Li
2
CO
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
2
Cl
2
F
4
12. 0.158 mol Au
0.159 mol Pt
0.288 mol Ag
13.0.234 mol C
6
H
5
OH
14. 3.8 g I
2
15. 1.00 10
22
atoms C
16. a. 0.0721 mol CaCl
2
55.49 mol H
2
O
b. 0.0721 mol Ca
2
0.144 mol Cl
17. a. 1.325 mol C
12
H
22
O
11
b. 7.762 mol NaCl
18. 0.400 mol ions
19. 4.75 mol atoms
20. a. 249 g H
2
O
b. 13.8 mol H
2
O
c. 36.1 mL H
2
O
d. 36.0 g H
2
O
21. The mass of a sugar molecule is much
greater than the mass of a water mole-
cule. 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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Holt Chemistry 324 Answer Key
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23. 0.14 mol O
2
24. a. 0.500 mol Ag
0.250 mol S
b. 0.157 mol Ag
2
S
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
2
H
3
O
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
2
O
b. 43.86% H
2
O
c. 20.70% H
2
O
d. 28.90% H
2
O
6. a. Ni(C
2
H
3
O
2
)
2
4H
2
O
23.58% Ni
b. Na
2
CrO
4
4H
2
O
22.22% Cr
c. Ce(SO
4
)
2
4H
2
O
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
2
O
14. 3.27 10
2
g S
EMPIRICAL FORMULAS
1. a. BaCl
2
b. BiO
3
H
3
or Bi(OH)
3
c. AlN
3
O
9
or Al(NO
3
)
3
d. ZnC
4
H
6
O
4
or Zn(CH
3
COO)
2
e. NiN
2
S
2
H
8
O
8
or Ni(NH
4
)
2
SO
4
f. C
2
HBr
3
O
2
or CBr
3
COOH
2. a. CuF
2
b. Ba(CN)
2
c. MnSO
4
3. a. NiI
2
b. MgN
2
O
6
or Mg(NO
3
)
2
c. MgS
2
O
3
, magnesium thiosulfate
d. K
2
SnO
3
, potassium stannate
4. a. As
2
S
3
b. Re
2
O
7
c. N
2
H
4
O
3
or NH
4
NO
3
d. Fe
2
Cr
3
O
12
or Fe
2
(CrO
4
)
3
e. C
5
H
9
N
3
f. C
6
H
5
F
2
N or C
6
H
3
F
2
NH
2
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Holt Chemistry 325 Answer Key
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