Chemical Calculations: Formula Masses,
Moles, and Chemical Equations
Chapter 6
Problem-Set Solutions
6.1
A formula mass is calculated by multiplying the atomic mass of each element by the number
of atoms of that element in the chemical formula, and then summing all of the atomic masses
of all the elements in the chemical formula.
a.
b.
c.
d.
[12(12.01) + 22(1.01) + 11(16.00)] amu = 342.34 amu
[7(12.01) + 16(1.01)] amu = 100.23 amu
[7(12.01) + 5(1.01) + 14.01 + 3(16.00) + 32.07] amu = 183.20 amu
[2(14.01) + 8(1.01) + 32.07 + 4(16.00)] amu = 132.17 amu
6.2
a.
b.
c.
d.
[20(12.01) + 30(1.01) + 16.00] amu = 286.50 amu
[14(12.01) + 9(1.01) + 35.45] amu = 212.68 amu
[8(12.01) + 10(1.01) + 4(14.01) + 2(16.00)] amu = 194.22 amu
[40.08 + 2(14.01) + 6(16.00)] amu = 164.10 amu
6.3
The chemist’s counting unit is the mole. A mole is 6.02 x 1023 objects.
a.
b.
c.
d.
1.00 mole of apples = 6.02 x 1023 apples
1.00 mole of elephants = 6.02 x 1023 elephants
1.00 mole of Zn atoms = 6.02 x 1023 Zn atoms
1.00 mole of CO2 molecules = 6.02 x 1023 CO2 molecules
6.4
a. 6.02 x 1023 oranges
c. 6.02 x 1023 Cu atoms
b. 6.02 x 1023 camels
d. 6.02 x 1023 CO molecules
6.5
Use a conversion factor derived from the definition of a mole. The equality is:
1 mole atoms = 6.02 x 1023 atoms.
6.02 x 10 23 atoms Fe 
23
a. 1.50 moles Fe x 
 = 9.03 x 10 atoms Fe
1
mole
Fe


6.02 x 10 23 atoms Ni
23
b. 1.50 moles Ni x 
 = 9.03 x 10 atoms Ni
1
mole
Ni


6.02 x 10 23 atoms C
23
c. 1.50 moles C x 
 = 9.03 x 10 atoms C
1 mole C


6.02 x 10 23 atoms Ne
23
d. 1.50 moles Ne x 
 = 9.03 x 10 atoms Ne
1 mole Ne


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353
354
6.6
Problem-Set Solutions Chapter 6
 6.02 x 1023 atoms Au 
23
a. 1.20 moles Au x 
 = 7.22 x 10 atoms Au
1 mole Au


 6.02 x 1023 atoms Ag 
23
b. 1.20 moles Ag x 
 = 7.22 x 10 atoms Ag
1 mole Ag


 6.02 x 1023 atoms Be 
23
c. 1.20 moles Be x 
 = 7.22 x 10 atoms Be
1 mole Be


 6.02 x 1023 atoms Si 
23
d. 1.20 moles Si x 
 = 7.22 x 10 atoms Si
1 mole Si


6.7
a.
b.
c.
d.
6.8
a. 0.300 mole
6.9
Molar mass is the mass in grams of a substance that is numerically equal to the substance’s
formula mass. Calculate the formula mass by adding together the atomic masses of the
elements in the compound.
a.
b.
c.
d.
0.200 mole Al atoms contains more moles (so more atoms) than 0.100 mole C atoms.
Avogadro’s number (1.00 mole) of C atoms has more atoms than 0.750 mole Al atoms.
1.50 moles Al atoms contain more atoms than 6.02 x 1023 atoms (1.00 mole) C atoms.
6.50 x 1023 C atoms contains more atoms than Avogadro’s number (6.02 x 1023)
of Al atoms.
b. 6.18 x 1023 atoms
c. 1.20 moles
d. 2.00 moles
[12.01 + 16.00] g = 28.01 g
[12.01 + 2(16.00)] g = 44.01 g
[22.99 + 35.45] g = 58.44 g
[12(12.01) + 22(1.01) + 11(16.00)] g = 342.34 g
6.10
a. [2(1.01) + 16.00] g = 18.0 g
c. [22.99 + 12.01 + 14.01] g = 49.0 g
b. [2(1.01) + 2(16.00)] g = 34.00 g
d. [39.10 + 12.01 + 14.01] g = 65.1 g
6.11
To solve these problems use a conversion factor relating formula mass of the substance to
moles of the substance. The equality will be:
Formula mass (g) substance = 1 mole substance
196.97 g Au 
a. 0.034 mole Au x 
 = 6.7 g Au
 1 mole Au 
107.87 g Ag 
b. 0.034 mole Ag x 
 = 3.7 g Ag
 1 mole Ag 
16.00 g O 
c. 3.00 moles O x 
 = 48.0 g O
1 mole O 
32.00 g O 2 
d. 3.00 moles O 2 x 
 = 96.0 g O2
1 mole O 2 
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Problem-Set Solutions Chapter 6
355
 63.55 g Cu 
 = 54 g Cu
6.12 a. 0.85 mole Cu x 
 1 mole Cu 
 58.69 g Ni 
1
b. 0.85 mole Ni x 
 = 5.0 x 10 g Ni
 1 mole Ni 
 14.01 g N 
c. 2.50 moles N x 
 = 35.0 g N
 1 mole N 
 28.02 g N 2 
d. 2.50 moles N 2 x 
 = 70.1 g N 2
 1 mole N 2 
6.13
Convert the given mass (5.00 g) to moles using the formula mass to form a conversion factor
relating 1 mole to its formula mass. (For example, 1 mole CO = 28.01 g CO)
1 mole CO 
a. 5.00 g CO x 
 = 0.179 mole CO
28.01 g CO
1 mole CO 2 
b. 5.00 g CO2 x 
 = 0.114 mole CO 2
44.01 g CO 2 
 1 mole B 4 H10 
c. 5.00 g B4 H10 x 
 = 0.0937 mole B 4 H10
53.34 g B 4 H10 
1 mole U 
d. 5.00 g U x 
 = 0.0210 mole U
238 g U 
6.14
 1 mole N2O 
a. 7.00 g N2O x 
 = 0.159 mole N2O
 44.02 g N2O 
 1 mole NO2 
b. 7.00 g NO2 x 
 = 0.152 mole NO2
 46.01 g NO2 
 1 mole P4 O10 
c. 7.00 g P4 O10 x 
 = 0.0247 mole P4 O10
 283.88 g P4 O10 
 1 mole V 
d. 7.00 g V x 
 = 0.137 mole V
 50.94 g V 
Copyright © Houghton Mifflin Company. All rights reserved.
356
6.15
Problem-Set Solutions Chapter 6
a. One mole H2SO4 contains 2 moles H atoms, 1 mole S atoms, and 4 moles O atoms. The
conversion factors derived from this statement are:
2 moles H
1 mole H 2 SO 4
,
1 mole H 2 SO 4
4 moles O
1 mole H 2 SO 4
2 moles H
,
1 mole S
,
1 mole H 2 SO 4
1 mole H 2 SO 4
1 mole S
,
1 mole H 2 SO 4
,
4 moles O
b. One mole POCl3 contains 1 mole P atoms, 1 mole O atoms, and 3 moles Cl atoms. The
factors derived from this statement are:
1 mole P
1 mole POCl 3
,
1 mole POCl 3
3 moles Cl
1 mole POCl 3
6.16 a.
1 mole P
,
,
1 mole O
1 mole POCl 3
,
1 mole POCl 3
1 mole O
,
1 mole POCl 3
3 moles Cl
1 mole HNO 3
1 mole HNO 3
1 mole H
1 mole N
,
,
,
,
1 mole HNO 3
1 mole H
1 mole HNO 3
1 mole N
3 moles O 1 mole HNO 3
,
1 mole HNO 3
3 moles O
b.
2 moles C
1 mole C2 H 4 Br2
4 moles H
1 mole C2 H 4 Br2
,
,
,
,
1 mole C2 H 4 Br2
2 moles C
1 mole C2 H 4 Br2
4 moles H
2 moles Br
1 mole C2 H 4 Br2
,
1 mole C2 H 4 Br2
2 moles Br
6.17
a. Use conversion factors relating moles S atoms and moles O atoms to moles SO2 molecules.
The equalities are: 1.00 mole S = 1 mole SO2; 2 moles O = 1 mole SO2
 1 mole S 
 = 2.00 moles S
 1 mole SO 2 
 2 moles O 
2.00 mole SO 2 x 
 = 4.00 moles O
 1 mole SO 2 
2.00 mole SO 2 x 
b. Use conversion factors relating moles S atoms and moles O atoms to moles SO3
molecules. The equalities are: 1.00 mole S = 1 mole SO3; 3 moles O = 1 mole SO3
 1 mole S 
 = 2.00 moles S
 1 mole SO 3 
 3 moles O 
2.00 mole SO 3 x 
 = 6.00 moles O
 1 mole SO 3 
2.00 mole SO 3 x 
Copyright © Houghton Mifflin Company. All rights reserved.
Problem-Set Solutions Chapter 6
c. Use conversion factors relating moles N atoms and moles H atoms to moles NH3
molecules. The equalities are: 1 mole N = 1 mole NH3; 3 moles H = 1 mole NH3
 1 mole N 
 = 3.00 moles N
 1 mole NH 3 
 3 moles H 
3.00 mole NH 3 x 
 = 9.00 moles H
 1 mole NH 3 
3.00 mole NH 3 x 
d. Use conversion factors relating moles N atoms and moles H atoms to moles N2H4
molecules. The equalities are: 2 moles N = 1 mole N2H4; 4 moles H = 1 mole N2H4
 2 mole N 
 = 6.00 moles N
 1 mole N 2 H 4 
 4 moles H 
3.00 mole N 2 H 4 x 
 = 12.0 moles H
 1 mole N 2 H 4 
3.00 mole N 2 H 4 x 
 1 mole N 
6.18 a. 4.00 moles NO2 x 
 = 4.00 moles N
 1 mole NO2 
 2 moles O 
4.00 moles NO2 x 
 = 8.00 moles O
 1 mole NO2 
 2 moles N 
b. 4.00 moles N 2 O x 
 = 8.00 moles N
 1 mole N 2 O 
 1 mole O 
4.00 moles N 2 O x 
 = 4.00 moles O
 1 mole N 2 O 
 2 moles H 
c. 7.00 moles H 2 O x 
 = 14.0 moles H
 1 mole H 2 O 
 1 mole O 
7.00 moles H 2 O x 
 = 7.00 moles O
 1 mole H 2 O 
 2 moles H 
d. 7.00 moles H 2 O2 x 
 = 14.0 moles H
 1 mole H 2 O2 
 2 moles O 
7.00 moles H 2 O2 x 
 = 14.0 moles O
 1 mole H 2 O2 
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357
358
6.19
Problem-Set Solutions Chapter 6
Find the total number of atoms in one molecule. Use this equality to form a conversion
factor.
a. There are 4 moles atoms (1 mole S atoms + 3 moles O atoms) in 1.00 mole SO3 molecules.
 4 moles atoms 
 = 16.0 moles atoms
 1.00 mole SO 3 
4.00 moles SO 3 x 
b. There are 7 moles atoms (2 moles H atoms + 1 mole S atoms + 4 moles O atoms) in
1.00 mole H2SO4 molecules.
 7 moles atoms 
 = 14.0 moles atoms
 1.00 mole H 2 SO 4 
2.00 moles H 2 SO 4 x 
c. There are 45 moles atoms (12 moles C atoms + 22 moles H atoms + 11 moles O atoms) in
1.00 mole C12H22O11 molecules.
 45 moles atoms

 = 45.0 moles atoms
 1.00 mole C12 H 22 O11 
1.00 mole C12 H 22 O11 x 
d. There are 5 moles atoms (1 mole Mg atoms + 2 moles O atoms + 2 moles H atoms) in 1.00
mole Mg(OH)2 molecules.


 = 15.0 moles atoms
 1.00 mole Mg OH 2 
3.00 mole Mg OH 2 x 
5 moles atoms
 6 moles atoms 
6.20 a. 3.00 moles N2O4 x 
 = 18.0 moles atoms
 1.00 mole N2O4 
 5 moles atoms 
b. 4.00 moles HNO3 x 
 = 20.0 moles atoms
 1.00 mole HNO3 
 9 moles atoms 
c. 0.500 moles C2 H6O x 
 = 4.50 moles atoms
 1.00 mole C2 H6O 
 11 moles atoms 
d. 5.00 moles (NH 4 )2S x 
 = 55.0 moles atoms
 1.00 mole (NH 4 )2S 
Copyright © Houghton Mifflin Company. All rights reserved.
Problem-Set Solutions Chapter 6
6.21
Convert the given mass of the element to moles using the atomic mass to form a conversion
factor relating 1 mole to its atomic mass. Use a second conversion factor based on the
definition of Avogadro’s number: 6.02 x 1023 atoms = 1 mole atoms
1 mole B  6.02 x 10 23 atoms B
23
a. 10.0 g B x 
 = 5.57 x 10 atoms B
 x 
10.81
g
B
1
mole
B

 

 1 mole Ca   6.02 x 10 23 atoms Ca 
23
 = 4.81 x 10 atoms Ca
 x
1 mole Ca
 40.08 g Ca  

b. 32.0 g Ca x 
1 mole Ne  6.02 x 10 23 atoms Ne
22
c. 2.0 g Ne x 
 = 6.0 x 10 atoms Ne
 x 
20.18
g
Ne
1
mole
Ne

 

1 mole N  6.02 x 10 23 atoms N 
23
d. 7.0 g N x 
 = 3.0 x 10 atoms N
 x 
1 mole N
14.01 g N  

6.22
359
 1 mole S   6.02 x 1023 atoms S 
23
a. 10.0 g S x 
 = 1.88 x 10 atoms S
 x
1 mole S
 32.07 g S  

 1 mole K   6.02 x 1023 atoms K 
23
b. 39.1 g K x 
 = 6.02 x 10 atoms K
 x
1 mole K
 39.10 g K  

 1 mole U   6.02 x 1023 atoms U 
21
c. 3.2 g U x 
 = 8.1 x 10 atoms U
 x
1 mole U
 238 g U  

 1 mole Be   6.02 x 1023 atoms Be 
23
d. 7.0 g Be x 
 = 4.7 x 10 atoms Be
 x
9.01
g
Be
1
mole
Be

 

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360
6.23
Problem-Set Solutions Chapter 6
In these problems, first convert atoms to moles using the definition of Avogadro’s number
(6.02 x 1023 atoms = 1 mole atoms). Then multiply by a second conversion factor changing
moles to grams of atoms (1 mole atoms = element’s atomic mass in grams).

 63.55 g Cu 
1 mole Cu
a. 6.02 x 10 23 atoms Cu x 
 x 
 = 63.6 g Cu
6.02 x 10 23 atoms Cu  1 mole Cu 

 63.55 g Cu 
1 mole Cu
b. 3.01 x 10 23 atoms Cu x 
 x 
 = 31.8 g Cu
6.02 x 10 23 atoms Cu 1 mole Cu 

 63.55 g Cu 
1 mole Cu
-20
c. 557 atoms Cu x 
 x 1 mole Cu  = 5.88 x 10 g Cu
23
6.02 x 10 atoms Cu  


 63.55 g Cu 
1 mole Cu
-22
d. 1 a tom Cu x 
 x 
 = 1.06 x 10 g Cu
23
6.02 x 10 atoms Cu 1 mole Cu 

  107.87 g Ag 
1 mole Ag
6.24 a. 6.02 x 1023 atoms Ag x 
 x
 = 108 g Ag
23
 6.02 x 10 atoms Ag   1 mole Ag 

  107.87 g Ag 
1 mole Ag
b. 3.01 x 1023 atoms Ag x 
 x
 = 53.9 g Ag
23
 6.02 x 10 atoms Ag   1 mole Ag 

  107.87 g Ag 
1 mole Ag
-16
c. 1.00 x 106 atoms Ag x 
 x
 = 1.79 x 10 g Ag
23
 6.02 x 10 atoms Ag   1 mole Ag 

  107.87 g Ag 
1 mole Ag
-22
d. 1 atom Ag x 
 x
 = 1.79 x 10 g Ag
23
 6.02 x 10 atoms Ag   1 mole Ag 
6.25
To convert grams to moles (parts a. and b.), multiply by a conversion factor derived from the
mass in grams of 1 mole (formula mass). To convert atoms to moles (parts c. and d.), use a
conversion factor derived from the definition of Avogadro’s number (1 mole = 6.02 x 1023).
1 mole He 
a. 10.0 g He x 
 = 2.50 mole s He
4.00 g He 
 1 mole N 2 O 
 = 0.227 moles N 2 O
 44.02 g N 2 O 
b. 10.0 g N 2 O x 


1 mole P
-14
c. 4.0 x 1010 atoms P x 
 = 6.6 x 10 mole P
23
6.02 x 10 atoms P 


1 mole Be
-14
d. 4.0 x 10 10 atoms Be x 
 = 6.6 x 10 mole Be
23
6.02 x 10 atoms Be
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361
Problem-Set Solutions Chapter 6
 1 mole N 
6.26 a. 25.0 g N x 
 = 1.78 moles N
 14.01 g N 
 1 mole Li 
b. 25.0 g Li x 
 = 3.60 moles Li
 6.94 g Li 


1 mole S
-8
c. 8.50 x 1015 atoms S x 
 = 1.41 x 10 mole S
23
 6.02 x 10 atoms S 


1 mole Cl
-8
d. 8.50 x 1015 atoms Cl x 
 = 1.41 x 10 mole Cl
23
 6.02 x 10 atoms Cl 
6.27
To change grams of molecules to atoms of S, we will use three conversion factors. 1) Change
grams of molecules to moles of molecules using the definition of formula mass. 2) Change
moles of molecules to moles of S in the molecule by determining the number of atoms of S in
the molecule. 3) Change moles of S to atoms of S using Avogadro’s number (1 mole
= 6.02 x 1023). In part d., we are given the number of moles of molecules, so the first
conversion factor is not needed.
1 mole H 2 SO4   1 mole S  6.02 x 10 23 atoms S 
a. 10.0 g H2 SO4 x 
 x 
 x 

1 mole S

98.09 g H 2 SO4  1 mole H 2 SO4  
= 6.14 x 10 22 atoms S
1 mole SO 3   1 mole S  6.02 x 10 23 atoms S 
b. 20.0 g SO 3 x 
 x 
 x 

1 mole S

80.07 g SO 3  1 mole SO 3  
= 1.50 x 10 23 atoms S
 1 mole Al 2 S3   3 moles S  6.02 x 10 23 atoms S
c. 30.0 g Al2 S3 x 
 x 
 x 

1 mole S

150.17 g Al 2 S3  1 mole Al2 S3  
= 3.61 x 10 23 atoms S
 2 moles S  6.02 x 10 23 atoms S
24
d. 2 moles S2 O x 
 = 2.41 x 10 atoms S
 x 
1 mole S
1 mole S 2 O  

Copyright © Houghton Mifflin Company. All rights reserved.
362
Problem-Set Solutions Chapter 6
 1 mole N 2 H 4   2 moles N   6.02 x 10 23 atoms N 
6.28 a. 10.0 g N 2 H 4 x 
 x
x

1 mole N

 32.06 g N 2 H 4   1 mole N 2 H 4  
= 3.76 x 10 23 atoms N
 1 mole HN 3   3 moles N   6.02 x 10 23 atoms N 
b. 20.0 g HN 3 x 
 x
x

1 mole N

 43.04 g HN 3   1 mole HN 3  
= 8.39 x 10 23 atoms N
 1 mole LiNO 3   1 mole N   6.02 x 10 23 atoms N 
c. 30.0 g LiNO 3 x 
 x
x

1 mole N

 68.95 g LiNO 3   1 mole LiNO 3  
= 2.62 x 10 23 atoms N
 2 moles N   6.02 x 10 23 atoms N 
d. 4.00 moles N 2 O5 x 
x

1 mole N

 1 mole N 2 O5  
= 4.82 x 10 24 atoms N
6.29
To calculate grams of S from molecules containing S atoms, use three conversion factors.
1) Change molecules of compound to moles of compound using Avogadro’s number (1 mole
= 6.02 x 1023). 2) Change moles of compound to moles S by determining the number of atoms
of S in each molecule of the compound. 3) Convert moles S to grams of S using the atomic
mass of sulfur. For parts c. and d., we are given moles of compound, so step 1 is not needed.

  2 moles S  32.07 g S 
1 mole S 2 O
a. 3.01 x 1023 molecules S 2 O x 
 x 
 x 

23
6.02 x 10 molecules S2 O 1 mole S 2 O  1 mole S 
= 32.1 g S

  4 moles S  32.07 g S 
1 mole S 4 N4
b. 3 molecules S 4 N4 x 
 x 
 x 

23
6.02 x 10 molecules S 4 N4  1 mole S 4 N4   1 mole S 
= 6.39 x 10 -22 g S
 1 mole S  32.07 g S 
c. 2.00 moles SO2 x 
 = 64.1 g S
 x 
1 mole SO 2  1 mole S 
 8 mole S  32.07 g S 
d. 4.50 moles S 8 x 
 = 1150 g S
 x 
1 mole S 8  1 mole S 
Copyright © Houghton Mifflin Company. All rights reserved.
363
Problem-Set Solutions Chapter 6
6.30

  3 moles O   16.00 g O 
1 mole SO 3
a. 4.50 x 10 22 molecules SO 3 x 
x
x

23
 6.02 x 10 molecules SO 3   1 mole SO 3   1 mole O 
= 3.59 g O

  10 moles O   16.00 g O 
1 mole P4 O10
b. 7 molecules P4 O10 x 
x
x

23
 6.02 x 10 molecules P4 O10   1 mole P4 O10   1 mole O 
= 1.86 x 10-21 g O
 4 moles O   16.00 g O 
c. 3.00 moles H2SO4 x 
x
 = 192 g O
 1 mole H2SO4   1 mole O 
 3 moles O   16.00 g O 
d. 1.50 moles O3 x 
x
 = 72.0 g O
 1 mole O3   1 mole O 
6.31
A balanced chemical equation has the same number of atoms of each element involved in the
reaction on each side of the equation.
a. Balanced chemical equation
b. Balanced chemical equation
c. The chemical equation in part c. is not balanced; there are different numbers of both S
atoms and O atoms on the two sides of the equation. The balanced chemical equation
should be: CS2 + 3O2  CO2 + 2SO2
d. Balanced chemical equation
6.32
a. not balanced
b. not balanced
c. balanced
d. balanced
6.33
To determine the number of atoms of each element on each side of the chemical equation,
multiply the number of atoms of the element in the molecule by the coefficient of the
molecule.
a. Reactant side: 2(2) = 4 N atoms, 3(2) = 6 O atoms;
Product side: 2(2) = 4 N atoms, 2(3) = 6 O atoms
b. Reactant side: 4(1) + 6(1) = 10 N atoms, 4(3) = 12 H atoms, 6(1) = 6 O atoms;
Product side: 5(2) = 10 N atoms, 6(2) = 12 H atoms, 6(1) = 6 O atoms
c. Reactant side: 1(1) = 1 P atoms, 1(3) = 3 Cl atoms, 3(2) = 6 H atoms;
Product side: 1(1) = 1 P atoms, 3(1) = 3 Cl atoms, 1(3) + 3(1) = 6 H atoms
d. Reactant side: 1(2) = 2 Al atoms, 1(3) = 3 O atoms, 6(1) = 6 H atoms, 6(1) = 6 Cl atoms;
Product side: 2(1) = 2 Al atoms, 3(1) = 3 O atoms, 3(2) = 6 H atoms, 2(3) = 6 Cl atoms
6.34
a. 4 Al, 6 O
b. 2 Na, 4 H, 2 O
6.35
To balance a chemical equation, examine the equation and pick one element to balance first.
Start with the compound that contains the greatest number of atoms, whether in the reactant or
product. Add coefficients where necessary to balance this element, then continue adding
coefficients to balance each of the other elements separately. As a final check, count the
number of atoms of each element on each side of the equation to make sure they are equal.
a. 2Na + 2H2O  2NaOH + H2
c. 2NaBr + Cl2  2NaCl + Br2
Copyright © Houghton Mifflin Company. All rights reserved.
c. 2 Co, 3 Hg, 6 Cl
d. 8 H, 1 S, 4 O, 2 N
b. 2Na + ZnSO4  Na2SO4 + Zn
d. 2ZnS + 3O2  2ZnO + 2SO2
364
Problem-Set Solutions Chapter 6
6.36
a. 2H2S + 3O2  2SO2 + 2H2O
c. 3IBr + 4NH3  3NH4Br + NI3
b. Ni + 2HCl  NiCl2 + H2
d. 2C2H6 + 7O2  4CO2 + 6H2O
6.37
In the following chemical equations a carbon-containing compound is oxidized with molecular
oxygen to form CO2 and H2O. It is usually convenient in this type of equation to begin by
balancing the hydrogen atoms, then the carbon atoms, and finally the oxygen atoms.
a. CH4 + 2O2  CO2 + 2H2O
c. C4H8O2 + 5O2  4CO2 + 4H2O
b. 2C6H6 + 15O2  12CO2 + 6H2O
d. C5H10O + 7O2  5CO2 + 5H2O
6.38
a. C2H4 + 3O2  2CO2 + 2H2O
c. C3H6O + 4O2  3CO2 + 3H2O
b. C6H12 + 9O2  6CO2 + 6H2O
d. 2C5H10O2 + 13O2  10CO2 + 10H2O
6.39
Use the general steps outlined in problem 6.35 to balance these chemical equations.
a. 3PbO + 2NH3  3Pb + N2 + 3H2O
b. 2Fe(OH)3 + 3H2SO4  Fe2(SO4)3 + 6H2O
6.40
a. SO2Cl2 + 8HI  H2S + 2H2O + 2HCl + 4I2
b. Na2CO3 + Mg(NO3)2  MgCO3 + 2NaNO3
6.41
The coefficients in a balanced chemical equation can be used to obtain several pairs of
conversion factors that can be used in solving problems. The coefficients in the balanced
chemical equation give the mole-to-mole ratios.
2 moles Ag 2 CO 3 2 moles Ag 2 CO 3 2 moles Ag 2 CO 3 4 moles Ag 4 moles Ag
,
,
,
,
,
4 moles Ag
2 moles CO2
1 mole O2
2 moles CO2 1 mole O2
2 moles CO2
1 mole O2
6.42
The other six are reciprocals of these six factors.
1 mole N2 H 4 1 mole N2 H 4 1 mole N2 H 4 2 moles H2O2 2 moles H2O2
1 mole N 2
,
,
,
,
,
2 moles H2O2 1 mole N2
4 moles H2O
1 mole N2
4 moles H2O 4 moles H 2O
The other six are reciprocals of these six factors.
Copyright © Houghton Mifflin Company. All rights reserved.
365
Problem-Set Solutions Chapter 6
6.43
The coefficients from a balanced chemical equation can be used to form conversion factors to
solve problems. In the problems below, the conversion factor is based on a mole-to-mole
ratio using the coefficient of the first reactant and the coefficient of the CO2 produced.
 7 moles CO 2 
a. 2.00 moles C 7 H16 x 
 = 14.0 mole s CO 2
1 mole C 7 H16 
 1 mole CO2 
 = 1.00 mole CO2
 2 moles HCl 
b. 2.00 moles HCl x 
 2 moles CO 2 
 = 4.00 moles CO 2
 1 mole Na 2 SO 4 
c. 2.00 moles Na 2 SO 4 x 
 1 mole CO 2 
 = 2.00 moles CO 2
 1 mole Fe 3O 4 
d. 2.00 moles Fe 3O 4 x 
6.44
 1 mole CO2 
a. 3.50 moles FeO x 
 = 3.50 moles CO2
 1 mole FeO 
 1 mole CO2 
b. 3.50 moles O2 x 
 = 1.17 moles CO2
 3 moles O2 
 16 moles CO2 
c. 3.50 moles C8 H18 x 
 = 28.0 moles CO2
 2 moles C8 H18 
 6 moles CO2 
d. 3.50 moles C6 H12O6 x 
 = 21.0 moles CO2
 1 mole C6 H12O6 
6.45
In the problem below, we can see from the road map in Figure 6.9 that the conversion of
grams of product to grams of reactant requires three conversion factors: 1) Use the molar mass
of the product to convert grams of product to moles of product. 2) Use the coefficients from
the balanced chemical equation to convert moles of product to moles of reactant. 3) Use the
molar mass of the reactant to convert moles of reactant to grams of reactant.
1 mole N 2  4 moles NH3 
a. 20.0 g N2 x 
 x 
 x
28.02 g N 2   2 moles N 2 
17.04 g NH3 
1 mole NH = 24.3 g NH3

3 
1 mole N 2  1 mole NH4 2 Cr2 O 7  252.10 g NH 4 2 Cr2 O7 
b. 20.0 g N 2 x 

 x 
 x 
1 mole N 2
28.02 g N 2  
  1 mole NH4 2 Cr2 O 7 
= 1.80 x 10 2 g NH4 2 Cr2 O 7
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366
Problem-Set Solutions Chapter 6
1 mole N 2  1 mole N 2 H 4  32.06 g N 2 H 4 
c. 20.0 g N2 x 
 x 
x 
= 22.9 g N 2 H4
28.02 g N 2   1 mole N 2  1 mole N 2 H4 
1 mole N 2  2 moles NH 3  17.04 g NH 3 
d. 20.0 g N 2 x 
 x 
 x 
 = 24.3 g NH3
28.02 g N 2   1 mole N 2  1 mole NH 3 
6.46
 1 mole H2O   1 mole N2 H 4   32.06 g N2 H 4 
a. 20.0 g H2O x 
 x
x
 = 8.90 g N 2 H 4
 18.02 g H2O   4 moles H2O   1 mole N2 H 4 
 1 mole H 2O   1 mole H 2O2   34.02 g H2O2 
b. 20.0 g H2O x 
 x
x
 = 18.9 g H2O2
 18.02 g H 2O   2 moles H2O   1 mole H2O2 
 1 mole H2O   2 moles HNO3   63.02 g HNO3 
2
c. 20.0 g H2O x 
 x
x
 = 1.40 x 10 g HNO3
18.02
g
H
O
1
mole
H
O
1
mole
HNO


 
2 
2
3 
 1 mole H2O   3 moles H 2   2.02 g H 2 
d. 20.0 g H 2O x 
 x
x
 = 2.24 g H 2
 18.02 g H2O   3 moles H2O   1 mole H2 
6.47
We can see from the road map in Figure 6.9 that the conversion of grams of CO2 to grams of
O2 requires three conversion factors: 1) Use the molar mass of CO2 to convert 3.50 g of CO2 to
moles of CO2. 2) Use the coefficients from the balanced chemical equation to convert moles
of CO2 to moles of O2. 3) Use the molar mass of O2 to convert moles of O2 to grams of O2.
1 mole CO 2   2 moles O2  32.00 g O 2 
3.50 g CO 2 x 
 x 
 x 
 = 5.09 g O2
44.01 g CO 2  1 mole CO2  1 mole O 2 
6.48
6.49
 1 mole W   3 moles H2   2.02 g H2 
1.00 g W x 
 = 0.0330 g H2
 x
x
 183.84 g W   1 mole W   1 mole H2 
Use the road map in Figure 6.9 to determine the conversion factors that will be needed. The
conversion of grams of CO to grams of O2 requires three conversion factors: 1) Use the molar
mass of CO to convert 25.0 g of CO to moles of CO. 2) Use the coefficients from the
balanced chemical equation to convert moles of CO to moles of O2. 3) Use the molar mass of
O2 to convert moles of O2 to grams of O2.
1 mole CO   1 mole O 2  32.00 g O 2 
25.0 g CO x 
 x 
 x 
 = 14.3 g O2
28.01 g CO  2 moles CO   1 mole O 2 
6.50
 1 mole H2O2   1 mole N2 H 4   32.06 g N2 H 4 
35.0 g H2O2 x 
 x
x
 = 16.5 g N 2 H 4
 34.02 g H2O2   2 moles H2O2   1 mole N 2 H 4 
Copyright © Houghton Mifflin Company. All rights reserved.
Problem-Set Solutions Chapter 6
6.51
367
Use the road map in Figure 6.9 to determine the conversion factors that will be needed. The
conversion of grams of SO2 to grams of H2O requires three conversion factors: 1) Use the
molar mass of SO2 to convert 10.0 g of SO2 to moles of SO2. 2) Use the coefficients from the
balanced chemical equation to convert moles of SO2 to moles of H2O. 3) Use the molar mass
of H2O to convert moles of H2O to grams of H2O.
1 mole SO 2  2 moles H 2 O  18.02 g H 2 O 
10.0 g SO 2 x 
 x 
 x 
= 5.63 g H 2O
64.07 g SO 2   1 mole SO 2  1 mole H 2 O 
6.52
6.53
 1 mole KHSO4   8 moles HCl   36.46 g HCl 
25.0 g KHSO4 x 
 x
x
 = 26.8 g HCl
 136.18 g KHSO4   2 moles KHSO4   1 mole HCl 
Formula mass is calculated by multiplying each atomic mass by the number of atoms of that
element in the chemical formula, and then summing all of the atomic masses of all the
elements in the chemical formula. Since we have been given the formula mass and since the
number of atoms of H (y) are unknown, we can set up an algebraic equation and solve for y.
[3(12.01) + y(1.01) + 32.07] amu = 76.18 amu
1.01y = 8.08
y = 8.00
6.54
a. 1.00 mole S8; it contains eight times as many atoms as 1.00 mole S.
b. 28.0 g Al; the molar mass of Al is 27.0 g, so 28.0 g of Al is more than 1.00 mole.
c. 30.0 g Mg; the molar mass of Mg is 24.3 g, so 30.0 g of Mg is more than 1.00 mole, and
the molar mass of Si is 28.1 g, so 28.1 g of Si is 1.00 mole.
d. 6.02 x1023 atoms He; this is 1.00 mole of He, and 2.00 g of Na is less than 1.00 mole of Na
since the molar mass of Na is 23.0 g.
6.55
Use the road map in Figure 6.9 to determine the conversion factors for these problems.
For example, in part a., use the following steps to obtain the answer:
(g Si) x (conversion factor using atomic mass of Si) = moles Si
(moles Si) x (conversion factor using mole-to-mole ratio, SiH4 to Si) = moles SiH4
a. g Si  moles Si  moles SiH4
 1 mole Si   1 mole SiH 4 
 = 0.03560 mole SiH 4
 x
 28.09 g Si   1 mole Si 
1.000 g Si x 
b. g Si  moles Si  moles SiO2  g SiO2
 1 mole Si   1 mole SiO 2   60.09 g SiO 2 
x
 = 2.139 g SiO 2
 x
 28.09 g Si   1 mole Si   1 mole SiO 2 
1.000 g Si x 
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368
Problem-Set Solutions Chapter 6
c. g Si  moles Si  moles (CH3)3SiCl  molecules (CH3)3SiCl
 1 mole Si   1 mole (CH 3 )3SiCl   6.022 x 10 23 molecules (CH 3 )3SiCl 
x

 x
 
1 mole Si
1 mole (CH 3 )3SiCl
 28.09 g Si  

1.000 g Si x 
= 2.144 x 10 22 molecules (CH 3 )3SiCl
d. g Si  moles Si  atoms Si
 1 mole Si   6.022 x 10 23 atoms Si 
22
 = 2.144 x 10 atoms Si
x
1 mole Si
 28.09 g Si  

1.000 g Si x 
6.56
6.57
 1 mole Si   28.09 g Si 
2.10 moles Ar x 
x
 = 59.0 g Si
 1 mole Ar   1 mole Si 
Because the oxygen is balanced with 22 atoms on each side of the equation, the compound
butyne contains only C and H. Write the balanced chemical equation with x and y substituted
for the numbers of C and H atoms.
2CxHy + 11O2  8CO2 + 6H2O
Write algebraic equations using the coefficients in the balanced chemical equation to balance
the C and H atoms:
Carbon balance:
Hydrogen balance:
2x = 8
2y = 6(2)
x=4
y=6
Butyne has the formula C4H6.
 1 mole (NH 4 )2 Cr2 O 7  

1 mole N 2
6.58 75.0 g (NH 4 )2 Cr2 O 7 x 
 x

 252 g (NH 4 )2 Cr2 O 7   1 mole (NH 4 )2 Cr2 O 7 
 28.0 g N 2 
x
 = 8.33 g N 2
 1 mole N 2 
 1 mole (NH 4 )2 Cr2 O 7  

1 mole H 2 O
75.0 g (NH 4 )2 Cr2 O 7 x 
 x

 252 g (NH 4 )2 Cr2 O 7   1 mole (NH 4 )2 Cr2 O 7 
 18.0 g H 2 O 
x
 = 21.4 g H 2 O
 1 mole H 2 O 
 1 mole (NH 4 )2 Cr2 O 7   4 moles Cr2 O 3 
75.0 g (NH 4 )2 Cr2 O 7 x 
 x

 252 g (NH 4 )2 Cr2 O 7   1 mole (NH 4 )2 Cr2 O 7 
 152 g Cr2 O 3 
x
 = 45.2 g Cr2 O 3
 1 mole Cr2 O 3 
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369
Problem-Set Solutions Chapter 6
6.59
Use the road map in Figure 6.9 to determine the conversion factors for the two parts of this
problem.
a. g Ag2S  moles Ag2S  moles Ag  g Ag
1 mole Ag 2 S   2 moles Ag   108 g Ag 
125 g Ag 2 S x 
 x 
 x 
= 109 g Ag
 248 g Ag 2 S  1 mole Ag 2 S  1 mole Ag 
b. g Ag2S  moles Ag2S  moles S  g S
1 mole Ag2 S   1 mole S  32.1 g S 
125 g Ag 2 S x 
 x 
 x 
= 16.2 g S
 248 g Ag 2 S  1 mole Ag 2 S  1 mole S 
6.60
6.61
 1 mole N 2   3 moles Be   9.01 g Be 
45.0 g N 2 x 
 x
x
 = 43.4 g Be
 28.0 g N 2   1 mole N2   1 mole Be 
The correct answer is d., 18.02 amu and 44.01 amu. Formula mass is the sum of the atomic
masses of all the atoms in the chemical formula of a substance.
H2O: 2(1.01) + 1(16.00) = 18.02
CO2: 1(12.01) + 2(16.00) = 44.01
6.62
b
6.63
Answer b., chemical formulas and atomic masses, is the correct answer. Molar mass is the
formula mass of a substance expressed in grams. To calculate the formula mass we need the
atomic masses of the individual atoms in the formula.
6.64
d
6.65
Answer a., 4.0 moles NH3, contains the most atoms: (4.0 moles) x (4 atoms) = 16 moles atoms
Answer b. contains (3.0 moles) x (4 atoms) = 12 moles atoms
Answer c. contains (6.0 moles) x (2 atoms) = 12 moles atoms
Answer d. contains (4.0 moles) x (3 atoms) = 12 moles atoms
6.66
b
6.67
The correct answer is b., 2, 1, 3.
6.68
b
6.69
The correct setup is b. The roadmap for solving the problem is:
2NH3  N2 + 3H2
g S4N4  moles S4N4  moles S  g S
Conversion factors:
1) Change g S4N4to moles S4N4 using 1 mole S4N4 = 184.32 g S4N4.
2) Change moles S4N4 to moles S using 1 mole S4N4 = 4 moles S
3) Change moles S to g S using 1 mole S = 32.07 g S
6.70
c
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