CHAPTER 3
Calculations with Chemical Formulas and
Equations
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
3.1. a. NO2 1 AW of N = 14.007 amu
2 AW of O = 2 15.999 = 31.998 amu
MW of NO2 = 46.005 = 46.0 amu (3 s.f.)
b. C6H12O6 6 AW of C = 6 12.011 = 72.066 amu
12 AW of H = 12 1.008 = 12.096 amu
6 AW of O = 6 15.999 = 95.994 amu
FW of C6H12O6 = 180.156 = 180. amu (3 s.f.)
c. NaOH 1 AW of Na = 22.98977 amu
1 AW of O = 15.999 amu
1 AW of H = 1.008 amu
MW of NaOH = 39.99677 = 40.0 amu (3 s.f.)
d. Mg(OH)2 1 AW of Mg = 24.305 amu
2 AW of O = 2 15.9994 = 31.998 amu
2 AW of H = 2 1.008 = 2.016 amu
FW of Mg(OH)2 = 58.319 = 58.3 amu (3 s.f.)
3.2. a. The molecular model represents a molecule made up of one S and three O. The chemical
formula is SO3. Calculating the formula weight by using the same approach as in Example
3.1 in the text yields 80.1 amu.
b. The molecular model represents one S, four O, and two H. The chemical formula is then
H2SO4. The formula weight is 98.1 amu.
65
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, 66 Chapter 3: Calculations with Chemical Formulas and Equations
3.3. a. The atomic weight of Ca = 40.078 amu; thus, the molar mass = 40.078 g/mol, and
1 mol Ca = 6.022 1023 Ca atoms.
40.078 g 1 mol
Mass of one Ca = = 6.6553 10−23
1 mol Ca 6.022 1023 atoms
= 6.655 10−23 g/atom
b. The molecular weight of C2H5OH, or C2H6O, = (2 12.01) + (6 1.008) + 16.00 = 46.068.
Its molar mass = 46.07 g/mol, and 1 mol = 6.022 1023 molecules of C2H6O.
46.07 g 1 mol
Mass of one C2H6O =
1 mol C2 H 6 O 6.022 1023 molecules
= 7.6503 10−23 = 7.650 10−23 g/molecule
3.4. The molar mass of H2O2 is 34.01 g/mol. Therefore,
34.01 g H 2O2
0.909 mol H 2O2 = 30.92 = 30.9 g H 2O 2
1 mol H 2O2
3.5. The molar mass of HNO3 is 63.01 g/mol. Therefore,
1 mol HNO3
28.5 g HNO3 = 0.4523 = 0.452 mol HNO3
63.01 g HNO3
3.6. Convert the mass of HCN from milligrams to grams. Then convert grams of HCN to moles of
HCN. Finally, convert moles of HCN to the number of HCN molecules.
1g 1 mol HCN 6.022 1023 HCN molecules
56 mg HCN
1000 mg 27.03 g HCN 1 mol HCN
= 1.248 1021 = 1.2 1021 HCN molecules
3.7. The molecular weight of NH4NO3 = 80.04; thus, its molar mass = 80.04 g/mol. Hence
28.014 g
Percent N = 100% = 35.00 = 35.0%
80.04 g
4.032 g
Percent H = 100% = 5.037 = 5.04%
80.04 g
47.997 g
Percent O = 100% = 59.97 = 60.0%
80.04 g
3.8. From the previous exercise, NH4NO3 is 35.0% N (fraction N = 0.350), so the mass of N in 48.5 g
of NH4NO3 is
48.5 g NH4NO3 (0.350 g N/1 g NH4NO3) = 16.975 = 17.0 g N
3.9. First, convert the mass of CO2 to moles of CO2. Next, convert this to moles of C (1 mol CO2 is
equivalent to 1 mol C). Finally, convert to mass of carbon, changing milligrams to grams first:
1 mol CO2 1 mol C 12.01 g C
5.80 10−3 g CO2 = 1.583 10−3 g C
44.01 g 1 mol CO2 1 mol C
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 3: Calculations with Chemical Formulas and Equations 67
Do the same series of calculations for water, noting that 1 mol H2O contains 2 mol H.
1 mol H 2 O 2 mol H 1.008 g H
1.58 10−3 g H2O = 1.767 10−4 g H
18.02 g 1 mol H 2 O 1 mol H
The mass percentages of C and H can be calculated using the masses from the previous
calculations:
1.583 mg
Percent C = 100% = 40.90 = 40.9% C
3.87 mg
0.1767 mg
Percent H = 100% = 4.5658 = 4.57% H
3.87 mg
The mass percentage of O can be determined by subtracting the sum of the above percentages
from 100%:
Percent O = 100.000% − (40.90 + 4.5658) = 54.5342 = 54.5% O
3.10. Convert the masses to moles that are proportional to the subscripts in the empirical formula:
1 mol S
33.4 g S = 1.042 mol S
32.06 g S
1 mol O
(83.5 − 33.4) g O = 3.131 mol O
16.00 g O
Next, obtain the smallest integers from the moles by dividing each by the smallest number of
moles:
3.131 mol O 1.042 mol S
For O: = 3.00 For S: = 1.00
1.042 mol S 1.042 mol S
The empirical formula is SO3.
3.11. For a 100.0-g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 26.2 g are O. Using the molar
masses, convert these masses to moles:
1 mol C
68.8 g C = 5.729 mol C
12.01 g C
1 mol H
5.0 g H = 4.96 mol H
1.008 g H
1 mol O
26.2 g O = 1.638 mol O
16.00 g O
These numbers are in the same ratio as the subscripts in the empirical formula. They must be
changed to integers. First, divide each one by the smallest number of moles:
5.729 4.96 1.638
For C: = 3.497 For H: = 3.03 For O: = 1.000
1.638 1.638 1.638
Rounding off, we obtain C3.5H3.0O1.0. Multiplying the numbers by 2 gives whole numbers, for an
empirical formula of C7H6O2.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 68 Chapter 3: Calculations with Chemical Formulas and Equations
3.12. For a 100.0-g sample of acetaldehyde, 54.5 g are C, 9.2 g are H, and 36.3 g are O. Using the
molar masses, convert these masses to moles:
1 mol C
54.5 g C = 4.537 mol C
12.01 g C
1 mol H
9.2 g H = 9.12 mol H
1.008 g H
1 mol O
36.3 g O = 2.268 mol O
16.00 g O
These numbers are in the same ratio as the subscripts in the empirical formula. They must be
changed to integers. First, divide each one by the smallest number of moles:
4.537 9.12 2.268
For C: = 2.000 For H: = 4.02 For O: = 1.000
2.268 2.268 2.268
Rounding off, we obtain C2H4O, the empirical formula, which is also the molecular formula.
3.13. H2 + Cl 2 2HCl
1 molec. (mol) H2 + 1 molec. (mol) Cl2 2 molec. (mol) HCl (molec., mole interp.)
2.016 g H2 + 70.90 g Cl2 2 36.46 g HCl (mass interp.)
3.14. Equation: Na + H2O 1/2H2 + NaOH, or 2Na + 2H2O H2 + 2NaOH.From this equation, one
mole of Na corresponds to one-half mole of H2, or two moles of Na corresponds to one mole of
H2. Therefore,
1 mol H 2 2 mol Na 22.99 g Na
7.81 g H2 = 178.1 = 178 g Na
2.016 g H 2 1 mol H 2 1 mol Na
3.15. Balanced equation: 2ZnS + 3O2 2ZnO + 2SO2
Convert grams of ZnS to moles of ZnS. Then determine the relationship between ZnS and O2
(2ZnS is equivalent to 3O2). Finally, convert to mass of O2.
1 mol ZnS 3 mol O2 32.00 g O2 1 kg
5.00 103g ZnS x
97.44 g ZnS 2 mol ZnS 1 mol O2 1000 g
= 2.463 = 2.46 kg O2
3.16. Balanced equation: 2HgO 2Hg + O2
Convert the mass of O2 to moles of O2. Using the fact that one mole of O2 is equivalent to two
moles of Hg, determine the number of moles of Hg, and convert to mass of Hg.
1 mol O2 2 mol Hg 200.59 g Hg
6.47 g O2 = 81.11 = 81.1 g Hg
32.00 g O2 1 mol O 2 1 mol Hg
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Calculations with Chemical Formulas and
Equations
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
3.1. a. NO2 1 AW of N = 14.007 amu
2 AW of O = 2 15.999 = 31.998 amu
MW of NO2 = 46.005 = 46.0 amu (3 s.f.)
b. C6H12O6 6 AW of C = 6 12.011 = 72.066 amu
12 AW of H = 12 1.008 = 12.096 amu
6 AW of O = 6 15.999 = 95.994 amu
FW of C6H12O6 = 180.156 = 180. amu (3 s.f.)
c. NaOH 1 AW of Na = 22.98977 amu
1 AW of O = 15.999 amu
1 AW of H = 1.008 amu
MW of NaOH = 39.99677 = 40.0 amu (3 s.f.)
d. Mg(OH)2 1 AW of Mg = 24.305 amu
2 AW of O = 2 15.9994 = 31.998 amu
2 AW of H = 2 1.008 = 2.016 amu
FW of Mg(OH)2 = 58.319 = 58.3 amu (3 s.f.)
3.2. a. The molecular model represents a molecule made up of one S and three O. The chemical
formula is SO3. Calculating the formula weight by using the same approach as in Example
3.1 in the text yields 80.1 amu.
b. The molecular model represents one S, four O, and two H. The chemical formula is then
H2SO4. The formula weight is 98.1 amu.
65
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 66 Chapter 3: Calculations with Chemical Formulas and Equations
3.3. a. The atomic weight of Ca = 40.078 amu; thus, the molar mass = 40.078 g/mol, and
1 mol Ca = 6.022 1023 Ca atoms.
40.078 g 1 mol
Mass of one Ca = = 6.6553 10−23
1 mol Ca 6.022 1023 atoms
= 6.655 10−23 g/atom
b. The molecular weight of C2H5OH, or C2H6O, = (2 12.01) + (6 1.008) + 16.00 = 46.068.
Its molar mass = 46.07 g/mol, and 1 mol = 6.022 1023 molecules of C2H6O.
46.07 g 1 mol
Mass of one C2H6O =
1 mol C2 H 6 O 6.022 1023 molecules
= 7.6503 10−23 = 7.650 10−23 g/molecule
3.4. The molar mass of H2O2 is 34.01 g/mol. Therefore,
34.01 g H 2O2
0.909 mol H 2O2 = 30.92 = 30.9 g H 2O 2
1 mol H 2O2
3.5. The molar mass of HNO3 is 63.01 g/mol. Therefore,
1 mol HNO3
28.5 g HNO3 = 0.4523 = 0.452 mol HNO3
63.01 g HNO3
3.6. Convert the mass of HCN from milligrams to grams. Then convert grams of HCN to moles of
HCN. Finally, convert moles of HCN to the number of HCN molecules.
1g 1 mol HCN 6.022 1023 HCN molecules
56 mg HCN
1000 mg 27.03 g HCN 1 mol HCN
= 1.248 1021 = 1.2 1021 HCN molecules
3.7. The molecular weight of NH4NO3 = 80.04; thus, its molar mass = 80.04 g/mol. Hence
28.014 g
Percent N = 100% = 35.00 = 35.0%
80.04 g
4.032 g
Percent H = 100% = 5.037 = 5.04%
80.04 g
47.997 g
Percent O = 100% = 59.97 = 60.0%
80.04 g
3.8. From the previous exercise, NH4NO3 is 35.0% N (fraction N = 0.350), so the mass of N in 48.5 g
of NH4NO3 is
48.5 g NH4NO3 (0.350 g N/1 g NH4NO3) = 16.975 = 17.0 g N
3.9. First, convert the mass of CO2 to moles of CO2. Next, convert this to moles of C (1 mol CO2 is
equivalent to 1 mol C). Finally, convert to mass of carbon, changing milligrams to grams first:
1 mol CO2 1 mol C 12.01 g C
5.80 10−3 g CO2 = 1.583 10−3 g C
44.01 g 1 mol CO2 1 mol C
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 3: Calculations with Chemical Formulas and Equations 67
Do the same series of calculations for water, noting that 1 mol H2O contains 2 mol H.
1 mol H 2 O 2 mol H 1.008 g H
1.58 10−3 g H2O = 1.767 10−4 g H
18.02 g 1 mol H 2 O 1 mol H
The mass percentages of C and H can be calculated using the masses from the previous
calculations:
1.583 mg
Percent C = 100% = 40.90 = 40.9% C
3.87 mg
0.1767 mg
Percent H = 100% = 4.5658 = 4.57% H
3.87 mg
The mass percentage of O can be determined by subtracting the sum of the above percentages
from 100%:
Percent O = 100.000% − (40.90 + 4.5658) = 54.5342 = 54.5% O
3.10. Convert the masses to moles that are proportional to the subscripts in the empirical formula:
1 mol S
33.4 g S = 1.042 mol S
32.06 g S
1 mol O
(83.5 − 33.4) g O = 3.131 mol O
16.00 g O
Next, obtain the smallest integers from the moles by dividing each by the smallest number of
moles:
3.131 mol O 1.042 mol S
For O: = 3.00 For S: = 1.00
1.042 mol S 1.042 mol S
The empirical formula is SO3.
3.11. For a 100.0-g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 26.2 g are O. Using the molar
masses, convert these masses to moles:
1 mol C
68.8 g C = 5.729 mol C
12.01 g C
1 mol H
5.0 g H = 4.96 mol H
1.008 g H
1 mol O
26.2 g O = 1.638 mol O
16.00 g O
These numbers are in the same ratio as the subscripts in the empirical formula. They must be
changed to integers. First, divide each one by the smallest number of moles:
5.729 4.96 1.638
For C: = 3.497 For H: = 3.03 For O: = 1.000
1.638 1.638 1.638
Rounding off, we obtain C3.5H3.0O1.0. Multiplying the numbers by 2 gives whole numbers, for an
empirical formula of C7H6O2.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 68 Chapter 3: Calculations with Chemical Formulas and Equations
3.12. For a 100.0-g sample of acetaldehyde, 54.5 g are C, 9.2 g are H, and 36.3 g are O. Using the
molar masses, convert these masses to moles:
1 mol C
54.5 g C = 4.537 mol C
12.01 g C
1 mol H
9.2 g H = 9.12 mol H
1.008 g H
1 mol O
36.3 g O = 2.268 mol O
16.00 g O
These numbers are in the same ratio as the subscripts in the empirical formula. They must be
changed to integers. First, divide each one by the smallest number of moles:
4.537 9.12 2.268
For C: = 2.000 For H: = 4.02 For O: = 1.000
2.268 2.268 2.268
Rounding off, we obtain C2H4O, the empirical formula, which is also the molecular formula.
3.13. H2 + Cl 2 2HCl
1 molec. (mol) H2 + 1 molec. (mol) Cl2 2 molec. (mol) HCl (molec., mole interp.)
2.016 g H2 + 70.90 g Cl2 2 36.46 g HCl (mass interp.)
3.14. Equation: Na + H2O 1/2H2 + NaOH, or 2Na + 2H2O H2 + 2NaOH.From this equation, one
mole of Na corresponds to one-half mole of H2, or two moles of Na corresponds to one mole of
H2. Therefore,
1 mol H 2 2 mol Na 22.99 g Na
7.81 g H2 = 178.1 = 178 g Na
2.016 g H 2 1 mol H 2 1 mol Na
3.15. Balanced equation: 2ZnS + 3O2 2ZnO + 2SO2
Convert grams of ZnS to moles of ZnS. Then determine the relationship between ZnS and O2
(2ZnS is equivalent to 3O2). Finally, convert to mass of O2.
1 mol ZnS 3 mol O2 32.00 g O2 1 kg
5.00 103g ZnS x
97.44 g ZnS 2 mol ZnS 1 mol O2 1000 g
= 2.463 = 2.46 kg O2
3.16. Balanced equation: 2HgO 2Hg + O2
Convert the mass of O2 to moles of O2. Using the fact that one mole of O2 is equivalent to two
moles of Hg, determine the number of moles of Hg, and convert to mass of Hg.
1 mol O2 2 mol Hg 200.59 g Hg
6.47 g O2 = 81.11 = 81.1 g Hg
32.00 g O2 1 mol O 2 1 mol Hg
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
