Ch 6 Book Answers

Chapter 6 Chemical Composition 

1. 100 washers 0.110 g 

1 washer 

= 11.0 g (assuming 100 washers is exact) 

100. g 1 washer 

0.110 g 

= 909 washers 

2. 500. g 1 cork 

1.63 g 

= 306.7 = 307 corks 

500. g 1 stopper 

4.31 g 

= 116 stoppers 

1 kg (1000 g) of corks contains (1000 g 1 cork ) = 613.49 = 613 corks 

1.63 g 

613 stoppers would weigh (613 stoppers 4.31 g ) = 2644 g = 2640 g 

1 stopper 

The ratio of the mass of a stopper to the mass of a cork is (4.31 g/1.63 g). So the mass of 

stoppers that contains the same number of stoppers as there are corks in 1000 g of corks is 

1000 g 4.31 g 

= 2644 g = 2640 g 

1.63 g 

3. The atomic mass unit (amu) is a unit of mass defined by scientists to more simply describe 

relative masses on an atomic or molecular scale. One amu is equivalent to 

1.66 10 –24 g. 

4. We use the average atomic mass of an element when performing calculations because the 

average mass takes into account the individual masses and relative abundance of all the 

isotopes of an element. 

5. a. 635 H atoms 

1.008 amu 

1 H atom 

= 640. amu 

b. 1.261 10 4 W atoms 

183.9 amu 

1 W atom = 2.319 106 amu 

c. 42 K atoms 

39.10 amu 

1 K atom 

= 1642 amu 

d. 7.213 10 23 N atoms 

14.01 amu 

1 N atom = 1.011 1025 amu 

e. 891 Fe atoms 

55.85 amu 

1 Fe atom = 4.976 104 amu 

World of Chemistry 37 Copyright Houghton Mifflin Company. 

All rights reserved.

38 Chapter 6 

6. a. 10.81 amu 1 B atom 

10.81 amu 

= 1.000 atom = 1 B atom 

b. 320.7 amu 1 S atom 

32.07 amu 

= 10 S atoms 

c. 19,691 amu 

d. 19,695 amu 

e. 3588.3 amu 

1 Au atom 

197.97 amu 

1 Xe atom 

131.3 amu 

1 Al atom 

26.98 amu 

= 100.00 Au atoms = 100 Au atoms 

= 150.0 Xe atoms = 150 Xe atoms 

= 133.0 Al atoms = 133 Al atoms 

7. 8274 amu 1 S atom 

32.07 amu 

5.213 10 24 S atoms 

= 258 S atoms 

32.07 amu 

1 S atom = 1.672 1026 amu 

8. 1.204 10 24 

9. Avogadro’s number (6.022 10 23 ) 

10. The ratio of the atomic mass of Ca to the atomic mass of Mg is (40.08 amu/24.31 amu), and 

the masses of calcium are given by 

12.16 g Mg 

24.31 g Mg 

40.08 amu 

24.31 amu 

40.08 amu 

24.31 amu 

= 20.05 g Ca 

= 40.08 g Ca 

11. The ratio of the atomic mass of Co to the atomic mass of F is (58.93 amu/19.00 amu), and the 

mass of cobalt is given by 

57.0 g F 

58.93 amu 

19.00 amu 

= 177 g Co 

12. 1 mol O = 16.00 g O = 6.02 10 23 O atoms 

1 O atom 

16.00 g O 

6.022 x 10 23 O atoms = 2.66 10–23 g O 

13. 0.50 mol O atoms 16.00 g O 

1 mol 

= 8.0 g O 

4 mol H atoms 1.008 g = 4 g H 

1 mol 

Half a mole of O atoms weighs more than 4 moles of H atoms. 

World of Chemistry 

Copyright Houghton Mifflin Company. 


Chemical Composition 39 

14. a. 26.2 g Au 

b. 41.5 g Ca 

1 mol Au 

197.0 g 

1 mol Ca 

40.08 g 

= 0.133 mol Au 

= 1.04 mol Ca 

c. 335 mg Ba 

1 g 

10 3 mg 

 

1 mol Ba 

137.3 g = 2.44 10–3 mol Ba 

d. 1.42 10 –3 g Pd 

1 mol Pd 

106.4 g = 1.33 10–5 mol Pd 

e. 3.05 10 –5 g Ni 1 g 

10 6 g 

f. 1.00 lb Fe 453.59 g 

1 lb 

g. 12.01 g C 1 mol C 

12.01 g 

 

1 mol Ni 

58.70 g = 5.20 10–13 mol Ni 

1 mol Fe 

= 8.12 mol Fe 

55.85 g 

= 1.000 mol C 

15. a. 2.00 mol Fe 55.85 g = 112 g Fe 

1 mol 

b. 0.521 mol Ni 58.70 g 

1 mol 

= 30.6 g Ni 

c. 1.23 10 –3 mol Pt 195.1 g = 0.240 g Pt 

1 mol 

d. 72.5 mol Pb 207.2 g 

1 mol 

e. 0.00102 mol Mg 24.31 g 

1 mol 

= 1.50 10 4 g Pb 

= 0.0248 g Mg 

f. 4.87 10 3 mol Al 26.98 g = 1.31 10 5 g Al 

1 mol 

g. 211.5 mol Li 6.941 g 

1 mol 

= 1468 g Li 

h. 1.72 10 –6 mol Na 22.99 g = 3.95 10 –5 g Na 

1 mol 

16. a. 0.00103 g Co 6.022 x 1023 Co atoms 

58.93 g Co 

= 1.05 10 19 Co atoms 

b. 0.00103 mol Co 6.022 x 1023 Co atoms 

1 mol 

= 6.20 10 20 Co atoms 





c. 2.75 g Co 

1 mol 

58.93 g Co 

= 0.0467 mol Co 

d. 5.99 10 21 Co atoms 

1 mol 

6.022 x 10 23 Co atoms 

= 0.00995 mol Co 

e. 4.23 mol Co 

58.93 g Co 

1 mol Co 

= 249 g Co 

f. 4.23 mol Co 6.022 x 1023 Co atoms 

1 mol Co 

= 2.55 10 24 Co atoms 

g. 4.23 g Co 6.022 x 1023 Co atoms 

58.93 g Co 

= 4.32 10 22 Co atoms 

17. molar mass 

18. adding together (summing) 

19. a. mass of 3 mol Na = 3 (22.99 g) = 68.97 g 

mass of 1 mol N = 1 (14.01 g) = 14.01 g 

molar mass of Na 3 N = 

82.98 g 

b. mass of 1 mol C = 12.01 g = 12.01 g 

mass of 2 mol S = 2 (32.07 g) = 64.14 g 

molar mass of CS 2 = 

76.15 g 

c. mass of 1 mol N = 14.01 g = 14.01 g 

mass of 4 mol H = 4 (1.008 g) = 4.032 g 

mass of 1 mol Br = 79.90 g = 79.90 g 

molar mass of NH 4 Br = 

97.942 g = 97.94 g 

d. mass of 2 mol C = 2 (12.01 g) = 24.02 g 

mass of 6 mol H = 6 (1.008 g) = 6.048 g 

mass of 1 mol O = 16.00 g = 16.00 g 

molar mass of C 2 H 6 O = 

46.068 g = 46.07 g 

e. mass of 2 mol H = 2 (1.008 g) = 2.016 g 

mass of 1 mol S = 32.07 g = 32.07 g 

mass of 3 mol O = 3 (16.00 g) = 48.00 g 

molar mass of H 2 SO 3 = 

82.086 g = 82.09 g 

f. mass of 2 mol H = 2 (1.008 g) = 2.016 g 

mass of 1 mol S = 32.07 g = 32.07 g 

mass of 4 mol O = 4 (16.00 g) = 64.00 g 

molar mass of H 2 SO 4 = 

98.086 g = 98.09 g 





20. a. mass of 1 mol Ba = 137.3 g = 137.3 g 

mass of 2 mol Cl = 2 (35.45 g) = 70.90 g 

mass of 8 mol O = 8 (16.00 g) = 128.0 g 

molar mass of Ba(ClO 4 ) 2 = 

336.2 g 

b. mass of 1 mol Mg = 24.31 g = 24.31 g 

mass of 1 mol S = 32.07 g = 32.07 g 

mass of 4 mol O = 4 (16.00 g) = 64.00 g 

molar mass of MgSO 4 = 

120.38 g 

c. mass of 1 mol Pb = 207.2 g = 207.2 g 


molar mass of PbCl 2 = 

278.1 g 

d. mass of 1 mol Cu = 63.55 g = 63.55 g 

mass of 2 mol N = 2 (14.01 g) = 28.02 g 

mass of 6 mol O = 6 (16.00 g) = 96.00 g 

molar mass of Cu(NO 3 ) 2 = 

187.57 g 

e. mass of 1 mol Sn = 118.7 g = 118.7 g 


molar mass of SnCl 4 = 

260.5 g 

f. mass of 6 mol C = 6 (12.01 g) = 72.06 g 

mass of 6 mol H = 6 (1.008 g) = 6.048 g 

mass of 1 mol O = 16.00 g = 16.00 g 

molar mass of C 6 H 6 O = 

94.11 g 

21. a. molar mass of SO 3 = 80.07 g 

1 g 

49.2 mg SO 3 

1000 mg 1 mol 

80.07 g 

= 6.14 10 –4 mol SO 3 

b. molar mass of PbO 2 = 239.2 g 

7.44 x 10 4 kg PbO 2 1000 g 

 

1 kg 

1 mol 

239.2 g 

= 3.11 10 5 mol PbO 2 

c. molar mass of CHCl 3 = 119.37 g 

59.1 g CHCl 3 1 mol 

119.37 g 

= 0.495 mol CHCl 3 

d. molar mass of C 2 H 3 Cl 3 = 133.39 g 

3.27 g C 2 H 3 Cl 3 1 g 

10 6 g 1 mol 

133.39 g 

= 2.45 10 –8 mol C 2 H 3 Cl 3 

e. molar mass of LiOH = 23.95 g 

4.01 g LiOH 1 mol 

23.95 g 

= 0.167 mol LiOH 





22. a. molar mass of NaH 2 PO 4 = 120.0 g 

4.26 10 –3 g NaH 2 PO 4 1 mol 

120.0 g 

= 3.55 10 –5 mol NaH 2 PO 4 

b. molar mass of CuCl = 99.00 g 

521 g CuCl 1 mol 

99.00 g 

= 5.26 mol CuCl 

c. molar mass of Fe = 55.85 g 

151 kg Fe 1000 g 

1 kg 

 

1 mol 

55.85 g 

= 2.70 10 3 mol Fe 

d. molar mass of SrF 2 = 125.6 g 

8.76 g SrF 2 1 mol 

125.6 g 

= 0.0697 mol SrF 2 

e. molar mass of Al = 26.98 g 

1.26 10 4 g Al 1 mol 

26.98 g 

= 467 mol Al 

23. a. molar mass of AlI 3 = 407.7 g 

1.50 mol AlI 3 407.7 g 

1 mol 

= 612 g AlI 3 

b. molar mass of C 6 H 6 = 78.11 g 

1.91 10 –3 mol C 6 H 6 78.11 g 

1 mol 

= 0.149 g C 6 H 6 

c. molar mass of C 6 H 12 O 6 = 180.2 g 

4.00 mol C 6 H 12 O 6 180.2 g 

1 mol 

= 721 g C 6 H 12 O 6 

d. molar mass of C 2 H 5 OH = 46.07 g 

4.56 10 5 mol C 2 H 5 OH 46.07 g 

1 mol 

= 2.10 10 7 g C 2 H 5 OH 

e. molar mass of Ca(NO 3 ) 2 = 164.1 g 

2.27 mol Ca(NO 3 ) 2 164.1 g 

1 mol 

= 373 g Ca(NO 3 ) 2 





24. a. molar mass of CO 2 = 44.01 g 

1.27 mmol 

1 mol 

10 3 mmol 

44.01 g 

1 mol 

= 0.0559 g CO 2 

b. molar mass of NCl 3 = 120.4 g 

4.12 10 3 mol NCl 3 120.4 g 

1 mol 

= 4.96 10 5 g NCl 3 

c. molar mass of NH 4 NO 3 = 80.05 g 

0.00451 mol NH 4 NO 3 80.05 g 

1 mol 

= 0.361 g NH 4 NO 3 

d. molar mass of H 2 O = 18.02 g 

18.0 mol H 2 O 18.02 g 

1 mol 

= 324 g H 2 O 

e. molar mass of CuSO 4 = 159.6 g 

62.7 mol CuSO 4 159.6 g = 1.00 10 4 g CuSO 4 

1 mol 

25. a. 6.37 mol CO 6.022 x 1023 molecules 

1 mol 

= 3.84 10 24 molecules CO 

b. molar mass of CO = 28.01 g 

6.37 g 1 mol 

28.01 g 

6.022 x 1023 molecules 

1 mol 

= 1.37 10 23 molecules CO 

c. molar mass of H 2 O = 18.02 g 

2.62 10 –6 g 6.022 x 1023 molecules 

18.02 g 

= 8.76 10 16 molecules H 2 O 

d. 2.62 10 –6 mol 6.022 x 1023 molecules 

1 mol 

= 1.58 10 18 molecules H 2 O 

e. molar mass of C 6 H 6 = 78.11 g 

5.23 g 6.022 x 1023 molecules 

= 4.03 10 22 molecules C 6 H 6 

78.11 g 





26. a. molar mass of Na 2 SO 4 = 142.1 g 

2.01 g Na 2 SO 4 1 mol Na 2 SO 4 

142.1 g 

b. molar mass of Na 2 SO 3 = 126.1 g 

 

1 mol S 

1 mol Na 2 

SO 4 

= 0.0141 mol S 

2.01 g Na 2 SO 3 1 mol Na 2 SO 3 

126.1 g 

 

1 mol S 

1 mol Na 2 

SO 3 

= 0.0159 mol S 

c. molar mass of Na 2 S = 78.05 g 

2.01 g Na 2 S 1 mol Na S 1 mol S 

2 

 

78.05 g 1 mol Na 2 

S 

= 0.0258 mol S 

d. molar mass of Na 2 S 2 O 3 = 158.1 g 

2.01 g Na 2 S 2 O 3 1 mol Na 2 S 2 O 3 

158.1 g 

 

2 mol S 

1 mol Na 2 

S 2 

O 3 

= 0.0254 mol S 

27. a. mass of Na present = 2 (22.99 g) = 45.98 g 

mass of S present = 32.07 g = 32.07 g 

mass of O present = 4 (16.00 g) = 64.00 g 

molar mass of Na 2 SO 4 = 

142.05 g 

% Na = 

45.98 g Na 

142.05 g 

100 = 32.37% Na 

% S = 32.07 g S 

142.05 g 

% O = 64.00 g O 

142.05 g 

100 = 22.58% S 

100 = 45.05% O 

b. mass of Na present = 2 (22.99 g) = 45.98 g 



molar mass of Na 2 SO 3 = 

126.05 g 

% Na = 

45.98 g Na 

126.05 g 

100 = 36.48% Na 

% S = 32.07 g S 

126.05 g 

100 = 25.44% S 

% O = 48.00 g O 

126.05 g 

100 = 38.08% O 





c. mass of Na present = 2 (22.99 g) = 45.98 g 


molar mass of Na 2 S = 

78.05 g 

% Na = 

45.98 g Na 

78.05 g 

100 = 58.91% Na 

% S = 32.07 g S 

78.05 g 

100 = 41.09% S 

d. mass of Na present = 2 (22.99 g) = 45.98 g 

mass of S present = 2 (32.07 g) = 64.14 g 


molar mass of Na 2 S 2 O 3 = 

158.12 g 

% Na = 

45.98 g Na 

158.12 g 

100 = 29.08% Na 

% S = 64.14 g S 

158.12 g 

% O = 48.00 g O 

158.12 g 

100 = 40.56% S 

100 = 30.36% O 

e. mass of K present = 3 (39.10 g) = 117.3 g 

mass of P present = 30.97 g = 30.97 g 


molar mass of K 3 PO 4 = 

212.3 g 

% K = 117.3 g K 

212.3 g 

% P = 30.97 g P 

212.3 g 

% O = 64.00 g O 

212.3 g 

100 = 55.25% K 

100 = 14.59% P 

100 = 30.15% O 

f. mass of K present = 2 (39.10 g) = 78.20 g 

mass of H present = 1.008 g = 1.008 g 



molar mass of K 2 HPO 4 = 

174.178 g = 174.18 g 

% K = 78.20 g K 

174.18 g 

% H = 1.008 g H 

174.18 g 

100 = 44.90% K 

100 = 0.5787% H 





% P = 30.97 g P 

174.18 g 

% O = 64.00 g O 

174.18 g 

100 = 17.78% P 

100 = 36.74% O 

g. mass of K present = 39.10 g = 39.10 g 

mass of H present = 2 (1.008 g) = 2.016 g 



molar mass of KH 2 PO 4 = 

136.09 g 

% K = 39.10 g K 

136.09 g 

% H = 2.016 g H 

136.09 g 

% P = 30.97 g P 

136.09 g 

% O = 64.00 g O 

136.09 g 

100 = 28.73% K 

100 = 1.481% H 

100 = 22.76% P 

100 = 47.03% O 

h. mass of K present = 3 (39.10) g = 117.3 g 


molar mass of K 3 P = 148.27 g = 148.3 g 

% K = 117.3 g K 

148.3 g 

% P = 30.97 g P 

148.3 g 

100 = 79.10% K 

100 = 20.88% P 

28. a. molar mass of CuBr 2 = 223.4 g 

63.55 g Cu 

% Cu = 100 = 28.45% Cu 

223.4 g 

b. molar mass of CuBr = 143.5 g 

63.55 g Cu 

% Cu = 100 = 44.29% Cu 

143.5 g 

c. molar mass of FeCl 2 = 126.75 g 

55.85 g Fe 

% Fe = 100 = 44.06% Fe 

126.75 g 





d. molar mass of FeCl 3 = 162.2 g 

55.85 g Fe 

% Fe = 100 = 34.43% Fe 

162.2 g 

e. molar mass of CoI 2 = 312.7 g 

58.93 g Co 

% Co = 100 = 18.85% Co 

312.7 g 

f. molar mass of CoI 3 = 439.6 g 

58.93 g Co 

% Co = 100 = 13.41% Co 

439.6 g 

g. molar mass of SnO = 134.7 g 

118.7 g Sn 

% Sn = 100 = 88.12% Sn 

134.7 g 

h. molar mass of SnO 2 = 150.7 g 

118.7 g Sn 

% Sn = 100 = 78.77% Sn 

150.7 g 

29. a. molar mass of C 6 H 10 O 4 = 146.1 g 

% C = 72.06 g C 

100 = 49.32% C 

146.1 g 

b. molar mass of NH 4 NO 3 = 80.05 g 

% N = 28.02 g N 

100 = 35.00% N 

80.05 g 

c. molar mass of C 8 H 10 N 4 O 2 = 194.2 g 

% C = 96.08 g C 

100 = 49.47% C 

194.2 g 

d. molar mass of ClO 2 = 67.45 g 

35.45 g Cl 

% Cl = 100 = 52.56% Cl 

67.45 g 

e. molar mass of C 6 H 11 OH = 100.2 g 

% C = 72.06 g C 

100 = 71.92% C 

100.2 g 

f. molar mass of C 6 H 12 O 6 = 180.2 g 

% C = 72.06 g C 

100 = 39.99% C 

180.2 g 





g. molar mass of C 20 H 42 = 282.5 g 

% C = 240.2 g C 

100 = 85.03% C 

282.5 g 

h. molar mass of C 2 H 5 OH = 46.07 g 

% C = 24.02 g C 

100 = 52.14% C 

46.07 g 

30. a. molar mass of NH 4 Cl = 53.49 g 

molar mass of NH + 4 ion = 18.04 g 

% NH + 4 = 18.04 g NH 4 

53.49 g 

b. molar mass of CuSO 4 = 159.62 

molar mass of Cu 2+ ion = 63.55 g 

+ 

100 = 33.73% NH 4 

+ 

% Cu 2+ = 

63.55 g Cu2+ 

159.62 g 

100 = 39.81% Cu 2+ 

c. molar mass of AuCl 3 = 303.4 g 

molar mass of Au 3+ ion = 197.0 g 

% Au 3+ = 

197.0 g Au3+ 

303.4 g 

100 = 64.93% Au 3+ 

d. molar mass of AgNO 3 = 169.9 g 

molar mass of Ag + ion = 107.9 g 

% Ag + = 

107.9 g Ag+ 

169.9 g 

100 = 63.51% Ag + 

31. To determine the empirical formula of a new compound, the composition of the compound 

by mass must be known. To determine the molecular formula of the compound, the molar 

mass of the compound must also be known. 

32. The empirical formula represents the smallest whole number ratio of the elements present in a 

compound. The molecular formula indicates the actual number of atoms of each element 

found in a molecule of the substance. 

33. a. NaO 

b. C 4 H 3 O 2 

c. C 12 H 12 N 2 O 3 is already the empirical formula 

d. C 2 H 3 Cl 





34. a. yes (each of these has empirical formula CH) 

b. no (the number of hydrogen atoms is wrong) 

c. yes (both have empirical formula NO 2 ) 

d. no (the number of hydrogen and oxygen atoms is wrong) 

35. 0.1929 g C 1 mol C 

12.01 g C 

= 0.01606 mol C 

0.01079 g H 1 mol H 

1.008 g H 

0.08566 g O 1 mol O 

16.00 g O 

= 0.01070 mol H 

= 0.005354 mol O 

0.1898 g Cl 

1 mol Cl 

35.45 g Cl 

= 0.005354 mol Cl 

Dividing each number of moles by the smallest number of moles gives: 

0.01606 mol C 

= 3.000 mol C 

0.005354 

0.01070 mol H 

= 1.999 mol H 

0.005354 

0.005354 mol O 

= 1.000 mol O 

0.005354 

0.005354 Cl 

= 1.000 mol Cl 

0.005354 

The empirical formula is C 3 H 2 OCl 

1 mol 

36. 2.514 g Ca 

= 0.06272 mol Ca 

40.08 g Ca 

The increase in mass represents the oxygen with which the calcium reacted: 

1.004 g O 1 mol O 

= 0.06275 mol O 

16.00 g O 

Since we have effectively the same number of moles of Ca and O, the empirical formula must 

be CaO. 

37. Consider having 100.0 g of the compound. Then the percentages of the elements present are 

numerically equal to their masses in grams. 

1 mol 

58.84 g Ba = 0.4286 mol Ba 

137.3 g Ba 

13.74 g S 1 mol 

= 0.4284 mol S 

32.07 g S 





27.43 g O 1 mol O 

= 1.714 mol O 

16.00 g O 

Dividing each number of moles by the smallest number of moles (0.4284 mol S) gives 

0.4286 mol Ba 

0.4284 

0.4284 mol S 

0.4284 

= 1.000 mol Ba 

= 1.000 mol S 

1.714 mol O 

= 4.001 mol O 

0.4284 

The empirical formula is BaSO 4 . 

38. The mass of chlorine involved in the reaction is 6.280 – 1.271 = 5.009 g Cl 

1.271 g Al 

5.009 g Cl 

1 mol Al 

26.98 g Al 

1 mol Cl 

34.45 g Cl 

= 0.04711 mol Al 

= 0.1413 mol Cl 

Dividing each of these numbers of moles by the smaller (0.04711 mol Al) shows that the 

empirical formula is AlCl 3 . 

39. Consider 100.0 g of the compound. 

55.06 g Co 

1 mol 

58.93 g Co 

= 0.9343 mol Co 

If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass. 

44.94 g S 1 mol 

= 1.401 mol S 

32.07 g S 

Dividing each number of moles by the smaller (0.9343 mol Co) gives 

0.9343 mol Co 

0.9343 

= 1.000 mol Co 

1.401 mol S 

= 1.500 mol S 

0.9343 

Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for 

the compound as Co 2 S 3 . 

40. 2.461 g Ca 

4.353 g Cl 

1 mol Ca 

40.08 g Ca 

1 mol Cl 

35.45 g Cl 

= 0.06140 mol Ca 

= 0.1228 mol Cl 

Dividing each of the number of moles by the smaller (0.06140 mol Ca) shows that the 

empirical formula is CaCl 2 . 





41. 10.00 g Cu 

1 mol 

63.55 g Cu 

= 0.1574 mol Cu 

2.52 g O 1 mol O 

= 0.158 mol O 

16.00 g O 

The numbers are almost equal: the empirical formula is CuO. 


1 mol Cu 

33.88 g Cu = 0.5331 mol Cu 

63.55 g Cu 

14.94 g N 1 mol N 

14.01 g N 

51.18 g O 1 mol O 

16.00 g O 

= 1.066 mol N 

= 3.199 mol O 

Dividing each number of moles by the smaller number of moles (0.5331 mol Cu) gives 

0.5331 mol Cu 

0.5331 

1.066 mol N 

0.5331 

= 1.000 mol Cu 

= 2.000 mol N 

3.199 mol O 

= 6.001 mol O 

0.5331 

The empirical formula is CuN 2 O 6 [i.e., Cu(NO 3 ) 2 ] 

43. Compound 1: Assume 100.0 g of the compound. 

1 mol Na 

83.12 g Na = 3.615 mol Na 

22.99 g Na 

16.88 g N 1 mol N 

= 1.205 mol Na 

14.01 g N 

Dividing each number of moles by the smaller (1.205 mol Na) indicates that the formula of 

Compound 1 is Na 3 N. 

Compound 2: Assume 100.0 g of the compound. 

1 mol Na 

35.36 g Na = 1.538 mol Na 

22.99 g Na 

64.64 g N 1 mol N 

14.01 g N 

= 4.614 mol N 

Dividing each number of moles by the smaller (1.538 mol Na) indicates that the formula of 

Compound 2 is NaN 3 . 





44. The empirical formula of a compound represents only the smallest whole number relationship 

between the number and type of atoms in a compound, whereas the molecular formula 

represents the actual number of atoms of each type in a true molecule of the substance. Many 

compounds (for example, H 2 O) may have the same empirical and molecular formulas. 

45. If only the empirical formula is known, the molar mass of the substance must be determined 

before the molecular formula can be calculated. 

46. empirical formula mass of CH 2 O = 30 g 

molar mass 

n = 

empirical formula mass = 90 g 

30 g = 3 

molecular formula is (CH 2 O) 3 = C 3 H 6 O 3 . 

47. empirical formula mass of CH 2 = 14 

molar mass 

n = 


14 g = 6 

molecular formula is (CH 2 ) 6 = C 6 H 12 . 

48. empirical formula mass of CH 4 O = 32.04 g 

molar mass 

n = 


32.04 g = 6 

molecular formula is (CH 4 O) 6 = C 6 H 24 O 6 . 


42.87 g C 1 mol C 

= 3.570 mol C 

12.01 g C 

3.598 g H 1 mol H 

1.008 g H 

28.55 g O 1 mol O 

16.00 g O 

25.00 g N 1 mol N 

14.01 g N 

= 3.569 mol H 

= 1.784 mol O 

= 1.784 mol N 

Dividing each number of moles by the smallest number of moles (1.784 mol O) gives 

3.570 mol C 

1.784 

= 2.001 mol C 

3.569 mol H 

1.784 

= 2.001 mol H 

1.784 mol O 

1.784 

= 1.000 mol O 





1.784 mol N 

= 1.000 mol N 

1.784 

The empirical formula of the compound is C 2 H 2 ON, empirical formula mass of 

C 2 H 2 ON = 56 

molar mass 

n = 


= 3 

56 g 

The molecular formula is (C 2 H 2 ON) 3 = C 6 H 6 O 3 N 3 . 


65.45 g C 1 mol C 

= 5.450 mol C 

12.01 g C 

5.492 g H 1 mol H 

= 5.448 mol H 

1.008 g H 

29.06 g O 1 mol O 

= 1.816 mol O 

16.00 g O 


5.450 mol C 

= 3.001 mol C 

1.816 

5.448 mol H 

= 3.000 mol H 

1.816 

1.816 mol O 

= 1.000 mol O 

1.816 

The empirical formula is C 3 H 3 O, and the empirical formula mass is approximately 55 g. 

molar mass 

n = 


= 2 

55 g 

The molecular formula is (C 3 H 3 O) 2 = C 6 H 6 O 2 . 

51. [1] c [6] d 

[2] e [7] a 

[3] j [8] g 

[4] h [9] i 

[5] b [10] f 

52. 5.00 g Al 0.185 mol 1.12 10 23 atoms 

0.140 g Fe 0.00250 mol 1.51 10 21 atoms 

2.7 10 2 g Cu 4.3 mol 2.6 10 24 atoms 

0.00250 g Mg 1.03 10 –4 mol 6.19 10 19 atoms 

0.062 g Na 2.7 10 –3 mol 1.6 10 21 atoms 

3.95 10 –18 g U 1.66 10 –20 mol 1.00 10 4 atoms 





53. 4.24 g 0.0543 mol 3.27 10 22 molec. 3.92 10 23 atoms 

4.04 g 0.224 mol 1.35 10 23 molec. 4.05 10 23 atoms 



126 g 6.99 mol 4.21 10 24 molec. 1.26 10 25 atoms 

0.267 g 0.00927 mol 5.58 x 10 21 molec. 3.35 x 10 22 atoms 

54. magnesium/nitrogen compound: 

mass of nitrogen contained = 1.2791 g – 0.9240 g = 0.3551 g N 

1 mol Mg 

0.9240 g Mg = 0.03801 mol Mg 

24.31 g Mg 

0.3551 g N 1 mol N 

14.01 g N 

= 0.02535 mol N 

Dividing each number of moles by the smaller number of moles gives 

0.03801 mol Mg 

0.02535 

= 1.499 mol Mg 

0.02535 mol N 

= 1.000 mol N 

0.02535 

Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg 3 N 2 . 

magnesium–oxygen compound: 

Consider 100.0 g of this compound. 

1 mol Mg 

60.31 g Mg = 2.481 mol Mg 

24.31 g Mg 

39.69 g O 1 mol O 

16.00 g O 

= 2.481 mol O 

Since the numbers of moles are the same, the compound contains the same relative number of 

Mg and O atoms: the empirical formula is MgO. 

55. For the first compound (restricted amount of oxygen): 

2.118 g Cu 

1 mol Cu 

63.54 g Cu 

= 0.03333 mol Cu 

0.2666 g O 1 mol O 

16.00 g O 

= 0.01666 mol O 

Since the number of moles of Cu (0.03333 mol) is twice the number of moles of 

O (0.01666 mol), the empirical formula is Cu 2 O. 





For the second compound (stream of pure oxygen): 

1 mol Cu 

2.118 g Cu = 0.03333 mol Cu 

63.54 g Cu 

0.5332 g O 1 mol O 

16.00 g O 

= 0.03333 mol O 

Since the numbers of moles are the same, the empirical formula is CuO. 

56. We need to find the subscripts for C a H b O c S d , where a:b:c:d is the mole ratio of C:H:O:S 

atoms. We are given the following relationships: 

b = 2a 

a = c 

b = 8d 

We can see that d will be the smallest subscript. Thus, let d = x and we get 

b = 8x 

2a = 8x or a = 4x 

since a = c, c = 4x 

Thus, we have C 4x H 8x O 4x S x . The empirical formula is C 4 H 8 O 4 S, which has a molar mass of 

about 152 g/mol [4(12.01) + 8(1.008) + 4(16.00) + 1(32.07)]. We are given that the molar 

mass of the compound is 152 g/mol. Thus, the molecular formula is C 4 H 8 O 4 S. 

57. 2.24 g Co 

55.85 g Fe 

58.93 g Co 

= 2.12 g Fe 

58. 2.24 g Fe 

58.93 g Co 

55.85 g Fe 

= 2.36 g Co 


1 mol Cu 

25.45 g Cu = 0.4005 mol Cu 

63.55 g Cu 

12.84 g S 1 mol S 

32.07 g S 

= 0.4004 mol S 

4.036 g H 1 mol H 

1.008 g H 

= 4.004 mol H 

57.67 g O 1 mol O 

16.00 g O 

= 3.604 mol O 





Dividing each number of moles by the smallest number of moles gives 

0.4005 mol Cu 

0.4004 

0.4004 mol S 

0.4004 

= 1.000 mol Cu 

= 1.000 mol S 

4.004 mol H 

0.4004 

= 10.00 mol H 

3.604 mol O 

= 9.001 mol O 

0.4004 

The empirical formula is CuSH 10 O 9 (which is usually written as CuSO 4 5H 2 O). 

60. 0.2990 g C 1 mol C 

12.01 g C 

= 0.02490 mol C 

0.05849 g H 1 mol H 

1.008 g H 

0. 2318 g N 1 mol N 

14.01 g N 

= 0.05803 mol H 

= 0.01655 mol N 

0.1328 g O 1 mol O 

16.00 g O 

= 0.008300 mol O 


0.02490 mol C 

0.008300 

0.05803 mol H 

0.008300 

0.01655 mol N 

0.008300 

= 3.000 mol C 

= 6.992 mol H 

= 1.994 mol N 

0.008300 mol O 

= 1.000 mol O 

0.008300 

The empirical formula is C 3 H 7 N 2 O. 

61. Mass of oxygen in compound = 4.33 g – 4.01 g = 0.32 g O 

4.01 g Hg 

1 mol Hg 

200.6 g Hg 

= 0.0200 mol Hg 

0.32 g O 1 mol O 

16.00 g O 

= 0.020 mol O 

Since the numbers of moles are equal, the empirical formula is HgO. 





62. Assume we have 100.0 g of the compound. 

1 mol Ba 

65.95 g Ba = 0.4803 mol Ba 

137.3 g Ba 

1 mol Cl 

34.05 g Cl = 0.9605 mol Cl 

35.45 g Cl 

Dividing each of these numbers of moles by the smaller number gives 

0.4803 mol Ba 

= 1.000 mol Ba 

0.4803 

0.9605 mol Cl 

= 2.000 mol Cl 

0.4803 

The empirical formula is then BaCl 2 . 

63. We need to find the subscripts for H x N y O z , where x:y:z is the mole ratio of H:N:O atoms. We 

are given that x = 4.0 moles H. 

To solve for y, we use 

56.0 g N 1 mol N 

= 4.00 moles N 

14.01 g N 

To solve for z, we use 

7.2 x 10 24 1 mol O 

atoms O 

= 12 moles O 

6.022 x 10 23 atoms 

The mole ratio is 4:4:12 or 1:1:3. The empirical formula is HNO 3 . 

64. For every 100.0 g of A 2 O, we have 63.7 g A and 36.3 g O. 

36.3 g O 1 mol O 

= 2.27 mol O 

16.00 g O 

The ratio between A and O is 2:1; with 2.27 mol O we must have 4.54 mol A. 

63.7 g A 

= 14.0 g/mol 

4.54 mol A 

Thus, A must be nitrogen. The compound is N 2 O. 

65. [1] g [6] i 

[2] c [7] f 

[3] b [8] h 

[4] a [9] e 

[5] j [10] d

Ch 6 Book Answers

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