Chapter 6 Chemical Composition - Francis Howell …fhsdfhhs.sharpschool.net/UserFiles/Servers/Server_999044/File/Staff...World of Chemistry 37 Copyright Houghton Mifflin Company. All

World of Chemistry 37 Copyright Houghton Mifflin Company. All rights reserved.

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

1.63 g) = 613.49 = 613 corks

613 stoppers would weigh (613 stoppers 4.31 g

1 stopper) = 2644 g = 2640 g

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

1.63 g = 2644 g = 2640 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. 1.261 104 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 1023 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

38 Chapter 6

World of Chemistry Copyright Houghton Mifflin Company. All rights reserved.

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 1 Au atom

197.97 amu = 100.00 Au atoms = 100 Au atoms

d. 19,695 amu 1 Xe atom

131.3 amu = 150.0 Xe atoms = 150 Xe atoms

e. 3588.3 amu 1 Al atom

26.98 amu = 133.0 Al atoms = 133 Al atoms

7. 8274 amu 1 S atom

32.07 amu = 258 S atoms

5.213 1024 S atoms 32.07 amu

1 S atom = 1.672 1026 amu

8. 1.204 1024

9. Avogadro’s number (6.022 1023)

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 40.08 amu

24.31 amu = 20.05 g Ca

24.31 g Mg 40.08 amu

24.31 amu = 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 1023 O atoms

1 O atom 16.00 g O

6.022 x 1023

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

1 mol = 4 g H

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

Chemical Composition 39


14. a. 26.2 g Au 1 mol Au

197.0 g = 0.133 mol Au

b. 41.5 g Ca 1 mol Ca

40.08 g = 1.04 mol Ca

c. 335 mg Ba 1 g

103 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

106

g

1 mol Ni

58.70 g = 5.20 10–13 mol Ni

f. 1.00 lb Fe 453.59 g

1 lb

1 mol Fe

55.85 g = 8.12 mol Fe

g. 12.01 g C 1 mol C

12.01 g = 1.000 mol C

15. a. 2.00 mol Fe 55.85 g

1 mol = 112 g Fe

b. 0.521 mol Ni 58.70 g

1 mol = 30.6 g Ni

c. 1.23 10–3 mol Pt 195.1 g

1 mol = 0.240 g Pt

d. 72.5 mol Pb 207.2 g

1 mol = 1.50 104 g Pb

e. 0.00102 mol Mg 24.31 g

1 mol = 0.0248 g Mg

f. 4.87 103 mol Al 26.98 g

1 mol = 1.31 105 g Al

g. 211.5 mol Li 6.941 g

1 mol = 1468 g Li

h. 1.72 10–6 mol Na 22.99 g

1 mol = 3.95 10–5 g Na

16. a. 0.00103 g Co 6.022 x 10

23 Co atoms

58.93 g Co = 1.05 1019 Co atoms

b. 0.00103 mol Co 6.022 x 10

23 Co atoms

1 mol = 6.20 1020 Co atoms

40 Chapter 6


c. 2.75 g Co 1 mol

58.93 g Co = 0.0467 mol Co

d. 5.99 1021 Co atoms 1 mol

6.022 x 1023

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 10

23 Co atoms

1 mol Co = 2.55 1024 Co atoms

g. 4.23 g Co 6.022 x 10

23 Co atoms

58.93 g Co = 4.32 1022 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 Na3N = 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 CS2 = 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 NH4Br = 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 C2H6O = 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 H2SO3 = 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 H2SO4 = 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(ClO4)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 MgSO4 = 120.38 g

c. mass of 1 mol Pb = 207.2 g = 207.2 g mass of 2 mol Cl = 2 (35.45 g) = 79.90 g molar mass of PbCl2 = 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(NO3)2 = 187.57 g

e. mass of 1 mol Sn = 118.7 g = 118.7 g mass of 4 mol Cl = 4 (35.45 g) = 141.80 g molar mass of SnCl4 = 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 C6H6O = 94.11 g

21. a. molar mass of SO3 = 80.07 g

49.2 mg SO3

1 g

1000 mg

1 mol

80.07 g = 6.14 10–4 mol SO3

b. molar mass of PbO2 = 239.2 g

7.44 x 104 kg PbO2

1000 g

1 kg

1 mol

239.2 g = 3.11 105 mol PbO2

c. molar mass of CHCl3 = 119.37 g

59.1 g CHCl3

1 mol

119.37 g = 0.495 mol CHCl3

d. molar mass of C2H3Cl3 = 133.39 g

3.27 g C2H3Cl3

1 g

106

g

1 mol

133.39 g = 2.45 10–8 mol C2H3Cl3

e. molar mass of LiOH = 23.95 g

4.01 g LiOH 1 mol

23.95 g = 0.167 mol LiOH

42 Chapter 6


22. a. molar mass of NaH2PO4 = 120.0 g

4.26 10–3 g NaH2PO4

1 mol

120.0 g = 3.55 10–5 mol NaH2PO4

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 = 2.70 103 mol Fe

d. molar mass of SrF2 = 125.6 g

8.76 g SrF2

1 mol

125.6 g = 0.0697 mol SrF2

e. molar mass of Al = 26.98 g

1.26 104 g Al 1 mol

26.98 g = 467 mol Al

23. a. molar mass of AlI3 = 407.7 g

1.50 mol AlI3

407.7 g

1 mol = 612 g AlI3

b. molar mass of C6H6 = 78.11 g

1.91 10–3 mol C6H6

78.11 g

1 mol = 0.149 g C6H6

c. molar mass of C6H12O6 = 180.2 g

4.00 mol C6H12O6

180.2 g

1 mol = 721 g C6H12O6

d. molar mass of C2H5OH = 46.07 g

4.56 105 mol C2H5OH46.07 g

1 mol = 2.10 107 g C2H5OH

e. molar mass of Ca(NO3)2 = 164.1 g

2.27 mol Ca(NO3)2

164.1 g

1 mol = 373 g Ca(NO3)2



24. a. molar mass of CO2 = 44.01 g

1.27 mmol 1 mol

103 mmol

44.01 g

1 mol = 0.0559 g CO2

b. molar mass of NCl3 = 120.4 g

4.12 103 mol NCl3

120.4 g

1 mol = 4.96 105 g NCl3

c. molar mass of NH4NO3 = 80.05 g

0.00451 mol NH4NO3

80.05 g

1 mol = 0.361 g NH4NO3

d. molar mass of H2O = 18.02 g

18.0 mol H2O18.02 g

1 mol = 324 g H2O

e. molar mass of CuSO4 = 159.6 g

62.7 mol CuSO4

159.6 g

1 mol = 1.00 104 g CuSO4

25. a. 6.37 mol CO 6.022 x 10

23 molecules

1 mol = 3.84 1024 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 1023 molecules CO

c. molar mass of H2O = 18.02 g

2.62 10–6 g 6.022 x 10

23 molecules

18.02 g = 8.76 1016 molecules H2O

d. 2.62 10–6 mol 6.022 x 10

23 molecules

1 mol = 1.58 1018 molecules H2O

e. molar mass of C6H6 = 78.11 g

5.23 g 6.022 x 10

23 molecules

78.11 g = 4.03 1022 molecules C6H6

44 Chapter 6


26. a. molar mass of Na2SO4 = 142.1 g

2.01 g Na2SO4

1 mol Na2SO

4

142.1 g

1 mol S

1 mol Na2SO

4

= 0.0141 mol S

b. molar mass of Na2SO3 = 126.1 g

2.01 g Na2SO3

1 mol Na2SO

3

126.1 g

1 mol S

1 mol Na2SO

3

= 0.0159 mol S

c. molar mass of Na2S = 78.05 g

2.01 g Na2S1 mol Na

2S

78.05 g

1 mol S

1 mol Na2S

= 0.0258 mol S

d. molar mass of Na2S2O3 = 158.1 g

2.01 g Na2S2O3

1 mol Na2S

2O

3

158.1 g

2 mol S

1 mol Na2S

2O

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 Na2SO4 = 142.05 g

% Na = 45.98 g Na

142.05 g 100 = 32.37% Na

% S = 32.07 g S

142.05 g 100 = 22.58% S

% O = 64.00 g O

142.05 g 100 = 45.05% O

b. 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 = 3 (16.00 g) = 48.00 g molar mass of Na2SO3 = 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 mass of S present = 32.07 g = 32.07 g molar mass of Na2S = 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 mass of O present = 3 (16.00 g) = 48.00 g molar mass of Na2S2O3 = 158.12 g

% Na = 45.98 g Na

158.12 g 100 = 29.08% Na

% S = 64.14 g S

158.12 g 100 = 40.56% S

% O = 48.00 g O

158.12 g 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 mass of O present = 4 (16.00 g) = 64.00 g molar mass of K3PO4 = 212.3 g

% K = 117.3 g K

212.3 g 100 = 55.25% K

% P = 30.97 g P

212.3 g 100 = 14.59% P

% O = 64.00 g O

212.3 g 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 mass of P present = 30.97 g = 30.97 g mass of O present = 4 (16.00 g) = 64.00 g molar mass of K2HPO4 = 174.178 g = 174.18 g

% K = 78.20 g K

174.18 g 100 = 44.90% K

% H = 1.008 g H

174.18 g 100 = 0.5787% H

46 Chapter 6


% P = 30.97 g P

174.18 g 100 = 17.78% P

% O = 64.00 g O

174.18 g 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 mass of P present = 30.97 g = 30.97 g mass of O present = 4 (16.00 g) = 64.00 g molar mass of KH2PO4 = 136.09 g

% K = 39.10 g K

136.09 g 100 = 28.73% K

% H = 2.016 g H

136.09 g 100 = 1.481% H

% P = 30.97 g P

136.09 g 100 = 22.76% P

% O = 64.00 g O

136.09 g 100 = 47.03% O

h. mass of K present = 3 (39.10) g = 117.3 g mass of P present = 30.97 g = 30.97 g molar mass of K3P = 148.27 g = 148.3 g

% K = 117.3 g K

148.3 g 100 = 79.10% K

% P = 30.97 g P

148.3 g 100 = 20.88% P

28. a. molar mass of CuBr2 = 223.4 g

% Cu = 63.55 g Cu

223.4 g 100 = 28.45% Cu

b. molar mass of CuBr = 143.5 g

% Cu = 63.55 g Cu

143.5 g 100 = 44.29% Cu

c. molar mass of FeCl2 = 126.75 g

% Fe = 55.85 g Fe

126.75 g 100 = 44.06% Fe



d. molar mass of FeCl3 = 162.2 g

% Fe = 55.85 g Fe

162.2 g 100 = 34.43% Fe

e. molar mass of CoI2 = 312.7 g

% Co = 58.93 g Co

312.7 g 100 = 18.85% Co

f. molar mass of CoI3 = 439.6 g

% Co = 58.93 g Co

439.6 g 100 = 13.41% Co

g. molar mass of SnO = 134.7 g

% Sn = 118.7 g Sn

134.7 g 100 = 88.12% Sn

h. molar mass of SnO2 = 150.7 g

% Sn = 118.7 g Sn

150.7 g 100 = 78.77% Sn

29. a. molar mass of C6H10O4 = 146.1 g

% C = 72.06 g C

146.1 g 100 = 49.32% C

b. molar mass of NH4NO3 = 80.05 g

% N = 28.02 g N

80.05 g 100 = 35.00% N

c. molar mass of C8H10N4O2 = 194.2 g

% C = 96.08 g C

194.2 g 100 = 49.47% C

d. molar mass of ClO2 = 67.45 g

% Cl = 35.45 g Cl

67.45 g 100 = 52.56% Cl

e. molar mass of C6H11OH = 100.2 g

% C = 72.06 g C

100.2 g 100 = 71.92% C

f. molar mass of C6H12O6 = 180.2 g

% C = 72.06 g C

180.2 g 100 = 39.99% C

48 Chapter 6


g. molar mass of C20H42 = 282.5 g

% C = 240.2 g C

282.5 g 100 = 85.03% C

h. molar mass of C2H5OH = 46.07 g

% C = 24.02 g C

46.07 g 100 = 52.14% C

30. a. molar mass of NH4Cl = 53.49 g molar mass of NH4

+ ion = 18.04 g

% NH4+ =

18.04 g NH4

+

53.49 g 100 = 33.73% NH4

+

b. molar mass of CuSO4 = 159.62

molar mass of Cu2+ ion = 63.55 g

% Cu2+ = 63.55 g Cu

2+

159.62 g 100 = 39.81% Cu2+

c. molar mass of AuCl3 = 303.4 g

molar mass of Au3+ ion = 197.0 g

% Au3+ =197.0 g Au

3+

303.4 g 100 = 64.93% Au3+

d. molar mass of AgNO3 = 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. C4H3O2

c. C12H12N2O3 is already the empirical formula

d. C2H3Cl



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 NO2)

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 = 0.01070 mol H

0.08566 g O 1 mol O

16.00 g O = 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

0.005354 = 3.000 mol C

0.01070 mol H

0.005354 = 1.999 mol H

0.005354 mol O

0.005354 = 1.000 mol O

0.005354 Cl

0.005354 = 1.000 mol Cl

The empirical formula is C3H2OCl

36. 2.514 g Ca 1 mol

40.08 g Ca = 0.06272 mol Ca

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

1.004 g O 1 mol O

16.00 g O = 0.06275 mol 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.

58.84 g Ba 1 mol

137.3 g Ba = 0.4286 mol Ba

13.74 g S 1 mol

32.07 g S = 0.4284 mol S

50 Chapter 6


27.43 g O 1 mol O

16.00 g O = 1.714 mol O

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

0.4286 mol Ba

0.4284 = 1.000 mol Ba

0.4284 mol S

0.4284 = 1.000 mol S

1.714 mol O

0.4284 = 4.001 mol O

The empirical formula is BaSO4.

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

1.271 g Al 1 mol Al

26.98 g Al = 0.04711 mol Al

5.009 g Cl 1 mol Cl

34.45 g Cl = 0.1413 mol Cl

Dividing each of these numbers of moles by the smaller (0.04711 mol Al) shows that the empirical formula is AlCl3.

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

32.07 g S = 1.401 mol 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

0.9343 = 1.500 mol S

Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for the compound as Co2S3.

40. 2.461 g Ca 1 mol Ca

40.08 g Ca = 0.06140 mol Ca

4.353 g Cl 1 mol Cl

35.45 g Cl = 0.1228 mol Cl

Dividing each of the number of moles by the smaller (0.06140 mol Ca) shows that the empirical formula is CaCl2.



41. 10.00 g Cu 1 mol

63.55 g Cu = 0.1574 mol Cu

2.52 g O 1 mol O

16.00 g O = 0.158 mol O

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


33.88 g Cu 1 mol Cu

63.55 g Cu = 0.5331 mol Cu

14.94 g N 1 mol N

14.01 g N = 1.066 mol N

51.18 g O 1 mol O

16.00 g O = 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 = 1.000 mol Cu

1.066 mol N

0.5331 = 2.000 mol N

3.199 mol O

0.5331 = 6.001 mol O

The empirical formula is CuN2O6 [i.e., Cu(NO3)2]

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

83.12 g Na 1 mol Na

22.99 g Na = 3.615 mol Na

16.88 g N 1 mol N

14.01 g N = 1.205 mol Na

Dividing each number of moles by the smaller (1.205 mol Na) indicates that the formula of Compound 1 is Na3N.

Compound 2: Assume 100.0 g of the compound.

35.36 g Na 1 mol Na

22.99 g Na = 1.538 mol 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 NaN3.

52 Chapter 6


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, H2O) 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 CH2O = 30 g

n = molar mass

empirical formula mass =

90 g

30 g = 3

molecular formula is (CH2O)3 = C3H6O3.

47. empirical formula mass of CH2 = 14

n = molar mass


84 g

14 g = 6

molecular formula is (CH2)6 = C6H12.

48. empirical formula mass of CH4O = 32.04 g

n = molar mass


192 g

32.04 g = 6

molecular formula is (CH4O)6 = C6H24O6.


42.87 g C 1 mol C

12.01 g C = 3.570 mol C

3.598 g H 1 mol H

1.008 g H = 3.569 mol H

28.55 g O 1 mol O

16.00 g O = 1.784 mol O

25.00 g N 1 mol N

14.01 g N = 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

1.784 = 1.000 mol N

The empirical formula of the compound is C2H2ON, empirical formula mass of

C2H2ON = 56

n = molar mass


168 g

56 g = 3

The molecular formula is (C2H2ON)3 = C6H6O3N3.


65.45 g C 1 mol C

12.01 g C = 5.450 mol C

5.492 g H 1 mol H

1.008 g H = 5.448 mol H

29.06 g O 1 mol O

16.00 g O = 1.816 mol O


5.450 mol C

1.816 = 3.001 mol C

5.448 mol H

1.816 = 3.000 mol H

1.816 mol O

1.816 = 1.000 mol O

The empirical formula is C3H3O, and the empirical formula mass is approximately 55 g.

n = molar mass


110 g

55 g = 2

The molecular formula is (C3H3O)2 = C6H6O2.

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 1023 atoms

0.140 g Fe 0.00250 mol 1.51 1021 atoms

2.7 102 g Cu 4.3 mol 2.6 1024 atoms

0.00250 g Mg 1.03 10–4 mol 6.19 1019 atoms

0.062 g Na 2.7 10–3 mol 1.6 1021 atoms

3.95 10–18 g U 1.66 10–20 mol 1.00 104 atoms

54 Chapter 6


53. 4.24 g 0.0543 mol 3.27 1022 molec. 3.92 1023 atoms

4.04 g 0.224 mol 1.35 1023 molec. 4.05 1023 atoms

1.98 g 0.0450 mol 2.71 1022 molec. 8.13 1022 atoms

45.9 g 1.26 mol 7.59 1023 molec. 1.52 1024 atoms

126 g 6.99 mol 4.21 1024 molec. 1.26 1025 atoms

0.267 g 0.00927 mol 5.58 x 1021 molec. 3.35 x 1022 atoms

54. magnesium/nitrogen compound:

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

0.9240 g Mg 1 mol Mg

24.31 g Mg = 0.03801 mol 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

0.02535 = 1.000 mol N

Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg3N2.

magnesium–oxygen compound:

Consider 100.0 g of this compound.

60.31 g Mg 1 mol Mg

24.31 g Mg = 2.481 mol 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 Cu2O.



For the second compound (stream of pure oxygen):

2.118 g Cu 1 mol Cu

63.54 g Cu = 0.03333 mol 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 CaHbOcSd, 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 C4xH8xO4xSx. The empirical formula is C4H8O4S, 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 C4H8O4S.

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


25.45 g Cu 1 mol Cu

63.55 g Cu = 0.4005 mol 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

56 Chapter 6


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

0.4005 mol Cu

0.4004 = 1.000 mol Cu

0.4004 mol S

0.4004 = 1.000 mol S

4.004 mol H

0.4004 = 10.00 mol H

3.604 mol O

0.4004 = 9.001 mol O

The empirical formula is CuSH10O9 (which is usually written as CuSO4 5H2O).

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 = 0.05803 mol H

0. 2318 g N 1 mol N

14.01 g N = 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 = 3.000 mol C

0.05803 mol H

0.008300 = 6.992 mol H

0.01655 mol N

0.008300 = 1.994 mol N

0.008300 mol O

0.008300 = 1.000 mol O

The empirical formula is C3H7N2O.

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.

65.95 g Ba 1 mol Ba

137.3 g Ba = 0.4803 mol Ba

34.05 g Cl 1 mol Cl

35.45 g Cl = 0.9605 mol Cl

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

0.4803 mol Ba

0.4803 = 1.000 mol Ba

0.9605 mol Cl

0.4803 = 2.000 mol Cl

The empirical formula is then BaCl2.

63. We need to find the subscripts for HxNyOz, 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

14.01 g N = 4.00 moles N

To solve for z, we use

7.2 x 1024 atoms O 1 mol O

6.022 x 1023

atoms = 12 moles O

The mole ratio is 4:4:12 or 1:1:3. The empirical formula is HNO3.

64. For every 100.0 g of A2O, we have 63.7 g A and 36.3 g O.

36.3 g O 1 mol O

16.00 g O = 2.27 mol 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

4.54 mol A = 14.0 g/mol

Thus, A must be nitrogen. The compound is N2O.

65. [1] g [6] i [2] c [7] f [3] b [8] h [4] a [9] e [5] j [10] d

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