QUANTITATIVE COMPOSITION OF COMPOUNDSfaculty.chemeketa.edu/lemme/Solutions/ch07.pdfCHAPTER 7...

C H A P T E R 7

QUANTITATIVE COMPOSITION OF COMPOUNDS

SOLUTIONS TO REVIEW QUESTIONS

1. A mole is an amount of substance containing the same number of particles as there are atoms in exactly

12 g of carbon-12.

It is Avogadro’s number (6.022� 1023) of anything (atoms, molecules, ping-pong balls, etc.).

2. A mole of gold (197.0 g) has a higher mass than a mole of potassium (39.10 g).

3. Both samples (Au and K) contain the same number of atoms. (6.022� 1023).

4. A mole of gold atoms contains more electrons than a mole of potassium atoms, as each Au atom has

79 e�, while each K atom has only 19 e�.

5. The molar mass of an element is the mass of one mole (or 6.022� 1023 atoms) of that element.

6. No. Avogadro’s number is a constant. The mole is defined as Avogadro’s number of C-12 atoms.

Changing the atomic mass to 50 amu would change only the size of the atomic mass unit, not Avogadro’s

number.

7. 6.022� 1023

8. There are Avogadro’s number of particles in one mole of substance.

9. (a) A mole of oxygen atoms (O) contains 6.022� 1023 atoms.

(b) A mole of oxygen molecules (O2) contains 6.022� 1023 molecules.

(c) A mole of oxygen molecules (O2) contains 1.204� 1024 atoms.

(d) A mole of oxygen atoms (O) has a mass of 16.00 grams.

(e) A mole of oxygen molecules (O2) has a mass of 32.00 grams.

10. 6.022� 1023 molecules in one molar mass of H2SO4.

4.215� 1024 atoms in one molar mass of H2SO4.

11. The molecular formula represents the total number of atoms of each element in a molecule. The

empirical formula represents the lowest number ratio of atoms of each element in a molecule.

12. Choosing 100.0 g of a compound allows us to simply drop the % sign and use grams instead of percent.

- 46 -

SOLUTIONS TO EXERCISES

1. Molar masses

(a) KBr 1 K 39.10 g

1 Br 79.90 g

119.0

(b) Na2SO4 2 Na 45.98 g

1 S 32.07 g

4 O 64.00 g

142.1 g

(c) Pb(NO3)2 1 Pb 207.2 g

2 N 28.02 g

6 O 96.00 g

331.2 g

(d) C2H5OH 2 C 24.02 g

6 H 6.048 g

1 O 16.00 g

46.07 g

(e) HC2H3O2 4 H 4.032 g

2 C 24.02 g

2 O 32.00 g

60.05 g

(f) Fe3O4 3 Fe 167.6 g

4 O 64.00 g

231.6 g

(g) C12H22O11 12 C 144.1 g

22 H 22.18 g

11 O 176.0 g

342.3 g

(h) Al2(SO4)3 2 Al 53.96 g

3 S 96.21 g

12 O 192.0 g

342.2 g

(i) (NH4)2HPO4 9 H 9.072 g

2 N 28.02 g

1 P 30.97 g

4 O 64.00 g

132.1 g

- 47 -

- Chapter 7 -

2. Molar masses

(a) NaOH 1 Na 22.99 g

1 O 16.00 g

1 H 1.008 g

40.00 g

(b) Ag2CO3 2 Ag 215.8 g

1 C 12.01 g

3 O 48.00 g

275.8 g

(c) Cr2O3 104.0 g

3 O 48.00 g

152.0 g

(d) (NH4)2CO3 2 N 28.02 g

8 H 8.064 g

1 C 12.01 g

3 O 48.00 g

96.09 g

(e) Mg(HCO3)2 1 Mg 24.31 g

2 H 2.016 g

2 C 24.02 g

6 O 96.00 g

146.3 g

(g) C6H5COOH 7 C 84.07 g

6 H 6.048 g

2 O 32.00 g

122.1 g

(g) C6H12O6 6 C 72.06 g

12 H 12.10 g

6 O 96.00 g

180.2 g

(h) K4Fe(CN)6 4 K 156.4 g

1 Fe 55.85 g

6 C 72.06 g

6 N 84.06 g

368.4 g

(i) BaCl2�2 H2O 1 Ba 137.3 g

2 Cl 70.90 g

4 H 4.032 g

2 O 32.00 g

244.2 g

- 48 -

- Chapter 7 -

2 Cr

3. Moles of atoms.

(a) 22:5 g Znð Þ 1 mol Zn

65:39 g Zn

� �¼ 0:344 mol Zn

(b) 0:688 gMgð Þ 1 mol Mg

24:31 gMg

� �¼ 2:83� 10�2 mol Mg

(c) 4:5� 1022atoms Cu� � 1 mol Cu

6:022� 1023 atoms Cu

� �¼ 7:5� 10�2mol Cu

(d) 382 g Coð Þ 1 mol Co

58:93 g Co

� �¼ 6:48 mol Co

(e) 0:055 g Snð Þ 1 mol Sn

118:7 g Sn

� �¼ 4:6� 10�4 mol Sn

(f) 8:5� 1024 molecules N2

� � 2 atoms N

1molecule N2

� �1 mol N atoms

6:022� 1023 atoms N

� �¼ 28 mol N atoms

4. Number of moles.

(a) 25:0 g NaOHð Þ 1 mol NaOH

40:00 g NaOH

� �¼ 0:625 mol NaOH

(b) 44:0 g Br2ð Þ 1 mol Br2

159:8 g Br2

� �¼ 0:275 mol Br2

(c) 0:684 gMgCl2ð Þ 1 mol MgCl2

95:21 gMgCl2

� �¼ 7:18� 10�3mol MgCl2

(d) 14:8 g CH3OHð Þ 1 mol CH3OH

32:04 g CH3OH

� �¼ 0:462 mol CH3OH

(e) 2:88 g Na2SO4ð Þ 1 mol Na2SO4

142:1 g Na2SO4

� �¼ 2:03� 10�2mol Na2SO4

(f) 4:20 lb ZnI2ð Þ 453:6 g

1 lb

� �1 mol ZnI2

319:2 g ZnI2

� �¼ 5:97 mol ZnI2

5. Number of grams.

(a) 0:550 mol Auð Þ 197:0 g Au

1 mol Au

� �¼ 108 g Au

(b) 15:8 mol H2Oð Þ 18:02 g H2O

mol H2O

� �¼ 285 g H2O

(c) 12:5 mol Cl2ð Þ 70:90 g Cl2mol Cl2

� �¼ 886 g Cl2

(d) 3:15 mol NH4NO3ð Þ 80:05 g NH4NO3

mol NH4NO3

� �¼ 252 g NH4NO3

- 49 -

- Chapter 7 -

6. Number of grams.

(a) 4:25� 10�4mol H2SO4

� � 98:09 g H2SO4

mol H2SO4

� �¼ 0:0417 g H2SO4

(b) 4:5� 1022 molecules CCl4� � 1 mol

6:022� 1023 molecules

� �153:8 g CCl4mol CCl4

� �¼ 11 g CCl4

(c) 0:00255 mol Tið Þ 47:87 g Ti

mol Ti

� �¼ 0:122 g Ti

(d) 1:5� 1016 atoms S� � 32:07 g S

6:022� 1023atoms S

� �¼ 8:0� 10�7g S

7. Number of molecules

(a) 2:5 mol S8ð Þ 6:022� 1023 molecules

mol

� �¼ 1:5� 1024 molecules S8

(b) 7:35 mol NH3ð Þ 6:022� 1023 molecules

mol

� �¼ 4:43� 1024 molecules NH3

(c) 17:5 g C2H5OHð Þ 6:022� 1023 molecules

46:07 g C2H5OH

� �¼ 2:29� 1023 molecules C2H5OH

(d) 255 g Cl2ð Þ 6:022� 1023 molecules

70:90 g Cl2

� �¼ 1:91� 1024 molecules Cl2

8. Number of molecules

(a) 9:6 mol C2H4ð Þ 6:022� 1023 molecules

mol

� �¼ 5:8� 1024 molecules C2H4

(b) 2:76 mol N2Oð Þ 6:022� 1023 molecules

mol

� �¼ 1:66� 1024 molecules N2O

(c) 23:2 g CH3OHð Þ 6:022� 1023 molecules

32:04 g CH3OH

� �¼ 4:36� 1023 molecules CH3OH

(d) 32:7 g CCl4ð Þ 6:022� 1023 molecules

153:8 g CCl4

� �¼ 1:28� 1023 molecules CCl4

9. Number of atoms

(a) 25 molecules P2O5ð Þ 7 atoms

1 molecules

� �¼ 1:8� 102 atoms

(b) 3:62 mol O2ð Þ 6:022� 1023 molecules

mol

� �2 atoms

1 molecule

� �¼ 4:36� 1024 atoms

(c) 12:2 mol CS2ð Þ 6:022� 1023 molecules

mol

� �3 atoms

1 molecule

� �¼ 2:20� 1025 atoms

- 50 -

- Chapter 7 -

(d) 1:25 g Nað Þ 6:022� 1023 atoms

22:99 g Na

� �¼ 3:27� 1022 atoms

(e) 2:7 g CO2ð Þ 6:022� 1023 molecules

44:01 g CO2

� �3 atoms

1 molecule

� �¼ 1:1� 1023 atoms

(f) 0:25 g CH4ð Þ 6:022� 1023 molecules

16:04 g CH4

� �5 atoms

1 molecule

� �¼ 4:7� 1022 atoms

10. (a) 2 molecules CH3COOHð Þ 8 atoms

1 molecule

� �¼ 16 atoms

(b) 0:75 mol C2H6ð Þ 6:022� 1023 molecules

mol

� �8 atoms

1 molecule

� �¼ 3:6� 1024 atoms

(c) 25 mol H2Oð Þ 6:022� 1023 molecules

mol

� �3 atoms

1 molecule

� �¼ 4:5� 1025 atoms

(d) 92:5 g Auð Þ 6:022� 1023 atoms

197:0 g Au

� �¼ 2:83� 1023 atoms

(e) 75 g PCl3ð Þ 6:022� 1023 molecules

137:3 g PCl3

� �4 atoms

1 molecule

� �¼ 1:3� 1024 atoms

(f) 15 g C6H12O6ð Þ 6:022� 1023 molecules

180:2 g C6H12O6

� �24 atoms

1 molecule

� �¼ 1:2� 1024 atoms

11. Number of grams.

(a) 1 atom Heð Þ 4:003 g

6:022� 1023 atoms

� �¼ 6:647� 10�24 g He

(b) 15 atoms Cð Þ 12:01 g

6:022� 1023 atoms

� �¼ 2:991� 10�22 g C

(c) 4 molecules N2O5ð Þ 108:0 g N2O5


� �¼ 7:175� 10�22 g N2O5

(d) 11 molecules C6H5NH2ð Þ 93:13 g C6H5NH2


� �¼ 1:701� 10�21 g C6H5NH2

12. (a) 1 atom Xeð Þ 131:3 g Xe

6:022� 1023 atoms

� �¼ 2:180� 10�22 g Xe

(b) 22 atoms Clð Þ 35:45 g Cl

6:022� 1023 atoms

� �¼ 1:295� 10�21 g Cl

- 51 -

- Chapter 7 -

(c) ð9 molecules CH3COOH60:05 g CH3COOH


� �¼ 8:975� 10�22 g CH3COOH

(d) 15 molecules C4H4O2 NH2ð Þ2� � 116:1 g C4H4O2 NH2ð Þ2


� �¼ 2:892� 10�21 g C4H4O2 NH2ð Þ2

13. (a) 25 kg CO2ð Þ 1000 g

kg

� �1 mol CO2

44:01 g CO2

� �¼ 5:7� 102 mol CO2

(b) 5 atoms Pbð Þ 1 mol Pb

6:022� 1023 atoms

� �¼ 8� 10�24 mol Pb

(c) 6 mol O2ð Þ 6:022� 1023 molecules O2

mol O2

� �2 atoms

1 molecule O2

� �¼ 7� 1024 atoms

(d) 25 molecules P4ð Þ 123:9 g P4

6:022� 1023 molecules P4

� �¼ 5:1� 10�21 g P4

14. (a) 275 atomsWð Þ 1 molW

6:022� 1023 atoms

� �¼ 4:57� 10�22 molW

(b) 95 mol H2Oð Þ 18:02 g H2O

mol H2O

� �1 kg

1000 g

� �¼ 1:7 kg H2O

(c) 12 molecules SO2ð Þ 64:07 g SO2

6:022� 1023 molecules SO2

� �¼ 1:3� 10�21 g SO2

(d) 25 mol Cl2ð Þ 6:022� 1023 molecules Cl2

mol Cl2

� �2 atoms

1 molecule

� �¼ 3:0� 1025 atoms

15. One molecule of tetraphosphorus decoxide (P4O10) contains:

(a) 1 molecule P4O10ð Þ 1 mol P4O10


� �¼ 1:661� 10�24mol P4O10

(b) 1:661� 10�24 mol P4O10

� � 283:9 g P4O10

mol

� �¼ 4:714� 10�22 g P4O10

(c) 1 molecule P4O10ð Þ 4 P atoms

1 molecule P4O10

� �¼ 4 atoms P

(d) 1 molecule P4O10ð Þ 10 O atoms

1 molecule P4O10

� �¼ 10 atoms O

(e) 1 molecule P4O10ð Þ 14 atoms

1 molecule P4O10

� �¼ 14 total atoms

- 52 -

- Chapter 7 -

16. 125 grams of disulfur decofluoride (S2F10) contains:

(a) 125 g S2F10ð Þ 1 mol S2F10

254:1 g S2F10

� �¼ 0:492 mol S2F10

(b) 0:492 mol S2F10ð Þ 6:022� 1023 molecules

mol S2F10

� �¼ 2:96� 1023 molecules S2F10

(c) 0:492 mol S2F10ð Þ 6:022� 1023 molecules

mol S2F10

� �12 atoms

1 molecule S2F10

� �¼ 3:56� 1024

(d) 0:492 mol S2F10ð Þ 6:022� 1023 molecules

mol S2F10

� �2 S atoms

1 molecule S2F10

� �¼ 5:93� 1023 atoms S

(e) 0:492 mol S2F10ð Þ 6:022� 1023 molecules

mol S2F10

� �10 F atoms

1 molecule S2F10

� �¼ 2:96� 1024 atoms F

17. Atoms of hydrogen in:

(a) 25 molecules C6H5CH3ð Þ 8 H atoms

molecule C6H5CH3

� �¼ 2:0� 102 atoms H

(b) 3:5 mol H2CO3ð Þ 6:022� 1023 molecules

mol H2CO3

� �2 H atoms

molecule H2CO3

� �¼ 4:2� 1024 atoms H

(c) 36 g CH3CH2OHð Þ 6:022� 1023 molecules

46:07 g CH3CH2OH

� �6 H atoms

molecule CH3CH2OH

� �¼ 2:8� 1024 atoms H

18. Atoms of hydrogen in:

(a) 23 molecules CH3CH2COOHð Þ 6 atoms H

1molecule CH3CH2COOH

� �¼ 1:4� 102 atoms H

(b) 7:4 mol H3PO4ð Þ 6:022� 1023 molecules

mol H3PO4

� �3 atoms H

1molecule H3PO4

� �¼ 1:3� 1025 atoms H

(c) 57 g C6H5ONH2ð Þ 6:022� 1023 molecules

109:1 g C6H5ONH2

� �7 atoms H

1molecule C6H5ONH2

� �¼ 2:2� 1024 atoms H

19. The number of grams of:

(a) silver in 25.0 g AgBr

25:0 g AgBrð Þ 107:9 g Ag

187:8 g AgBr

� �¼ 14:4 g Ag

(b) nitrogen in 6.34 mol (NH4)3PO4

6:34 mol NH4ð Þ3PO4

� � 42:03 g N

mol NH4ð Þ3PO4

� �¼ 266 g N

- 53 -

- Chapter 7 -

(c) oxygen in 8.45� 1022 molecules SO3

The conversion is: molecules SO3�!mol SO3�! g O

8:45� 1022 molecules SO3

� � 1 mol


� �48:00 g O

mol SO3

� �¼ 6:74 g O

20. The number of grams of:

(a) chlorine in 5.00 g PbCl2

5:00 g PbCl2ð Þ 70:90 g Cl

278:1 g PbCl2

� �¼ 1:27 g Cl

(b) hydrogen in 4.50 g H2SO4

4:50 g H2SO4ð Þ 2:016 g H

98:09 g H2SO4

� �¼ 9:25� 10�2g H

(c) hydrogen in 5.45� 1022 molecules NH3

The conversion is: molecules NH3�!moles NH3�! g H

5:45� 1022 molecules NH3

� � 1 mol


� �3:024 g H

mol NH3

� �¼ 0:274 g H

21. Percent composition

(a) NaBr Na 22.99 g 22:99 g

102:9 g

� �100ð Þ ¼ 22:34%Na

Br 79.90 g

102.9 g

79:90 g

102:9 g

� �100ð Þ ¼ 77:65% Br

(b) KHCO3 K 39.10 g 39:10 g

100:1 g

� �100ð Þ ¼ 39:06%K

H 1.008 g

3 O 48.00 g

C 12.01 g 1:008 g

100:1 g

� �100ð Þ ¼ 1:007%H100.1 g

12:01 g

100:1 g

� �100ð Þ ¼ 12:00% C

48:00 g

100:1 g

� �100ð Þ ¼ 47:95%O

- 54 -

- Chapter 7 -

(c) FeCl3 Fe 55.85 g 55:85 g

162:3 g

� �100ð Þ ¼ 34:41% Fe

3 Cl 106.4 g

162.3 g

106:4 g

162:3 g

� �100ð Þ ¼ 65:56%Cl

(d) SiCl4 Si 28.09 g 28:09 g

169:9 g

� �100ð Þ ¼ 16:53% Si

4 Cl 141.8 g

169.9 g

141:8 g

169:9 g

� �100ð Þ ¼ 83:46%Cl

(e) Al2(SO4)3 2 Al 53.96 g 53:96 g

342:2 g

� �100ð Þ ¼ 15:77%Al

3 S 96.21 g

12 O 192.0 g 96:21 g

342:2 g

� �100ð Þ ¼ 28:12% S342.2 g

192:0 g

342:2 g

� �100ð Þ ¼ 56:11%O

(e) AgNO3 Ag 107.9 g 107:9 g

169:9 g

� �100ð Þ ¼ 63:51%Ag

N 14.01 g

3 O 48.00 g 14:01 g

169:9 g

� �100ð Þ ¼ 8:246%N169.9 g

48:00 g

169:9 g

� �100ð Þ ¼ 28:25%O

22. Percent composition

(a) ZnCl2 Zn 65.39 g 65:39 g

136:3 g

� �100ð Þ ¼ 47:98% Zn

2 Cl 70.90 g

136.3 g

70:90 g

136:3 g

� �100ð Þ ¼ 52:02%Cl

(b) NH4C2H3O2 N 14.01 g 14:01 g

77:09 g

� �100ð Þ ¼ 18:17%N

7 H 7.056 g

2 C 24.02 g

2 O 32.00 g 7:056 g

77:09 g

� �100ð Þ ¼ 9:153%H77.09 g

24:02 g

77:09 g

� �100ð Þ ¼ 31:16% C

32:00 g

77:09 g

� �100ð Þ ¼ 41:51%O

- 55 -

- Chapter 7 -

(c) MgP2O7 Mg 24.31 g 24:31 g

198:3 g

� �100ð Þ ¼ 12:26%Mg

2 P 61.94 g

7 O 112.0 g 61:94 g

198:3 g

� �100ð Þ ¼ 31:24% P198.3 g

112:0 g

198:3 g

� �100ð Þ ¼ 56:48%O

(d) (NH4)2SO4 2 N 28.02 g 28:02 g

132:2 g

� �100ð Þ ¼ 21:20%N

8 H 8.064 g

S 32.07 g

4 O 64.00 g 8:064 g

132:2 g

� �100ð Þ ¼ 6:100%H132.2 g

32:07 g

132:2 g

� �100ð Þ ¼ 24:26% S

64:00 g

132:2 g

� �100ð Þ ¼ 48:41%O

(e) Fe(NO3)3 Fe 55.85 g 55:85 g

241:9 g

� �100ð Þ ¼ 23:09% Fe

3 N 42.03 g

9 O 144.0 g

241.9 g 42:03 g

241:9 g

� �100ð Þ ¼ 17:37%N

144:0 g

241:9 g

� �100ð Þ ¼ 59:53%O

(f) ICl3 I 126.9 g 126:9 g

233:3 g

� �100ð Þ ¼ 54:39% I

3 Cl 106.4 g

233.3 g 106:4 g

233:3 g

� �100ð Þ ¼ 45:61%Cl

23. Percent of iron

(a) FeO Fe 55.85 g 55:85 g

71:85 g

� �100ð Þ ¼ 77:73% Fe

O 16.00 g

71.85 g

(b) Fe2O3 2 Fe 111.7 g 111:7 g

159:7 g

� �100ð Þ ¼ 69:94% Fe

3 O 48.00 g

159.7 g

(c) Fe3O4 3 Fe 167.6 g 167:6 g

231:6 g

� �100ð Þ ¼ 72:37% Fe

4 O 64.00 g

231.6 g

- 56 -

- Chapter 7 -

(d) K4Fe(CN)6 Fe 55.85 g 55:85 g

368:4 g

� �100ð Þ ¼ 15:16% Fe

4 K 156.4 g

6 C 72.06 g

6 N 84.06 g

368.4 g

24. Percent chlorine

(a) KCl K 39.10 g 35:45 g

74:55 g

� �100ð Þ ¼ 47:55% Cl

Cl 35.45 g

74.55 g

(b) BaCl2 Ba 137.3 g 70:90 g

208:2 g

� �100ð Þ ¼ 34:05%Cl

2 Cl 70.90 g

208.2 g

(c) SiCl4 Si 28.09 g 141:8 g

169:9 g

� �100ð Þ ¼ 83:46%Cl

4 Cl 141.8 g

169.9 g

(d) LiCl Li 6.941 g 35:45 g

42:39 g

� �100ð Þ ¼ 83:63% Cl

Cl 35.45 g

42.39 g

Highest % Cl is LiCl; lowest % Cl is in BaCl2

25. Percent composition of an oxide

25.75 g oxide 6:68 g

25:75 g

� �100ð Þ ¼ 25:9%N26.68 g N

19.07 g O19:02 g

25:75 g

� �100ð Þ ¼ 74:06%O

26. Percent Composition of an alcohol.

35.75 g alcohol 18:64 g

35:75 g

� �100ð Þ ¼ 52:14%C�18.64 g C

24.70 g H4:70 g

35:75 g

� �100ð Þ ¼ 13:1%H

12.41 g O

12:41 g

35:75 g

� �100ð Þ ¼ 34:71%O

27. (a) H2O has the higher percent Hydrogen

(b) N2O3 has the lower percent Nitrogen

(c) Both have the same percent Oxygen

- 57 -

- Chapter 7 -

28. (a) KClO3 (Because a K atom has more mass than a Na atom.)

(b) KHSO4 (Because a H atom has less mass than a K atom.)

(c) Na2CrO4 (Because only one Cr atom is present.)

29. Empirical formulas from percent composition.

(a) Step 1. Express each element as grams/100 g material.

63:6%N ¼ 63:6 g N=100 g material

36:4%O ¼ 36:4 g O=100 g material

Step 2. Calculate the relative moles of each element.

63:6 g Nð Þ 1 mol N

14:01 g N

� �¼ 4:54 mol N

36:4 g Oð Þ 1 mol O

16:00 g O

� �¼ 2:28 mol O

Step 3. Chance these moles to whole numbers by dividing each by the smaller number.

4:54 mol N

2:28¼ 1:99 mol N

2:28 mol O

2:28¼ 1:00 mol O

The simplest ratio of N:O is 2:1. The empirical formula, therefore, is N2O.

(b) 46.7% N, 53.3% O


14:01 g N

� �¼ 3:33 mol N

3:33 mol N

3:33¼ 1:00 mol N

53:3 g Nð Þ 1 mol O

16:00 g O

� �¼ 3:33 mol O

3:33 mol O

3:33¼ 1:00 mol O

The empirical formula is NO.

(c) 25.9% N, 71.4% O


14:01 g N

� �¼ 1:85 mol N

1:85 mol N

1:85¼ 1:00 mol N


16:00 g O

� �¼ 4:63 mol O

4:63 mol O

1:85¼ 2:5 mol O

Since these values are not whole numbers, multiply each by 2 to change them to whole numbers.

1:00 mol Nð Þ 2ð Þ ¼ 2:00 mol N; 2:5 mol Oð Þ 2ð Þ ¼ 5:00 mol O

The empirical formula is N2O5.

- 58 -

- Chapter 7 -

(d) 43.4% Na, 11.3% C, 45.3% O

43:4 g Nað Þ 1 mol Na

22:99 g Na

� �¼ 1:89 mol Na

1:89 mol Na

0:941¼ 2:01 mol Na

11:3 g Cð Þ 1 mol C

12:01 g C

� �¼ 0:941 mol C

0:941 mol C

0:941¼ 1:00 mol C


16:00 g O

� �¼ 2:83 mol O

2:83 mol O

0:941¼ 3:00 mol O

The empirical formula is Na2CO3.

(e) 18.8% Na, 29.0% Cl, 52.3% O

18:8 g Nað Þ 1 mol Na

22:99 g Na

� �¼ 0:818 mol Na

0:818 mol Na

0:818¼ 1:00 mol Na

29:0 g Clð Þ 1 mol Cl

35:45 g Cl

� �¼ 0:818 mol Cl

0:818 mol Cl

0:818¼ 1:00 mol Cl


16:00 g O

� �¼ 3:27 mol O

3:27 mol O

0:818¼ 4:00 mol O

The empirical formula is NaClO4.

(f) 72.02% Mn, 27.98% O

72:02 gMnð Þ 1 mol Mn

54:94 gMn

� �¼ 1:311 mol Mn

1:311 mol Mn

1:311¼ 1:000 mol Mn


16:00 g O

� �¼ 1:749 mol O

1:749 mol O

1:311¼ 1:334 mol O

Multiply both values by 3 to give whole numbers.

1:000 mol Mnð Þ 3ð Þ ¼ 3:000 mol Mn; 1:334 mol Oð Þ 3ð Þ ¼ 4:002 mol O

The empirical formula is Mn3O4.

30. Empirical formulas from percent composition.

(a) 64.1% Cu, 35.9% Cl

64:1 g Cuð Þ 1 mol Cu

63:55 g Cu

� �¼ 1:01 mol Cu

1:01 mol Cu

1:01¼ 1:00 mol Cu


35:45 g Cl

� �¼ 1:01 mol Cl

1:01 mol Cl

1:01¼ 1:00 mol Cl

The empirical formula is CuCl.

- 59 -

- Chapter 7 -

(b) 47.2% Cu, 52.8% Cl


63:55 g Cu

� �¼ 0:743 mol Cu

0:743 mol Cu

0:743¼ 1:00 mol Cu


35:45 g Cl

� �¼ 1:49 mol Cl

1:49 mol Cl

0:743¼ 2:01 mol Cl

The empirical formula is CuCl2.

(c) 51.9% Cr, 48.1% S

51:9 g Crð Þ 1 mol Cr

52:00 g Cr

� �¼ 0:998 mol Cr

0:998 mol Cr

0:998¼ 1:00 mol Cr

48:1 g Sð Þ 1 mol S

32:07 g S

� �¼ 1:50 mol S

1:50 mol S

0:998¼ 1:50 mol S

Multiply both values by 2 to give whole numbers.

1:00 mol Crð Þ 2ð Þ ¼ 2:00 mol Cr; 1:50 mol Sð Þ 2ð Þ ¼ 3:00 mol S

The empirical formula is Cr2S3.

(d) 55.3% K, 14.6% P, 30.1% O

55:3 g Kð Þ 1 mol K

39:10 g K

� �¼ 1:41 mol K

1:41 mol K

0:471¼ 2:99 mol K

14:6 g Pð Þ 1 mol P

30:97 g P

� �¼ 0:471 mol P

1:471 mol P

0:471¼ 1:00 mol P


16:00 g O

� �¼ 1:88 mol O

1:88 mol O

0:471¼ 3:99 mol O

The empirical formula is K3PO4.

(e) 38.9% Ba, 29.4% Cr, 31.7% O

38:9 g Bað Þ 1 mol Ba

137:3 g Ba

� �¼ 0:283 mol Ba

0:283 mol Ba

0:283¼ 1:00 mol Ba

29:4 g Crð Þ 1 mol Cr

52:00 g Cr

� �¼ 0:565 mol Cr

0:565 mol Cr

0:283¼ 2:00 mol Cr


16:00 g O

� �¼ 1:98 mol O

1:98 mol O

0:283¼ 7:00 mol O

The empirical formula is BaCr2O7.

- 60 -

- Chapter 7 -

(f) 3.99% P, 82.3% Br, 13.7% Cl


30:97 g P

� �¼ 0:129 mol P

0:129 mol P

0:129¼ 1:00 mol P

82:3 g Brð Þ 1 mol Br

79:90 g Br

� �¼ 1:03 mol Br

1:03 mol Br

0:129¼ 7:98 mol Br


35:45 g Cl

� �¼ 0:386 mol Cl

0:386 mol Cl

0:129¼ 2:99 mol Cl

The empirical formula is PBr8Cl3.

31. Empirical formula:

(a) 26:08 g Znð Þ 1 mol Zn

65:39 g

� �¼ 0:3988 mol Zn

0:3988 mol Zn

0:3988¼ 1:00 mol Zn


12:01 g

� �¼ 0:399 mol C

0:399 mol C

0:3988¼ 1:00 mol C


16:00 g

� �¼ 1:196 mol O

1:196 mol O

0:3988¼ 2:999 mol O

The empirical formula is ZnCO3

(b) 150.0 g compound�57:66 g C�7:26 g H85:1 g Cl


12:01 g C

� �¼ 4:801 mol C

4:801 mol C

2:40¼ 2:000 mol C

7:26 g Hð Þ 1 mol H

1:008 g H

� �¼ 7:20 mol H

7:20 mol H

2:40¼ 3:00 mol H


35:45 g

� �¼ 2:40 mol Cl

2:40 mol Cl

2:40¼ 1:00 mol Cl

The empirical formula is C2H3Cl

- 61 -

- Chapter 7 -

(c) 75.0 g Oxide � 42.0 gV¼ 33.0 g O

42:0 g Vð Þ 1 mol V

50:94 g V

� �¼ 0:824 mol V

0:824 mol V

0:824¼ 1:00 mol V


16:00 g O

� �¼ 2:06 mol O

2:06 mol O

0:824¼ 2:50 mol O

Multiplying both by 2 gives the empirical formula V2O5

(d) 67:35 g Nið Þ 1 mol Ni

58:69 g Ni

� �¼ 1:148 mol Ni

1:148 mol Ni

0:7649¼ 1:501 mol Ni


16:00 g O

� �¼ 3:060 mol O

3:060 mol O

0:7649¼ 4:001 mol O


30:97 g P

� �¼ 0:7649 mol P

0:7649 mol P

0:7649¼ 1:000 mol P

Multiplying all by 2 gives the empirical formula Ni3O8P2

32. Empirical formula

(a) 55:08 g Cð Þ 1 mol C

12:01 g C

� �¼ 4:586 mol C

4:586 mol C

0:7643 mol¼ 6:000 mol C


1:008 g H

� �¼ 3:82 mol H

3:82 mol H

0:7643¼ 5:00 mol H

61:07 g Brð Þ 1 mol Br

79:90 g Br

� �¼ 0:7643 mol Br

0:7643 mol Br

0:7643¼ 1:000 mol Br

The empirical formula is C6H5Br

(b) 65.2 g compound � 36.8 g Ag � 12.1 g Cl¼ 16.3 g O

36:8 g Agð Þ 1 mol Ag

107:9 g Ag

� �¼ 0:341 mol Ag

0:341 mol Ag

0:341¼ 1:00 mol Ag


35:45 g Cl

� �¼ 0:341 mol Cl

0:341mol Cl

0:341¼ 1:00 mol Cl


16:00 g O

� �¼ 1:02 mol O

1:02 mol O

0:341¼ 2:99 mol O

The empirical formula is AgClO3

- 62 -

- Chapter 7 -

(c) 25.25 g sulfide � 12.99 gV¼ 12.26 g S

12:99 g Vð Þ 1 mol V

50:94 g V

� �¼ 0:2550 mol V

0:2550 mol V

0:2550¼ 1:000 mol V


32:07 g S

� �¼ 0:3823 mol S

0:3823 mol S

0:2550¼ 1:499 mol S

Multiplying both by 2 gives the empirical formula V2S3

(d) 38:0 g Znð Þ 1 mol Zn

65:39 g

� �¼ 0:581 mol Zn

0:581 mol Zn

0:387¼ 1:50 mol Zn


30:97 g

� �¼ 0:387 mol P

0:387 mol P

0:387¼ 1:00 mol P

Multiplying both by 2 gives the empirical formula Zn3P2

33. 15.267 g sulfide � 12.272 g Au¼ 2.995 g S

12:272 g Auð Þ 1 mol Au

197:09 g Au

� �¼ 0:06229 mol Au

0:06229 mol Au

0:06229¼ 1:000 mol Au


32:07 g S

� �¼ 0:09339 mol S

0:09339 mol S

0:06229¼ 1:499 mol S

Multiplying both by 2 gives the empirical formula Au2S3

34. 10.724 g oxide � 7.143 g Ti¼ 3.581 g O

7:143 g Tið Þ 1 mol Ti

47:88 g Ti

� �¼ 0:1492 mol Ti

0:1492 mol Ti

0:1492¼ 1:000 mol Ti


16:00 g O

� �¼ 0:2238 mol O

0:2238 mol O

0:1492¼ 1:500 mol O

Multiplying both by 2 gives the empirical formula Ti2O3


2.775 g oxide � 2.465 g Cu¼ 0.310 g O


63:55 g Cu

� �¼ 0:03879 mol Cu

0:03879 mol Cu

0:0194¼ 2:00 mol Cu


16:00 g O

� �¼ 0:0194 mol O

0:0194 mol O

0:0194¼ 1:00 mol O

The empirical formula is Cu2O.

- 63 -

- Chapter 7 -


5.276 g compound � 3.898 g Hg¼ 1.378 g Cl

3:898 g Hgð Þ 1 mol Hg

200:6 g Hg

� �¼ 0:01943 mol Hg

0:01943 mol Hg

0:01943¼ 1:000 mol Hg


35:45 g Cl

� �¼ 0:03887 mol Cl

0:03887 mol Cl

0:01943¼ 2:001 mol Cl

The empirical formula is HgCl2.

37. Empirical and molecular formulas of benzoyl peroxide.

69.42% C, 4.16% H, 26.42% O; molar mass¼ 242 g


12:01 g C

� �¼ 5:780 mol C

5:780 mol C

1:651¼ 3:501 mol C


1:008 g H

� �¼ 4:13 mol H

4:13 mol H

1:651¼ 2:50 mol H


16:00 g O

� �¼ 1:651 mol O

1:651 mol O

1:651¼ 1:000 mol O

Multiplying all by 2 gives the empirical formula C7H5O2. The empirical formula mass is 121 g.

molar mass

empirical formula mass¼ 242 g

121 g¼ 2

The molecular formula is twice that of the empirical formula.

Molecular formula is (C7H5O2)2¼C14H10O4

38. Empirical and Molecular formulas of dixanthogen.

29.73% C, 4.16% H, 13.20% O, 52.91% S; molar mass¼ 242.4 g


12:01 g C

� �¼ 2:475 mol C

2:475 mol C

0:8250¼ 3:000 mol C


1:008 g H

� �¼ 4:13 mol H

4:13 mol H

0:8250¼ 5:01 mol H


16:00 g O

� �¼ 0:8250 mol O

0:8250 mol O

0:8250¼ 1:000 mol O


32:07 g S

� �¼ 1:650 mol S

1:650 mol S

0:8250¼ 2:000 mol S

- 64 -

- Chapter 7 -

The empirical formula is C3H5OS2. The empirical formula mass is 121.2 g.

molar mass

empirical formula mass¼ 242:4 g

121:2 g¼ 2

The molecular formula is twice that of the empirical formula

Molecular formula is (C3H5OS2)2¼C6H10O2S4

39. Molecular formula of ethanedioic acid

26.7% C, 2.24% H, 71.1% O; molar mass¼ 90.04

26:7 g C1mol C

12:01 g C

� �¼ 2:22 mol C

2:22 mol C

2:2¼ 1:0 mol C

2:2 g H1mol H

1:008 g H

� �¼ 2:2 mol H

2:2 mol H

2:2¼ 1:0 mol H

71:1 g O1mol O

16:00 g O

� �¼ 4:44 mol O

4:44 mol O

2:2¼ 2:0 mol O

The empirical formula is CHO2, making the empirical formula mass 45.02 g.

molar mass

mass of empirical formula¼ 90:04 g

45:02 g¼ 2


Molecular formula¼ (CHO2)2¼C2H2O4

40. Molecular formula of butyric acid

54.5% C, 9.2% H, 36.3% O; molar mass¼ 88.11


12:01 g C

� �¼ 4:54 mol C

4:54 mol C

2:27¼ 2:00 mol C


1:008 g H

� �¼ 9:1 mol H

9:1 mol H

2:27¼ 4:0 mol H


16:00 g O

� �¼ 2:27 mol O

2:27 mol O

2:27¼ 1:0 mol O

The empirical formula is C2H4O, making the empirical formula mass 44.05 g.

molar mass

mass of empirical formula¼ 88:11 g

44:05 g¼ 2


Molecular formula¼ (C2H4O)2¼C4H8O2

- 65 -

- Chapter 7 -

41. % nitrogen ¼ 12:04 g

39:54 g100ð Þ ¼ 30:45%

% oxygen ¼ 39:54 g� 12:04 g

39:54 g100ð Þ ¼ 69:55%

empirical formula: moles of nitrogen ¼ 12:04 g N

14:01 g=mol¼ 0:8594 mol N

moles of oxygen ¼ 27:50 g O

16:00 g=mol O¼ 1:719 mol O

relative number of nitrogen atoms ¼ 0:8594 mol

0:8594 mol¼ 1:000

relative number of oxygen atoms ¼ 1:719 mol

0:8594 mol¼ 2:00

empirical formula ¼ NO2

molecular formula: (molar mass of NO2) x¼ 92.02 g, 46.01 x¼ 92.02, x¼ 2

The molecular formula is twice the empirical formula.

molecular formula¼N2O4

42. Total mass of Cþ Hþ O ¼ 30:21 gþ 5:08 gþ 40:24 g ¼ 75:53 g

% carbon ¼ 30:21 g

75:53 g100ð Þ ¼ 40:0%

% hydrogen ¼ 5:08 g

75:53 g100ð Þ ¼ 6:73%

% oxygen ¼ 40:24 g

75:53 g100ð Þ ¼ 53:3%

empirical formula: moles of carbon ¼ 30:21 g C

12:01 g=mol¼ 2:515 mol C

moles of hydrogen ¼ 5:080 g H

1:008 g=mol¼ 5:03 mol H

moles of oxygen ¼ 40:24 g O

16:00 g=mol¼ 2:515 mol O

relative number of carbon atoms ¼ 2:515 mol

2:515 mol¼ 1:000

relative number of hydrogen atoms ¼ 5:03 mol

2:515 mol¼ 2:00

relative number of oxygen atoms ¼ 2:515 mol

2:515 mol¼ 1:000

empirical formula ¼ CH2O

molecular formula: (molar mass of CH2O) x¼ 180.18 g=mol,

30:03 g=molð Þx ¼ 180:18 g=mol,

x ¼ 180:18 g=mol

30:03 g=mol¼ 6

The molecular formula is six times the empirical formula.

molecular formula¼C6H12O6

- 66 -

- Chapter 7 -

43. What is compound XYZ3

X: 0:4004ð Þ 100:09 gð Þ ¼ 40:08 g calciumð ÞY: 0:1200ð Þ 100:09 gð Þ ¼ 12:01 g carbonð ÞZ: 0:4796ð Þ 100:09 gð Þ ¼ 48:00 g;

48:00 g

3¼ 16:00 g oxygenð Þ

Elements determined from atomic masses in the periodic table.

XYZ3¼CaCO3

44. What is compound X2(YZ3)3

X: 0:1912ð Þ 282:23 gð Þ ¼ 53:96 g

2¼ 26:98 g aluminumð Þ

Y: 0:2986ð Þ 282:23 gð Þ ¼ 84:27 g

3¼ 28:09 g siliconð Þ

Z: 0:5102ð Þ 282:23 gð Þ ¼ 143:99 g


Elements determined from atomic masses in the periodic table.

X2(YZ3)3¼Al2(SiO3)3

45. 0:350 mol P4ð Þ 6:022� 1023molecules

mol

� �4 atoms P

molecule P4

� �¼ 8:43� 1023 atoms P

46. 10:0 g Kð Þ 1 mol K

39:10 g K

� �1 mol Na

1 mol K

� �22:99 g Na

mol Na

� �¼ 5:88 g Na

47. 1:79� 10�23 g=atom� �

6:022� 1023 atoms=molar mass� � ¼ 10:8 g=molar mass

48. 5 lb C12H22O11ð Þ 453:6 g

1 lb

� �6:022� 1023 molecules

342:3 g

� �¼ 4� 1024 molecules

49. 6:022� 1023 sheets� � 4:60 cm

500 sheets

� �1 m

100 cm

� �¼ 5:54� 1019m

50.6:022� 1023dollars

6:1� 109people

� �¼ 9:9� 1013dollars=person

51. The conversion is: mi3�! ft3�! in:3�! cm3�! drops

(a) 1 mi3� � 5280 ft

mile

� �312:0 in:

ft

� �32:54 cm

inch

� �320 drops

1:0 cm3

� �¼ 8� 1016 drops

(b) 6:022� 1023drops� � 1 mi3

8� 1016drops

� �¼ 8� 106mi3

- 67 -

- Chapter 7 -

52. 1 mol Ag¼ 107.9 g Ag

(a) 107:9 g Agð Þ 1 cm3

10:5 g

� �¼ 10:3 cm3 volume of cubeð Þ

(b) 10:3 cm3 ¼ volume of cube ¼ sideð Þ3

side ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10:3 cm33

p¼ 2:18 cm

53. (a) Determine the molar mass of each compound.

CO2, 44.01 g; O2, 32.00 g; H2O, 18.02 g; CH3OH, 32.04 g. The 1.00 gram sample with the lowest

molar mass will contain the most molecules. Thus, H2O will contain the most molecules.

(b) 1:00 g H2Oð Þ 1 mol

18:02 g

� �3ð Þ 6:022� 1023atoms� �

mol

� �¼ 1:00� 1023atoms

1:00 g CH3OHð Þ 1 mol

32:04 g

� �6ð Þ 6:022� 1023atoms� �

mol

� �¼ 1:13� 1023atoms

1:00 g CO2ð Þ 1 mol

44:01 g

� �3ð Þ 6:022� 1023atoms� �

mol

� �¼ 4:10� 1022atoms

1:00 g O2ð Þ 1 mol

32:00 g

� �2ð Þ 6:022� 1023atoms� �

mol

� �¼ 3:76� 1022atoms

The 1.00 g sample of CH3OH contains the most atoms

54. 1 mol Fe2S3¼ 207.9 g Fe2S3¼ 6.022� 1023 formula units

6:022� 1023atoms� � 1 formula unit

5 atoms

� �207:9 g Fe2S3

6:022� 1023formula units

� �¼ 41:58 g Fe2S3

55. The conversion is g P�!mol P�!mol Ca�! g Ca


30:97 g P

� �3 mol Ca

2 mol P

� �40:08 g Ca

1 mol Ca

� �¼ 1:94 g Ca

1.94 g Ca combines with 1.00 g P.

56. Grams of Fe per ton of ore that contains 5% FeSO4.

The conversion is: ton�! lb�! g�! g FeSO4�! g Fe

1:0 tonð Þton

453:6 g

lb

� �0:05 FeSO4ð Þ 55:85 g Fe

151:9 g FeSO4

� �4

1.0 ton of iron ore contains 2� 104 g Fe.

57. From the formula, 2 Li (13.88 g) combine with 1 S (32.07 g).

13:88 g Li

32:07 g S

� �20:0 g Sð Þ ¼ 8:66 g Li

- 68 -

- Chapter 7 -

2000 lb��

¼ 1.7� 10 g Fe

58. (a) HgCO3 Hg 200.6 g 200:6 g Hg

260:6 g

� �100ð Þ ¼ 76:98%Hg

C 12.01 g

3 O 48.00 g

260.6 g

(b) Ca(ClO3)2 6 O 96.00 g 96:00 g O

207:0 g

� �100ð Þ ¼ 46:38%O

2 Cl 70.90 g

Ca 40.08 g

207.0 g

(c) C10H14N2 2 N 28.02 g 28:02 g N

162:6 g

� �100ð Þ ¼ 17:27%N

10 C 120.1 g

14 H 14.11 g

162.2 g

(d) C55H72MgN4O5

Mg 24.31 g 24:31 gMg

893:5 g

� �100ð Þ ¼ 2:721%Mg

55 C 660.55 g

72 H 72.58 g

4 N 56.04 g

5 O 80.00 g

893.5 g

59. According to the formula, 1 mol (65.39 g) Zn combines with 1 mol (32.07 g) S.

19:5 g Znð Þ 32:07 g S

65:39 g Zn

� �¼ 9:56 g S

19.5 g Zn require 9.56 g S for complete reaction. Therefore, there is not sufficient S present (9.40 g) to

react with the Zn.

60. Percent composition of C21H28O3

21 C 252.2 g 252:2 g C

328:4 g

� �100ð Þ ¼ 76:80%C

28 H 28.22 g

3 O 48.00 g 28:22 g H

328:4 g

� �100ð Þ ¼ 8:593%H

328.4 g

48:00 g O

328:4 g

� �100ð Þ ¼ 14:62%O

- 69 -

- Chapter 7 -

61. Percent composition of C17H21NO �HCl17 C 204.2 g 204:2 g C

291:8 g

� �100ð Þ ¼ 69:98%C

22 H 22.18 g

N 14.01 g 22:18 g H

291:8 g

� �100ð Þ ¼ 7:60%H

O 16.00 g

Cl 35.45 g 14:01 g N

291:8 g

� �100ð Þ ¼ 4:80%N291.8 g

16:00 g O

291:8 g

� �100ð Þ ¼ 5:48%O

35:45 g Cl

291:8 g

� �100ð Þ ¼ 12:15% Cl

62. Percent composition of sucrose

12 C 144.1 g 144:1 g C

342:3 g

� �100ð Þ ¼ 42:10% C

22 H 22.18 g

11 O 176.0 g 22:18 g H

342:3 g

� �100ð Þ ¼ 6:480%H342.3 g

176:0 g O

342:3 g

� �100ð Þ ¼ 51:42%O

63. Molecular formula of aspirin

60.0% C, 4.48% H, 35.5% O; molar mass of aspirin¼ 180.2


12:01 g C

� �¼ 5:00 mol C

5:00 mol C

2:22¼ 2:25 mol C


1:008 g H

� �¼ 4:44 mol H

4:44 mol H

2:22¼ 2:00 mol H


16:00 g O

� �¼ 2:22 mol O

2:22 mol O

2:22¼ 1:00 mol O

Multiplying each by 4 give the empirical formula C9H8O4. The empirical formula mass is 180.2 g. Since

the empirical formula mass equals the molar mass, the molecular formula is the same as the empirical

formula, C9H8O4.

64. Calculate the percent oxygen in Al2(SO4)3.

2 Al 53.96 g 192:0 g

342:2 g

� �100ð Þ ¼ 56:11%O

3 S 96.21 g

12 O 192.0 g Now take 56:11% of 8:50 g342.2 g 8:50 g Al2 SO4ð Þ3

� �0:5611ð Þ ¼ 4:77 g O

- 70 -

- Chapter 7 -

65. Empirical formula of gallium arsenide; 48.2% Ga, 51.8% As

48:2 g Gað Þ 1 mol Ga

69:72 g Ga

� �¼ 0:691 mol Ga

0:691 mol Ga

0:691¼ 1:00 mol Ga

51:8 g Asð Þ 1 mol As

74:92 g As

� �¼ 0:691 mol As

0:691 mol As

0:691¼ 1:00 mol As

The empirical formula is GaAs.

66. Empirical formula of calcium tartrate; 25.5% C, 2.1% H, 21.3% Ca, 51.0% O.


12:01 g C

� �¼ 2:12 mol C

2:212 mol C

0:531¼ 3:99 mol C


1:008 g H

� �¼ 2:1 mol H

2:1 mol H

0:531¼ 4:0 mol H

21:2 g Cað Þ 1 mol C

40:08 g Ca

� �¼ 0:531 mol Ca

0:529 mol Ca

0:531¼ 1:00 mol Ca


16:00 g O

� �¼ 3:19 mol O

3:19 mol O

0:531¼ 6:01 mol O

The empirical formula is C4H4CaO6

67. (a) 7.79% C, 92.21% Cl


12:01 g C

� �¼ 0:649 mol C

0:649 mol C

0:649¼ 1:00 mol C


35:45 g Cl

� �¼ 2:601 mol Cl

2:601 mol Cl

0:649¼ 4:01 mol Cl

The empirical formula is CCl4. The empirical formula mass is 153.8 which equals the molar mass,

therefore the molecular formula is CCl4.

(b) 10.13% C, 89.87% Cl


12:01 g C

� �¼ 0:8435 mol C

0:8435 mol C

0:8435¼ 1:000 mol C


35:45 g Cl

� �¼ 2:535 mol Cl

2:535 mol Cl

0:8435¼ 3:005 mol Cl

The empirical formula is CCl3. The empirical formula mass is 118.4 g.

molar mass


118:4 g¼ 1:999


Molecular formula¼C2Cl6.

- 71 -

- Chapter 7 -

(c) 25.26% C, 74.74% Cl


12:01 g C

� �¼ 2:103 mol C

2:103 mol C

2:103¼ 1:000 mol C


35:45 g Cl

� �¼ 2:108 mol Cl

2:103 mol Cl

2:108¼ 1:002 mol Cl

The empirical formula is CCl. The empirical formula mass is 47.46 g.

molar mass


47:46 g¼ 6:000

The molecular formula is six times that of the empirical formula.

Molecular formula¼C6Cl6.

(d) 11.25% C, 88.75% Cl


12:01 g C

� �¼ 0:9367 mol C

0:9367 mol C

0:9367¼ 1:000 mol C


35:45 g Cl

� �¼ 2:504 mol Cl

2:504 mol Cl

0:9367¼ 2:673 mol Cl

Multiplying each by 3 give the empirical formula C3Cl8. The empirical formula mass is 319.6.

Since the molar mass is also 319.6 the molecular formula is C3Cl8.

68. The conversion is: s�!min�! hr�! day�! yr

6:022� 1023s� � 1 min

60 s

� �1 hr

60 min

� �1 day

24 hr

� �1 year

365 days

� �¼ 1:910� 1016years

69. The conversion is: g�!mol�! atom


63:55 g Cu

� �6:022� 1023atoms

mol

� �¼ 2:4� 1022atoms Cu

70. The conversion is: molecules�!mol�! g 1 trillion ¼ 1012

1000:� 1012molecules C3H8O3

� � 1 mol

6:022� 1023molecules

� �92:09 g C3H8O3

mol C3H8O3

� �¼ 1:529� 10�7g C3H8O3

71. 6:1� 109people� � 1 mol people

6:022� 1023people

� �¼ 1:0� 10�14mol of people

- 72 -

- Chapter 7 -

72. Empirical formula 23.3% Co, 25.3% Mo, 51.4% Cl

23:3 g Coð Þ 1 mol Co

58:93 g Co

� �¼ 0:935 mol Co

0:395 mol Co

0:264¼ 1:50 mol Co

25:3 gMoð Þ 1 mol Mo

95:94 gMo

� �¼ 0:264 mol Mo

0:264 mol Mo

0:264¼ 1:00 mol Mo


35:45 g Cl

� �¼ 1:45 mol Cl

1:45 mol Cl

0:264¼ 5:49 mol Cl

Multiplying by 2 gives the empirical formula Co3Mo2Cl11.

73. The conversion is: g Al�!mol Al�!molMg�! gMg

18 g Alð Þ 1 mol Al

26:98 g Al

� �2 mol Mg

1 mol Al

� �24:31 gMg

molMg

� �¼ 32 gMg

74. (10.0 g compound) (0.177)¼ 1.77 g N


14:01 g N

� �¼ 0:126 mol N

3:8� 1023atoms H� � 1 mol

6:022� 1023atoms

� �¼ 0:63 mol H

To determine the mol C, first find grams H and subtract the grams of H and N from the grams of the

sample.

0:63 mol Hð Þ 1:008 g H

mol H

� �¼ 0:64 g H

10:0 g sample

�1:77 g N

�0:64 g H7:6 g C


12:01 g C

� �¼ 0:63 mol C

Now determine the empirical formula from the moles of C, H, and N.

N0:126 mol N

0:126¼ 1:00 mol N

H0:63 mol H

0:126¼ 5:0 mol H

C0:63 mol C

0:126¼ 5:0 mol C

The empirical formula is C5H5N

- 73 -

- Chapter 7 -

75. Let x¼molar mass of A2O

0:400x ¼ 16:00 g O Since A2O has only one mol of O atomsð Þx ¼ 40:0 g O=mol A2O

40:0 ¼ 16:00þ 2y y ¼ molar mass of A

40:0� 16:00 ¼ 2y

12:0g

mol¼ y

Look in the periodic table for the element that has 12.0 g=mol.

The element is carbon. The mystery element is carbon.

76. (a) CH2O (divide the molecular formula by 6)

(b) C4H9 (divide the molecular formula by 2)

(c) CH2O (divide the molecular formula by 3)

(d) C25H52 (divide the molecular formula by 1)

(e) C6H2Cl2O (divide the molecular formula by 2)

77. First determine the element in compound A(BC)3:

A: 0:3459ð Þ 78:01 gð Þ ¼ 26:98 g aluminumð ÞB: 0:6153ð Þ 78:01 gð Þ ¼ 48:00 g


C: 0:0388ð Þ 78:01 gð Þ ¼ 3:03 g

3¼ 1:01 g hydrogenð Þ

Element determined from atomic masses in the periodic table.

A(BC)3¼Al(OH)3Then compound A2B3¼Al2O3 with a molar mass of

2 26:98 gð Þ þ 3 16:00 gð Þ ¼ 102:0 g

%Al ¼ 2 26:98 gð Þ102:0 g

100ð Þ ¼ 52:90%

%O ¼ 3 16:00ð Þ102:0

100ð Þ ¼ 47:06%

78. (a) Percent composition of the original unknown compound.

Convert g CO2 to g C and g H2O to g H

4:776 g CO2ð Þ 12:01 g C

44:01 g CO2

� �¼ 1:303 g C

2:934 g H2Oð Þ 2:016 g H

18:02 g H2O

� �¼ 0:3282 g H

- 74 -

- Chapter 7 -

2.500 g compound �1.303 g C �0.3282 g H¼ 0.869 g O

1:303 g C

2:500 g

� �100ð Þ ¼ 52:12%C

0:328 g H

2:500 g

� �100ð Þ ¼ 13:13%H

0:869 g O

2:500 g

� �100ð Þ ¼ 34:76%O

(b) Empirical formula of unknown compound; 52.12% C, 13.13% H, 34.76% O.


12:01 g

� �¼ 4:340

4:340 mol C

2:173¼ 1:997 mol C


1:008 g H

� �¼ 13:03

13:03 mol H

2:173¼ 5:996 mol H

34:76 g Oð Þ 1 mol C

16:00 g

� �¼ 2:173

2:173 mol O

2:173¼ 1:000 mol O

The empirical formula is C2H6O

- 75 -

- Chapter 7 -

QUANTITATIVE COMPOSITION OF COMPOUNDSfaculty.chemeketa.edu/lemme/Solutions/ch07.pdfCHAPTER 7...

Documents

Transcript of QUANTITATIVE COMPOSITION OF COMPOUNDSfaculty.chemeketa.edu/lemme/Solutions/ch07.pdfCHAPTER 7...