QUANTITATIVE COMPOSITION OF COMPOUNDSfaculty.chemeketa.edu/lemme/CH...

CHAPTER 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 areatoms in exactly 12 g of carbon-12.

It is Avogadro’s number 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.

4. A mole of gold atoms contains more electrons than a mole of potassium atoms, as eachAu atom has while each K atom has only

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

6. No. Avogadro’s number is a constant. The mole is defined as Avogadro’s number of C-12atoms. Changing the atomic mass to 50 amu would change only the size of the atomicmass unit, not Avogadro’s number.

7.

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

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

(b) A mole of oxygen molecules contains molecules.

(c) A mole of oxygen molecules contains atoms.

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

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

10. molecules in one molar mass of atoms in one molar mass of

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

H2SO4.4.215 * 1024H2SO4.6.022 * 1023

(O2)

1.204 : 1024(O2)

6.022 : 1023(O2)

6.022 : 1023

6.022 * 1023

6.022 * 1023

19 e-.79 e-,

16.022 * 10232.

16.022 * 10232

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CHAPTER 7

SOLUTIONS TO EXERCISES

1. Molar masses

(a) KBr

(b)

(c)

(d)

(e)

(f) 167.6 g

64.00 g

231.6 g

3

4

Fe

O

Fe3O4

4.032 g

24.02 g

32.00 g

60.05 g

4

2

2

H

C

O

HC2H3O2

24.02 g

6.048 g

16.00 g

46.07 g

2

6

1

C

H

O

C2H5OH

207.2 g

28.02 g

96.00 g

331.2 g

1

2

6

Pb

N

O

Pb(NO3)2

45.98 g

32.07 g

64.00 g

142.1 g

2

1

4

Na

S

O

Na2SO4

39.10 g

79.90 g

119.0 g

1

1

K

Br

- 55 -


(g)

(h)

(i)

2. Molar masses

(a) NaOH

(b)

(c)

(d) 28.02 g

8.064 g

12.01 g

48.00 g

96.09 g

2

8

1

3

N

H

C

O

(NH4)2CO3

104.0 g

48.00 g

152.0 g

2

3

Cr

O

Cr2O3

215.8 g

12.01 g

48.00 g

275.8 g

2

1

3

Ag

C

O

Ag2CO3

22.99 g

16.00 g

1.008 g

40.00 g

1

1

1

Na

O

H

9.072 g

28.02 g

30.97 g

64.00 g

132.1 g

9

2

1

4

H

N

P

O

(NH4)2HPO4

53.96 g

96.21 g

192.0 g

342.2 g

2

3

12

Al

S

O

Al2(SO4)3

144.1 g

22.18 g

176.0 g

342.3 g

12

22

11

C

H

O

C12H22O11

- 56 -

- Chapter 7 -


(e)

(f)

(g)

(h)

(i)

3. Moles of atoms.

(a)

(b) 10.688 g Mg2a 1 mol Mg

24.31 g Mgb = 2.83 * 10-2 mol Mg

122.5 g Zn2a 1 mol Zn

65.39 g Znb = 0.344 mol Zn

137.3 g

70.90 g

4.032 g

32.00 g

244.2 g

1

2

4

2

Ba

Cl

H

O

BaCl2# 2 H2O

156.4 g

55.85 g

72.06 g

84.06 g

368.4 g

4

1

6

6

K

Fe

C

N

K4Fe(CN)6

72.06 g

12.10 g

96.00 g

180.2 g

6

12

6

C

H

O

C6H12O6

84.07 g

6.048 g

32.00 g

122.1 g

7

6

2

C

H

O

C6H5COOH

24.31 g

2.016 g

24.02 g

96.00 g

146.3 g

1

2

2

6

Mg

H

C

O

Mg(HCO3)2

- 57 -

- Chapter 7 -


(c)

(d)

(e)

(f)

4. Number of moles.

(a)

(b)

(c)

(d)

(e)

(f)

5. Number of grams.

(a)

(b)

(c)

(d) 13.15 mol NH4NO32a80.05 g NH4NO3

mol NH4NO3b = 252 g NH4NO3

112.5 mol Cl22a70.90 g Cl2mol Cl2

b = 886 g Cl2

115.8 mol H2O2a18.02 g H2O

mol H2Ob = 285 g H2O

10.550 mol Au2a197.0 g Au

1 mol Aub = 108 g Au

14.20 lb ZnI22a453.6 g

1 lbb a 1 mol ZnI2

319.2 g ZnI2b = 5.97 mol ZnI2

12.88 g Na2SO42a 1 mol Na2SO4

142.1 g Na2SO4b = 2.03 * 10-2 mol Na2SO4

114.8 g CH3OH2a 1 mol CH3OH

32.04 g CH3OHb = 0.462 mol CH3OH

10.684 g MgCl22a 1 mol MgCl295.21 g MgCl2

b = 7.18 * 10-3 mol MgCl2

144.0 g Br22a 1 mol Br2

159.8 g Br2b = 0.275 mol Br2

125.0 g NaOH2a 1 mol NaOH

40.00 g NaOHb = 0.625 mol NaOH

= 28 mol N atoms

18.5 * 1024 molecules N22a 2 atoms N

1 molecule N2b a 1 mol N atoms

6.022 * 1023 atoms Nb

10.055 g Sn2a 1 mol Sn

118.7 g Snb = 4.6 * 10-4 mol Sn

1382 g Co2a 1 mol Co

58.93 g Cob = 6.48 mol Co

14.5 * 1022 atoms Cu2a 1 mol Cu

6.022 * 1023 atoms Cub = 7.5 * 10-2 mol Cu

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- Chapter 7 -

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6. Number of grams.

(a)

(b)

(c)

(d)

7. Number of molecules.

(a)

(b)

(c)

(d)

8. (a)

(b)

(c)

(d)

9. Number of atoms.

(a)

(b) (25.0 mol Ag) a6.022 * 1023atoms

mol b = 1.5 * 1025 atoms Ag

(11 molecules C2H5OH) a 9 atoms

1 molecule b = 99 atoms C2H5OH

1100. g CH42a6.022 * 1023 molecules CH4

16.04 g CH4b = 3.75 * 1024 molecules CH4

112.0 g CO22a6.022 * 1023 molecules CO2

44.01 g CO2b = 1.64 * 1023 molecules CO2

10.27 mol C2H6O2a6.022 * 1023 molecules

molb = 1.6 * 1023 molecules C2H6O

11.75 mol Cl22a6.022 * 1023 molecules

molb = 1.05 * 1024 molecules Cl2

11000. g HCl2a6.022 * 1023 molecules HCl

36.46 g HClb = 1.652 * 1025 molecules HCl

116.0 g CH42a6.022 * 1023 molecules CH4

16.04 g CH4b = 6.01 * 1023 molecules CH4

10.56 mol C6H62a6.022 * 1023 molecules

molb = 3.4 * 1023 molecules C6H6

11.26 mol O22a6.022 * 1023 molecules

molb = 7.59 * 1023 molecules O2

11.5 * 1016 atoms S2a 32.07 g S

6.022 * 1023 atoms Sb = 8.0 * 10-7 g S

10.00255 mol Ti2a47.87 g Ti

mol Tib = 0.122 g Ti

= 11 g CCl4

14.5 * 1022 molecules CCl42a 1 mol

6.022 * 1023 moleculesb a153.8 g CCl4

mol CCl4b

14.25 * 10-4 mol H2SO42a98.09 g H2SO4

mol H2SO4b = 0.0417 g H2SO4

- 59 -

- Chapter 7 -


(c)

(d)

10. Number of atoms.

(a)

(b)

(c)

(d)

11. Number of grams.

(a)

(b)

(c)

(d)

12. (a)

(b)

(c) 11 molecule NH32a 17.03 g NH3

6.022 * 1023 molecules NH3b = 2.828 * 10-23 g NH3

11 atom U2a 238.0 g U

6.022 * 1023 atoms Ub = 3.952 * 10-22 g U

11 atom Au2a 197.0 g Au

6.022 * 1023 atoms Aub = 3.271 * 10-22 g Au

= 3.771 * 10-22 g C3H5(NO3)3

[1 molecule C3H5(NO3)3]a 227.1 g C3H5(NO3)3

6.022 * 1023 molecules C3H5(NO3)3b

11 molecule H2O2a 18.02 g H2O

6.022 * 1023 molecules H2O b = 2.992 * 10-23 g H2O

11 atom Ag2a 107.9 g Ag

6.022 * 1023 atoms Agb = 1.792 * 10-22 g Ag

11 atom Pb2a 207.2 g Pb

6.022 * 1023 atoms Pbb = 3.441 * 10-22 g Pb

(15.2 g U) a6.022 * 1023atoms U

238.0 g Ub = 3.85 * 1022 atoms U

(75.2 g BF3) a6.022 * 1023 molecules BF3

67.81 g BF3b a 4 atoms

1 moleculesb = 2.67 * 1024 atoms BF3

(10.0 mol Au) a6.022 * 1023atoms

molb = 6.02 * 1024 atoms Au

(18 molecules N2O5) a 7 atoms

1 moleculeb = 1.3 * 102 atoms N2O5

= 1.83 * 1024 atom CHCl3

(72.5 g CHCl3) a6.022 * 1023 molecules

119.4b a 5 atoms

1 moleculeb

(0.0986 g Xe) a6.022 * 1023atoms

131.3 gb = 4.52 * 1020 atoms Xe

- 60 -

- Chapter 7 -


(d)

13. (a)

(b)

(c)

(d)

14. (a)

(b)

(c)

(d)

15. One mole of carbon disulfide contains:

(a)

(b)

(c)

(d) 16.022 * 1023 atoms2 + 11.204 * 1024 atoms2 = 1.806 * 1024 total atoms

16.022 * 1023 molecules of CS22¢ 2 S atoms

1 molecule CS2≤ = 1.204 * 1024 S atoms

16.022 * 1023 molecules of CS22¢ 1 C atom

1 molecule CS2≤ = 6.022 * 1023 C atoms

6.022 * 1023 molecules of CS2

(CS2)

1485 mL Br22a3.12 g

mLb a 1 mol Br2

159.8 g Br2b = 9.47 mol Br2

142.4 g Mg2a6.022 * 1023 atoms

24.31 gb = 1.05 * 1024 atoms Mg

12.50 kg NaCl2a1000 g

kgb a 1 mol NaCl

58.44 g NaClb = 42.8 mol NaCl

128.4 g S2a 1 mol S

32.07 g Sb = 0.886 mol S

15000 molecules CO22a 1 mol

6.022 * 1023 moleculesb = 8 * 10-21 mol CO2

110. atoms C2a 1 mol

6.022 * 1023 atomsb = 1.7 * 10-23 mol C

1125 mol Au2a197.0 g Au

mol Aub a 1 kg

1000 gb = 24.6 kg Au

18.66 mol Cu2a63.55 g Cu

mol Cub = 550. g Cu

= 1.795 * 10-22 g C6H4(NH2)2

[1 molecule C6H4(NH2)2]a 108.1 g C6H4(NH2)2

6.022 * 1023 molecules C6H4(NH2)2b

- 61 -

- Chapter 7 -


16. One mole of ammonia contains

(a)

(b)

(c)

(d)

17. Atoms of oxygen in:

(a)

(b)

(c)

18. Atoms of oxygen in:

(a)

(b)

(c)

19. The number of grams of:

(a) silver in 25.0 g AgBr

125.0 g AgBr2a 107.9 g Ag

187.8 g AgBrb = 14.4 g Ag

15.0 * 1018 molecules H2O2¢ 1 atom O

molecule H2O≤ = 5.0 * 1018 atoms O

= 5.46 * 1024 atoms O

1255 g MgCO32a 1 mol MgCO3

84.32 g MgCO3b ¢ 3 mol O

mol MgCO3≤ ¢ 6.022 * 1023 atoms

mol≤

15.0 mol MnO22¢ 2 mol O

mol MnO2≤ ¢6.022 * 1023 atoms

mol≤ = 6.0 * 1024 atoms O

16.00 * 1022 molecules C6H12O62¢ 6 atoms O

molecule C6H12O6≤ = 3.60 * 1023 atoms O

10.622 mol MgO2a 1 mol O

mol MgOb ¢6.022 * 1023 atoms

mol≤ = 3.75 * 1023 atoms O

116.0 g O22a 1 mol O2

32.00 g O2b ¢ 2 mol O

1 mol O2≤ ¢6.022 * 1023 atoms

mol≤ = 6.02 * 1023 atoms O

16.022 * 1023 atoms2 + 11.807 * 1024 atoms2 = 2.409 * 1024 total atoms

16.022 * 1023 molecules of NH32¢ 3 H atoms

molecule NH3≤ = 1.807 * 1024 H atoms

16.022 * 1023 molecules of NH32¢ 1 N atom

molecule NH3≤ = 6.022 * 1023 N atoms

6.022 * 1023 molecules of NH3

(NH3)

- 62 -

- Chapter 7 -


(b) nitrogen in 6.34 mol

(c) oxygen in molecules

The conversion is: molecules

20. The number of grams of:

(a) chlorine in 5.00 g

(b) hydrogen in 4.50 g

(c) hydrogen in molecules

The conversion is: molecules

21. Percent composition

(a) NaBr

(b)

a 1.008 g

100.1 gb11002 = 1.007% H

a 39.10 g

100.1 gb11002 = 39.06% K

39.10 g

1.008 g

48.00 g

12.01 g

100.1 g

K

H

3 O

C

KHCO3

a 79.90 g

102.9 gb11002 = 77.65% Br

a 22.99 g

102.9 gb11002 = 22.34% Na

22.99 g

79.90 g

102.9 g

Na

Br

15.45 * 1022 molecules NH32¢ 1 mol

6.022 * 1023 molecules≤ ¢3.024 g H

mol NH3≤ = 0.274 g H

NH3 ¡ moles NH3 ¡ g H

NH35.45 * 1022

14.50 g H2SO42¢ 2.016 g H

98.09 g H2SO4≤ = 9.25 * 10-2 g H

H2SO4

15.00 g PbCl22¢ 70.90 g Cl

278.1 g PbCl2≤ = 1.27 g Cl

PbCl2

18.45 * 1022 molecules SO32¢ 1 mol

6.022 * 1023 molecules≤ ¢48.00 g O

mol SO3≤ = 6.74 g O

SO3 ¡ mol SO3 ¡ g O

SO38.45 * 1022

16.34 mol (NH4)3PO42¢ 42.03 g N

mol (NH4)3PO4≤ = 266 g N

(NH4)3PO4

- 63 -

- Chapter 7 -


(c)

(d)

(e)

(f)

a 48.00 g

169.9 gb11002 = 28.25% O

a 14.01 g

169.9 gb11002 = 8.246% N

a 107.9 g

169.9 gb11002 = 63.51% Ag

107.9 g

14.01 g

48.00 g

169.9 g

Ag

N

3 O

AgNO3

a 192.0 g

342.2 gb11002 = 56.11% O

a 96.21 g

342.2 gb11002 = 28.12% S

a 53.96 g

342.2 gb11002 = 15.77% Al

53.96 g

96.21 g

192.0 g

342.2 g

2 Al

3 S

12 O

Al2(SO4)3

a 141.8 g

169.9 gb11002 = 83.46% Cl

a 28.09 g

169.9 gb11002 = 16.53% Si

28.09 g

141.8 g

169.9 g

Si

4 Cl

SiCl4

a 106.4 g

162.3 gb11002 = 65.56% Cl

a 55.85 g

162.3 gb11002 = 34.41% Fe

55.85 g

106.4 g

162.3 g

Fe

3 Cl

FeCl3

a 48.00 g

100.1 gb11002 = 47.95% O

a 12.01 g

100.1 gb11002 = 12.00% C

- 64 -

- Chapter 7 -


22. Percent composition

(a)

(b)

(c)

(d)

a 64.00 g

132.2 gb11002 = 48.41% O

a 32.07 g

132.2 gb11002 = 24.26% S

a 8.064 g

132.2 gb11002 = 6.100% H

a 28.02 g

132.2 gb11002 = 21.20% N

28.02 g

8.064 g

32.07 g

64.00 g

132.2 g

2 N

8 H

S

4 O

(NH4)2SO4

a 112.0 g

198.3 gb11002 = 56.48% O

a 61.94 g

198.3 gb11002 = 31.24% P

a 24.31 g

198.3 gb11002 = 12.26% Mg

24.31 g

61.94 g

112.0 g

198.3 g

Mg

2 P

7 O

MgP2O7

a 32.00 g

77.09 gb11002 = 41.51% O

a 24.02 g

77.09 gb11002 = 31.16% C

a 7.056 g

77.09 gb11002 = 9.153% H

a 14.01 g

77.09 gb11002 = 18.17% N

14.01 g

7.056 g

24.02 g

32.00 g

77.09 g

N

7 H

2 C

2 O

NH4C2H3O2

a 70.90 g

136.3 gb11002 = 52.02% Cl

a 65.39 g

136.3 gb11002 = 47.98% Zn

65.39 g

70.90 g

136.3 g

Zn

2 Cl

ZnCl2

- 65 -

- Chapter 7 -


(e)

(f)

23. Percent of iron

(a) FeO

(b)

(c)

(d) a 55.85 g

368.4 gb11002 = 15.16% Fe

55.85 g

156.4 g

72.06 g

84.06 g

368.4 g

Fe

4 K

6 C

6 N

K4Fe(CN)6

a 167.6 g

231.6 gb11002 = 72.37% Fe

167.6 g

64.00 g

231.6 g

3 Fe

4 O

Fe3O4

a 111.7 g

159.7 gb11002 = 69.94% Fe

111.7 g

48.00 g

159.7 g

2 Fe

3 O

Fe2O3

a 55.85 g

71.85 gb11002 = 77.73% Fe

55.85 g

16.00 g

71.85 g

Fe

O

a 106.4 g

233.3 gb11002 = 45.61% Cl

a 126.9 g

233.3 gb11002 = 54.39% I

126.9 g

106.4 g

233.3 g

I

3 Cl

ICI3

a 144.0 g

241.9 gb11002 = 59.53% O

a 42.03 g

241.9 gb11002 = 17.37% N

a 55.85 g

241.9 gb11002 = 23.09% Fe

55.85 g

42.03 g

144.0 g

241.9 g

Fe

3 N

9 O

Fe(NO3)3

- 66 -

- Chapter 7 -


24. Percent chlorine

(a) KCl

(b)

(c)

(d) LiCl

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

25. Percent composition of an oxide

26. Percent composition of ethylene chloride

27. (a) (It has 2 H atoms/O atom)

(b) (It has only 0.67 N atom/O atom)

(c) equal (Both have 20 atoms/N atom)

N2O3

H2O

a17.75 g

24.75 gb11002 = 71.72% Cl

a 1.00 g

24.75 gb11002 = 4.04% H

a 6.00 g

24.75 gb11002 = 24.2% C

6.00 g C

1.00 g H

17.75 g Cl

24.75 g total

a 8.00 g

14.20 gb11002 = 56.3% O

a 6.20 g

14.20 gb11002 = 43.7% P

14.20 g oxide

-6.20 g P

8.00 g oxygen

BaCl2

a 35.45 g

42.39 gb11002 = 83.63% Cl

6.941 g

35.45 g

42.39 g

Li

Cl

a 141.8 g

169.9 gb11002 = 83.46% Cl

28.09 g

141.8 g

169.9 g

Si

4 Cl

SiCl4

a 70.90 g

208.2 gb11002 = 34.05% Cl

137.3 g

70.90 g

208.2 g

Ba

2 Cl

BaCl2

a 35.45 g

74.55 gb11002 = 47.55% Cl

39.10 g

35.45 g

74.55 g

K

Cl

- 67 -

- Chapter 7 -


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

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

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

29. Empirical Formulas. Change all percents to grams.

(a)

This is a ratio of 1 mol H to 1 mol O.

The empirical formula is HO

(b)

This is a ratio of 1 mol Zn to 1 mol O.

The empirical formula is ZnO

(c)

The empirical formula is

(d)

The empirical formula is NH4Cl

1.869 mol Cl

1.869= 1.000 mol Cl

mol ratios: 1.869 mol N

1.869= 1 mol N

7.49 mol H

1.869= 4.00 mol H

66.26 g Cl

35.45 g>mol= 1.869 mol Cl

26.19 g N

14.01 g>mol= 1.869 mol N

7.55 g H

1.008 g>mol= 7.49 mol H

Fe(ClO)2

1.260 mol Cl

0.6299= 2.000 mol Cl

mol ratios =0.6299 Fe

0.6299= 1.000 mol Fe

1.260 mol O

0.6299= 2.000 mol O

20.16 g O

16.00 g O>mol= 1.260 mol O

35.18 g Fe

55.85 g>mol= 0.6299 mol Fe

44.66 g Cl

35.45 g>mol= 1.260 mol Cl

80.34 g Zn

65.39 g>mol= 1.228 mol Zn

19.66 g O

16.00 g>mol= 1.229 mol O

5.94 gH

1.008 g>mol= 5.89 mol H

94.06 g O

16.00 g>mol= 5.879 mol O

Na2CrO4

KHSO4

KClO3

- 68 -

- Chapter 7 -


30. Empirical formulas: Change all percents to grams.

(a)

This is a ratio of 1 mol K to 1 mol Br.

The empirical formula is KBr

(b)


(c)

The empirical formula is or

(d)

The empirical formula is H2SO4

4.078 mol O

1.019= 4.002 mol O

mole ratios: 2.04 mol H

1.019= 2.00 mol H

1.019 mol S

1.019= 1.000 mol S

65.25 g O

16.00 g>mol= 4.078 mol O

2.06 g H

1.008 g>mol= 2.07 mol H

32.69 g S

32.07 g>mol= 1.019 mol S

Ca(OH)2CaO2H2

2.699 mol O

1.35= 1.999 mol O

mole ratios: 1.350 mol Ca

1.350= 1.000 mol Ca

2.70 mol H

1.350= 2.00 mol H

43.18 g O

16.00 g>mol= 2.699 mol O

54.09 g Ca

40.08 g>mol= 1.350 mol Ca

2.72 g H

1.008 g>mol= 2.70 mol H

Ag NO3

1.766 mol O

0.5885= 3.000 mol O

mole ratios:0.5885 mol Ag

0.5885= 1.000 mol Ag

0.5889 mol N

0.5885= 1.001 mol N

28.25 g O

16.00 g>mol= 1.766 mol O

63.50 g Ag

107.9 g>mol= 0.5885 mol Ag

8.25 g N

14.01 g>mol= 0.5889 mol N

32.86 g K

39.10 g>mol= 0.8404 mol K

67.14 g Br

79.90 g>mol= 0.8403 mol Br

- 69 -

- Chapter 7 -


31. Empirical formulas from percent composition.

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

Step 2. Calculate the relative moles of each element.

Step 3. Change these moles to whole numbers by dividing each by the smallernumber.

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

(b) 46.7% N, 53.3% O

The empirical formula is NO.

(c) 25.9% N, 71.4% O

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

The empirical formula is N2O5.11.00 mol N2122 = 2.00 mol N; 12.5 mol O2122 = 5.00 mol O

174.1 g O2a 1 mol O

16.00 g Ob = 4.63 mol O

4.63 mol O

1.85= 2.5 mol O

125.9 g N2a 1 mol N

14.01 g Nb = 1.85 mol N

1.85 mol N

1.85= 1.00 mol N

153.3 g O2a 1 mol O

16.00 g Ob = 3.33 mol O

3.33 mol O

3.33= 1.00 mol O

146.7 g N2a 1 mol N

14.01 g Nb = 3.33 mol N

3.33 mol N

3.33= 1.00 mol N

N2O.

2.28 mol O

2.28= 1.00 mol O

4.54 mol N

2.28= 1.99 mol N

136.4 g O2a 1 mol O

16.00 g Ob = 2.28 mol O

163.6 g N2a 1 mol N

14.01 g Nb = 4.54 mol N

36.4% O = 36.4 g O>100 g material

63.6% N = 63.6 g N>100 g material

>

- 70 -

- Chapter 7 -


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


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


(f) 72.02% Mn, 27.98% O

Multiply both values by 3 to give whole numbers.


32. Empirical formulas from percent composition.

(a) 64.1% Cu, 35.9% Cl

The empirical formula is CuCl.

135.9 g Cl2a 1 mol Cl

35.45 g Clb = 1.01 mol Cl

1.01 mol Cl

1.01= 1.00 mol Cl

164.1 g Cu2a 1 mol Cu

63.55 g Cub = 1.01 mol Cu

1.01 mol Cu

1.01= 1.00 mol Cu

Mn3O4.

11.000 mol Mn2132 = 3.000 mol Mn; 11.334 mol O2132 = 4.002 mol O

127.98 g O2a 1 mol O

16.00 g Ob = 1.749 mol O

1.749 mol O

1.311= 1.334 mol O

172.02 g Mn2a 1 mol Mn

54.94 g Mnb = 1.311 mol Mn

1.311 mol Mn

1.311= 1.000 mol Mn

NaClO4.

152.3 g O2a 1 mol O

16.00 g Ob = 3.27 mol O

3.27 mol O

0.818= 4.00 mol O


35.45 g Clb = 0.818 mol Cl

0.818 mol Cl

0.818= 1.00 mol Cl

118.8 g Na2a 1 mol Na

22.99 g Nab = 0.818 mol Na

0.818 mol Na

0.818= 1.00 mol Na

Na2CO3.

145.3 g O2a 1 mol O

16.00 g Ob = 2.83 mol O

2.83 mol O

0.941= 3.00 mol O

111.3 g C2a 1 mol C

12.01 g Cb = 0.941 mol C

0.941 mol C

0.941= 1.00 mol C

143.4 g Na2a 1 mol Na

22.99 g Nab = 1.89 mol Na

1.89 mol Na

0.941= 2.01 mol Na

- 71 -

- Chapter 7 -


(b) 47.2% Cu, 52.8% Cl


(c) 51.9% Cr, 48.1% S

Multiply both values by 2 to give whole numbers.


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


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

The empirical formula is BaCr2O7.

131.7 g O2a 1 mol O

16.00 g Ob = 1.98 mol O

1.98 mol O

0.283= 7.00 mol O

129.4 g Cr2a 1 mol Cr

52.00 g Crb = 0.565 mol Cr

0.565 mol Cr

0.283= 2.00 mol Cr

138.9 g Ba2a 1 mol Ba

137.3 g Bab = 0.283 mol Ba

0.283 mol Ba

0.283= 1.00 mol Ba

K3PO4.

130.1 g O2a 1 mol O

16.00 g Ob = 1.88 mol O

1.88 mol O

0.471= 3.99 mol O

114.6 g P2a 1 mol P

30.97 g Pb = 0.471 mol P

0.471 mol P

0.471= 1.00 mol P

155.3 g K2a 1 mol K

39.10 g Kb = 1.41 mol K

1.41 mol K

0.471= 2.99 mol K

Cr2S3.

11.50 mol S2122 = 3.00 mol S11.00 mol Cr2122 = 2.00 mol Cr;

148.1 g S2a 1 mol S

32.07 g Sb = 1.50 mol S

1.50 mol S

0.998= 1.50 mol S

151.9 g Cr2a 1 mol Cr

52.00 g Crb = 0.998 mol Cr

0.998 mol Cr

0.998= 1.00 mol Cr

CuCl2 .


35.45 g Clb = 1.49 mol Cl

1.49 mol Cl

0.743= 2.01 mol Cl


63.55 g Cub = 0.743 mol Cu

0.743 mol Cu

0.743= 1.00 mol Cu

- 72 -

- Chapter 7 -


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


33. Empirical formula



Multiplying both by 2 gives the empirical formula




13.898 g Hg2a 1 mol Hg

200.6 g Hgb = 0.01943 mol Hg

0.01943 mol Hg

0.01943= 1.000 mol Hg

5.276 g compound - 3.898 g Hg = 1.378 g Cl

Cu2O.

10.310 g O2a 1 mol O

16.00 g Ob = 0.0194 mol O

0.0194 mol O

0.0194= 1.00 mol O


63.55 g Cub = 0.03879 mol Cu

0.03879 mol Cu

0.0194= 2.00 mol Cu

2.775 g oxide - 2.465 g Cu = 0.310 g O

V2O5.

12.400 g O2a 1 mol O

16.00 g Ob = 0.1500 mol O

0.1500 mol O

0.0600= 2.50 mol O

13.054 g V2a 1 mol V

50.94 g Vb = 0.0600 mol V

0.0600 mol V

0.0600= 1.00 mol V

5.454 g product - 3.054 g V = 2.400 g O

SnO2.

11.077 g O2a 1 mol O

16.00 g Ob = 0.0673 mol O

0.0673 mol O

0.0337= 2.00 mol O

13.996 g Sn2a 1 mol Sn

118.7 g Snb = 0.0337 mol Sn

0.0337 mol Sn

0.0337= 1.00 mol Sn

PBr8Cl3 .


35.45 g Clb = 0.386 mol Cl

0.386 mol Cl

0.129= 2.99 mol Cl

182.3 g Br2a 1 mol Br

79.90 g Brb = 1.03 mol Br

1.03 mol Br

0.129= 7.98 mol Br

13.99 g P2a 1 mol P

30.97 g Pb = 0.129 mol P

0.129 mol P

0.129= 1.00 mol P

- 73 -

- Chapter 7 -



37. Molecular formula of hydroquinone

65.45% C, 5.45% H, 29.09% O;

The empirical formula is making the empirical formula mass 55.05 g.

The molecular formula is twice that of the empirical formula.

38. Molecular formula of fructose40.0% C, 6.7% H, 53.3% O;

The empirical formula is making the empirical formula mass 33.03 g.

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

39. Molecular formula of ethanedioic acid26.7% C, 2.24% H, 71.1% O;

26.7 g Ca 1 mol C

12.01 g Cb = 2.22 mol C

2.22 mol C

2.2= 1.0 mol C

molar mass = 90.04

Molecular formula = (CH2O)6 = C6H12O6

molar mass

mass of empirical formula=

180.1 g

33.03 g= 5.994

CH2O

153.3 g O2a 1 mol O

16.00 g Ob = 3.33 mol O

3.33 mol O

3.33= 1.00 mol O

16.7 g H2a 1 mol H

1.008 g Hb = 6.6 mol H

6.6 mol H

3.33= 2.0 mol H

140.0 g C2a 1 mol C

12.01 g Cb = 3.33 mol C

3.33 mol C

3.33= 1.00 mol C

molar mass = 180.1

Molecular formula = (C3H3O)2 = C6H6O2

molar mass


110.1 g

55.05 g= 2

C3H3O

129.09 g O2a 1 mol O

16.00 g Ob = 1.818 mol O

1.818 mol O

1.818= 1.000 mol O

15.45 g H2a 1 mol H

1.008 g Hb = 5.41 mol H

5.41 mol H

1.818= 2.98 mol H

165.45 g C2a 1 mol C

12.01 g Cb = 5.450 mol C

5.450 mol C

1.818= 2.998 mol C

molar mass = 110.1

HgCl2 .


35.45 g Clb = 0.03887 mol Cl

0.03887 mol Cl

0.01943= 2.001 mol Cl

- 74 -

- Chapter 7 -


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


40. Molecular formula of butyric acid54.5% C, 9.2% H, 36.3% O;

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


41.

relative number of nitrogen atoms =0.8594 mol

0.8594 mol= 1.000

moles of oxygen =27.50 g O

16.00 g>mol O= 1.719 mol O

empirical formula: moles of nitrogen =12.04 g N

14.01 g>mol= 0.8594 mol N

% oxygen =39.54 g - 12.04 g

39.54 g 11002 = 69.55%

% nitrogen =12.04 g

39.54 g 11002 = 30.45%

Molecular formula = (C2H4O)2 = C4H8O2

molar mass


88.11 g

44.05 g= 2

C2H4O

136.3 g O2a 1 mol O

16.00 g Ob = 2.27 mol O

2.27 mol O

2.27= 1.0 mol O

19.2 g H2a 1 mol H

1.008 g Hb = 9.1 mol H

9.1 mol H

2.27= 4.0 mol H

154.5 g C2a 1 mol C

12.01 g Cb = 4.54 mol C

4.54 mol C

2.27= 2.00 mol C

molar mass = 88.11

Molecular formula = (CHO2)2 = C2H2O4

molar mass


90.04 g

45.02 g= 2

CHO2

71.1 g Oa 1 mol O

16.00 g Ob = 4.44 mol O

4.44 mol O

2.2= 2.0 mol O

2.2 g Ha 1 mol H

1.008 g Hb = 2.2 mol H

2.2 mol H

2.2= 1.0 mol H

- 75 -

- Chapter 7 -


relative number of oxygen

molecular formula:

The molecular formula is twice the empirical formula.

42. Total mass of

empirical formula:

molecular formula:

The molecular formula is six times the empirical formula.

molecular formula = C6H12O6

x =180.18 g>mol

30.03 g>mol= 6

130.03 g>mol2x = 180.18 g>mol,

1molar mass of CH2O2x = 180.18 g>mol,

empirical formula = CH2O

relative number of oxygen atoms =2.515 mol

2.515 mol= 1.000

relative number of hydrogen atoms =5.03 mol

2.515 mol= 2.00

relative number of carbon atoms =2.515 mol

2.515 mol= 1.000

moles of oxygen =40.24 g O

16.00 g>mol= 2.515 mol O

moles of hydrogen =5.080 g H

1.008 g>mol= 5.03 mol H

moles of carbon =30.21 g C

12.01 g>mol= 2.515 mol C

% oxygen =40.24 g

75.53 g 11002 = 53.3%

% hydrogen =5.08 g

75.53 g 11002 = 6.73%

% carbon =30.21 g

75.53 g 11002 = 40.0%

C + H + O = 30.21 g + 5.08 g + 40.24 g = 75.53 g

molecular formula = N2O4

1molar mass of NO22x = 92.02 g, 46.01x = 92.02, x = 2

empirical formula = NO2

atoms =1.719 mol

0.8594 mol= 2.000

- 76 -

- Chapter 7 -


43. What is compound

Elements determined from atomic masses in the periodic table.

44. What is compound


45.

46.

47.

48.

49.

50. ¢6.022 * 1023 dollars

6.1 * 109 people≤ = 9.9 * 1013 dollars>person

16.022 * 1023 sheets2a 4.60 cm

500 sheetsb a 1 m

100 cmb = 5.54 * 1019 m

15 lb C12H22O112¢453.6 g

1 lb≤ ¢6.022 * 1023 molecules

342.3 g≤ = 4 * 1024 molecules

11.79 * 10-23 g>atom216.022 * 1023 atoms>molar mass2 = 10.8 g>molar mass

110.0 g K2a 1 mol K

39.10 g Kb a1 mol Na

1 mol Kb a22.99 g Na

mol Nab = 5.88 g Na

10.350 mol P42¢6.022 * 1023 molecules

mol≤ ¢ 4 atoms P

molecule P4≤ = 8.43 * 1023 atoms P

X2(YZ3)3 = Al2(SiO3)3

Z: 10.510221282.23 g2 =143.99 g

9= 16.00 g (oxygen)

Y: 10.298621282.23 g2 =84.27 g

3= 28.09 g (silicon)

X: 10.191221282.23 g2 =53.96 g

2= 26.98 g (aluminum)

X2(YZ3)3

XYZ3 = CaCO3

Z: 10.479621100.09 g2 = 48.00 g; 48.00 g

3= 16.00 g (oxygen)

Y: 10.120021100.09 g2 = 12.01 g (carbon)

X: 10.400421100.09 g2 = 40.08 g (calcium)XYZ3

- 77 -

- Chapter 7 -


51. The conversion is:

(a)

(b)

52.

(a)

(b)

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

The 1.00 gram samplewith the lowest molar mass will contain the most molecules. Thus, willcontain the most molecules.

(b)

The 1.00 g sample of contains the most atoms

54. formula units

16.022 * 1023 atoms2a1 formula unit

5 atomsb ¢ 207.9 g Fe2S3

6.022 * 1023 formula units≤ = 41.58 g Fe2S3

1 mol Fe2S3 = 207.9 g Fe2S3 = 6.022 * 1023

CH3OH

11.00 g O22a 1 mol

32.00 gb ¢ 12216.022 * 1023 atoms2

mol≤ = 3.76 * 1022 atoms

11.00 g CO22a 1 mol

44.01 gb ¢ 13216.022 * 1023 atoms2

mol≤ = 4.10 * 1022 atoms

11.00 g CH3OH2a 1 mol

32.04 gb ¢ 16216.022 * 1023 atoms2

mol≤ = 1.13 * 1023 atoms

11.00 g H2O2a 1 mol

18.02 gb ¢ 13216.022 * 1023 atoms2

mol≤ = 1.00 * 1023 atoms

H2OH2O, 18.02 g; CH3OH, 32.04 g.CO2, 44.01 g; O2, 32.00 g;

side = 23 10.3 cm3 = 2.18 cm

10.3 cm3 = volume of cube = (side)3

1107.9 g Ag2¢ 1 cm3

10.5 g≤ = 10.3 cm3 (volume of cube)

1 mol Ag = 107.9 g Ag

16.022 * 1023 drops2¢ 1 mi3

8 * 1016 drops≤ = 8 * 106 mi3

11 mi32a5280 ft

mileb3a12.0 in.

ftb3a2.54 cm

inchb3a20 drops

1.0 cm3 b = 8 * 1016 drops

mi3 ¡ ft3 ¡ in.3 ¡ cm3 ¡ drops

- 78 -

- Chapter 7 -


55. The conversion is

1.94 g Ca combines with 1.00 g P.

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

The conversion is:

1.0 ton of iron ore contains

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

58. (a)

(b)

(c) 28.02 g

120.1 g

14.11 g

162.2 g

2 N

10 C

14 H

C10H14N2

96.00 g

70.90 g

40.08 g

207.0 g

6 O

2 Cl

Ca

Ca(ClO3)2

200.6 g

12.01 g

48.00 g

260.6 g

Hg

C

3 O

HgCO3

a13.88 g Li

32.07 g Sb120.0 g S2 = 8.66 g Li

2 * 104 g Fe.

11.0 ton2a2000 lb

tonb a453.6 g

lbb10.05 FeSO42¢ 55.85 g Fe

151.9 g FeSO4≤ = 2 * 104 g Fe

ton ¡ lb ¡ g ¡ g FeSO4 ¡ g Fe

FeSO4.

11.00 g P2a 1 mol P

30.97 g Pb a 3 mol Ca

2 mol Pb a 40.08 g Ca

1 mol Cab = 1.94 g Ca

g P ¡ mol P ¡ mol Ca ¡ g Ca

- 79 -

- Chapter 7 -

a200.6 g Hg

260.6 gb11002 = 76.98% Hg

a 96.00 g O

207.0 gb11002 = 46.38% O

a28.02 g N

162.2 gb11002 = 17.27% N


(d)

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

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

60. Percent composition of

61. Percent composition of

a 35.45g Cl

291.8 gb(100) = 12.15% Cl

a204.2 g C

291.8 gb11002 = 69.98% C

a22.18 g H

291.8 gb(100) = 7.60% H

a14.01 g N

291.8 gb(100) = 4.80% N

a16.00 g O

291.8 gb(100) = 5.48% O

204.2 g

22.18 g

14.01 g

16.00 g

35.45 g

291.8 g

17 C

22 H

N

O

Cl

C17H21NO•HCl

a252.2 g C

328.4 g b11002 = 76.80% C

a28.22 g H

328.4 gb(100) = 8.593% H

a48.00 g O

328.4 gb (100) = 14.62% O

252.2 g

28.22 g

48.00 g

328.4 g

21 C

28 H

3 O

C21H28O3

119.5 g Zn2a 32.07 g S

65.39 g Znb = 9.56 g S

a 24.31 g Mg

893.5 gb11002 = 2.721% Mg

24.31 g

660.55 g

72.58 g

56.04 g

80.00 g

893.5 g

Mg

55 C

72 H

4 N

5 O

C55H72MgN4O5

- 80 -

- Chapter 7 -


62. Percent composition of sucrose

63. Molecular formula of aspirin

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

Multiplying each by 4 give the empirical formula The empirical formula massis 180.2 g. Since the empirical formula mass equals the molar mass, the molecularformula is the same as the empirical formula,

64. Calculate the percent oxygen in

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

The empirical formula is GaAs.

151.8 g As2a 1 mol As

74.92 g Asb = 0.691 mol As

0.691 mol As

0.691= 1.00 mol As

148.2 g Ga2a 1 mol Ga

69.72 g Gab = 0.691 mol Ga

0.691 mol Ga

0.691= 1.00 mol Ga

a192.0 g

342.2 gb11002 = 56.11% O

Now take 56.11% of 8.50 g

18.50 g Al2(SO4)3210.56112 = 4.77 g O

53.96 g

96.21 g

192.0 g

342.2 g

2 Al

3 S

12 O

Al2(SO4)3 .

C9H8O4.

C9H8O4.

135.5 g O2a 1 mol O

16.00 g Ob = 2.22 mol O

2.22 mol O

2.22= 1.00 mol O

14.48 g H2a 1 mol H

1.008 g Hb = 4.44 mol H

4.44 mol H

2.22= 2.00 mol H

160.0 g C2a 1 mol C

12.01 g Cb = 5.00 mol C

5.00 mol C

2.22= 2.25 mol C

aspirin = 180.2

a144.1 g C

342.3 gb11002 = 42.10% C

a22.18 g H

342.3 gb(100) = 6.480% H

a176.0 g O

342.3 gb (100) = 51.42% O

144.1 g

22.18 g

176.0 g

342.3 g

12 C

22 H

11 O

- 81 -

- Chapter 7 -


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

The empirical formula is C4H4CaO6

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

The empirical formula is The empirical formula mass is 153.8 which equalsthe molar mass, therefore the molecular formula is

(b) 10.13% C, 89.87% Cl

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

The molecular formula is twice that of the empirical formula. Molecular

(c) 25.26% C, 74.74% Cl


35.45 g Clb = 2.108 mol Cl

2.103 mol Cl

2.108= 1.002 mol Cl

125.26 g C2a 1 mol C

12.01 g Cb = 2.103 mol C

2.103 mol C

2.103= 1.000 mol C

formula = C2Cl6 .

molar mass

empirical formula mass=

236.7 g

118.4 g= 1.999

CCl3 .


35.45 g Clb = 2.535 mol Cl

2.535 mol Cl

0.8435= 3.005 mol Cl

110.13 g C2a 1 mol C

12.01 g Cb = 0.8435 mol C

0.8435 mol C

0.8435= 1.000 mol C

CCl4 .CCl4 .


35.45 g Clb = 2.601 mol Cl

2.601 mol Cl

0.649= 4.01 mol Cl

17.79 g C2a 1 mol C

12.01 g Cb = 0.649 mol C

0.649 mol C

0.649= 1.00 mol C

151.0 g O2a 1 mol O

16.00 g Ob = 3.19 mol O

3.19 mol O

0.531= 6.01 mol O

121.2 g Ca2a 1 mol Ca

40.08 g Cab = 0.531 mol Ca

0.529 mol Ca

0.531= 1.00 mol Ca

12.1 g H2a 1 mol H

1.008 g Hb = 2.1 mol H

2.1 mol H

0.531= 4.0 mol H

125.5 g C2a 1 mol C

12.01 g Cb = 2.12 mol C

2.212 mol C

0.531= 3.99 mol C

- 82 -

- Chapter 7 -


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

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

(d) 11.25% C, 88.75% Cl

Multiplying each by 3 give the empirical formula The empirical formulamass is 319.6. Since the molar mass is also 319.6 the molecular formula is




71.


23.3% Co, 25.3% Mo, 51.4% Cl

123.3 g Co2a 1 mol Co

58.93 g Cob = 0.395 mol Co

0.395 mol Co

0.264= 1.50 mol Co

16.1 * 109 people2¢ 1 mol people

6.022 * 1023 people≤ = 1.0 * 10-14 mol of people

= 1.529 * 10-7 g C3H8O3

11000. * 1012 molecules C3H8O32¢ 1 mol

6.022 * 1023 molecules≤ a 92.09 g C3H8O3

mol C3H8O3b

1 trillion = 1012molecules ¡ mol ¡ g


63.55 g Cub ¢6.022 * 1023 atoms

mol≤ = 2.4 * 1022 atoms Cu

g ¡ mol ¡ atom

16.022 * 1023 s2a 1 min

60 sb a 1 hr

60 minb a1 day

24 hrb a 1 year

365 daysb = 1.910 * 1016 years

s ¡ min ¡ hr ¡ day ¡ yr

C3Cl8 .C3Cl8 .


35.45 g Clb = 2.504 mol Cl

2.504 mol Cl

0.9367= 2.673 mol Cl

111.25 g C2a 1 mol C

12.01 g Cb = 0.9367 mol C

0.9367 mol C

0.9367= 1.000 mol C

formula = C6Cl6 .

molar mass

empirical formula mass=

284.8 g

47.46 g= 6.000

- 83 -

- Chapter 7 -


Multiplying by 2 gives the empirical formula


74.

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

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

The empirical formula is C5H5N

C 0.63 mol C

0.126= 5.0 mol C

H 0.63 mol H

0.126= 5.0 mol H

N 0.126 mol N

0.126= 1.00 mol N

17.6 g C2a 1 mol C

12.01 g Cb = 0.63 mol C

10.0 g sample

-1.77 g N

-0.64 g H

7.6 g C

10.63 mol H2a1.008 g H

mol Hb = 0.64 g H

13.8 * 1023 atoms H2¢ 1 mol

6.022 * 1023 atoms≤ = 0.63 mol H

11.77 g N2a 1 mol N

14.01 g Nb = 0.126 mol N

110.0 g compound210.1772 = 1.77 g N

118 g Al2a 1 mol Al

26.98 g Alb a2 mol Mg

1 mol Alb a24.31 g Mg

mol Mgb = 32 g Mg

g Al ¡ mol Al ¡ mol Mg ¡ g Mg

Co3Mo2Cl11 .


35.45 g Clb = 1.45 mol Cl

1.45 mol Cl

0.264= 5.49 mol Cl

125.3 g Mo2a 1 mol Mo

95.94 g Mob = 0.264 mol Mo

0.264 mol Mo

0.264= 1.00 mol Mo

- 84 -

- Chapter 7 -


75. Let

Look in the periodic table for the element that has The element is carbon. The mystery element is carbon.

76. (a) (divide the molecular formula by 6)(b) (divide the molecular formula by 2)(c) (divide the molecular formula by 3)(d) (divide the molecular formula by 1)(e) (divide the molecular formula by 2)

77. First determine the elements in compound


Then compound with a molar mass of

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

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

(2.934 g H2O)a 2.016 g H

18.02 g H2Ob = 0.3282 g H

(4.776 g CO2)a 12.01 g C

44.01 g CO2b = 1.303 g C

% O =3116.002

103.0 11002 = 47.06%

% Al =2126.98 g2

102.0 g 11002 = 52.90%

2126.98 g2 + 3116.00 g2 = 102.0 gA2B3 = Al2O3

A(BC)3 = Al(OH)3

C: 10.03882178.01 g2 =3.03 g

3= 1.01 g (hydrogen)

B: (0.6153)(78.01 g) =48.00 g

3= 16.00 g (oxygen)

A: 10.34592178.01 g2 = 26.98 g (aluminum)

A(BC)3 :

C6H2Cl2OC25H52

CH2OC4H9

CH2O

12.0 g>mol.

12.0

g

mol= y

40.0 - 16.00 = 2y

40.0 = 16.00 + 2y y = molar mass of A

x = 40.0 g O>mol A2O

0.400x = 16.00 g O (Since A2O has only one mol of O atoms)

x = molar mass of A2O

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- Chapter 7 -


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

The empirical formula is C2H6O

(34.76 g O)a 1 mol O

16.00 g Ob = 2.173

2.173 mol O

2.173= 1.000 mol O

(13.13 g H)a 1 mol H

1.008 g Hb = 13.03

13.03 mol H

2.173= 5.996 mol H

(52.12 g C)a1 mol C

12.01 gb = 4.340

4.340 mol C

2.173= 1.997 mol C

a0.869 g O

2.500 gb (100) = 34.76% O

a0.328 g H

2.500 gb (100) = 13.13% H

a1.303 g C

2.500 gb(100) = 52.12% C

2.500 g compound -1.303 g C -0.3282 g H = 0.869 g O

- 86 -

- Chapter 7 -


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