Chapter 11 Solutions to HW

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Solutions Manual Chemistry: Matter and Change • Chapter 11 135

The MoleThe Mole

Section 11.1 Measuring Matter pages 309–312

Practice Problemspages 311, 312

1. Determine the number of atoms in 2.50 mol Zn.

2.50 mol Zn �

� 1.51 � 1024 atoms of Zn

2. Given 3.25 mol AgNO3, determine the numberof formula units.

3.25 mol AgNO3 �

� 1.96 � 1024 formula units of AgNO3

3. Calculate the number of molecules in 11.5 molH2O.

11.5 mol H2O �

� 6.92 � 1024 molecules of H2O

4. How many moles contain each of the following?

a. 5.75 � 1024 atoms Al

5.75 � 1024 atoms Al �

� 9.55 mol Al

b. 3.75 � 1024 molecules CO2

3.75 � 1024 molecules CO2 �

� 6.23 mol CO2

c. 3.58 � 1023 formula units ZnCl23.58 � 1023 formula units ZnCl2 �

� 0.595 mol ZnCl2

d. 2.50 � 1020 atoms Fe

2.50 � 1020 atoms Fe �

� 4.15 � 10�4 mol Fe

Section 11.1 Assessment page 312

5. How is a mole similar to a dozen?

The mole is a unit for counting 6.02 � 1023

representative particles. The dozen is used tocount 12 items.

6. What is the relationship between Avogadro’snumber and one mole?

One mole contains Avogadro’s number (6.02 � 1023) of representative particles.

7. Explain how you can convert from the numberof representative particles of a substance tomoles of that substance.

Multiply the number of representative particles by1 mol/6.02 � 1023 representative particles.

8. Explain why chemists use the mole.

Chemists use the mole because it is a convenientway of knowing how many representativeparticle are in a sample.

9. Thinking Critically Arrange the followingfrom the smallest number of representativeparticles to the largest number of representativeparticles: 1.25 � 1025 atoms Zn; 3.56 mol Fe;6.78 � 1022 molecules glucose (C6H12O6).

From smallest to largest: 6.78 � 1022 moleculesglucose, 2.14 � 1024 atoms Fe, 1.25 � 1025 atomsZn.

10. Using Numbers Determine the number ofrepresentative particles in each of the followingand identify the representative particle: 11.5 molAg; 18.0 mol H2O; 0.150 mol NaCl.

11.5 mol Ag �

� 6.92 � 1024 atoms Ag

18.0 mol H2O �

� 1.08 � 1025 molecules H2O

0.150 mol NaCl �

� 9.03 � 1022 formula units NaCl

6.02 � 1023 formula units NaCl��

1 mol NaCl

6.02 � 1023 molecules H2O��

1 mol H2O

6.02 � 1023 atoms Ag��

1 mol Ag

1 mol��6.02 � 1023 atoms

1 mol��6.02 � 1023 formula units

1 mol��6.02 � 1023 molecules

1 mol��6.02 � 1023 atoms

6.02 � 1023 molecules��

1 mol

6.02 � 1023 formula units��

1 mol

6.02 � 1023 atoms��

1 mol

SOLUTIONS MANUALCHAPTER 11

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136 Chemistry: Matter and Change • Chapter 11 Solutions Manual

Section 11.2 Mass and the Mole pages 313–319

Practice Problemspages 316, 318

11. Determine the mass in grams of each of the following.

a. 3.57 mol Al

3.57 mol Al � � 96.3 g Al

b. 42.6 mol Si

42.6 mol Si � � 1.20 � 103 g Si

c. 3.45 mol Co

3.45 mol Co � � 203 g Co

d. 2.45 mol Zn

2.45 mol Zn � � 1.60 � 102 g Zn

12. Determine the number of moles in each of thefollowing.

a. 25.5 g Ag

25.5 g Ag � � 0.236 mol Ag

b. 300.0 g S

300.0 g S � � 9.355 mol S

c. 125 g Zn

125 g Zn � � 1.91 mol Zn

d. 1.00 kg Fe

1.00 kg Fe � � �

17.9 mol Fe

13. How many atoms are in each of the followingsamples?

a. 55.2 g Li

55.2 g Li � �

� 4.79 � 1024 atoms Li

b. 0.230 g Pb

0.230 g Pb � �

� 6.68 � 1020 atoms Pb

c. 11.5 g Hg

11.5 g Hg � �

� 3.45 � 1022 atoms Hg

d. 45.6 g Si

45.6 g Si � �

� 9.77 � 1023 atoms Si

e. 0.120 kg Ti

0.120 kg Ti � � �

� 1.51 � 1024 atoms Ti

14. What is the mass in grams of each of the following?

a. 6.02 � 1024 atoms Bi

6.02 � 1024 atoms Bi � �

� 2.09 � 103 g Bi

b. 1.00 � 1024 atoms Mn

1.00 � 1024 atoms Mn � �

� 91.3 g Mn54.94 g Mn��1 mol Mn

1 mol Mn��6.02 � 1023 atoms

209.0 g Bi��1 mol Bi

1 mol Bi��6.02 � 1023 atoms

6.02 � 1023 atoms��

1 mol

1 mol Ti��47.87 g Ti

1000 g Ti��

1 kg Ti

6.02 � 1023 atoms��

1 mol

1 mol Si��28.09 g Si

6.02 � 1023 atoms��

1 mol

1 mol Hg��200.6 g Hg

6.02 � 1023 atoms��

1 mol

1 mol Pb��207.2 g Pb

6.02 � 1023 atoms��

1 mol1 mol Li��6.94 g Li

1 mol Fe��55.85 g Fe

1000 g Fe��

1 kg Fe

1 mol Zn��65.38 g Zn

1 mol S��32.07 g S

1 mol Ag��107.9 g Ag

65.38 g Zn��1 mol Zn

58.93 g Co��1 mol Co

28.09 g Si��1 mol Si

26.98 g Al��1 mol Al


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c. 3.40 � 1022 atoms He

3.40 � 1022 atoms He � �

� 0.226 g He

d. 1.50 � 1015 atoms N

1.50 � 1015 atoms N � �

� 3.49 � 10�8 g N

e. 1.50 � 1015 atoms U

1.50 � 1015 atoms U � �

� 5.93 � 10�7 g U

Section 11.2 Assessmentpage 319

15. Explain what is meant by molar mass.

Molar mass is the mass in grams of one mole ofany pure substance.

16. What conversion factor should be used to convert from mass to moles? Moles to mass?

Mass-to-mole conversions use the conversionfactor 1 mol/number of grams. Moles-to-massconversions use the conversion factor number ofgrams/1 mol.

17. Explain the steps needed to convert the mass of an element to the number of atoms of the element.

Multiply the mass by the inverse of molar mass,and then multiply by Avogadro’s number.

18. Thinking Critically The mass of a single atomis usually given in the unit amu. Would it bepossible to express the mass of a single atom ingrams? Explain.

Because one mole is equal to 6.02 � 1023 atomsand the molar mass of an element is equal to onemole, the mass of a single atom can be calculatedby dividing the mass of one mole by Avogadro’snumber.

19. Sequencing Arrange the following in order ofmass from the smallest mass to the largest: 1.0mol Ar, 3.0 � 1024 atoms Ne, 20 g Kr.

1.0 mol Ar � � 39.95 g Ar

3.0 � 1024 atoms Ne � �

� 101 g Ne

20 g Kr, 1.0 mol Ar, 3.0 � 1024 atoms Ne

Section 11.3 Moles ofCompounds pages 320–327

Practice Problemspages 321–324, 326

20. Determine the number of moles of chloride ionsin 2.50 mol ZnCl2.

2.50 mol ZnCl2 � � 5.00 mol Cl�

21. Calculate the number of moles of each elementin 1.25 mol glucose (C6H12O6).

1.25 mol C6H12O6 � � 7.50 mol C

1.25 mol C6H12O6 � � 15.0 mol H

1.25 mol C6H12O6 � � 7.50 mol O

22. Determine the number of moles of sulfate ionspresent in 3.00 mol iron(III) sulfate (Fe2(SO4)3).

3.00 mol Fe2(SO4)3 �

� 9.00 mol SO42�

23. How many moles of oxygen atoms are presentin 5.00 mol diphosphorus pentoxide (P2O5)?

5.00 mol P2O5 � � 25.0 mol O5 mol O��1 mol P2O5

3 mol SO42�

��1 mol Fe2(SO4)3

6 mol O��1 mol C6H12O6

12 mol H��1 mol C6H12O6

6 mol C��1 mol C6H12O6

2 mol Cl��1 mol ZnCl2

20.18 g Ne��1 mol Ne

1 mol Ne��6.02 � 1023 atoms Ne

39.95 g Ar��1 mol Ar

238.0 g U��1 mol U

1 mol U��6.02 � 1023 atoms

14.01 g N��1 mol N

1 mol N��6.02 � 1023 atoms

4.003 g He��1 mol He

1 mol He��6.02 � 1023 atoms


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24. Calculate the number of moles of hydrogenatoms in 11.5 mol water.

11.5 mol H2O � � 23.0 mol H

25. Determine the molar mass of each of the following ionic compounds: NaOH, CaCl2,KC2H3O2, Sr(NO3)2, and (NH4)3PO4.

NaOH

1 mol Na � � 22.99 g

1 mol O � � 16.00 g

1 mol H � � 1.008 g

molar mass NaOH � 40.00 g/mol

CaCl2

1 mol Ca � � 40.08 g

2 mol Cl � � 70.90 g

molar mass CaCl2 � 110.98 g/mol

KC2H3O2

1 mol K � � 39.10 g

2 mol C � � 24.02 g

3 mol H � � 3.024 g

2 mol O � � 32.00 g

molar mass KC2H3O2 � 98.14 g/mol

Sr(NO3)2

1 mol Sr � � 87.62 g

2 mol N � � 28.02 g

6 mol O � � 96.00 g

molar mass Sr(NO3)2 � 211.64 g/mol

(NH4)3PO4

3 mol N � � 42.03 g

12 mol H � � 12.096 g

1 mol P � � 30.97 g

4 mol O � � 64.00 g

molar mass (NH4)3PO4 � 149.10 g/mol

26. Calculate the molar mass of each of the following molecular compounds: C2H5OH,C12H22O11, HCN, CCl4, and H2O.

C2H5OH

2 mol C � � 24.02 g

6 mol H � � 6.048 g

1 mol O � � 16.00 g

molar mass C2H5OH � 46.07 g/mol

C12H22O11

12 mol C � � 144.12 g

22 mol H � � 22.176 g

11 mol O � � 176.00 g

molar mass C12H22O11 � 342.30 g/mol

HCN

1 mol H � � 1.008 g

1 mol C � � 12.01 g

1 mol N � � 14.01 g

molar mass HCN � 27.03 g/mol

CCl4

1 mol C � � 12.01 g

4 mol Cl � � 141.80 g

molar mass CCl4 � 153.81 g/mol

35.45 g Cl��1 mol Cl

12.01 g C��1 mol C

14.01 g N��1 mol N

12.01 g C��1 mol C

1.008 g H��1 mol H

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

16.00 g O��1 mol O

30.97 g P��1 mol P

1.008 g H��1 mol H

14.01 g N��1 mol N

16.00 g O��1 mol O

14.01 g N��1 mol N

87.62 g Sr��1 mol Sr

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

39.10 g K��1 mol K


40.08 g Ca��1 mol Ca

1.008 g H��1 mol H

16.00 g O��1 mol O

22.99 g Na��1 mol Na

2 mol H��1 mol H2O


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H2O

2 mol H � � 2.016 g

1 mol O � � 16.00 g

molar mass H2O � 18.02 g/mol

27. What is the mass of 3.25 moles of sulfuric acid(H2SO4)?

Step 1: Find the molar mass of H2SO4.

2 mol H � � 2.016 g

1 mol S � � 32.07 g

4 mol O � � 64.00 g

molar mass H2SO4 � 98.09 g/mol

Step 2: Make mole 0 mass conversion.

3.25 mol H2SO4 �

� 319 g H2SO4

28. What is the mass of 4.35 � 10�2 moles of zincchloride (ZnCl2)?

Step 1: Find the molar mass of ZnCl2.

1 mol Zn � � 65.38 g

2 mol Cl � � 70.90 g

molar mass ZnCl2 � 136.28 g/mol


4.35 � 10�2 mol ZnCl2 �

� 5.93 g ZnCl2

29. How many grams of potassium permanganateare in 2.55 moles?

Step 1: Find the molar mass of KMnO4.

1 mol K � � 39.10 g

1 mol Mn � � 54.94 g

4 mol O � � 64.00 g

molar mass KMnO4 � 158.04 g/mol


2.55 mol KMnO4 �

� 403 g KMnO4

30. Determine the number of moles present in eachof the following.

a. 22.6 g AgNO3

Step 1: Find the molar mass of AgNO3.

1 mol Ag � � 107.9 g

1 mol N � � 14.01 g

3 mol O � � 48.00 g

molar mass AgNO3 � 169.9 g/mol

Step 2: Make mass 0 mole conversion.

22.6 g AgNO3 �

� 0.133 mol AgNO3

b. 6.50 g ZnSO4

Step 1: Find the molar mass of ZnSO4.

1 mol Zn � � 65.39 g

1 mol S � � 32.07 g

4 mol O � � 64.00 g

molar mass ZnSO4 � 161.46 g/mol


6.50 g ZnSO4 �

� 0.0403 mol ZnSO4

1 mol ZnSO4��161.46 g ZnSO4

16.00 g O��1 mol O

32.07 g S��1 mol S


1 mol AgNO��169.9 g AgNO3

16.00 g O��1 mol O

14.01 g N��1 mol N

107.9 g Ag��1 mol Ag

158.04 g KMnO4��1 mol KMnO4

16.00 g O��1 mol O

54.94 g Mn��1 mol Mn

39.10 g K��1 mol K

136.28 g ZnCl2��1 mol ZnCl2



98.09 g H2SO4��1 mol H2SO4

16.00 g O��1 mol O

32.07 g S��1 mol S

1.008 g H��1 mol H

16.00 g O��1 mol O

1.008 g H��1 mol H


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c. 35.0 g HCl

Step 1: Find the molar mass of HCl.

1 mol H � � 1.008 g

1 mol Cl � � 35.45 g

molar mass HCl � 36.46 g/mol


35.0 g HCl � � 0.960 mol HCl

d. 25.0 g Fe2O3

Step 1: Find the molar mass of Fe2O3.

2 mol Fe � � 111.70 g

3 mol O � � 48.00 g

molar mass Fe2O3 � 159.70 g/mol


25.0 g Fe2O3 �

� 0.157 mol Fe2O3

e. 254 g PbCl4Step 1: Find the molar mass of PbCl4.

1 mol Pb � � 207.2 g

4 mol Cl � � 141.80 g

molar mass PbCl4 � 349.0 g/mol


254 g PbCl4 �

� 0.728 mol PbCl4

31. A sample of silver chromate (Ag2CrO4) has amass of 25.8 g.

Step 1: Find the molar mass of Ag2CrO4.

2 mol Ag � � 215.8 g

1 mol Cr � � 52.00 g

4 mol O � � 64.00 g

molar mass Ag2CrO4 � 331.8 g/mol


25.8 g Ag2CrO4 �

� 0.0778 mol Ag2CrO4

Step 3: Make mole 0 formula unit conversion.

0.0778 mol Ag2CrO4 �

� 4.68 � 1022 formula units Ag2CrO4

a. How many Ag� ions are present?

4.68 � 1022 formula units Ag2CrO4 �

� 9.36 � 1022 Ag� ions

b. How many CrO42� ions are present?

4.68 � 1022 formula units Ag2CrO4 �

� 4.68 � 1022 CrO42� ions

c. What is the mass in grams of one formulaunit of silver chromate?

�

� 5.51 � 10�22 g Ag2CrO4/formula unit

32. What mass of sodium chloride contains 4.59 � 1024 formula units?

Step 1: Find the number of moles of NaCl.

4.59 � 1024 formula units NaCl �

� 7.62 mol NaCl1 mol NaCl

��6.02 � 1023 formula units

1 mol Ag2CrO4��6.02 � 1023 formula units

331.8 g Ag2CrO4��1 mol Ag2CrO4

1 CrO42� ion

��1 formula unit Ag2CrO4

2 Ag� ions��1 formula unit Ag2CrO4


1 mol

1 mol Ag2CrO4��331.8 g Ag2CrO4

16.00 g O��1 mol O

52.00 g Cr��1 mol Cr

107.9 g Ag��1 mol Ag

1 mol PbCl4��349.0 g PbCl4


207.2 g Pb��1 mol Pb

1 mol Fe2O3��159.70 g Fe2O3

16.00 g O��1 mol O

55.85 g Fe��1 mol Fe

1 mol HCl��36.46 g HCl


1.008 g H��1 mol H


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Step 2: Find the molar mass of NaCl.

1 mol Na � � 22.99 g

1 mol Cl � � 35.45 g

molar mass NaCl � 58.44 g/mol


7.62 mol NaCl � � 445 g NaCl

33. A sample of ethanol (C2H5OH) has a mass of45.6 g.

Step 1: Find the molar mass of C2H5OH.

2 mol C � � 24.02 g

6 mol H � � 6.048 g

1 mol O � � 16.00 g

molar mass C2H5OH � 46.07 g/mol


45.6 g C2H5OH �

� 0.990 mol C2H5OH

Step 3: Make mole 0 molecule conversion.

0.990 mol C2H5OH � �

5.96 � 1023 molecules C2H5OH

a. How many carbon atoms does the samplecontain?

5.96 � 1023 molecules C2H5OH �

� 1.19 � 1024 C atoms

b. How many hydrogen atoms are present?


� 3.58 � 1024 H atoms

c. How many oxygen atoms are present?


� 5.96 � 1023 O atoms

34. A sample of sodium sulfite (Na2SO3) has amass of 2.25 g.

Step 1: Find the molar mass of Na2SO3

2 mol Na � � 45.98 g

1 mol S � � 32.07 g

3 mol O � � 48.00 g

molar mass Na2SO3 � 126.05 g/mol

Step 2: Make mass 0 mole conversion

2.25 g Na2SO3 �

� 0.0179 mol Na2SO3

Step 3: Make mole 0 formula unit conversion

0.0179 mol Na2SO3 �

� 1.08 � 1022 formula units Na2SO3

a. How many Na� ions are present?

1.08 � 1022 formula units Na2SO3 �

� 2.16 � 1022 Na� ions

b. How many SO32� ions are present?

1.08 � 1022 formula units Na2SO3 �

� 1.08 � 1022 SO32� ions

c. What is the mass in grams of one formulaunit of Na2SO3?

�

� 2.09 � 10�22 g Na2SO3/formula unit

1 mol Na2SO3��6.02 � 1023 formula units

126.08 g Na2SO3��1 mol Na2SO3

1 SO32� ions

��1 formula unit Na2SO3

2 Na� ions��1 formula unit Na2SO3


1 mol

1 mol Na2SO3��126.05 g Na2SO3

16.00 g O��1 mol O

32.07 g S��1 mol S


1 O atom��1 molecule C2H5OH

6 H atoms��1 molecule C2H5OH

2 C atoms��1 molecule C2H5OH

6.02 � 1023 molecules��

1 mol

1 mol C2H5OH��46.07 g C2H5OH

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

58.44 g NaCl��1 mol NaCl




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35. A sample of carbon dioxide has a mass of 52.0 g.

Step 1: Find the molar mass of CO2.

1 mol C � � 12.01 g

2 mol O � � 32.00 g

molar mass CO2 � 44.01 g/mol


52.0 g CO2 � � 1.18 mol CO2

Step 3: Make mole 0 molecule conversion.

1.18 mol CO2 �

� 7.11 � 1023 molecules CO2

a. How many carbon atoms are present?


� 7.11 � 1023 C atoms

b. How many oxygen atoms are present?


� 1.42 � 1024 O atoms

c. What is the mass in grams of one moleculeof CO2?

�

� 7.31 � 10�23 g CO2/molecule


36. Describe how you can determine the molarmass of a compound.

Multiply the mass of one mole of each element bythe ratio of that element to one mole of thecompound. Add the resulting masses.

37. What three conversion factors are often used inmole conversions?

,

,

38. Explain how you can determine the number ofatoms or ions in a given mass of a compound.

Convert the mass to moles, multiply the numberof moles by the ratio of the number of atoms orions to one mole, multiply by Avogadro’s number.

39. If you know the mass in grams of a molecule ofa substance, could you obtain the mass of amole of that substance? Explain.

Yes, multiply the mass in grams of the moleculeby Avogadro’s number.

40. Thinking Critically Design a bar graph thatwill show the number of moles of each elementpresent in 500 g dioxin (C12H4Cl4O2), apowerful poison.

12 mol C � � 144.12 g C

4 mol H � � 4.032 g H

4 mol Cl � � 141.80 g Cl

2 mol O � � 32.00 g O

molar mass � 321.96 g/mol

16.00 g O��1 mol O


1.008 g H��1 mol H

12.01 g C��1 mol C

0369

121518212427

C H Cl O

Moles of Elements in 500 g Dioxin

Element

Nu

mb

er o

f m

ole

s

number of atoms of element��

1 mol compound

6.02 � 1023 representative particles��

1 mol

number of grams��

1 mol

1 mol��6.02 � 1023 molecules

44.01 g CO2��1 mol CO2

2 O atoms��1 molecule CO2

1 C atom��1 molecule CO2

6.02 � 1023 molecules��

1 mol

1 mol CO2��44.01 g CO2

16.00 g O��1 mol O

12.01 g C��1 mol C


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500 g C12H4Cl4O2 �

� 2 mol C12H4Cl4O2

2 mol C12H4Cl4O2 � � 24 mol C

2 mol C12H4Cl4O2 � � 8 mol H

2 mol C12H4Cl4O2 � � 8 mol Cl

2 mol C12H4Cl4O2 � � 4 mol O

41. Applying Concepts The recommended dailyallowance of calcium is 1000 mg of Ca2� ions.Calcium carbonate is used to supply the calciumin vitamin tablets. How many moles of calciumions does 1000 mg represent? How many molesof calcium carbonate are needed to supply therequired amount of calcium ions? What mass ofcalcium carbonate must each tablet contain?

1 mol Ca � � 40.08 g Ca

1 mol C � � 12.01 g C

3 mol O � � 48.00 g O


1000 mg Ca2� � �

� 0.02 mol Ca2�

0.02 mol Ca2� � �

� 2 g CaCO3

Section 11.4 Empirical andMolecular Formulas pages 328–337

Practice Problemspages 331, 333, 335, 337

42. Determine the percent by mass of each elementin calcium chloride.

Steps 1 and 2: Assume 1 mole; calculate molarmass of CaCl2.

1 mol Ca � � 40.08 g

2 mol Cl � � 70.90 g

molar mass CaCl2 � 110.98 g/mol

Step 3: Determine percent by mass of eachelement.

percent Ca � � 100

� 36.11% Ca

percent Cl � � 100

� 63.89% Cl

43. Calculate the percent composition of sodiumsulfate.

Steps 1 and 2: Assume 1 mole; calculate molarmass of Na2SO4.

2 mol Na � � 45.98 g

1 mol S � � 32.07 g

4 mol O � � 64.00 g

molar mass Na2SO4 � 142.05 g/mol


percent Na � � 100

� 32.37% Na

percent S � � 100

� 22.58% S

32.07 g S��142.05 g Na2SO4

45.98 g Na��142.05 g Na2SO4

16.00 g O��1 mol O

32.07 g S��1 mol S


70.90 g Cl��110.98 g CaCl2

40.08 g Ca��110.98 g CaCl2



100.09 g CaCO3��1 mol CaCO3

1 mol CaCO3��1 mol Ca2�

1 mol Ca2��40.08 g Ca2�

1 g Ca2�

��103 mg Ca2�

16.00 g O��1 mol O

12.01 g C��1 mol C


2 mol O��1 mol C12H4Cl4O2

4 mol Cl��1 mol C12H4Cl4O2

4 mol H��1 mol C12H4Cl4O2

12 mol C��1 mol C12H4Cl4O2

1 mol C12H4Cl4O2��321.96 g C12H4Cl4O2


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percent O � � 100

� 45.05% O

44. Which has the larger percent by mass of sulfur,H2SO3 or H2S2O8?

Steps 1 and 2: Assume 1 mole; calculate molarmass of H2SO3.

2 mol H � � 2.016 g

1 mol S � � 32.06 g

3 mol O � � 48.00 g

molar mass H2SO3 � 82.08 g/mol

Step 3: Determine percent by mass of S.

percent S � � 100 � 39.06% S

Repeat steps 1 and 2 for H2S2O8. Assume 1 mole;calculate molar mass of H2S2O8.

2 mol H � � 2.016 g

2 mol S � � 64.12 g

8 mol O � � 128.00 g

molar mass H2S2O8 � 194.14 g/mol

Step 3: Determine percent by mass of S.

percent S � � 100

� 33.03% S

H2SO3 has a larger percent by mass of S.

45. What is the percent composition of phosphoricacid (H3PO4)?

Steps 1 and 2: Assume 1 mole; calculate molarmass of H3PO4.

3 mol H � � 3.024 g

1 mol P � � 30.97 g

4 mol O � � 64.00 g

molar mass H3PO4 � 97.99 g/mol


percent H � � 100 � 3.08% H

percent P � � 100 � 31.61% P

percent O � � 100 � 65.31% O

46. A blue solid is found to contain 36.84%nitrogen and 63.16% oxygen. What is theempirical formula for this solid?

Step 1: Assume 100 g sample; calculate moles ofeach element.

36.84 g N � � 2.630 mol N

63.16 g O � � 3.948 mol O

Step 2: Calculate mole ratios.

� �

� �

The simplest ratio is 1 mol N: 1.5 mol O.

Step 3: Convert decimal fraction to whole number.

In this case, multiply by 2 because 1.5 � 2 � 3.Therefore, the empirical formula is N2O3.

1.5 mol O��

1 mol N1.500 mol O��1.000 mol N

3.948 mol O��2.630 mol N

1 mol N�1 mol N

1.000 mol N��1.000 mol N

2.630 mol N��2.630 mol N

1 mol O��16.00 g O

1 mol N��14.01 g N

64.00 g O��97.99 g H3PO4

30.97 g P��97.99 g H3PO4

3.024 g H��97.99 g H3PO4

16.00 g O��1 mol O

30.97 g P��1 mol P

1.008 g H��1 mol H

64.12 g S��194.14 g H2S2O8

16.00 g O��1 mol O

32.06 g S��1 mol S

1.008 g H��1 mol H

32.06 g S��82.08 g H2SO3

16.00 g O��1 mol O

32.06 g S��1 mol S

1.008 g H��1 mol H

64.00 g O��142.05 g Na2SO4


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47. Determine the empirical formula for acompound that contains 35.98% aluminum and64.02% sulfur.


35.98 g Al � � 1.334 mol Al

64.02 g S � � 1.996 mol S


� �

� �

The simplest ratio is 1 mol Al: 1.5 mol S.


In this case, multiply by 2 because 1.5 � 2 � 3.Therefore, the empirical formula is Al2S3.

48. Propane is a hydrocarbon, a compoundcomposed only of carbon and hydrogen. It is81.82% carbon and 18.18% hydrogen. What isthe empirical formula?


81.82 g C � � 6.813 mol C

18.18 g H � � 18.04 mol H


� �

� �

The simplest ratio is 1 mol: 2.65 mol H.


In this case, multiply by 3 because 2.65 � 3 � 7.95 � 8. Therefore, the empirical formula isC3H8.

49. The chemical analysis of aspirin indicates thatthe molecule is 60.00% carbon, 4.44%hydrogen, and 35.56% oxygen. Determine theempirical formula for aspirin.


60.00 g C � � 5.00 mol C

4.44 g H � � 4.40 mol H

35.56 g O � �2.22 mol O


� �

� �

� �

The simplest ratio is 2.25 mol C: 2 mol H: 1 mol O.


In this case, multiply by 4 because 2.25 � 4 � 9.Therefore, the empirical formula is C9H8O4.

50. What is the empirical formula for a compoundthat contains 10.89% magnesium, 31.77% chlo-rine, and 57.34% oxygen?


10.89 g Mg � � 0.4480 mol Mg

31.77 g Cl � � 0.8962 mol Cl

57.34 g O � � 3.584 mol O


� �

� �

� �8 mol O��1 mol Mg

7.999 mol O��1.000 mol Mg

3.584 mol O��0.4480 mol Mg

2 mol Cl��1 mol Mg

2.000 mol Cl��1.000 mol Mg

0.8962 mol Cl��0.4480 mol Mg

1 mol Mg��1 mol Mg

1.000 mol Mg��1.000 mol Mg

0.4480 mol Mg��0.4480 mol Mg

1 mol O��16.00 g O

1 mol Cl��35.45 g Cl

1 mol Mg��24.31 g Mg

1 mol O�1 mol O

1.00 mol O��1.00 mol O

2.22 mol O��2.22 mol O

2 mol H�1 mol O

1.98 mol H��1.00 mol O

4.40 mol H��2.22 mol O

2.25 mol C��

1 mol O2.25 mol C��1.00 mol O

5.00 mol C��2.22 mol O

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C

2.65 mol H��

1 mol C2.649 mol H��1.000 mol C

18.04 mol H��6.813 mol C

1 mol C��

1 mol C1.000 mol C��1.000 mol C

6.813 mol C��6.813 mol C

1 mol H��1.008 g H

1 mol C��12.01 g C

1.5 mol S��1 mol Al

1.500 mol S��1.000 mol Al

1.996 mol S��1.334 mol Al

1 mol Al��1 mol Al

1.000 mol Al��1.000 mol Al

1.334 mol Al��1.334 mol Al

1 mol S��32.06 g S

1 mol Al��26.98 g Al


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The empirical formula is MgCl2O8 or Mg(ClO4)2.

The simplest ratio is 1 mol Mg: 2 mol Cl: 8 mol O.

51. Analysis of a chemical used in photographicdeveloping fluid indicates a chemical composi-tion of 65.45% C, 5.45% H, and 29.09% O. Themolar mass is found to be 110.0 g/mol.Determine the molecular formula.


65.45 g C � � 5.450 mol C

5.45 g H � � 5.41 mol H

29.09 g O � � 1.818 mol O


� �

� �

� �

The simplest ratio is 1 mol: 2.65 mol H.

Therefore, the empirical formula is C3H3O.

Step 3: Calculate the molar mass of the empiricalformula.

3 mol C � � 36.03 g

3 mol H � � 3.024 g

1 mol O � � 16.00 g

molar mass C3H3O � 55.05 g/mol

Step 4: Determine whole number multiplier.

� 1.998, or 2

The molecular formula is C6H6O2.

52. A compound was found to contain 49.98 g carbon and 10.47 g hydrogen. The molar massof the compound is 58.12 g/mol. Determine the molecular formula.


49.98 g C � � 4.162 mol C

10.47 g H � � 10.39 mol H


� �

� �

The simplest ratio is 1 mol C: 2.5 mol H.

Because 2.5 � 2 � 5, the empirical formula isC2H5.


2 mol C � � 24.02 g

5 mol H � � 5.040 g

molar mass C2H5 � 29.06 g/mol


� 2.000

The molecular formula is C4H10.

53. A colorless liquid composed of 46.68% nitrogenand 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula?


46.68 g N � � 3.332 mol N

53.32 g O � � 3.333 mol O


� �1 mol N�1 mol N

1.000 mol N��1.000 mol N

3.332 mol N��3.332 mol N

1 mol O��16.00 g O

1 mol N��14.01 g N

58.12 g/mol��29.06 g/mol

1.008 g H��1 mol H

12.01 g C��1 mol C

2.5 mol H��

1 mol C2.50 mol H��1.000 mol C

10.39 mol H��4.162 mol C

1 mol C�1 mol C

1.000 mol C��1.000 mol C

4.162 mol C��4.162 mol C

1 mol H��1.008 g H

1 mol C��12.01 g C

110.0 g/mol��55.05 g/mol

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

1 mol O�1 mol O

1.000 mol O��1.000 mol O

1.818 mol O��1.818 mol O

3 mol H�1 mol O

2.97 mol H��1.00 mol O

5.41 mol H��1.818 mol O

3 mol C�1 mol O

3.000 mol C��1.000 mol O

5.450 mol C��1.818 mol O

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C


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� �

The simplest ratio is 1 mol N: 1 mol O.

The empirical formula is NO.


1 mol N � � 14.01 g

1 mol O � � 16.00 g

molar mass NO � 30.01 g/mol


� 2.000

The molecular formula is N2O2.

54. When an oxide of potassium is decomposed,19.55 g K and 4.00 g O are obtained. What isthe empirical formula for the compound?

Step 1: Calculate moles of each element.

19.55 g K � � 0.5000 mol K

4.00 g O � � 0.250 mol O


� �

� �

The simplest ratio is 2 mol K: 1 mol O.

The empirical formula is K2O.

55. Analysis of a compound composed of iron andoxygen yields 174.86 g Fe and 75.14 g O. Whatis the empirical formula for this compound?

Step 1: Calculate moles of each element

174.86 g Fe � � 3.131 mol Fe

75.14 g O � � 4.696 mol O


� �

� �

The simplest ratio is 1 mol Fe: 1.5 mol O.

Because 1.5 � 2 � 3, the empirical formula isFe2O3.

56. The pain reliever morphine contains 17.900 g C,1.680 g H, 4.225 g O, and 1.228 g N. Determinethe empirical formula.


17.900 g C � � 1.490 mol C

1.680 g H � � 1.667 mol H

4.225 g O � � 0.2641 mol O

1.228 g N � � 0.087 65 mol N


� �

� �

� �

� �

The simplest ratio is 17 mol C: 19 mol H: 3 molO: 1 mol N.

The empirical formula is C17H19O3N.

57. An oxide of aluminum contains 0.545 g Al and0.485 g O. Find the empirical formula for theoxide.


0.545 g Al � � 0.0202 mol Al

0.485 g O � � 0.0303 mol O1 mol O��6.00 g O

1 mol Al��26.98 g Al

3 mol O�1 mol N

3.013 mol O��1.000 mol N

0.2641 mol O��0.08765 mol N

19 mol H��1 mol N

19.02 mol H��1.000 mol N

1.667 mol H��0.08765 mol N

17 mol C��1 mol N

17.00 mol C��1.000 mol N

1.490 mol C��0.08765 mol N

1 mol N�1 mol N

1.000 mol N��1.000 mol N

0.08765 mol N��0.08765 mol N

1 mol N��14.01 g N

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C

1.5 mol O��1 mol Fe

1.500 mol O��1.000 mol Fe

4.696 mol O��3.131 mol Fe

1 mol Fe��1 mol Fe

1.000 mol Fe��1.000 mol Fe

3.131 mol Fe��3.131 mol Fe

1 mol O��16.00 g O

1 mol Fe��55.85 g Fe

1 mol O�1 mol O

1.00 mol O��1.00 mol O

0.250 mol O��0.250 mol O

2 mol K�1 mol O

2.00 mol K��1.00 mol O

0.5000 mol K��0.250 mol O

1 mol O��16.00 g O

1 mol K��39.10 g K

60.01 g/mol��30.01 g/mol

16.00 g O��1 mol O

14.01 g N��1 mol N

1 mol O�1 mol N

1.000 mol O��1.000 mol N

3.333 mol O��3.332 mol N


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� �

� �

The simplest ratio is 1 mol Al: 1.5 mol O.

Because 1.5 � 2 � 3, the empirical formula isAl2O3.


58. Explain how percent composition data for acompound are related to the masses of theelements in the compound.

Percent composition is numerically equal to themass in grams of each element in a 100.0-gsample.

59. What is the difference between an empiricalformula and a molecular formula?

The empirical formula is the simplest whole-number ratio of the atoms in a compound. Themolecular formula is the actual ratio of atoms inthe compound.

60. Explain how you can find the mole ratio in achemical compound.

The mole ratio is determined by calculating themoles of each element in the compound anddividing each number of moles by the smallestnumber of moles. It is sometimes necessary tomultiply the ratio by an integer to obtain wholenumbers.

61. Thinking Critically An analysis for copper wasperformed on two pure solids. One solid wasfound to contain 43.0% copper; the othercontained 32.0% copper. Could these solids besamples of the same copper-containingcompound? Explain your answer.

No, a compound will always have the samechemical analysis. Therefore, if two solid havedifferent percents of copper, they are differentcompounds.

62. Inferring Hematite (Fe2O3) and magnetite(Fe3O4) are two ores used as sources of iron.Which ore provides the greater percent of ironper kilogram?

2 mol Fe � � 111.70 g Fe

3 mol O � � 48.00 g O


3 mol Fe � � 167.55 g Fe

4 mol O � � 64.00 g O


percent by mass � � 100

� 69.94% Fe in Fe2O3


� 72.36% Fe in Fe3O4

Hematite is 69.94% Fe, magnetite is 72.36% Fe.Magnetite contains a greater percentage of ironper kilogram than hematite.

Section 11.5 The Formula for aHydratepages 338–341

Practice Problemspage 340

63. A hydrate is found to have the following percentcomposition: 48.8% MgSO4 and 51.2% H2O.What is the formula and name for this hydrate?

Step 1: Assume 100 g sample; calculate moles ofeach component.

48.8 g MgSO4 �

� 0.405 mol MgSO4

51.2 g H2O � � 2.84 mol H2O1 mol H2O��18.02 g H2O

1 mol MgSO4��120.38 g MgSO4

167.55 g Fe��231.55 g Fe3O4

111.70 g Fe��159.70 g Fe2O3

16.00 g O��1 mol O


16.00 g O��1 mol O


1.5 mol O��1 mol Al

1.50 mol O��1.00 mol Al

0.0303 mol O��0.0202 mol Al

1 mol Al��1 mol Al

1.00 mol Al��1.00 mol Al

0.0202 mol Al��0.0202 mol Al


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�

�

�

�

The formula of the hydrate is MgSO4�7H2O. Itsname is magnesium sulfate heptahydrate.

64. Figure 11-13 shows a common hydrate ofcobalt(II) chloride. If 11.75 g of this hydrate isheated, 9.25 g of anhydrous cobalt chlorideremains. What is the formula and name for thishydrate?

Step 1: Calculate the mass of water driven off.

mass of hydrated compound � mass ofanhydrous compound remaining � 11.75 g CoCl2�xH2O � 9.25 g CoCl2� 2.50 g H2O

Step 2: Calculate moles of each component.

9.25 g CoCl2 �

� 0.0712 mol CoCl2

2.50 g H2O � � 0.139 mol H2O


�

�

�

�

The formula of the hydrate is CoCl2�2H2O. Itsname is cobalt(II) chloride dihydrate.

Section 11.5 Assessment page 341

65. What is a hydrate? What does its name indicateabout its composition?

A hydrate is a compound that has a specificnumber of water molecules associated with it.The name indicates the name of the compoundand the number of water molecules associatedwith each formula unit of compound.

66. Describe the experimental procedure for deter-mining the formula for a hydrate. Explain thereason for each step.

Mass an empty crucible. Add some hydrate andremass. Heat the crucible to drive out the water.Cool and remass. Determine the moles of theanhydrous compound. Subtract the mass of thecrucible after heating from the mass of the cruciblewith the hydrate. The difference is the mass of thewater lost. Determine the moles of water.Determine the simplest whole-number ratio ofmoles of water to moles of anhydrous compound,which will yield the formula of the hydrate.

67. Name the compound having the formulaSrCl2�6H2O.

strontium chloride hexahydrate

68. Thinking Critically Explain how the hydrate illus-trated in Figure 11-13 might be used as a means ofroughly determining the probability of rain.

The hydrate is pink in moist air.

69. Sequencing Arrange these hydrates in order ofincreasing percent water content:MgSO4�7H2O, Ba(OH)2�8H2O, CoCl2�6H2O.

1 mol Mg � � 24.31 g Mg

1 mol S � � 32.07 g S

11 mol O � � 176.00 g O

14 mol H � � 14.112 g H

molar mass MgSO4�7H2O � 246.49 g

1.008 g H��1 mol H

16.00 g O��1 mol O

32.07 g S��1 mol S

24.31 g Mg��1 mol Mg

2 mol H2O��1 mol CoCl2

1.95 mol H2O��1.00 mol CoCl2

0.139 mol H2O��0.0712 mol CoCl2

1 mol CoCl2��1 mol CoCl2

1.00 mol CoCl2��1.00 mol CoCl2

0.0712 mol CoCl2��0.0712 mol CoCl2

1 mol H2O��18.02 g H2O

1 mol CoCl2��129.83 g CoCl2

7 mol H2O��1 mol MgSO4

7.01 mol H2O��1.00 mol MgSO4

2.84 mol H2O��0.405 mol MgSO4

1 mol MgSO4��1 mol MgSO4

1.00 mol MgSO4��1.00 mol MgSO4

0.405 mol MgSO4��0.405 mol MgSO4


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1 mol Ba � � 137.33 g Ba

10 mol O � � 160.00 g O

18 mol H � � 18.144 g H

molar mass Ba(OH)2�8H2O � 315.47 g

1 mol Co � � 58.93 g Co

2 mol Cl � � 70.90 g Cl

6 mol O � � 96.00 g O

12 mol H � � 12.096 g H

molar mass CoCl2�6H2O � 237.93 g

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H

molar mass H2O � 18.02 g

� 100

� 51.17% H2O in MgSO4�7H2O

� 100

� 45.70% H2O in Ba(OH)2�8H2O

� 100

� 45.44% H2O in CoCl2�6H2O

CoCl2�6H2O; Ba(OH)2�8H2O; MgSO4�7H2O

Chapter 11 Assessment pages 346–350

Concept Mapping70. Complete this concept map by placing in each

box the conversion factor needed to convertfrom each measure of matter to the next.

1.

2.

3.

4.

Mastering Concepts71. Why is the mole an important unit to chemists?

(11.1)

A mole allows a chemist to accurately measurethe number of atoms, molecules, or formula unitsin a substance.

72. How is a mole similar to a dozen? (11.1)

A mole, like a dozen, contains a specific numberof items.

73. What is the numerical value of Avogadro’s number? (11.1)

6.02 � 1023

74. What is molar mass? (11.2)

Molar mass is the mass in grams of one mole ofany element or compound.


1 mol

1 mol��6.02 � 1023 representative particles


1 mol

1 mol��number of grams

MolesMoles

Number ofparticles

Mass

1.

2. 3.

4.

6(18.02 g H2O)��237.93 g CoCl2�6H2O

8(18.02 g H2O)��315.47 g Ba(OH)2�8H2O

7(18.02 g H2O)��246.49 g MgSO4�7H2O

1.008 g H��1 mol H

16.00 g O��1 mol O

1.008 g H��1 mol H

16.00 g O��1 mol O



1.008 g H��1 mol H

16.00 g O��1 mol O

137.33 g Ba��

1 mol Ba


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75. Which contains more atoms, a mole of silveratoms or a mole of gold atoms? Explain youranswer. (11.2)

They both contain the same number of atomsbecause a mole of anything contains 6.02 � 1023

representative particles.

76. Discuss the relationships that exist between themole, molar mass, and Avogadro’s number. (11.2)

Molar mass is the mass in grams of one mole ofany pure substance. Avogadro’s number is thenumber of representative particles in one mole.The mass of 6.02 � 1023 representative particlesof a substance is the molar mass of the substance.

77. Which has a greater mass, a mole of silveratoms or a mole of gold atoms? Explain youranswer. (11.2)

The molar mass of Au is 197.0 g/mol; the molarmass of Ag is 107.9 g/mol. Thus, a mole of Au hasa greater mass.

78. Explain the difference between atomic mass(amu) and molar mass (gram). (11.2)

Atomic mass (amu) is the mass of an individualparticle (atom, molecule). Molar mass (grams) isthe mass of a mole of particles.

79. If you divide the molar mass of an element byAvogadro’s number, what is the meaning of thequotient? (11.2)

the mass of one atom of the element

80. List three conversion factors used in molar conversions. (11.3)

,

,

,

81. What information is provided by the formula forpotassium chromate (K2CrO4)? (11.3)

One mole of K2CrO4 contains two moles of K�

ions and one mole of CrO42� ions.

82. Which of the following molecules contains themost moles of carbon atoms per mole of thecompound: ascorbic acid (C6H8O6), glycerin(C3H8O3), or vanillin (C8H8O3)? Explain. (11.3)

The formula for vanillin (C8H8O3) shows there areeight carbon atoms per molecule, more thanascorbic acid or glycerin.

83. Explain what is meant by percent composition.(11.4)

Percent composition is the percent by mass ofeach element in a compound.

84. What is the difference between an empiricalformula and a molecular formula? Use anexample to illustrate your answer. (11.4)

An empirical formula is the smallest whole-number ratio of elements that make up acompound (CH). A molecular formula specifies theactual number of atoms of each element in onemolecule or formula unit of the substance (C6H6).

85. Do all pure samples of a given compound havethe same percent composition? Explain. (11.4)

Yes, for every pure substance, the percent bymass of each element is the same regardless ofthe size of the sample.

86. What information must a chemist obtain inorder to determine the empirical formula of anunknown compound? (11.4)

the percent composition of the compound

87. What is a hydrated compound? Use an exampleto illustrate your answer. (11.5)

A hydrated compound is a compound that has aspecific number of water molecules associatedwith its atoms, for example, Na2CO3�10H2O andCuSO4�5H2O.

88. Explain how hydrates are named. (11.5)

First, name the compound. Then, add a prefix(mono- di- tri-) that indicates how many watermolecules are associated with one mole ofcompound.


1 mol

1 mol��6.02 � 1023 representative particles


1 mol

1 mol��number of grams


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Mastering Problems

Mole-Particle Conversions (11.1)Level 1

89. Determine the number of representative parti-cles in each of the following.

a. 0.250 mol silver

0.250 mol Ag �

� 1.51 � 1023 atoms Ag

b. 8.56 � 10�3 mol sodium chloride

8.56 � 10�3 mol NaCl �


c. 35.3 mol carbon dioxide

35.3 mol CO2 �

� 2.13 � 1025 molecules CO2

d. 0.425 mol nitrogen (N2)

0.425 mol N2 �

� 2.56 � 1023 molecules N2

90. Determine the number of moles in each of thefollowing.

a. 3.25 � 1020 atoms lead

3.25 � 1020 atoms Pb �

� 5.39 � 10�4 mol Pb

b. 4.96 � 1024 molecules glucose

4.96 � 1024 molecules glucose �

� 8.24 mol of glucose

c. 1.56 � 1023 formula units sodium hydroxide

1.56 � 1023 formula units NaOH �

� 2.59 � 10�1 mol NaOH

d. 1.25 � 1025 copper(II) ions

1.25 � 1025 Cu2� ions �

� 2.08 � 101 mol Cu2� ions

91. Make the following conversions.

a. 1.51 � 1015 atoms Si to mol Si

1.51 � 1015 atoms Si �

� 2.51 � 10�9 mol Si

b. 4.25 � 10�2 mol H2SO4 to moleculesH2SO4

4.25 � 10�2 mol H2SO4 �

� 2.56 � 1022 molecules H2SO4

c. 8.95 � 1025 molecules CCl4 to mol CCl48.95 � 1025 molecules CCl4 �

� 1.49 � 102 mol CCl4

d. 5.90 mol Ca to atoms Ca

5.90 mol Ca �

� 3.55 � 1024 atoms Ca

92. How many molecules are contained in each ofthe following?

a. 1.35 mol carbon disulfide (CS2)

1.35 mol CS2 �

� 8.13 � 1023 molecules CS2

6.02 � 1023 molecules CS2��1 mol CS2

6.02 � 1023 atoms Ca��

1 mol Ca

1 mol CCl4��6.02 � 1023 molecules CCl4

6.02 � 1023 molecules H2SO4��1 mol H2SO4

1 mol Si��6.02 � 1023 atoms Si

1 mol Cu2� ions��6.02 � 1023 Cu2� ions

1 mol NaOH��6.02 � 1023 formula units NaOH

1 mol glucose��6.02 � 1023 molecules glucose

1 mol Pb��6.02 � 1023 atoms Pb

6.02 � 1023 molecules N2��1 mol N2

6.02 � 1023 molecules CO2��1 mol CO2


1 mol NaCl

6.02 � 1023 atoms Ag��

1 mol Ag


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b. 0.254 mol diarsenic trioxide (As2O3)

0.254 mol As2O3 �

� 1.53 � 1023 molecules As2O3

c. 1.25 mol water

1.25 mol H2O �


d. 150.0 mol HCl

150.0 mol HCl �

� 9.030 � 1025 molecules HCl

93. How many moles contain each of the following?

a. 1.25 � 1015 molecules carbon dioxide


� 2.08 � 10�9 mol CO2

b. 3.59 � 1021 formula units sodium nitrate

3.59 � 1021 formula units NaNO3 �

� 5.96 � 10�3 mol NaNO3

c. 2.89 � 1027 formula units calcium carbonate

2.89 � 1027 formula units CaCO3 �

� 4.80 � 103 mol CaCO3

Level 2

94. A bracelet containing 0.200 mol of metal atomsis 75% gold. How many particles of gold atomsare in the bracelet?

0.200 mol Au � 0.75 � 0.150 mol Au

0.150 mol Au �

� 9.03 � 1022 atoms Au

95. A solution containing 0.250 mol Cu2� ions isadded to another solution containing 0.130 molCa2� ions. What is the total number of metalions in the combined solution?

0.250 mol Ca2� ions � 0.130 mol Ca2� ion � 0.380 mol metal ions

0.380 mol metal ions �

� 2.29 � 1023 Cu2� and Ca2� ions

96. If a snowflake contains 1.9 � 1018 molecules of water, how many moles of water does it contain?

1.9 � 1018 molecules H2O �

� 3.2 � 10�6 mol H2O

97. If you could count two atoms every second,how long would it take you to count a mole ofatoms? Assume that you counted continually 24 hours every day. How does the time youcalculated compare with the age of Earth, whichis estimated to be 4.5 � 109 years old?

6.02 � 1023 particles � � 3.01 � 1023 s

3.01 � 1023 s � � �

� 2.1 � 106

� 9.5 � 1015 yr; about 2 million times longer

Mole-Mass Conversions (11.2)Level 1

98. Calculate the mass of the following.

a. 5.22 mol He

5.22 mol He � � 20.9 g He

b. 0.0455 mol Ni

0.0455 mol Ni � � 2.67 g Ni58.69 g Ni��1 mol Ni

4.00 g He��1 mol He

9.54 � 1015 y��

4.5 � 109 y

1 y�365 day

1 day�24 h

1 h�3600 s

1 s��2 particles

1 mol H2O��6.02 � 1023 molecules H2O

6.02 � 1023 mol metal ions��

1 mol metal ions

6.02 � 1023 atoms Au��

1 mol Au

1 mol CaCO3��6.02 � 1023 formula units CaCO3

1 mol NaNO3��6.02 � 1023 formula units NaNO3

1 mol CO2��6.02 � 1023 molecules CO2

6.02 � 1023 molecules HCl��

1 mol HCl

6.02 � 1023 molecules H2O ��

1 mol H2O

6.02 � 1023 molecules As2O3��1 mol As2O3


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c. 2.22 mol Ti

2.22 mol Ti � � 106 g Ti

d. 0.00566 mol Ge

0.00566 mol Ge � � 0.411 g Ge

99. Make the following conversions.

a. 3.50 mol Li to g Li

3.50 mol Li � � 24.3 g Li

b. 7.65 g Co to mol Co

7.65 g Co � � 0.130 mol Co

c. 5.62 g Kr to mol Kr

5.62 g Kr � � 0.0671 mol Kr

d. 0.0550 mol As to g As

0.0550 mol As � � 4.12 g As

Level 2

100. Determine the mass in grams of the following.

a. 1.33 � 1022 mol Sb

1.33 � 1022 mol Sb �

� 1.62 � 1024 g Sb

b. 4.75 � 1014 mol Pt

4.75 � 1014 mol Pt �

� 9.27 � 1016 g Pt

c. 1.22 � 1023 mol Ag

1.22 � 1023 mol Ag �

� 1.32 � 1025 g Ag

d. 9.85 � 1024 mol Cr

9.85 � 1024 mol Cr �

� 5.12 � 1026 g Cr

Particle-Mass Conversions (11.2)Level 1

101. Convert the following to mass in grams.

a. 4.22 � 1015 atoms U

4.22 � 1015 atoms U � �

� 1.67 � 10�6 g U

b. 8.65 � 1025 atoms H

8.65 � 1025 atoms H �

�

� 145 g H

c. l.25 � 1022 atoms O

1.25 � 1022 atoms O �

�

� 0.332 g O

d. 4.44 � 1023 atoms Pb

4.44 � 1023 atoms Pb �

�

� 153 g Pb

102. A sensitive balance can detect masses of 1 � 10�8 g. How many atoms of silver wouldbe in a sample having this mass?

1 � 10�8 g Ag � �

� 5 � 1013 atoms Ag

103. Calculate the number of atoms in each of the following.

a. 25.8 g Hg

25.8 g Hg � �

� 7.74 � 1022 atoms Hg

6.02 � 1023 atoms Hg��

1 mol Hg

1 mol Hg��200.59 g Hg

6.02 � 1023 atoms Ag��

1 mol Ag

1 mol Ag��107.87 g Ag



16.00 g O��1 mol O

1 mol O��6.02 � 1023 atoms O

1.008 g H��1 mol H

1 mol H��6.02 � 1023 atoms H

283.03 g U��

1 mol U

1 mol U��6.02 � 1023 atoms U

52.00 g Cr��1 mol Cr

107.87 g Ag��

1 mol Ag

195.08 g Pt��

1 mol Pt

121.76 g Sb��

1 mol Sb

74.92 g As��1 mol As

1 mol Kr��83.80 g Kr

1 mol Co��58.93 g Co

6.94 g Li��1 mol Li

72.61 g Ge��1 mol Ge

47.87 g Ti��1 mol Ti


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b. 0.0340 g Zn

0.0340 g Zn � �

� 3.13 � 1020 atoms Zn

c. 150 g Ar

150 g Ar � �

� 2.3 � 1024 atoms Ar

d. 0.124 g Mg

0.124 g Mg � �

� 3.07 � 1021 atoms Mg

104. Which has more atoms, 10.0 g of carbon or10.0 g of calcium? How many atoms does each have?

10.0 g C � �

� 5.01 � 1023 atoms C

10.0 g Ca � �

� 1.50 � 1023 atoms Ca

10.0 g C contains more atoms.

105. Which has more atoms, 10.0 moles of carbonor 10.0 moles of calcium? How many doeseach have?

One mole of any substance contains 6.02 � 1023

representative particles. Thus, 10.0 moles ofcarbon and 10.0 moles of calcium contain thesame number or atoms.

10.0 mol �

� 6.02 � 1024 atoms

Level 2

106. A mixture contains 0.250 mol Fe and 1.20 g C.What is the total number of atoms in the mixture?

0.250 mol Fe �

� 1.51 � 1023 atoms Fe

1.20 g C � �

� 0.601 � 1023 atoms C

1.51 � 1023 atoms Fe � 0.601 � 1023 atoms C � 2.11 � 1023 total atoms

Chemical Formulas (11.3)Level 1

107. In the formula for sodium phosphate(Na3PO4), how many moles of sodium are represented? How many moles of phosphorus?How many moles of oxygen?

3 mol Na, 1 mol P, 4 mol O

108. How many moles of oxygen atoms arecontained in the following?

a. 2.50 mol KMnO4

2.50 mol KMnO4 �

� 10.0 mol O

b. 45.9 mol CO2

45.9 mol CO2 � � 91.8 mol O

c. 1.25 � 10�2 mol CuSO4�5H2O

1.25 � 10�2 mol CuSO4�5H2O �

� 0.113 mol O9 mol O

��1 mol CuSO4�5H2O

2 mol O��1 mol CO2

4 mol O��1 mol KMnO4

6.02 � 1023 atoms C��

1 mol C1 mol C��12.01 g C

6.02 � 1023 atoms Fe��

1 mol Fe

6.02 � 1023 atoms��

1 mol

6.02 � 1023 atoms Ca��

1 mol Ca1 mol Ca��40.08 g Ca

6.02 � 1023 atoms C��

1 mol C1 mol C��12.01 g C

6.02 � 1023 atoms Mg��

1 mol Mg

1 mol Mg��24.31 g Mg

6.02 � 1023 atoms Ar��

1 mol Ar

1 mol Ar��39.95 g Ar

6.02 � 1023 atoms Zn��

1 mol Zn



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Level 2

109. The graph shows the numbers of atoms of eachelement in a compound. What is its formula?What is its molar mass?

CaC4H6O4 (calcium acetate)

1 mol Ca � � 40.08 g Ca

4 mol C � � 48.04 g C

6 mol H � � 6.048 g H

4 mol O � � 64.00 g O


110. How many carbon tetrachloride molecules arein 3.00 mol carbon tetrachloride (CCl4)? Howmany carbon atoms? How many chlorineatoms? How many total atoms?

3.00 mol CCl4 �

� 1.81 � 1024 molecules CCl4

1.81 � 1024 molecules CCl4 �

� 1.81 � 1024 atoms C

1.81 � 1024 molecules CCl4 �

� 7.24 � 1024 atoms Cl

1.81 � 1024 atoms C � 7.24 � 1024 atoms Cl � 9.05 � 1024 total atoms

Molar Mass (11.3)Level 1

111. Determine the molar mass of each of the following.

a. nitric acid (HNO3)

1 mol H � � 1.008 g H

1 mol N � � 14.01 g N

3 mol O � � 48.00 g O

molar mass � 63.02 g/mol HNO3

b. ammonium nitrate (NH4NO3)

2 mol N � � 28.02 g N

4 mol H � � 4.032 g H

3 mol O � � 48.00 g O

molar mass � 80.05 g/mol NH4NO3

c. zinc oxide (ZnO)

1 mol Zn � � 65.39 g Zn

1 mol O � � 16.00 g O

molar mass � 81.39 g/mol ZnO

d. cobalt chloride (CoCl2)

1 mol Co � � 58.93 g Co

2 mol Cl � � 70.90 g Cl

molar mass � 129.83 g/mol CoCl2



16.00 g O��1 mol O


16.00 g O��1 mol O

1.008 g H��1 mol H

14.01 g N��1 mol N

16.00 g O��1 mol O

14.01 g N��1 mol N

1.008 g H��1 mol H

4 mol atoms Cl��1 molecule CCl4

1 mol atoms C��1 molecule CCl4

6.02 � 1023 molecules CCl4��1 mol CCl4

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C


Ca C H OAto

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654321

Atoms

The Composition ofa Compound


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112. Calculate the molar mass of each of the following.

a. ascorbic acid (C6H8O6)

6 mol C � � 72.06 g C

8 mol H � � 8.064 g H

6 mol O � � 96.00 g O

molar mass � 176.12 g/mol C6H8O6

b. sulfuric acid (H2SO4)

2 mol H � � 2.016 g H

1 mol S � � 32.07 g S

4 mol O � � 64.00 g O

molar mass � 98.09 g/mol H2SO4

c. silver nitrate (AgNO3)

1 mol Ag � � 107.87 g Ag

1 mol N � � 14.01 g N

3 mol O � � 48.00 g O

molar mass � 169.88 g/mol AgNO3

d. saccharin (C7H5NO3S)

7 mol C � � 84.07 g C

5 mol H � � 5.040 g H

1 mol N � � 14.01 g N

3 mol O � � 48.00 g O

1 mol S � � 32.07 g S

molar mass � 183.19 g/mol C7H5NO3S

113. Determine the molar mass of allyl sulfide, thecompound responsible for the smell of garlic.The chemical formula of allyl sulfide is(C3H5)2S.

6 mol C � � 72.06 g C

10 mol H � � 10.080 g H

1 mol S � � 32.07 g S

molar mass � 114.21 g/mol (C3H5)2S

Mass-Mole Conversions (11.3)Level 1

114. How many moles are in 100.0 g of each of the following compounds?

a. dinitrogen oxide (N2O)

2 mol N � � 28.02 g N

1 mol O � � 16.00 g O


100.0 g N2O � � 2.27 mol N2O

b. methanol (CH3OH)

1 mol C � � 12.01 g C

4 mol H � � 4.032 g H

1 mol O � � 16.00 g O


100.0 g CH3OH �

� 3.12 mol CH3OH

1 mol CH3OH��32.04 g CH3OH

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

1 mol N2O��44.02 g N2O

16.00 g O��1 mol O

14.01 g N��1 mol N

32.07 g S��1 mol S

1.008 g H��1 mol H

12.01 g C��1 mol C

32.07 g S��1 mol S

16.00 g O��1 mol O

14.01 g N��1 mol N

1.008 g H��1 mol H

12.01 g C��1 mol C

16.00 g O��1 mol O

14.01 g N��1 mol N

107.87 g Ag��

1 mol Ag

16.00 g O��1 mol O

32.07 g S��1 mol S

1.008 g H��1 mol H

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C


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115. What is the mass of each of the following?

a. 4.50 � 10�2 mol CuCl2

1 mol Cu � � 63.55 g Cu

2 mol Cl � � 70.90 g Cl


4.50 � 10�2 mol CuCl2 �

� 6.05 g CuCl2

b. 1.25 � 102 mol Ca(OH)2

1 mol Ca � � 40.08 g Ca

2 mol H � � 2.016 g H

2 mol O � � 32.00 g O


1.25 � 102 mol Ca(OH)2 �

� 9.26 � 102 g Ca(OH)2

116. Determine the number of moles in each of the following.

a. 1.25 � 102 g Na2S

2 mol Na � � 45.98 g Na

1 mol S � � 32.07 g S


1.25 � 102 g Na2S �

� 1.60 mol Na2S

b. 0.145 g H2S

2 mol H � � 2.016 g H

1 mol S � � 32.07 g S


0.145 g H2S �

� 4.25 � 10�3 mol H2S

Level 2

117. Benzoyl peroxide is a substance used as anacne medicine. What is the mass in grams of3.50 � 10�2 moles of benzoyl peroxide(C14H10O4)?

14 mol C � � 168.14 g C

10 mol H � � 10.080 g H

4 mol O � � 64.00 g O


3.50 � 10�2 mol �

� 8.48 g benzoyl peroxide

118. Hydrofluoric acid is a substance used to etchglass. Determine the mass of 4.95 � 1025 HFmolecules.

1 mol H � � 1.008 g H

1 mol F � � 19.00 g F


4.95 � 1025 molecules HF �

�

� 1650 g HF

119. How many moles of aluminum ions are in 45.0 g of aluminum oxide?

2 mol Al � � 53.96 g Al

3 mol O � � 48.00 g O


45.0 g Al2O3 � �

� 0.883 mol Al3� ions2 mol Al3� ions��

1 mol Al2O3

1 mol Al2O3��101.96 g Al2O3

16.00 g O��1 mol O


20.01 g HF��1 mol HF

1 mol HF��6.02 � 1023 molecules HF

19.00 g F��1 mol F

1.008 g H��1 mol H

242.22 g��

1 mol

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

1 mol H2S��34.09 g H2S

32.07 g S��1 mol S

1.008 g H��1 mol H

1 mol Na2S��78.05 g Na2S

32.07 g S��1 mol S


74.10 g Ca(OH)2��1 mol Ca(OH)2

16.00 g O��1 mol O

1.008 g H��1 mol H


134.35 g CuCl2��1 mol CuCl2


63.55 g Cu��1 mol Cu


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120. How many moles of ions are in the following?

a. 0.0200 g AgNO3

1 mol Ag � � 107.87 g Ag

1 mol N � � 14.01 g N

3 mol O � � 48.00 g O


0.0200 g AgNO3 � �

� 2.35 � 10�4 mol ions

b. 0.100 mol K2CrO4

0.100 mol K2CrO4 �

� 0.300 mol ions

c. 0.500 g Ba(OH)2

1 mol Ba � � 137.33 g Ba

2 mol O � � 32.00 g O

2 mol H � � 2.016 g H


0.500 g Ba(OH)2 � �

� 8.75 � 10�3 mol ions

d. 1.00 � 10�9 mol Na2CO3

1.00 � 10�9 mol Na2CO3 �

� 3.00 � 10�9 mol ions

Mass-Particle Conversions (11.3)Level 1

121. Calculate the values that will complete the table.

Silver acetate:

1 mol Ag � � 107.87 g Ag

2 mol C � � 24.02 g C

3 mol H � � 3.024 g H

2 mol O � � 32.00 g O


2.50 mol AgC2H3O2 �

� 417 g AgC2H3O2

2.50 mol AgC2H3O2 �

� 1.51 � 1024 formula units

Glucose:

6 mol C � � 72.06 g C

12 mol H � � 12.10 g H

6 mol O � � 96.00 g O


16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C


1 mol AgC2H3O2

166.91 g AgC2H3O2��1 mol AgC2H3O2

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

107.87 g Ag��

1 mol Ag

3 mol ions��1 mol Na2CO3

3 mol ions��1 mol Ba(OH)2

1 mol Ba(OH)2��171.35 g Ba(OH)2

1.008 g H��1 mol H

16.00 g O��1 mol O

137.33 g Ba��

1 mol Ba

3 mol ions��1 mol K2CrO4

2 mol ions��1 mol AgNO3

1 mol AgNO3��169.88 g AgNO3

16.00 g O��1 mol O

14.01 g N��1 mol N

107.87 g Ag��

1 mol Ag


Moles, Mass, and Representative Particles

Compound Number Mass(g) Representativeof moles particles

Silver acetate 2.50 417 1.51 � 1024

Ag(C2H3O2)

Glucose 1.798 324.0 1.082 � 1024

C6H12O6

Benzene 0.009 39 0.733 5.65 � 1021

C6H6

Lead(II) sulfide 0.4178 100.0 2.516 � 1023

PbS

Table 11-2

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324.0 g glucose �

� 1.798 mol glucose

1.798 mol glucose �

� 1.082 � 1024 molecules

Benzene:

6 mol C � � 72.06 g C

6 mol H � � 6.048 g H


5.65 � 1021 molecules benzene �

� 9.39 � 10�3 mol benzene

9.39 � 10�3 mol benzene �

� 0.733 g benzene

Lead(II) sulfide:

1 mol Pb � � 207.2 g Pb

1 mol S � � 32.07 g S


100.0 g PbS � � 0.4178 mol PbS

0.4178 mol PbS �

� 2.516 � 1023 formula units

122. How many formula units are present in 500.0 glead(II) chloride?

1 mol Pb � � 207.2 g Pb

2 mol Cl � � 70.90 g Cl


500.0 g PbCl2 � � 1.798 mol PbCl2

1.798 mol PbCl2 �

� 1.082 � 1024 formula units PbCl2

123. Determine the number of atoms in 3.50 g gold.

3.50 g Au � �

� 1.07 � 1022 atoms Au

124. Calculate the mass of 3.62 � 1024 moleculesof glucose (C6H12O6).

From solution #121, the molar mass of glucoseequals 180.16 g/mol.

3.62 � 1024 molecules glucose �

�

� 1080 g glucose

125. Determine the number of molecules of ethanol(C2H5OH) in 47.0 g.

2 mol C � � 24.02 g C

6 mol H � � 6.048 g H

1 mol O � � 16.00 g O


47.0 g ethanol � �

� 6.14 � 1023 molecules ethanol

126. What mass of iron(III) chloride contains 2.35 � 1023 chloride ions?

1 mol Fe � � 55.85 g Fe

3 mol Cl � � 106.35 g Cl


2.35 � 1023 chloride ions �

�

�

� 21.1 g FeCl3

162.20 g FeCl3��1 mol FeCl3

1 mol FeCl3��3 mol chloride ions

1 mol chloride ions��6.02 � 1023 chloride ions



6.02 � 1023 molecules ethanol��

1 mol ethanol

1 mol ethanol��46.07 g ethanol

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

180.16 g glucose��

1 mol glucose

1 mol glucose��6.02 � 1023 molecules glucose

6.02 � 1023 atoms Au��

1 mol Au

1 mol Au��196.97 g Au


1 mol PbCl2

1 mol PbCl2��278.1 g PbCl2




1 mol PbS

1 mol PbS��239.3 g PbS

32.07 g S��1 mol S


78.11 g benzene��1 mol benzene

1 mol benzene��6.02 � 1023 molecules benzene

1.008 g H��1 mol H

12.01 g C��1 mol C

6.02 � 1023 molecules��

1 mol glucose

1 mol glucose��180.16 g glucose


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127. How many moles of iron can be recoveredfrom 100.0 kg Fe3O4?

3 mol Fe � � 167.55 g Fe

4 mol O � � 64.00 g O


100.0 kg Fe3O4 �

� 1.000 � 105 g Fe3O4

1.000 � 105 g Fe3O4 �

� � 1296 mol Fe

128. The mass of an electron is 9.11 � 10�28 g.What is the mass of a mole of electrons?

�

� 5.48 � 10�4 g/mol electrons

129. Vinegar is 5.0% acetic acid (CH3COOH). Howmany molecules of acetic acid are present in25.0 g vinegar?

2 mol C � � 24.02 g C

4 mol H � � 4.032 g H

2 mol O � � 32.00 g O


0.050 � 25.0 g vinegar � 1.25 g CH3COOH

1.25 g CH3COOH �

�

� 1.25 � 1022 molecules CH3COOH

130. The density of lead is 11.3 g/cm3. Calculatethe volume of one mole lead.

1 mol Pb � � � 18.3 cm3

131. Calculate the moles of aluminum ions presentin 250.0 g aluminum oxide (Al2O3).

From solution # 119, the molar mass ofaluminum oxide equals 101.96 g/mol.

250.0 g Al2O3 � �

� 4.90 mol Al3�

Level 2

132. Determine the number of chloride ions in10.75 g of magnesium chloride.

1 mol Mg � � 24.31 g

2 mol Cl � � 70.90 g


10.75 g MgCl2 � �

�

� 1.37 � 1023 ions Cl�

133. Acetaminophen, a common aspirin substitute,has the formula C8H9NO2. Determine the number of molecules of acetaminophen in a 500 mg tablet.

8 mol C � � 96.08 g C

9 mol H � � 9.072 g H

1 mol N � � 14.01 g N

2 mol O � � 32.00 g O


500 mg C8H9NO2 � �

� 0.003 mol C8H9NO2

0.003 mol C8H9NO2 �

� 2 � 1021 molecules C8H9NO2

6.02 � 1023 molecules C8H9NO2��1 mol C8H9NO2

1 mol C8H9NO2��151.16 g C8H9NO2

1 g C8H9NO2��103 mg C8H9NO2

16.00 g O��1 mol O

14.01 g N��1 mol N

1.008 g H��1 mol H

12.01 g C��1 mol C

6.02 � 1023 ions MgCl2��1 mol Cl�

2 mol Cl��1 mol MgCl2

1 mol MgCl2��94.31 g MgCl2



2 mol Al3��1 mol Al2O3

1 mol Al2O3��101.96 g Al2O3

1 cm3��11.3 g Pb


6.02 � 1023 molecules CH3COOH��

1 mol CH3COOH

1 mol CH3COOH��60.05 g CH3COOH

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

6.02 � 1023 electrons��

1 mol electrons9.11 � 10�28 g��

1 electron

3 mol Fe��1 mol Fe3O4

1 mol Fe3O4��231.55 g Fe3O4

103 g Fe3O4��1 kg Fe3O4

16.00 g O��1 mol O



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134. Calculate the number of sodium ions present in25.0 g sodium chloride.

1 mol Na � � 22.99 g Na

1 mol Cl � � 35.45 g Cl


25.0 g NaCl � �

� 2.58 � 1023 ions Na�

135. Determine the number of oxygen atoms present in 25.0 g carbon dioxide.

1 mol C � � 12.01 g C

2 mol O � � 32.00 g O


25.0 g CO2 � � 0.568 mol CO2

0.568 mol CO2 �

� 6.84 � 1023 atoms O

Percent Composition (11.4)

136. Express the composition of each of thefollowing as the mass percent of its elements(percent composition).

a. sucrose (C12H22O11)

12 mol C � � 144.12 g C

22 mol H � � 22.18 g H

11 mol O � � 176.00 g O



� 42.10% C


� 6.480% H


� 51.42% O

b. magnetite (Fe3O4)

3 mol Fe � � 167.55 g Fe

4 mol O � � 64.00 g O



� 72.360% Fe


� 27.64% O

c. aluminum sulfate (Al2(SO4)3)

2 mol Al � � 53.96 g Al

3 mol S � � 96.21 g S

12 mol O � � 192.00 g O



� 15.77% Al


� 28.12% S

percent by mass = � 100

� 56.11% O

137. Which of the following iron compounds contain the greatest percentage of iron: pyrite(FeS2), hematite (Fe2O3), or siderite (FeCO3)?

1 mol Fe � � 55.85 g Fe

2 mol S � � 64.14 g S

molar mass FeS2 � 119.99 g/mol

32.07 g S��1 mol S


96.21 g��342.17 g

96.21 g��342.17 g

53.96 g��342.17 g

16.00 g O��1 mol O

32.07 g S��1 mol S


64.00 g��231.55 g

167.55 g��231.55 g

16.00 g O��1 mol O


176.00 g��342.30 g

22.18 g��342.30 g

144.12 g��342.30 g

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

2(6.02 � 1023 atoms O)��

1 mol CO2

1 mol CO2��44.01 g CO2

16.00 g O��1 mol O

12.01 g C��1 mol C

6.02 � 1023 ions Na��

1 mol NaCl

1 mol NaCl��58.44 g NaCl




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2 mol Fe � � 111.70 g Fe

3 mol O � � 48.00 g O


1 mol Fe � � 55.85 g Fe

1 mol C � � 12.01 g C

3 mol O � � 96.00 g O

molar mass FeCO3 � 152.95 g/mol

percent mass � � 100

� 46.55% in FeS2


� 69.944% in Fe2O3


� 36.51% in FeCO3

Hematite, with 69.944% Fe, has the greatestpercentage of iron.

138. Determine the empirical formula for each ofthe following compounds.

a. ethylene (C2H4)

Both subscripts can be divided by two. The empirical formula is CH2.

b. ascorbic acid (C6H8O6)

All subscripts can be divided by two. The empirical formula is C3H4O3.

c. naphthalene (C10H8)

Both subscripts can be divided by two. The empirical formula is C5H4.

139. Caffeine, a stimulant found in coffee, has thechemical formula C8H10N4O2.

a. Calculate the molar mass of caffeine.

8 mol C � � 96.08 g C

10 mol H � � 10.080 g H

4 mol N � � 56.04 g N

2 mol O � � 32.00 g O


b. Determine the percent composition ofcaffeine.


� 49.47% C


� 5.191% H


� 28.86% N


� 16.48% O

140. Which of the titanium-containing mineralsrutile (TiO2) or ilmenite (FeTiO3) has thelarger percent of titanium?

1 mol Ti � � 47.87 g Ti

2 mol O � � 32.00 g O

molar mass TiO2 � 79.87 g/mol

1 mol Fe � � 55.85 g Fe

1 mol Ti � � 47.87 g Ti

3 mol O � � 48.00 g O

molar mass FeTiO3 � 151.72 g/mol

16.00 g O��1 mol O



16.00 g O��1 mol O


32.00 g��194.20 g

56.04 g��194.20 g

10.08 g��194.20 g

96.08 g��194.20 g

16.00 g O��1 mol O

14.01 g N��1 mol N

1.008 g H��1 mol H

12.01 g C��1 mol C

55.85 g��152.95 g

111.70 g��159.70 g

55.85 g��119.99 g

16.00 g O��1 mol O

12.01 g C��1 mol C


16.00 g O��1 mol O



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� 59.93% Ti in TiO2


� 31.55% Ti in FeTiO3

TiO2 has the greater percent by mass of titanium.

141. Vitamin E, found in many plants, is thought toretard the aging process in humans. Theformula for vitamin E is C29H50O2. What is thepercent composition of vitamin E?

29 mol C � � 348.29 g C

50 mol H � � 50.400 g H

2 mol O � � 32.00 g O


percent by mass � � 100 � 80.87% C

percent by mass � � 100 � 11.70% H

percent by mass � � 100 � 7.430% O

142. Aspartame, an artificial sweetener, has theformula C14H18N2O5. Determine the percentcomposition of aspartame.

14 mol C � � 168.14 g C

18 mol H � � 18.144 g H

2 mol N � � 28.02 g N

5 mol O � � 80.00 g O


percent by mass � � 100 � 57.13% C

percent by mass � � 100 � 6.165% H

percent by mass � � 100 � 9.521% N

percent by mass � � 100 � 27.18% O

Empirical and Molecular Formulas (11.4)

143. The hydrocarbon used in the manufacture offoam plastics is called styrene. Analysis ofstyrene indicates the compound is 92.25% Cand 7.75% H and has a molar mass of 104 g/mol. Determine the molecular formulafor styrene.

92.25 g C � � 7.681 mol C

7.75 g H � � 7.69 mol H

� 1.00

� 1.00

The empirical formula is CH.

1 mol C � � 12.01 g C

1 mol H � � 1.008 g H

molar mass CH � 13.02 g/mol

n � � 8.00

Multiply the subscripts in the empirical formulaby 8 to obtain the molecular formula C8H8.

144. Monosodium glutamate (MSG) is sometimesadded to food to enhance flavor. Analysisdetermined this compound to be 35.5% C,4.77% H, 8.29% N, 13.6% Na, and 37.9% O.What is the empirical formula for MSG?

35.5 g C � � 2.96 mol C

4.77 g H � � 4.73 mol H

8.29 g N � � 0.592 mol N1 mol N��14.01 g N

1 mol H��1.008 g H

1 mol C��12.01 g C

104 g/mol��13.02 g/mol

1.008 g H��1 mol H

12.01 g C��1 mol C

7.69 mol H��

7.69

7.681 mol C��

7.69

1 mol H��1.008 g H

1 mol C��12.01 g C

80.00 g��294.30 g

28.02 g��294.30 g

18.144 g��294.30 g

168.14 g��294.30 g

16.00 g O��1 mol O

14.01 g N��1 mol N

1.008 g H��1 mol H

12.01 g C��1 mol C

32.00 g��430.69 g

50.40 g��430.69 g

348.29 g��430.69 g

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

47.87 g��151.72 g

47.87 g�79.87 g


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13.6 g Na � � 0.592 mol Na

37.9 g O � � 2.37 mol O

� 5.00 mol C

� 8.00 mol H

� 1.00 mol N

� 1.00 mol Na

� 4.00 mol O

The simplest whole-number ratio is 5.00 mol C:8.00 mol H: 1.00 mol N: 1.00 mol Na: 4.00 mol O.The empirical formula is C5H8NO4Na.

145. Determine the molecular formula foribuprofen, a common headache remedy.Analysis of ibuprofen yields a molar mass of 206 g/mol and a percent composition of75.7% C, 8.80% H and 15.5% O.

75.7 g C � � 6.30 mol C

8.80 g H � � 8.73 mol H

15.5 g O � � 0.969 mol O

= 6.50 mol C

= 9.01 mol H

= 1.00 mol O

The simplest ratio is 6.50 mol C: 9.01 mol H: 1.00mol O. Multiply the simplest ratio by two toobtain the simplest whole-number ratio. Theempirical formula is C13H18O2.

13 mol C � � 156.13 g C

18 mol H � � 18.144 g H

2 mol O � � 32.00 g O

molar mass C13H18O2 � 206.27 g/mol

The molar mass C13H18O2 equals the molar massof ibuprofen. Thus, the molecular formula is thesame as the empirical formula C13H18O2.

146. Vanadium oxide is used as an industrial cata-lyst. The percent composition of this oxide is56.0% vanadium and 44.0% oxygen. Determinethe empirical formula for vanadium oxide.

56.0 g V � � 1.10 mol V

44.0 g O � � 2.75 mol O

� 1.00 mol V

� 2.50 mol O

The simplest ratio is 1.00 mol V: 2.50 mol O.Multiply the simplest ratio by two to obtain thesimplest whole-number ratio. The empiricalformula is V2O5.

147. What is the empirical formula of a compoundthat contains 10.52 g Ni, 4.38 g C, and 5.10 g N?

10.52 g Ni � � 0.1792 mol Ni

4.38 g C � � 0.3470 mol C

5.10 g N � � 0.3640 mol N

� 1.000 mol Ni

� 1.936 mol C

� 2.031 mol N

The simplest ratio is 1.00 mol Ni: 1.936 mol C:2.031 mol N. The empirical formula is Ni(CN)2.

0.3640 mol N��

0.1792

0.3470 mol C��

0.1792

0.1792 mol Ni��

0.1792

1 mol N��14.01 g N

1 mol C��12.01 g C

1 mol Ni��58.69 g Ni

2.75 mol O��

1.10

1.10 mol V��

1.10

1 mol O��16.00 g O

1 mol V��50.94 g V

16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

0.969 mol O��

0.969

8.73 mol H��

0.969

6.30 mol C��

0.969

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C

2.37 mol O��

0.592

0.592 mol Na��

0.592

0.592 mol N��

0.592

4.73 mol H��

0.592

2.96 mol C��

0.592

1 mol O��16.00 g O

1 mol Na��22.99 g Na


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148. The Statue of Liberty turns green in airbecause of the formation of two coppercompounds, Cu3(OH)4SO4 and Cu4(OH)6SO4.Determine the mass percent of copper in thesecompounds.

Cu3(OH)4SO4

3 mol Cu � � 190.65 g Cu

8 mol O � � 128.00 g O

4 mol H � � 4.032 g H

1 mol S � � 32.07 g S

molar mass Cu3(OH)4SO4 � 354.75 g/mol

percent by mass � � 100 � 53.74%

Cu4(OH)6SO4

4 mol Cu � � 254.20 g Cu

10 mol O � � 160.00 g O

6 mol H � � 6.048 g H

1 mol S � � 32.07 g S

molar mass Cu4(OH)6SO4 � 452.32 g/mol

percent by mass � � 100 � 56.20%

149. Analysis of a compound containing chlorineand lead reveals that the compound is 59.37%lead. The molar mass of the compound is349.0 g/mol. What is the empirical formula forthe chloride? What is the molecular formula?

percent Pb � percent Cl � 100%

percent Cl � 100% � percent Pb � 100% � 59.37% � 40.63%

59.37 g Pb � � 0.2865 mol Pb

40.63 g Cl � � 1.146 mol Cl

� 1.000 mol Pb

� 4.000 mol Cl

The simplest ratio is 1.000 mol Pb: 4.000 mol Cl.The empirical formula is PbCl4.

1 mol Pb � � 207.2 g Pb

4 mol Cl � � 141.80 g Cl

molar mass PbCl4 � 349.00 g/mol

The molar mass based on the empirical formulaequals the molar mass of the compound. Themolecular formula is PbCl4.

150. Glycerol is a thick, sweet liquid obtained as abyproduct of the manufacture of soap. Its percent composition is 39.12% carbon,8.75% hydrogen, and 52.12% oxygen. Themolar mass is 92.11 g/mol. What is the molecular formula for glycerol?

39.12 g C � � 3.257 mol C

8.75 g H � � 8.681 mol H

52.12 g O � � 3.258 mol O

� 1.000 mol C

� 2.665 mol H

� 1.000 mol O

The simplest whole-number ratio is 1.000 mol C:2.655 mol H: 1.000 mol O. To obtain the simplestwhole-number ratio, multiply each number bythree. The empirical formula is C3H8O3.

3 mol C � � 36.03 g C

8 mol H � � 8.064 g H

3 mol O � � 48.00 g O


16.00 g O��1 mol O

1.008 g H��1 mol H

12.01 g C��1 mol C

3.258 mol O��

3.257

8.681 mol H��

3.257

3.257 mol C��

3.257

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C



1.146 mol Cl��

0.2865

0.2865 mol Pb��

0.2865

1 mol Cl��35.45 g Cl

1 mol Pb��207.2 g Pb

254.20 g��452.32 g

32.07 g S��1 mol S

1.008 g H��1 mol H

16.00 g O��1 mol O


190.65 g��338.75 g

32.07 g S��1 mol S

1.008 g H��1 mol H

16.00 g O��1 mol O



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The molar mass based on the empirical formulaequals the molar mass of the compound. Themolecular formula is C3H8O3.

The Formula for a Hydrate (11.5)Level 1

151. Determine the mass percent of anhydroussodium carbonate (Na2CO3) and water insodium carbonate decahydrate(Na2CO3�10H2O).

2 mol Na � � 45.98 g Na

1 mol C � � 12.01 g C

13 mol O � � 208.00 g O

20 mol H � � 20.16 g H

molar mass Na2CO3�10H2O � 286.15 g/mol

2 mol Na � � 45.98 g Na

1 mol C � � 12.01 g C

3 mol O � � 48.00 g O

molar mass Na2CO3 � 105.99 g/mol

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H


� 100 � 37.03% Na2CO3

� 100 � 62.97% H2O

152. What is the formula and name for a hydrate thatis 85.3% barium chloride and 14.7% water?

1 mol Ba � � 137.33 g Ba

2 mol Cl � � 70.90 g Cl

molar mass BaCl2 � 208.23 g/mol

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H


85.3 g BaCl2 � � 0.410 mol BaCl2

14.7 g H2O � � 0.816 mol H2O

n � � 2.00

BaCl2�2H2O, barium chloride dihydrate

153. Gypsum is hydrated calcium sulfate. A 4.89-gsample of this hydrate was heated, and afterthe water was driven off, 3.87 g anhydrouscalcium sulfate remained. Determine theformula of this hydrate and name thecompound.

mass of hydrated CaSO4 � mass of CaSO4 �

mass of H2O

mass of H2O � mass of hydrated CaSO4 � massof H2O � 4.89 g � 3.87 g � 1.02 g

1 mol Ca � � 40.08 g Ca

1 mol S � � 32.07 g S

4 mol O � � 64.00 g O

molar mass CaSO4 � 136.15 g/mol

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H


3.87 g CaSO4 �

� 0.0284 mol CaSO4

1.02 g H2O � � 0.0566 mol H2O

n � � 2.00

CaSO4�2H2O, calcium sulfate dihydrate

0.0566 mol��0.0284 mol

1 mol H2O ��18.02 g H2O

1 mol CaSO4��136.15 g CaSO4

1.008 g H��1 mol H

16.00 g O��1 mol O

16.00 g O��1 mol O

32.07 g S��1 mol S


0.816 mol��0.410 mol

1 mol H2O��18.02 g H2O

1 mol BaCl2��208.23 g BaCl2

1.008 g H��1 mol H

16.00 g O��1 mol O


137.33 g Ba��

1 mol Ba

10(18.02 g)��

286.15 g

105.99 g��286.15 g

1.008 g H��1 mol H

16.00 g O��1 mol O

16.00 g O��1 mol O

12.01 g C��1 mol C


1.008 g H��1 mol H

16.00 g O��1 mol O

12.01 g C��1 mol C



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154. The table shows data from an experiment todetermine the formulas of hydrated bariumchloride. Determine the formula for thehydrate and its name.

initial mass of hydrate � (mass of hydrate �crucible) � (mass of empty crucible) � 31.35 g �21.30 g � 10.05 g

mass of anhydrous solid � (mass after heating 5 min) � (mass of empty crucible) � 29.87 g �21.30 g � 8.57 g

mass of water � (initial mass of hydrate) �(mass of anhydrous solid) � 10.05 g � 8.57 g � 1.48 g

1 mol Ba � � 137.33 g Ba

2 mol Cl � � 70.90 g Cl

molar mass BaCl2 � 208.23 g/mol

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H


8.57 g BaCl2 �

� 0.0412 mol BaCl2

1.48 g H2O � � 0.0821 mol H2O

n � � 2.00

BaCl2�2H2O, barium chloride dihydrate

155. A 1.628-g sample of a hydrate of magnesiumiodide is heated until its mass is reduced to1.072 g and all water has been removed. Whatis the formula of the hydrate?

mass of water � mass of hydrate � mass ofanhydrous solid � 1.628 g � 1.072 g � 0.556 g

1 mol Mg � � 24.31 g Mg

2 mol I � � 253.80 g I

molar mass MgI2 � 278.11 g/mol

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H


1.072 g MgI2 �

� 0.003855 mol MgI2

0.556 g H2O � � 0.0309 mol H2O

n � � 8.02

MgI2�8H2O

156. Hydrated sodium tetraborate (Na2B4O7�xH2O)is commonly called borax. Chemical analysisindicates that this hydrate is 52.8% sodiumtetraborate and 47.2% water. Determine theformula and name the hydrate.

2 mol Na � � 45.98 g Na

4 mol B � � 48.04 g B

7 mol O � � 112.00 g O

molar mass Na2B4O7 � 206.02 g/mol

1 mol O � � 16.00 g O

2 mol H � � 2.016 g H


1.008 g H��1 mol H

16.00 g O��1 mol O

16.00 g O��1 mol O

12.01 g B��1 mol B


0.0309 mol��0.003855 mol

1 mol H2O��18.02 g H2O

1 mol MgI2��278.11 g MgI2

1.008 g H��1 mol H

16.00 g O��1 mol O

126.90 g I��

1 mol I


0.0821 mol��0.0412 mol

1 mol H2O��18.02 g H2O

1 mol BaCl2��208.23 g BaCl2

1.008 g H��1 mol H

16.00 g O��1 mol O


137.33 g B��1 mol Ba


Data for BaCl2�xH20

Mass of empty crucible 21.30 g

Mass of hydrate � crucible 31.35 g

Initial mass of hydrate 10.05 g

Mass after heating 5 min 29.87 g

Mass of anhydrous solid 8.57 g

Table 11-3

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52.8 g Na2B4O7 �

� 0.256 mol Na2B4O7

47.2 g H2O � � 2.61 mol H2O

n � � 10.2

Na2B4O7�10H2O, sodium tetraborate decahydrate

Mixed Review

Sharpen your problem-solving skills by answeringthe following.

157. Determine the following:

a. the number of representative particles in3.75 g Zn

3.75 g Zn � �

� 3.45 � 1022 atoms Zn

b. the mass of 4.32 � 1022 atoms Ag

4.32 � 1022 atoms Ag �

�

� 7.74 g Ag

c. the number of sodium ions is 25.0 g ofNa2O.

25.0 g Na2O � �

�

� 4.86 � 1023 ions Na�

158. Which of the following has the greatest number of oxygen atoms?

a. 17.63 g CO2

1 mol C � � 12.01 g C

2 mol O � � 32.00 g O


17.64 g CO2 � � 0.4008 mol CO2

0.4008 mol CO2 � � 0.8016 mol O

0.8016 mol O �

� 4.82 � 1023 atoms O

b. 3.21 � 1022 molecules CH3OH

3.21 � 1022 molecules CH3OH �

� 3.21 � 1022 atoms O

c. 0.250 mol C6H12O6

0.250 mol C6H12O6 �

� 1.5 mol O

1.5 mol O �

� 9.0 � 1023 atoms O

159. Which of the following compounds has thegreatest percent of oxygen by mass: TiO2,Fe2O3, Al2O3?

1 mol Ti � � 47.87 g Ti

2 mol O � � 32.00 g O

molar mass TiO2 � 79.87 g/mol

2 mol Fe � � 111.70 g Fe

3 mol O � � 48.00 g O


2 mol Al � � 53.96 g Al

3 mol O � � 48.00 g O

molar mass Al2O3 � 54.44 g/mol


� 40.07% O in TiO2

32.00 g O��79.87 g TiO2

16.00 g O��1 mol O


16.00 g O��1 mol O


16.00 g O��1 mol O


6.02 � 1023 atoms O��

1 mol O

6 mol O��1 mol C6H12O6

1 atom O��1 molecule CH3OH

6.02 � 1023 atoms O��

1 mol O

2 mol O��1 mol CO2

1 mol CO2��44.01 g CO2

16.00 g O��1 mol O

12.01 g C��1 mol C

6.02 � 1023 ions Na��

1 mol Na�2 mol Na��1 mol Na2O

1 mol Na2O��61.98 g Na2O

107.87 g Ag��

1 mol Ag1 mol Ag

��6.02 � 1023 atoms Ag

6.02 � 1023 atoms Zn��

1 mol Zn


2.61 mol��0.256 mol

1 mol H2O��18.02 g H2O

1 mol Na2B4O7��206.02 g Na2B4O7


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� 30.06% O in Fe2O3


� 87.99% O in Al2O3

Al2O3 has the greatest percent of oxygen by mass.

160. Naphthalene, commonly known as moth balls, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of napthaleneis 128 g/mol. Determine the empirical andmolecular formulas for naphthalene.

93.7 g C � � 7.80 mol C

6.3 g H � � 6.2 mol H

� 1.25 mol C

� 1.0 mol H

The simplest ratio is 1.25 mol C : 1.0 mol H.Multiply both subscripts by four to obtain thesimplest whole-number ratio. The empiricalformula is C5H4.

5 mol C � � 60.05 g C

4 mol H � � 4.032 g H


n � � 2.00

Molecular formula is C10H8.

161. Which of the following molecular formulas arealso empirical formulas: ethyl ether (C4H10O),aspirin (C9H8O4), butyl dichloride (C4H8Cl2),glucose (C6H12O6).

The subscripts in C4H10O and C9H8O4 representsimplest whole-number ratios.

162. Calculate each of the following:

a. the number of moles in 15.5 g Na2SO4

2 mol Na � � 45.98 g Na

1 mol S � � 32.07 g S

4 mol O � � 64.00 g O


15.5 g Na2SO4 �

� 0.109 mol Na2SO4

b. the number of formula units in 0.255 molNaCl

0.255 mol NaCl �


c. the mass in grams of 0.775 mol SF6

1 mol S � � 32.07 g S

6 mol F � � 114.00 g F


0.775 mol SF6 � � 113 g SF6

d. the number of Cl� ions in 14.5 g MgCl2.

1 mol Mg � � 24.31 g Mg

2 mol Cl � � 70.90 g Cl


14.5 g MgCl2 �

� 0.152 mol MgCl2

0.152 mol MgCl2 �

� 1.83 � 1023 ions Cl�

2(6.02 � 1023 ion Cl�)��

1 mol MgCl2

1 mol MgCl2��95.21 g MgCl2



146.07 g SF6��1 mol SF6

19.00 g F��1 mol F

32.07 g S��1 mol S


1 mol NaCl

1 mol Na2SO4��142.05 g Na2SO4

16.00 g O��1 mol O

32.07 g S��1 mol S


128 g/mol��64.09 g/mol

1.008 g H��1 mol H

12.01 g C��1 mol C

6.2 mol H��

6.2

7.80 mol C��

6.2

1 mol H��1.008 g H

1 mol C��12.01 g C

48.00 g O��54.44 g Al2O3

48.00 g O��159.70 g Fe2O3


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163. The graph shows the percent composition of a compound containing carbon, hydrogen,oxygen, and nitrogen. How many grams ofeach element are present in 100 g of thecompound?

52.42 g O, 22.95 g N, 19.68 g C, 4.96 g H

164. A party balloon was filled with 9.80 � 1022

atoms of helium. After 24 hours, 45% of thehelium had escaped. How many atoms ofhelium remained?

percent remaining � 100% � percent thatescaped � 100% � 45% � 55%

0.55 � 9.80 � 1022 atoms He � 5.39 � 1022 atoms He

165. Tetrafluoroethylene, which is used in the production of Teflon, is composed of 24.0%carbon and 76.0% fluorine and has a molarmass of 100.0 g/mol. Determine the empiricaland molecular formulas of this compound.

24.0 g C � � 2.00 mol C

76.0 g F � � 4.00 mol F

� 1.00 mol C

� 2.00 mol F

The simplest ratio is 1.00 mol C : 2.00 mol F. The empirical formula is CF2.

1 mol C � � 12.01 g C

2 mol F � � 38.00 g F

molar mass CF2 � 50.01 g/mol

n � � 2.000

The molar mass of the compound is two times greater than the molar mass of theempirical formula. Multiply the subscripts inthe empirical formula by two to obtain themolecular formula C2F4.

166. Calculate the mass in grams of one atom oflead.

1 atom Pb � �

� 3.44 � 10�22 g Pb

167. Diamond is a naturally occurring form of carbon. If you have a 0.25-carat diamond,how many carbon atoms are present? (1 carat � 0.200 g)

0.25 carat � � 5 � 10�2 g C

5 � 10�2 g C � �

� 2.5 � 1021 atoms C

168. How many molecules of isooctane (C8H18) are present in 1.00 L? (density of isooctane �0.680 g/mL)

8 mol C � � 96.08 g C

18 mol H � � 18.18 g H


1.00 L � �

� 6.80 � 102 g isooctane

0.680 g isooctane��

1 mL103 mL�

1 L

1.008 g H��1 mol H

12.01 g C��1 mol C

6.02 � 1023 atoms C��

1 mol C

1 mol C��12.01 g C

0.200 g C��

1 carat

207.02 g Pb��

1 mol Pb


100.0 g/mol��50.01 g/mol

38.00 g F��1 mol F

12.01 g C��1 mol C

4.00 mol F��

2.00

2.00 mol C��

2.00

1 mol F��19.00 g F

1 mol C��12.01 g C

H4.96%

C19.68%

N22.95%

O52.42%


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6.80 � 102 g isooctane �

� 5.95 mol isooctane

5.95 mol isooctane �

� 3.58 � 1024 molecules isooctane

169. Calculate the number of molecules of water ina swimming pool that is 40.0 m in length, 20.0m in width, and 5.00 m in depth. Assume thatthe density of water is 1.00 g/cm3.

volume � length � width � height

volume � 40.0 m � 20.0 m � 5.00 m � 4.00 � 103 m3

4.00 � 103 m3 � �

� 4.00 � 109 g H2O

4.00 � 1032 g H2O � �


Thinking Critically170. Analyze and Conclude A mining company

has two possible sources of copper: chalcopy-rite (CuFeS2) and chalcocite (Cu2S). If themining conditions and the extraction of copperfrom the ore were identical for each of theores, which ore would yield the greater quan-tity of copper? Explain your answer.

Chalcopyrite (CuFeS2) is 34.6% copper by mass(determined from percent composition) andchalcocite (Cu2S) is 79.9% copper by mass.Chalcocite would yield the greater quantity ofcopper because the ore has the greaterpercentage copper by mass.

171. Designing an Experiment Design an experi-ment that can be used to determine the amountof water in alum (KAl(SO4)2�xH2O).

Determine and record the mass of an emptyevaporating dish. Add about 2 g of the hydrate.Measure and record the mass. Heat theevaporating dish gently for 5 minutes, andstrongly for another 5 minutes to evaporate allthe water. Allow the dish to cool, and measureand record the mass. Determine the masses of theanhydrous solid and the water lost. Calculate thenumber of moles of anhydrous compound andwater. Determine the ratio of moles of water tomoles of anhydrous compound. Use the whole-number ratio of the moles as the coefficient ofH2O in the formula.

172. Concept Mapping Design a concept map thatillustrates the mole concept. Include moles,Avogadro’s number, molar mass, number ofparticles, percent composition, empiricalformula, and molecular formula.

Concept maps will vary.

173. Communicating If you use the expression “a mole of nitrogen,” is it perfectly clear whatyou mean, or is there more than one way tointerpret the expression? Explain. How couldyou change the expression to make it moreprecise?

The phrase could be interpreted to mean a moleof nitrogen atoms or a mole of nitrogenmolecules. Specify atoms or molecules.

Writing in Chemistry174. Octane ratings are used to identify certain

grades of gasoline. Research octane rating andprepare a pamphlet for consumers identifyingthe different types of gasoline, the advantagesof each, and when each grade is used.

Student pamphlets will vary but should discusshow octane ratings are found. Octane rating of100% means pure octane. Octane rating of 90%means 90% octane and 10% heptane.

6.02 � 1023 molecules H2O��

1 mol H2O

1 mol H2O��18.02 g H2O

1 g H2O��

1 cm3

(102 cm)3

��1 m3

6.02 � 1023 molecules isooctane��

1 mol isooctane

1 mol isooctane��114.26 g isooctane


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175. Research the life of the Italian chemistAmedeo Avogadro (1776–1856) and how hiswork led scientists to the number of particlesin a mole.

Students should mention Avogadro’s hypothesis.Avogadro formulated his hypothesis as anexplanation for earlier works by Gay-Lussac andRitter. His ideas were rejected by chemists of hisday but were revived later by the Italian chemistStanislao Cannizzaro. Avogadro died beforeseeing his ideas accepted.

Cumulative Review

Refresh your understanding of previous chapters byanswering the following.

176. Express the following answers with the correctnumber of significant figures. (Chapter 2)

a. 18.23 � 456.7

�438.5

b. 4.233 � 0.0131

323

c. (82.44 � 4.92) � 0.125

406

177. Distinguish between atomic number and massnumber. How do these two numbers comparefor isotopes of an element? (Chapter 4)

Atomic number equals the number of protons.Mass number equals the number of protons plusthe number of neutrons. Two isotopes of anelement will have the same atomic number butdifferent mass numbers.

178. Write balanced equations for the followingreactions. (Chapter 10)

a. Magnesium metal and water combine toform solid magnesium hydroxide andhydrogen gas.

Mg(s) � 2H2O(l) 0 Mg(OH)2(s) � H2(g)

b. Dinitrogen tetroxide gas decomposes intonitrogen dioxide gas.

N2O4(g) 0 2NO2(g)

c. Aqueous solutions of sulfuric acid andpotassium hydroxide undergo a doublereplacement reaction.

H2SO4(aq) � 2KOH(aq) 0 K2SO4(aq) � 2H2O(l)

179. How can you tell if a chemical equation is balanced? (Chapter 10)

There are equal numbers of each kind of atomon both sides of the equation.

Standardized Test PracticeChapter 11page 351

Interpreting Graphs Use the graph to answer questions 1–4.

1. Acetaldehyde and butanoic acid must have thesame _____ .

a. molecular formulab. empirical formula c. molar massd. chemical properties

Acetaldehyde and butanoic acid must have thesame empirical formula because they have thesame number of grams and therefore the samenumber of moles of each element in a 100.0 gsample.

b

Compound name

Percent Composition of Some Organic CompoundsPe

rcen

t b

y m

ass

0

10

20

30

40

50

60

Ethanol Formaldehyde Acetaldehyde Butanoic acid

%C%H%O

52.2

13.0

34.840.0

6.7 9.1

36.4

54.5

9.1

36.4

53.3 54.5


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2. If the molar mass of butanoic acid is 88.1 g/mol,then what is its molecular formula?

a. CH2Ob. C2H4Oc. C6HO2d. C4H8O2

Determine the empirical formula. Assume a100.0 g sample.

54.5 g C � � 4.54 mol C

9.1 g H � � 9.03 mol H

36.4 g O � � 2.28 mol O

Determine the mole ratios.

� 1.99 � 2 mol C

� 3.96 � 4 mol H

� 1.00 � 1 mol O

The empirical formula for butanoic acid is C2H4O.

The mass of the empirical formula is 44.06 g/mol.

� 2.00

The molecular formula for butanoic acid � (C2H4O)2 � C4H8O2.

d

3. What is the empirical formula of ethanol?

a. C4HO3b. C52H13O35c. C2H6O d. C4H13O2

52.2 g C � � 4.35 mol C

13.0 g H � � 12.9 mol H

34.8 g O � � 1 mol O

Determine the mole ratios.

� 2.00 mol C � 2 mol C

� 5.92 mol H � 6 mol H

� 1.00 mol O � 1 mol O

The empirical formula for ethanol is C2H6O.

c

4. The empirical formula of formaldehyde is thesame as its molecular formula. How manygrams are in 2.000 moles of formaldehyde?

a. 30.00 g b. 60.06 g c. 182.0 gd. 200.0 g

Determine the empirical formula forformaldahyde.

40.0 g C � � 3.33 mol C

6.7 g H � � 6.65 mol H

53.3 g O � � 3.33 mol O

� 1.00 mol C � 1 mol C

� 2.00 mol H � 2 mol H

� 1.00 mol O � 1 mol O

The empirical formula for formaldahyde is CH2O.Molar mass of CH2O � 30.03 g/mol

� 2.000 mol � 60.06 g formaldahyde

b

30.03 g�1 mol

3.33 mol O��

3.33

6.65 mol H��

3.33

3.33 mol C��

3.33

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C

2.18 mol O��

2.18

12.9 mol H��

2.18

4.35 mol C��

2.18

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C

88.1 g/mol��44.06 g/mol

2.28 mol O��

2.28

9.03 mol H��

2.28

4.54 mol C��

2.28

1 mol O��16.00 g O

1 mol H��1.008 g H

1 mol C��12.01 g C


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5. A mole is all of the following EXCEPT _____ .

a. the atomic or molar mass of an element orcompound

b. Avogadro’s number of molecules of acompound

c. the number of atoms in exactly 12 g of pure 12C

d. the SI measurement unit for the amount of asubstance

A mole is not a mass so a is the answer.

a

6. How many atoms are in 0.625 moles of Ge(atomic mass � 72.59 amu)?

a. 2.73 � 1025

b. 6.99 � 1025

c. 3.76 � 1023

d. 9.63 � 1023

0.625 mol Ge �

� 3.76 � 1023 atoms Ge

c

7. The molar mass of fluorapatite (Ca5(PO4)3F) is _____ .

a. 314 g/molb. 344 g/molc. 442 g/mold. 504 g/mol

5 mol Ca � � 200.40 g Ca

3 mol P � � 92.91 g P

12 mol O � � 192.00 g O

1 mol F � � 19.00 g F


d

8. How many moles of cobalt(III) titanate(Co2TiO4) are in 7.13 g of the compound?

a. 2.39 � 101 mol b. 3.14 � 10�2 molc. 3.22 � 101 mold. 4.17 � 10�2 mol

Find the molar mass of Co2TiO4.

2 mol Co � � 117.86 g Co

1 mol Ti � � 44.88 g Ti

4 mol O � � 64.00 g O

Molar mass Co2TiO4 � 226.74 g/mol

7.13 g Co2TiO4 �

� 3.14 � 10�2 mol

b

1 mol Co2TiO4��227 g Co2TiO4

16.00 g O��1 mol O



19.00 g F��1 mol F

16.00 g O��1 mol O

30.97 g P��1 mol P


6.02 � 1023 atoms Ge��

1 mol Ge


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9. Magnesium sulfate (MgSO4) is often added towater-insoluble liquid products of chemicalreactions to remove any unwanted water.MgSO4 readily absorbs water to form twodifferent hydrates. One of these hydrates isfound to contain 13.0% H2O and 87.0%MgSO4. What is the name of this hydrate?

a. magnesium sulfate monohydrate b. magnesium sulfate dihydratec. magnesium sulfate hexahydrated. magnesium sulfate heptahydrate

Assume a 100.0 g sample of the hydrate. Then, thesample contains 87.0 g MgSO4 and 13.0 g H2O.

Molar mass MgSO4 � 120.38 g/mol

Molar mass H2O � 18.02 g/mol

87.0 g MgSO4 �

� 0.723 mol MgSO4

13.0 g H2O � � 0.721 mol H2O

� �

The hydrate is MgSO4�H2O, magnesium sulfatemonohydrate

a

10. The mass of one molecule of barium hexafluo-rosilicate (BaSiF6) is _____ .

a. 1.68 � 1026 g b. 2.16 � 1021 gc. 4.64 � 10�22 g d. 6.02 � 10�23 g

Determine the molar mass of BaSiF6.

1 mol Ba � � 137.33 g Ba

1 mol Si � � 28.09 g Si

6 mol F � � 114.00 g F

Molar mass � 279.42 g/mol BaSiF6

�

� 4.64 � 10�22 g/molecule BaSiF6

c

1 mol BaSiF6��6.02 � 1023 molecules BaSiF6

279.42 g BaSiF6��1 mol BaSiF6

19.00 g F��1 mol F

28.09 g Si��1 mol Si

137.33 g Ba��

1 mol Ba

1�1

0.997�

10.721 mol H2O

��0.723 mol MgSO4

1 mol H2O��18.02 g H2O

1 mol MgSO4��120.38 g MgSO4


Chapter 11 Solutions to HW

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Transcript of Chapter 11 Solutions to HW