Chapter 11 Solutions to HW
Comments
Description
11CHAPTER SOLUTIONS MANUAL The Mole Section 11.1 Measuring Matter pages 309–312 Section 11.1 Assessment page 312 5. How is a mole similar to a dozen? Practice Problems pages 311, 312 The mole is a unit for counting 6.02 1023 representative particles. The dozen is used to count 12 items. 1. Determine the number of atoms in 2.50 mol Zn. 6.02 1023 atoms 2.50 mol Zn 1 mol 1.51 1024 atoms of Zn 2. Given 3.25 mol AgNO3, determine the number of formula units. 6.02 1023 formula units 3.25 mol AgNO3 1 mol 1.96 1024 formula units of AgNO3 3. Calculate the number of molecules in 11.5 mol H2O. Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. number and one mole? One mole contains Avogadro’s number (6.02 1023) of representative particles. 7. Explain how you can convert from the number of representative particles of a substance to moles of that substance. Multiply the number of representative particles by 1 mol/6.02 1023 representative particles. 8. Explain why chemists use the mole. 6.02 1023 molecules 11.5 mol H2O 1 mol 6.92 1024 molecules of H2O 4. How many moles contain each of the following? a. 5.75 1024 atoms Al 1 mol 5.75 1024 atoms Al 6.02 1023 atoms 9.55 mol Al b. 3.75 3.75 6. What is the relationship between Avogadro’s 1024 1024 molecules CO2 molecules CO2 1 mol 6.23 mol CO2 6.02 1023 molecules c. 3.58 1023 formula units ZnCl2 3.58 1023 formula units ZnCl2 1 mol 0.595 mol ZnCl2 6.02 1023 formula units d. 2.50 1020 atoms Fe 1 mol 2.50 1020 atoms Fe 6.02 1023 atoms 4.15 104 mol Fe Chemists use the mole because it is a convenient way of knowing how many representative particle are in a sample. 9. Thinking Critically Arrange the following from the smallest number of representative particles to the largest number of representative particles: 1.25 1025 atoms Zn; 3.56 mol Fe; 6.78 1022 molecules glucose (C6H12O6). From smallest to largest: 6.78 1022 molecules glucose, 2.14 1024 atoms Fe, 1.25 1025 atoms Zn. 10. Using Numbers Determine the number of representative particles in each of the following and identify the representative particle: 11.5 mol Ag; 18.0 mol H2O; 0.150 mol NaCl. 6.02 1023 atoms Ag 11.5 mol Ag 1 mol Ag 6.92 1024 atoms Ag 6.02 1023 molecules H2O 18.0 mol H2O 1 mol H2O 1.08 1025 molecules H2O 6.02 1023 formula units NaCl 0.150 mol NaCl 1 mol NaCl 9.03 1022 formula units NaCl Solutions Manual Chemistry: Matter and Change • Chapter 11 135 11 Section 11.2 Mass and the Mole pages 313–319 Practice Problems SOLUTIONS MANUAL 13. How many atoms are in each of the following samples? a. 55.2 g Li pages 316, 318 6.02 1023 atoms 1 mol Li 55.2 g Li 6.94 g Li 1 mol 11. Determine the mass in grams of each of the 4.79 1024 atoms Li following. a. 3.57 mol Al 26.98 g Al 3.57 mol Al 96.3 g Al 1 mol Al b. 42.6 mol Si 28.09 g Si 42.6 mol Si 1.20 103 g Si 1 mol Si c. 3.45 mol Co 58.93 g Co 3.45 mol Co 203 g Co 1 mol Co d. 2.45 mol Zn 65.38 g Zn 2.45 mol Zn 1.60 102 g Zn 1 mol Zn 12. Determine the number of moles in each of the following. a. 25.5 g Ag 1 mol Ag 25.5 g Ag 0.236 mol Ag 107.9 g Ag b. 300.0 g S 1 mol S 300.0 g S 9.355 mol S 32.07 g S c. 125 g Zn 1 mol Zn 125 g Zn 1.91 mol Zn 65.38 g Zn d. 1.00 kg Fe 1000 g Fe 1 mol Fe 1.00 kg Fe 55.85 g Fe 1 kg Fe 17.9 mol Fe b. 0.230 g Pb 1 mol Pb 0.230 g Pb 207.2 g Pb 6.02 1023 atoms 6.68 1020 atoms Pb 1 mol c. 11.5 g Hg 1 mol Hg 11.5 g Hg 200.6 g Hg 6.02 1023 atoms 3.45 1022 atoms Hg 1 mol d. 45.6 g Si 1 mol Si 45.6 g Si 28.09 g Si 6.02 1023 atoms 9.77 1023 atoms Si 1 mol e. 0.120 kg Ti 1000 g Ti 1 mol Ti 0.120 kg Ti 47.87 g Ti 1 kg Ti 6.02 1023 atoms 1.51 1024 atoms Ti 1 mol 14. What is the mass in grams of each of the following? a. 6.02 1024 atoms Bi 1 mol Bi 6.02 1024 atoms Bi 6.02 1023 atoms 209.0 g Bi 2.09 103 g Bi 1 mol Bi b. 1.00 1024 atoms Mn 1 mol Mn 1.00 1024 atoms Mn 6.02 1023 atoms 54.94 g Mn 91.3 g Mn 1 mol Mn 136 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. CHAPTER 11 CHAPTER c. 3.40 1022 atoms He 3.40 1022 1 mol He atoms He 6.02 1023 atoms 4.003 g He 0.226 g He 1 mol He d. 1.50 1015 atoms N SOLUTIONS MANUAL 19. Sequencing Arrange the following in order of mass from the smallest mass to the largest: 1.0 mol Ar, 3.0 1024 atoms Ne, 20 g Kr. 39.95 g Ar 1.0 mol Ar 39.95 g Ar 1 mol Ar 1 mol Ne 3.0 1024 atoms Ne 6.02 1023 atoms Ne 1 mol N 1.50 1015 atoms N 6.02 1023 atoms 20.18 g Ne 101 g Ne 1 mol Ne 14.01 g N 3.49 108 g N 1 mol N 20 g Kr, 1.0 mol Ar, 3.0 1024 atoms Ne e. 1.50 1015 atoms U 1.50 1015 1 mol U atoms U 6.02 1023 atoms 238.0 g U 5.93 107 g U 1 mol U Section 11.3 Moles of Compounds pages 320–327 Practice Problems pages 321–324, 326 Section 11.2 Assessment in 2.50 mol ZnCl2. 15. Explain what is meant by molar mass. 2 mol Cl 2.50 mol ZnCl2 5.00 mol Cl 1 mol ZnCl2 Molar mass is the mass in grams of one mole of any pure substance. Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. 20. Determine the number of moles of chloride ions page 319 16. What conversion factor should be used to convert from mass to moles? Moles to mass? Mass-to-mole conversions use the conversion factor 1 mol/number of grams. Moles-to-mass conversions use the conversion factor number of grams/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 atom is usually given in the unit amu. Would it be possible to express the mass of a single atom in grams? Explain. Because one mole is equal to 6.02 1023 atoms and the molar mass of an element is equal to one mole, the mass of a single atom can be calculated by dividing the mass of one mole by Avogadro’s number. Solutions Manual 21. Calculate the number of moles of each element in 1.25 mol glucose (C6H12O6). 6 mol C 1.25 mol C6H12O6 7.50 mol C 1 mol C6H12O6 12 mol H 1.25 mol C6H12O6 15.0 mol H 1 mol C6H12O6 6 mol O 1.25 mol C6H12O6 7.50 mol O 1 mol C6H12O6 22. Determine the number of moles of sulfate ions present in 3.00 mol iron(III) sulfate (Fe2(SO4)3). 3 mol SO42 3.00 mol Fe2(SO4)3 1 mol Fe2(SO4)3 9.00 mol SO42 23. How many moles of oxygen atoms are present in 5.00 mol diphosphorus pentoxide (P2O5)? 5 mol O 5.00 mol P2O5 25.0 mol O 1 mol P2O5 Chemistry: Matter and Change • Chapter 11 137 11 SOLUTIONS MANUAL 24. Calculate the number of moles of hydrogen atoms in 11.5 mol water. 14.01 g N 3 mol N 42.03 g 1 mol N 2 mol H 11.5 mol H2O 23.0 mol H 1 mol H2O 1.008 g H 12 mol H 12.096 g 1 mol H 25. Determine the molar mass of each of the following ionic compounds: NaOH, CaCl2, KC2H3O2, Sr(NO3)2, and (NH4)3PO4. NaOH 22.99 g Na 1 mol Na 22.99 g 1 mol Na 30.97 g P 1 mol P 1 mol P 16.00 g O 4 mol O 64.00 g 1 mol O molar mass (NH4)3PO4 149.10 g/mol 26. Calculate the molar mass of each of the 16.00 g O 1 mol O 1 mol O 16.00 g 1.008 g H 1 mol H 1 mol H following molecular compounds: C2H5OH, C12H22O11, HCN, CCl4, and H2O. 1.008 g C2H5OH molar mass NaOH 40.00 g/mol CaCl2 40.08 g Ca 1 mol Ca 1 mol Ca 40.08 g 35.45 g Cl 2 mol Cl 1 mol Cl 70.90 g molar mass CaCl2 110.98 g/mol KC2H3O2 39.10 g K 1 mol K 1 mol K 39.10 g 12.01 g C 2 mol C 1 mol C 24.02 g 1.008 g H 3 mol H 3.024 g 1 mol H 16.00 g O 2 mol O 32.00 g 1 mol O molar mass KC2H3O2 98.14 g/mol Sr(NO3)2 87.62 g Sr 1 mol Sr 87.62 g 1 mol Sr 14.01 g N 2 mol N 28.02 g 1 mol N 16.00 g O 6 mol O 96.00 g 1 mol O molar mass Sr(NO3)2 211.64 g/mol (NH4)3PO4 12.01 g C 2 mol C 24.02 g 1 mol C 1.008 g H 6 mol H 6.048 g 1 mol H 16.00 g O 1 mol O 16.00 g 1 mol O molar mass C2H5OH Chemistry: Matter and Change • Chapter 11 46.07 g/mol C12H22O11 12.01 g C 12 mol C 1 mol C 144.12 g 1.008 g H 22 mol H 22.176 g 1 mol H 16.00 g O 11 mol O 176.00 g 1 mol O molar mass C12H22O11 342.30 g/mol HCN 1.008 g H 1 mol H 1.008 g 1 mol H 12.01 g C 1 mol C 1 mol C 12.01 g 14.01 g N 1 mol N 14.01 g 1 mol N molar mass HCN 27.03 g/mol CCl4 12.01 g C 1 mol C 1 mol C 12.01 g 35.45 g Cl 4 mol Cl 141.80 g 1 mol Cl molar mass CCl4 138 30.97 g 153.81 g/mol Solutions Manual Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc. CHAPTER 65.90 g molar mass ZnCl2 136.38 g Zn 1 mol Zn 65.0403 mol ZnSO4 54.39 g Zn 1 mol Zn 65.016 g 1 mol H 16.93 g ZnCl2 29.38 g 1 mol Zn 35. 27.008 g H 2 mol H 2.55 mol KMnO4 1 mol KMnO4 403 g KMnO4 30.28 g/mol Step 2: Make mole 0 mass conversion. a.02 g/mol (H2SO4)? Step 1: Find the molar mass of H2SO4.9 g 1 mol Ag 16.46 g/mol Step 2: Make mass 0 mole conversion. a division of the McGraw-Hill Companies.00 g 1 mol O 28.55 moles? Step 1: Find the molar mass of KMnO4.50 g ZnSO4 Step 1: Find the molar mass of ZnSO4. 32. 158.07 g S 1 mol S 1 mol S 32.00 g molar mass ZnSO4 161. What is the mass of 3. 98.45 g Cl 2 mol Cl 1 mol Cl 70.94 g 1 mol Mn Solutions Manual Chemistry: Matter and Change • Chapter 11 139 . 39.25 mol H2SO4 1 mol H2SO4 319 g H2SO4 Copyright © Glencoe/McGraw-Hill. 98.00 g O 4 mol O 1 mol O 64.9 g/mol Step 2: Make mass 0 mole conversion.94 g Mn 1 mol Mn 54. Inc.04 g/mol Step 1: Find the molar mass of AgNO3.35 158.00 g O 4 mol O 1 mol O moles of zinc 14.008 g H 2 mol H 2.04 g KMnO4 2.016 g 1 mol H molar mass H2SO4 64.00 g 1 mol O molar mass H2O 18. How many grams of potassium permanganate are in 2. 0.07 g 102 molar mass KMnO4 of the following.28 g ZnCl2 4.01 g N 1 mol N 1 mol N 14.9 g Ag 1 mol Ag 107.10 g b. 136. What is the mass of 4.07 g S 1 mol S 1 mol S 16. 6.00 g O 4 mol O 64.00 g O 3 mol O 1 mol O 48.25 moles of sulfuric acid 32.39 g 1 mol Zn 32. 1 mol ZnSO4 6.9 g AgNO3 Step 1: Find the molar mass of ZnCl2.35 102 mol ZnCl2 1 mol ZnCl2 5. 22.50 g ZnSO4 161.09 g/mol Step 2: Make mole 0 mass conversion.6 g AgNO3 169.00 g molar mass AgNO3 169.07 g 16.00 g O 1 mol O 16.133 mol AgNO3 65.00 g Step 2: Make mole 0 mass conversion. chloride (ZnCl2)? 1 mol AgNO 22.10 g K 1 mol K 1 mol K 39.09 g H2SO4 3.01 g 16. Determine the number of moles present in each 107.46 g ZnSO4 0.CHAPTER 11 SOLUTIONS MANUAL H2O 1.6 g AgNO3 1. 0778 mol Ag2CrO4 Step 1: Find the molar mass of Fe2O3.59 1024 formula units NaCl 1 mol NaCl 7.8 g/mol Step 2: Make mass 0 mole conversion. 1. 35.70 g 1 mol Fe 16.0 g/mol Step 2: Make mass 0 mole conversion.70 g/mol Step 2: Make mass 0 mole conversion.02 1023 formula units Solutions Manual Copyright © Glencoe/McGraw-Hill.960 mol HCl 36.8 g Ag2CrO4 1 mol Ag2CrO4 1 mol Ag2CrO4 6.00 g molar mass Ag2CrO4 331.80 g molar mass PbCl4 349.8 g Ag2CrO4 0.008 g 35.00 g 16.45 g Cl 4 mol Cl 1 mol Cl 141. a division of the McGraw-Hill Companies.51 1022 g Ag2CrO4/formula unit 32.008 g H 1 mol H 1 mol H 1. 1 mol PbCl4 254 g PbCl4 349.00 g O 4 mol O 1 mol O 64.157 mol Fe2O3 6.36 1022 Ag ions b.0 g Fe2O3 52.728 mol PbCl4 31.0 g Fe2O3 159.02 1023 formula units 5. 0. 55. 1 mol HCl 35.68 1022 formula units Ag2CrO4 2 Ag ions 1 formula unit Ag2CrO4 9.2 g 1 mol Pb 35. 254 g PbCl4 Step 1: Find the molar mass of PbCl4.59 1024 formula units? Step 1: Find the number of moles of NaCl. What mass of sodium chloride contains 4.62 mol NaCl 6. Step 1: Find the molar mass of Ag2CrO4.70 g Fe2O3 0.0 g HCl 0.0778 mol Ag2CrO4 Step 3: Make mole 0 formula unit conversion. 1 mol Fe2O3 25.0 g HCl Step 1: Find the molar mass of HCl.9 g Ag 2 mol Ag 215. A sample of silver chromate (Ag2CrO4) has a mass of 25.8 g Ag2CrO4 331. 207.00 g Cr 1 mol Cr 1 mol Cr Chemistry: Matter and Change • Chapter 11 1 CrO42 ion 1 formula unit Ag2CrO4 4. How many CrO42 ions are present? 4. What is the mass in grams of one formula unit of silver chromate? 331.2 g Pb 1 mol Pb 207. CHAPTER .46 g HCl d.02 1023 formula units 1 mol 4.68 1022 formula units Ag2CrO4 e.0 g PbCl4 0. How many Ag ions are present? 4. 107. Inc.8 g.45 g Cl 1 mol Cl 35.45 g 1 mol Cl molar mass HCl 36. 4. 1 mol Ag2CrO4 25.68 1022 formula units Ag2CrO4 a.8 g 1 mol Ag 140 52.46 g/mol Step 2: Make mass 0 mole conversion.00 g O 3 mol O 1 mol O 48.85 g Fe 2 mol Fe 111.00 g molar mass Fe2O3 159.68 1022 CrO42 ions c. 25.11 SOLUTIONS MANUAL c. 96 1023 molecules C2H5OH 1 SO32 ions 1 formula unit Na2SO3 2 C atoms 1.16 1022 Na ions b. How many oxygen atoms are present? 5.58 1024 H atoms 1 molecule C2H5OH c.048 g 1 mol H 16.00 g O 3 mol O 1 mol O 48.00 g 1 mol O molar mass C2H5OH 46.09 1022 g Na2SO3/formula unit Chemistry: Matter and Change • Chapter 11 141 .02 1023 molecules 0.07 g C2H5OH 0. molar mass Na2SO3 126.25 g Na2SO3 126.01 g C 2 mol C 24.25 g.08 1022 SO32 ions c. What is the mass in grams of one formula unit of Na2SO3? 126.990 mol C2H5OH Step 3: Make mole 0 molecule conversion. 22.44 g/mol Step 3: Make mole 0 mass conversion. How many SO32 ions are present? contain? 1.96 1023 molecules C2H5OH 6 H atoms 3.96 1023 molecules C2H5OH a.62 mol NaCl 445 g NaCl 1 mol NaCl 34.98 g 1 mol Na 32.07 g S 1 mol S 1 mol S 32.08 g Na2SO3 1 mol Na2SO3 1 mol Na2SO3 6. 6.45 g molar mass NaCl 58.6 g C2H5OH 46.19 1024 C atoms 1 molecule C2H5OH b. A sample of ethanol (C2H5OH) has a mass of 12. 16. Step 1: Find the molar mass of Na2SO3 22.008 g H 6 mol H 6. a division of the McGraw-Hill Companies.990 mol C2H5OH 1 mol 5.08 1022 formula units Na2SO3 5. How many Na ions are present? 1.44 g NaCl 7.02 g 1 mol C 1.05 g Na2SO3 0.08 1022 formula units Na2SO3 2 Na ions 1 formula unit Na2SO3 2. Copyright © Glencoe/McGraw-Hill.99 g Na 1 mol Na 22.07 g/mol Step 2: Make mass 0 mole conversion. A sample of sodium sulfite (Na2SO3) has a mass of 2. 1 O atom 5.00 g O 1 mol O 16. Inc.96 1023 molecules C2H5OH Solutions Manual 1.07 g 45.0179 mol Na2SO3 Step 3: Make mole 0 formula unit conversion 0.CHAPTER 11 SOLUTIONS MANUAL Step 2: Find the molar mass of NaCl.99 g 1 mol Na 35.99 g Na 2 mol Na 45.6 g.0179 mol Na2SO3 6.00 g Step 1: Find the molar mass of C2H5OH. How many carbon atoms does the sample Step 2: Make mass 0 mole conversion 1 mol Na2SO3 2.02 1023 formula units 2. 58.05 g/mol 33.02 1023 formula units 1 mol 1.08 1022 formula units Na2SO3 a.96 1023 O atoms 1 molecule C2H5OH 1 mol C2H5OH 45. How many hydrogen atoms are present? 5.45 g Cl 1 mol Cl 1 mol Cl 35. What is the mass in grams of one molecule of CO2? 44.11 1023 C atoms 1 molecule CO2 b. c.11 1023 molecules CO2 a. Step 3: Make mole 0 molecule conversion.45 g Cl 4 mol Cl 141. multiply by Avogadro’s number. Add the resulting masses. A sample of carbon dioxide has a mass of 52. 6. Multiply the mass of one mole of each element by the ratio of that element to one mole of the compound. a division of the McGraw-Hill Companies. Explain how you can determine the number of atoms or ions in a given mass of a compound. H Cl Element O 12.12 g C 1 mol C 1. What three conversion factors are often used in mole conversions? number of grams .01 g C 1 mol C 12.11 1023 molecules CO2 38.00 g O 2 mol O 32. 1 mol 6. Describe how you can determine the molar mass of a compound.02 1023 molecules 1 mol CO2 7.01 g CO2 .0 g CO2 44.032 g H 35.01 g 1 mol C 16.02 1023 representative particles . 1 mol CO2 1.80 g Cl 1 mol Cl 16.02 1023 molecules 1. Thinking Critically Design a bar graph that will show the number of moles of each element present in 500 g dioxin (C12H4Cl4O2).3 Assessment page 327 Number of moles 2 O atoms 1.01 g CO2 1 mol 6. Inc.0 g. Yes.96 g/mol Solutions Manual Copyright © Glencoe/McGraw-Hill.18 mol CO2 52. a powerful poison. How many oxygen atoms are present? 7.11 1023 molecules CO2 1 C atom 7.18 mol CO2 1 mol 7. 39.01 g/mol 37.00 g O 2 mol O 32.00 g O 1 mol O molar mass 142 Chemistry: Matter and Change • Chapter 11 321. If you know the mass in grams of a molecule of a substance. 12. could you obtain the mass of a mole of that substance? Explain.008 g H 4 mol H 1 mol H 4. How many carbon atoms are present? 7.42 1024 O atoms 1 molecule CO2 27 24 21 18 15 12 9 6 3 0 Moles of Elements in 500 g Dioxin C 36.01 g C 12 mol C 144.31 1023 g CO2/molecule Section 11.00 g 1 mol O molar mass CO2 44. 40. multiply the number of moles by the ratio of the number of atoms or ions to one mole. 1 mol number of atoms of element 1 mol compound Step 2: Make mass 0 mole conversion. Convert the mass to moles. multiply the mass in grams of the molecule by Avogadro’s number.CHAPTER 11 SOLUTIONS MANUAL 35. Step 1: Find the molar mass of CO2. 05 g/mol Step 3: Determine percent by mass of each element. Calculate the percent composition of sodium sulfate.07 g S percent S 100 142.90 g Cl percent Cl 100 110. Applying Concepts The recommended daily allowance of calcium is 1000 mg of Ca2 ions. 333. 40.98 g 1 mol Na 32. 40.99 g Na 2 mol Na 45.37% Na 32.08 g 35.01 g C 1 mol C 1 mol C 12.96 g C12H4Cl4O2 2 mol C12H4Cl4O2 pages 328–337 12 mol C 2 mol C12H4Cl4O2 24 mol C 1 mol C12H4Cl4O2 4 mol H 2 mol C12H4Cl4O2 8 mol H 1 mol C12H4Cl4O2 4 mol Cl 2 mol C12H4Cl4O2 8 mol Cl 1 mol C12H4Cl4O2 2 mol O 2 mol C12H4Cl4O2 4 mol O 1 mol C12H4Cl4O2 41.89% Cl 43.98 g CaCl2 63. 45. a division of the McGraw-Hill Companies.CHAPTER 11 SOLUTIONS MANUAL 1 mol C12H4Cl4O2 500 g C12H4Cl4O2 321.09 g/mol 1 g Ca2 1 mol Ca2 1000 mg Ca2 3 2 10 mg Ca 40.09 g CaCO3 0.00 g molar mass Na2SO4 142.01 g C 16.98 g Na percent Na 100 142.08 g Ca percent Ca 100 110. 335. Calcium carbonate is used to supply the calcium in vitamin tablets.00 g O 4 mol O 1 mol O 64.11% Ca 70.4 Empirical and Molecular Formulas Practice Problems pages 331.05 g Na2SO4 32. 337 42.05 g Na2SO4 22. Inc.08 g Ca 1 mol Ca 12. 40.98 g CaCl2 36.98 g/mol Step 3: Determine percent by mass of each element. Determine the percent by mass of each element in calcium chloride.58% S Solutions Manual Chemistry: Matter and Change • Chapter 11 143 .02 mol Ca2 1 mol CaCO3 1 mol Ca2 2 g CaCO3 Section 11.02 mol Ca2 1 mol CaCO3 100. calculate molar mass of Na2SO4. How many moles of calcium ions does 1000 mg represent? How many moles of calcium carbonate are needed to supply the required amount of calcium ions? What mass of calcium carbonate must each tablet contain? Copyright © Glencoe/McGraw-Hill.00 g O molar mass 100.08 g Ca 1 mol Ca 1 mol Ca 40.00 g O 3 mol O 1 mol O 48. Steps 1 and 2: Assume 1 mole.08 g Ca2 0.45 g Cl 2 mol Cl 1 mol Cl 70.07 g S 1 mol S 1 mol S 32.90 g molar mass CaCl2 110. calculate molar mass of CaCl2. 22. Steps 1 and 2: Assume 1 mole.08 g Ca 1 mol Ca 40.07 g 16. 84 g N 2. calculate molar mass of H2S2O8.00 g O 4 mol O 64.016 g 1 mol H 32.01 g N 1 mol O 63.08% H 97.630 mol N 1.948 mol O 1.31% O 97.000 mol N 1 mol N The simplest ratio is 1 mol N: 1. 144 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. a division of the McGraw-Hill Companies. 64.99 g H3PO4 46.97 g 1.630 mol N 1.630 mol N 14. Step 3: Convert decimal fraction to whole number.06 g S 2 mol S 1 mol S 64. In this case. Inc.11 SOLUTIONS MANUAL 64.05 g Na2SO4 45. 2.00 g O percent O 100 142.008 g H 2 mol H 2.024 g 1 mol H Steps 1 and 2: Assume 1 mole. calculate moles of each element.14 g/mol Step 3: Determine percent by mass of S.00 g 1 mol O molar mass H2S2O8 194.016 g 16.08 g/mol Step 3: Determine percent by mass of S. multiply by 2 because 1.12 g 16. 1 mol N 36. the empirical formula is N2O3.99 g H3PO4 30.06 g S 1 mol S 1 mol S 32.16% oxygen.12 g S percent S 100 194. Which has the larger percent by mass of sulfur. calculate molar mass of H3PO4.5 mol O 3. 30. 45.000 mol N 1 mol N 2.00 g O 3 mol O 1 mol O 48. calculate molar mass of H2SO3.00 g molar mass H2SO3 82. 3.97 g P 1 mol P 1 mol P 30.05% O 44.61% P 97.00 g O percent O 100 65. What is the empirical formula for this solid? Step 1: Assume 100 g sample. H2SO3 or H2S2O8? 1. Assume 1 mole.16 g O 3.06% S 82.008 g H 3 mol H 3.948 mol O 16.008 g H 2 mol H 1 mol H 2.84% nitrogen and 63.97 g P percent P 100 31. What is the percent composition of phosphoric acid (H3PO4)? Steps 1 and 2: Assume 1 mole. 97.99 g H3PO4 64.630 mol N 1.08 g H2SO3 Repeat steps 1 and 2 for H2S2O8.000 mol N 1 mol N 1.500 mol O 2.06 g molar mass H3PO4 16. Therefore.14 g H2S2O8 33.03% S H2SO3 has a larger percent by mass of S.99 g/mol Step 3: Determine percent by mass of each element.5 2 3.00 g O 8 mol O 128. CHAPTER .5 mol O. 32.024 g H percent H 100 3.06 g S percent S 100 39.00 g 1 mol O 32. A blue solid is found to contain 36. 1.00 g O Step 2: Calculate mole ratios. 1 mol Al 35.000 mol Mg 1 mol Mg 8 mol O 3.06 g S Step 2: Calculate mole ratios.01 g C that contains 10. the empirical formula is C3H8.44% hydrogen.500 mol S 1. multiply by 4 because 2.01 g C 1 mol H 4.999 mol O 0.8962 mol Cl 35.00 mol O 1 mol O 1 mol O 2.4480 mol Mg 1. Step 1: Assume 100 g sample.996 mol S 32. Step 2: Calculate mole ratios.34 g O 3.34% oxygen? 1 mol H 18.008 g H 1 mol O 35.02 g S 1.18% hydrogen. Step 3: Convert decimal fraction to whole number. calculate moles of each element.22 mol O 1. the empirical formula is C9H8O4.98 g Al 1.22 mol O 1.5 mol S 1. Step 3: Convert decimal fraction to whole number. the empirical formula is Al2S3.18 g H 18.77% chlorine.31 g Mg 1 mol Cl 31. It is 81. 0.CHAPTER 11 47.000 mol Mg 1 mol Mg Solutions Manual Chemistry: Matter and Change • Chapter 11 145 .77 g Cl 0.44 g H 4.584 mol O 7. composed only of carbon and hydrogen.00 mol C 2.22 mol O 1. 1 mol Mg 10.00 g C 5.5 2 3.000 mol Al 1 mol Al 1. calculate moles of each element. calculate moles of each element.334 mol Al 26.000 mol Cl 2 mol Cl 0. 50. a compound Copyright © Glencoe/McGraw-Hill.813 mol C 12.4480 mol Mg 1.65 3 7.22 mol O 16.00 g O Step 2: Calculate mole ratios.56% oxygen.813 mol C 1. What is the empirical formula for a compound 1 mol C 81. What is the empirical formula? Step 1: Assume 100 g sample. and 57. Therefore.45 g Cl 1 mol O 57.334 mol Al 1. 4.40 mol H 1. calculate moles of each element.649 mol H 6.00 mol O 1 mol O 4. 5. 1 mol C 60. Therefore.00% carbon. Step 1: Assume 100 g sample.4480 mol Mg 1.000 mol C 1 mol C 6.000 mol Mg 1 mol Mg 0. The chemical analysis of aspirin indicates that the molecule is 60. In this case. multiply by 3 because 2.813 mol C 1.00 mol O 2.000 mol C 1 mol C 2.04 mol H 1.40 mol H 1. Propane is a hydrocarbon.008 g H Step 1: Assume 100 g sample.000 mol Mg 1 mol Mg 0. Determine the empirical formula for aspirin.65 mol H.25 mol C 2.4480 mol Mg 24.98 g Al 1 mol S 64.996 mol S 1.82 g C 6.02% sulfur.334 mol Al 1.000 mol Al 1 mol Al 1. In this case.5 mol S.56 g O 2.82% carbon and 18. multiply by 2 because 1.813 mol C 1.22 mol O 1.000 mol C 1 mol C The simplest ratio is 1 mol: 2. 1. a division of the McGraw-Hill Companies. In this case.95 8. Inc. 48. Therefore.00 mol C 12.98 mol H 2 mol H 2.25 mol C: 2 mol H: 1 mol O.00 g O Step 2: Calculate mole ratios.89 g Mg 0.65 mol H 18.25 4 9.04 mol H 2.4480 mol Mg 1. Determine the empirical formula for a compound that contains 35. and 35. Step 3: Convert decimal fraction to whole number.98% aluminum and 64.334 mol Al 1. SOLUTIONS MANUAL 49.25 mol C 2.000 mol Al 1 mol Al The simplest ratio is 1 mol Al: 1.89% magnesium. 31.00 mol O 1 mol O The simplest ratio is 2.584 mol O 16.8962 mol Cl 2. 6. 53. 110. a division of the McGraw-Hill Companies. calculate moles of each element.333 mol O 16.008 g H 3 mol H 3.32 g O 3.97 mol H 3 mol H 1.68 g N 3. Determine the molecular formula. CHAPTER .000 mol O 1 mol O The simplest ratio is 1 mol: 2.00 g O 1 mol O 16.000 mol C 1 mol C Step 2: Calculate mole ratios.818 mol O 16. 1 mol C 65.98 g C 4.5 mol H 4.01 g C 2 mol C 24. A compound was found to contain 49.11 SOLUTIONS MANUAL The empirical formula is MgCl2O8 or Mg(ClO4)2. What is the molecular formula? Step 1: Assume 100 g sample.162 mol C 1. Inc. The molar mass of the compound is 58. Therefore. the empirical formula is C3H3O.00 g O Step 2: Calculate mole ratios. 1 mol O 53.0 g/mol 1. or 2 55.332 mol N 1.45% C.06 g/mol Step 4: Determine whole number multiplier.41 mol H 2. 1 mol H 5. Analysis of a chemical used in photographic developing fluid indicates a chemical composition of 65. the empirical formula is C2H5.41 mol H 1.332 mol N 1. calculate moles of each element.998.00 g 1 mol O molar mass C3H3O 55.12 g/mol 2.162 mol C 1.5 mol H.008 g H 5 mol H 5.450 mol C 3.818 mol O 1. Determine the molecular formula. 58.12 g/mol.818 mol O 1.818 mol O 1. Because 2. 1 mol C 49.000 mol C 3 mol C 1.000 mol N 1 mol N 3.000 mol O 1 mol O 5.05 g/mol Step 4: Determine whole number multiplier. Step 3: Calculate the molar mass of the empirical formula.000 mol N 1 mol N 146 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.5 2 5.32% oxygen has a molar mass of 60.50 mol H 2.02 g 1 mol C 1.45 g C 5. Step 1: Assume 100 g sample. 3.39 mol H 2.0 g/mol.01 g C 1 mol H 10. Step 1: Assume 100 g sample. The simplest ratio is 1 mol C: 2.162 mol C 12.000 29. Step 3: Calculate the molar mass of the empirical formula. and 29.68% nitrogen and 53.008 g H 4.008 g H Step 2: Calculate mole ratios. 5.01 g C 3 mol C 36.000 mol O 1.47 g H 10.01 g C 52.47 g hydrogen.09 g O 1. A colorless liquid composed of 46.01 g/mol.01 g N The molecular formula is C6H6O2.000 mol C 1 mol C 4. 12.040 g 1 mol H molar mass C2H5 29. 12.00 g O 10.818 mol O 1.45% H.00 mol O 1 mol O 1 mol O 1.332 mol N 14.450 mol C 12. The molar mass is found to be 110.05 g/mol 1 mol N 46. calculate moles of each element. 5.09% O. The simplest ratio is 1 mol Mg: 2 mol Cl: 8 mol O.39 mol H 1.024 g 1 mol H 16.06 g/mol The molecular formula is C4H10. 51.98 g carbon and 10.162 mol C 1.000 mol C 1 mol C 1 mol O 29.45 g H 5.65 mol H.03 g 1 mol C 1. 500 mol O 3.696 mol O 1. 3.250 mol O 16. The pain reliever morphine contains 17.5 2 3.000 mol N 1 mol N The simplest ratio is 1 mol N: 1 mol O.000 mol Fe 1 mol Fe 3.333 mol O 1. The empirical formula is NO.08765 mol N 1. Find the empirical formula for the oxide. 14. 54. a division of the McGraw-Hill Companies.900 g C 1.00 mol O 0.131 mol Fe 1.131 mol Fe 1.000 mol N 1 mol N The simplest ratio is 17 mol C: 19 mol H: 3 mol O: 1 mol N.00 g O Step 2: Calculate mole ratios. 57.5000 mol K 39. the empirical formula is Fe2O3.00 mol O 1 mol O The simplest ratio is 2 mol K: 1 mol O. and 1. Step 2: Calculate mole ratios.131 mol Fe 55.00 g 1 mol O 56.000 mol N 1 mol N 0.000 mol O 3.0303 mol O 6.000 30. 55.00 mol K 2 mol K 0.667 mol H 19. 4.00 g O 1 mol O 16.485 g O.680 g H 1. molar mass NO 30.696 mol O 16.228 g N 0.5000 mol K 2. Step 2: Calculate mole ratios.250 mol O 1. What is the empirical formula for this compound? Step 1: Calculate moles of each element 0.250 mol O 1.228 g N.2641 mol O 16.00 mol O 1 mol O 1 mol O 0.225 g O 0.225 g O.013 mol O 3 mol O 0. 1. Step 3: Calculate the molar mass of the empirical formula. 0.08765 mol N 1. Analysis of a compound composed of iron and oxygen yields 174.00 g O Solutions Manual Chemistry: Matter and Change • Chapter 11 147 .08765 mol N 1.10 g K 1 mol O 4.01 g N 1 mol N 14.98 g Al 1 mol O 75.490 mol C 17.000 mol N 1 mol N 1.5 mol O.000 mol Fe 1 mol Fe The simplest ratio is 1 mol Fe: 1.01 g/mol 2. 1 mol Fe 174.01 g N Copyright © Glencoe/McGraw-Hill. 1 mol K 19.00 g O 0.545 g Al 0.667 mol H 1.485 g O 0.2641 mol O 3.131 mol Fe 1. Step 1: Calculate moles of each element.02 mol H 19 mol H 0. The empirical formula is K2O.86 g Fe 3.000 mol Fe 1 mol Fe 1.01 g/mol Step 4: Determine whole number multiplier. Inc. 16.55 g K and 4.01 g/mol The molecular formula is N2O2.00 g O 1 mol N 1. 60.14 g O.14 g O 4. An oxide of aluminum contains 0.01 g C 1 mol H 1. When an oxide of potassium is decomposed.55 g K 0. The empirical formula is C17H19O3N.86 g Fe and 75.332 mol N 1.00 g O are obtained.00 g O 1 mol O 0.08765 mol N 1.545 g Al and 0.5 mol O 4.008 g H 1 mol O 4.08765 mol N 1.900 g C. What is the empirical formula for the compound? Step 1: Calculate moles of each element. 1 mol C 17. Determine the empirical formula.000 mol N 1 mol N 1.250 mol O 1.680 g H.85 g Fe 1 mol Al 0.0202 mol Al 26.01 g 1 mol N Because 1. Step 1: Calculate moles of each element.087 65 mol N 14.490 mol C 12.000 mol N 1 mol N 0.CHAPTER 11 SOLUTIONS MANUAL 1 mol O 3. 19.00 mol C 17 mol C 0. Section 11. Magnetite contains a greater percentage of iron per kilogram than hematite.55 g Fe percent by mass 100 231.4 Assessment page 337 58.02 g H2O Solutions Manual Copyright © Glencoe/McGraw-Hill.5 mol O 0. Because 1.0-g sample.00 g O molar mass Fe3O4 231.70 g Fe2O3 69. Inc. Explain how you can find the mole ratio in a chemical compound. Percent composition is numerically equal to the mass in grams of each element in a 100.8% MgSO4 and 51. The mole ratio is determined by calculating the moles of each element in the compound and dividing each number of moles by the smallest number of moles. 61.55 g Fe 1 mol Fe 16.85 g Fe 3 mol Fe 167. Thinking Critically An analysis for copper was performed on two pure solids. calculate moles of each component. magnetite is 72. Therefore. SOLUTIONS MANUAL 62. 0.2 g H2O 2. It is sometimes necessary to multiply the ratio by an integer to obtain whole numbers.5 mol O.70 g Fe 1 mol Fe 16.0303 mol O 1.0% copper.00 mol Al 1 mol Al 0.94% Fe in Fe2O3 formula and a molecular formula? 167. What is the formula and name for this hydrate? Step 1: Assume 100 g sample.2% H2O. Explain how percent composition data for a compound are related to the masses of the elements in the compound.0202 mol Al 1. Which ore provides the greater percent of iron per kilogram? 55.84 mol H2O 18.00 mol Al 1 mol Al The simplest ratio is 1 mol Al: 1.00 g O 4 mol O 1 mol O 64. A hydrate is found to have the following percent composition: 48.94% Fe. What is the difference between an empirical 72.5 The Formula for a Hydrate pages 338–341 Practice Problems page 340 63. CHAPTER . 148 Chemistry: Matter and Change • Chapter 11 Section 11.0% copper.5 2 3.00 mol Al 1 mol Al 1.36% Fe in Fe3O4 60.55 g Fe3O4 The empirical formula is the simplest wholenumber ratio of the atoms in a compound. the empirical formula is Al2O3. Hematite is 69. a division of the McGraw-Hill Companies. One solid was found to contain 43. No.8 g MgSO4 120.50 mol O 0.55 g/mol 111. Could these solids be samples of the same copper-containing compound? Explain your answer.70 g/mol 55. a compound will always have the same chemical analysis.405 mol MgSO4 1 mol H2O 51.70 g Fe percent by mass 100 159.38 g MgSO4 0. if two solid have different percents of copper.00 g O 3 mol O 1 mol O 48. 59. The molecular formula is the actual ratio of atoms in the compound.85 g Fe 2 mol Fe 111. Inferring Hematite (Fe2O3) and magnetite (Fe3O4) are two ores used as sources of iron. they are different compounds.36% Fe.11 Step 2: Calculate mole ratios.00 g O molar mass Fe2O3 159. the other contained 32. 1 mol MgSO4 48.0202 mol Al 1.0202 mol Al 1. If 11. The difference is the mass of the water lost. 64.95 mol H2O 0.01 mol H2O 0.50 g H2O Copyright © Glencoe/McGraw-Hill. Add some hydrate and remass. Cool and remass.139 mol H2O 1.0712 mol CoCl2 1. 0.139 mol H2O 18.CHAPTER 11 Step 2: Calculate mole ratios. a division of the McGraw-Hill Companies.008 g H 14 mol H 1 mol H 14. Sequencing Arrange these hydrates in order of increasing percent water content: MgSO47H2O.00 mol CoCl2 32.49 g Solutions Manual Chemistry: Matter and Change • Chapter 11 149 .5 Assessment page 341 65. The name indicates the name of the compound and the number of water molecules associated with each formula unit of compound. Its name is magnesium sulfate heptahydrate.07 g S 2 mol H2O 1 mol CoCl2 16.00 mol MgSO4 7 mol H2O 1 mol MgSO4 The formula of the hydrate is MgSO47H2O.25 g CoCl2 2. strontium chloride hexahydrate 68.0712 mol CoCl2 1.00 mol CoCl2 1 mol CoCl2 1 mol CoCl2 69.84 mol H2O 7. Determine the moles of the anhydrous compound. Step 3: Calculate mole ratios. What is the formula and name for this hydrate? Step 1: Calculate the mass of water driven off. Subtract the mass of the crucible after heating from the mass of the crucible with the hydrate. Determine the moles of water.31 g Mg 1 mol Mg 24.0712 mol CoCl2 1 mol H2O 2. mass of hydrated compound mass of anhydrous compound remaining 11.00 g O 11 mol O 1 mol O 176.405 mol MgSO4 1.02 g H2O SOLUTIONS MANUAL Section 11. 0. CoCl26H2O.0712 mol CoCl2 1.50 g H2O 0.31 g Mg 1 mol Mg 0.00 g O 1. which will yield the formula of the hydrate.07 g S 1 mol S 1 mol S 32. Describe the experimental procedure for deter- mining the formula for a hydrate. 67. What is a hydrate? What does its name indicate about its composition? A hydrate is a compound that has a specific number of water molecules associated with it. 66. 24. The hydrate is pink in moist air. Thinking Critically Explain how the hydrate illus- trated in Figure 11-13 might be used as a means of roughly determining the probability of rain. Explain the reason for each step.25 g CoCl2 129.25 g of anhydrous cobalt chloride remains. Inc. 1 mol CoCl2 9. 9.405 mol MgSO4 1. Ba(OH)28H2O. molar mass MgSO47H2O 246. Step 2: Calculate moles of each component.75 g CoCl2xH2O 9.83 g CoCl2 0. Its name is cobalt(II) chloride dihydrate.75 g of this hydrate is heated.00 mol MgSO4 1 mol MgSO4 1 mol MgSO4 2. Figure 11-13 shows a common hydrate of cobalt(II) chloride. Name the compound having the formula SrCl26H2O.00 mol MgSO4 0. Determine the simplest whole-number ratio of moles of water to moles of anhydrous compound. Mass an empty crucible. Heat the crucible to drive out the water.405 mol MgSO4 1.112 g H The formula of the hydrate is CoCl22H2O.00 mol CoCl2 0. 1) 6. 3.00 g O 10 mol O 1 mol O 160.00 g O 6 mol O 1 mol O 96. Complete this concept map by placing in each box the conversion factor needed to convert from each measure of matter to the next. Ba(OH)28H2O. 150 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.00 g O 1.02 g 7(18.70% H2O in Ba(OH)28H2O Number of particles 1 mol 1.1) CoCl26H2O. Inc. a division of the McGraw-Hill Companies. 58. number? (11.49 g MgSO47H2O 51.93 g Co 1 mol Co 1 mol Co 58. Moles Moles 2.016 g H 1 mol 3.47 g Chapter 11 Assessment pages 346–350 Concept Mapping 70.47 g Ba(OH)28H2O 45. or formula units in a substance. 1 mol 1. MgSO47H2O 73.02 g H2O) 100 246.33 g Ba 1 mol Ba 1 mol Ba 137. How is a mole similar to a dozen? (11. CHAPTER .93 g Co 35.90 g Cl 16. 6(18. Mass 1.1) A mole allows a chemist to accurately measure the number of atoms.93 g CoCl26H2O 72.096 g H molar mass CoCl26H2O 237.44% H2O in CoCl26H2O A mole.02 g H2O) 100 237.008 g H 2 mol H 1 mol H 2. molecules.144 g H molar mass Ba(OH)28H2O 315.2) Molar mass is the mass in grams of one mole of any element or compound.00 g O 1 mol O 1 mol O 16.11 SOLUTIONS MANUAL 137.02 1023 representative particles 2. 4.33 g Ba 16.93 g 16. 6. like a dozen. What is the numerical value of Avogadro’s 45.02 1023 representative particles molar mass H2O 18.00 g O 1. number of grams number of grams 4.008 g H 18 mol H 1 mol H 18. contains a specific number of items. What is molar mass? (11. 1 mol Mastering Concepts 71. Why is the mole an important unit to chemists? (11.45 g Cl 2 mol Cl 1 mol Cl 70.17% H2O in MgSO47H2O 8(18.02 1023 74.00 g O 6.02 g H2O) 100 315.008 g H 12 mol H 1 mol H 12. 2) the same percent composition? Explain. (11.4) the mass of one atom of the element the percent composition of the compound 80.2) order to determine the empirical formula of an unknown compound? (11. A molecular formula specifies the actual number of atoms of each element in one molecule or formula unit of the substance (C6H6). Copyright © Glencoe/McGraw-Hill. a mole of Au has a greater mass. for example.0 g/mol. Na2CO310H2O and CuSO45H2O. Molar mass (grams) is the mass of a mole of particles.3) to illustrate your answer. 76. Explain the difference between atomic mass SOLUTIONS MANUAL 82. Then. more than ascorbic acid or glycerin. (11. 85.2) Molar mass is the mass in grams of one mole of any pure substance. If you divide the molar mass of an element by 86.5) 1 mol .2) The molar mass of Au is 197. Thus. number of grams A hydrated compound is a compound that has a specific number of water molecules associated with its atoms. add a prefix (mono. (11. Do all pure samples of a given compound have (amu) and molar mass (gram).3) One mole of K2CrO4 contains two moles of K ions and one mole of CrO42 ions. 81. glycerin (C3H8O3). What is the difference between an empirical formula and a molecular formula? Use an example to illustrate your answer. (11. and Avogadro’s number. (11. (11. (11. a mole of silver atoms or a mole of gold atoms? Explain your answer. Which contains more atoms. what is the meaning of the quotient? (11. Inc. 6.02 1023 representative particles . The mass of 6.02 1023 representative particles number of grams 1 mol 88. Yes. 78.di. 77. List three conversion factors used in molar 87. What information is provided by the formula for potassium chromate (K2CrO4)? (11. Solutions Manual Chemistry: Matter and Change • Chapter 11 151 . a division of the McGraw-Hill Companies. Which has a greater mass. Discuss the relationships that exist between the mole.4) An empirical formula is the smallest wholenumber ratio of elements that make up a compound (CH). Explain how hydrates are named.3) The formula for vanillin (C8H8O3) shows there are eight carbon atoms per molecule. 79. name the compound. What is a hydrated compound? Use an example conversions. the molar mass of Ag is 107. (11. 6. 1 mol 1 mol . What information must a chemist obtain in Avogadro’s number.02 1023 representative particles. the percent by mass of each element is the same regardless of the size of the sample. (11.02 1023 representative particles of a substance is the molar mass of the substance. molecule).5) First. (11.4) Atomic mass (amu) is the mass of an individual particle (atom. molar mass.CHAPTER 11 75. Avogadro’s number is the number of representative particles in one mole.tri-) that indicates how many water molecules are associated with one mole of compound. 83. Explain what is meant by percent composition. Which of the following molecules contains the most moles of carbon atoms per mole of the compound: ascorbic acid (C6H8O6). 84. a mole of silver atoms or a mole of gold atoms? Explain your answer.2) They both contain the same number of atoms because a mole of anything contains 6. for every pure substance.9 g/mol. or vanillin (C8H8O3)? Explain. (11.4) Percent composition is the percent by mass of each element in a compound. 90 mol Ca to atoms Ca 6. 0. 5. 1.51 1015 atoms Si to mol Si 1 mol Si 1.02 1023 molecules glucose 8.425 mol N2 1 mol N2 1023 2.35 mol carbon disulfide (CS2) 6. a division of the McGraw-Hill Companies. How many molecules are contained in each of the following? a.3 mol CO2 1 mol CO2 1023 2.56 1023 formula units sodium hydroxide Mastering Problems Mole-Particle Conversions (11.250 mol silver 6. 1.425 mol nitrogen (N2) molecules N2 6.02 atoms Ag 0. a. 35. 0.02 molecules CO2 35.02 1023 formula units NaCl 1 mol NaCl 5. 1.51 1015 atoms Si 6.13 1023 molecules CS2 152 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.02 1023 atoms Si 2. CHAPTER .02 1023 Cu2 ions 2. 4. Determine the number of representative particles in each of the following.15 1021 formula units NaCl c.02 1023 molecules CCl4 1.25 102 mol H2SO4 6.56 1023 formula units NaOH 1 mol NaOH 6.25 1025 Cu2 ions 1 mol Cu2 ions 6.02 1023 atoms Pb 5.25 1020 atoms Pb 6. 8.25 1025 copper(II) ions 1.51 109 mol Si b.49 102 mol CCl4 d. a.08 101 mol Cu2 ions 91.25 1020 atoms lead 1 mol Pb 3. 1.250 mol Ag 1 mol Ag 1023 1.02 1023 molecules CS2 1. 4.1) Level 1 89. a.11 SOLUTIONS MANUAL c. Inc.56 1022 molecules H2SO4 c. 3.35 mol CS2 1 mol CS2 8.95 1025 molecules CCl4 to mol CCl4 8.59 101 mol NaOH d.56 103 mol sodium chloride 8.56 1023 molecules N2 90.90 mol Ca 1 mol Ca 3.96 1024 molecules glucose 1 mol glucose 6. 8. Make the following conversions.55 1024 atoms Ca 92.95 1025 molecules CCl4 1 mol CCl4 6.56 103 mol NaCl 6.02 0.24 mol of glucose 1.51 1023 atoms Ag b.13 1025 molecules CO2 d.3 mol carbon dioxide 6.02 1023 molecules H2SO4 1 mol H2SO4 2.39 104 mol Pb b.96 1024 molecules glucose 4.02 1023 atoms Ca 5.02 1023 formula units NaOH 2.25 102 mol H2SO4 to molecules H2SO4 4. Determine the number of moles in each of the following. 0 mol HCl 1 mol HCl of water. 2.53 1023 molecules H2O 2.02 1023 mol metal ions 1 mol metal ions 7.02 1023 formula units CaCO3 4. A bracelet containing 0.2 106 mol H2O 97.0455 mol Ni 2.02 1023 molecules HCl 150.CHAPTER 11 SOLUTIONS MANUAL b. 150.9 1018 molecules 6.01 1023 s 2 particles 5. How many moles contain each of the following? a. How many particles of gold atoms are in the bracelet? 0.25 mol water added to another solution containing 0. how long would it take you to count a mole of atoms? Assume that you counted continually 24 hours every day. Calculate the mass of the following.01 1023 s 3600 s 24 h 365 day c.59 1021 formula units sodium nitrate Copyright © Glencoe/McGraw-Hill. a.25 1015 molecules CO2 1 mol CO2 6.89 1027 formula units calcium carbonate 2.02 atoms Au 0.9 g He 1 mol He b.54 1015 y 2.00 g He 5.2) Level 1 98.75 0.02 1023 molecules CO2 2.67 g Ni 1 mol Ni 9.53 1023 molecules As2O3 c. Inc.150 mol Au 1 mol Au 1023 9. A solution containing 0.80 103 mol CaCO3 Level 2 94.130 mol Ca2 ions.200 mol of metal atoms is 75% gold.380 mol metal ions 0.96 103 mol NaNO3 1y 1 day 1h 3. 0.030 1025 molecules HCl 1.380 mol metal ions 6.22 mol He 20. If a snowflake contains 1.254 mol diarsenic trioxide (As2O3) 0.130 mol Ca2 ion 0.02 1023 molecules As2O3 1 mol As2O3 1. how many moles of water does it contain? 9.0 mol HCl 96. 0. 1.22 mol He 4.02 1023 particles 3.03 1022 atoms Au Solutions Manual Chemistry: Matter and Change • Chapter 11 153 .29 1023 Cu2 and Ca2 ions 1023 d.5 109 years old? 1 mol NaNO3 6. If you could count two atoms every second. which is estimated to be 4.1 106 4.200 mol Au 0. How does the time you calculated compare with the age of Earth.25 mol H2O 1 mol H2O 6.254 mol As2O3 6. What is the total number of metal ions in the combined solution? 0.89 1027 formula units CaCO3 1 mol CaCO3 6.69 g Ni 0.5 1015 yr.08 109 mol CO2 b.150 mol Au 6.02 1023 formula units NaNO3 1s 6.0455 mol Ni 58. 1. 3.250 mol Cu2 ions is 3.02 1023 molecules H2O 3.25 1015 molecules carbon dioxide 1.9 1018 molecules H2O 93. 5. about 2 million times longer Mole-Mass Conversions (11.250 mol Ca2 ions 0.59 1021 formula units NaNO3 1 mol H2O 6.02 molecules H2O 1.5 109 y 9. 95. a division of the McGraw-Hill Companies. 85 1024 mol Cr 1 mol Cr Particle-Mass Conversions (11. 3.65 g Co 0.02 1023 atoms U 283.22 1023 mol Ag 1.85 1024 mol Cr 52.25 1022 atoms O 1. Calculate the number of atoms in each of the following.75 1014 mol Pt 1 mol Pt 9. CHAPTER . 0.332 g O d. How many atoms of silver would be in a sample having this mass? 1 mol Ag 1 108 g Ag 107. Inc.80 g Kr d.8 g Hg 1 mol Hg 25.02 1023 atoms H 145 g H c. 0.0671 mol Kr 83.2 g Pb 1 mol Pb 1 mol Pb 6.62 1024 g Sb b. 1.00566 mol Ge 0.130 mol Co 58.25 1022 atoms O 16.3 g Li 1 mol Li b.87 g Ag 6.65 g Co to mol Co 1 mol Co 7.59 g Hg 6. 4.12 1026 g Cr 154 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.50 mol Li 24.76 g Sb 1. a. 4.22 mol Ti 47.03 g U 1.92 g As 0. a. Convert the following to mass in grams. 9. 4. Determine the mass in grams of the following.02 1023 atoms O 0.22 1015 atoms U 6.12 g As 1 mol As Level 2 100.11 SOLUTIONS MANUAL c. A sensitive balance can detect masses of 1 108 g.02 1023 atoms Ag 5 1013 atoms Ag 1 mol Ag 103.00 g Cr 9.22 1015 atoms U 1 mol U 4. 7. 1.411 g Ge 1 mol Ge 99.00566 mol Ge 72.74 1022 atoms Hg 5. a. a division of the McGraw-Hill Companies.87 g Ti 2.8 g Hg 200.33 1022 mol Sb 121.61 g Ge 0.0550 mol As to g As 74. 8.93 g Co c. a.44 1023 atoms Pb 207.00 g O 1 mol O 1 mol O 6.008 g H 1 mol H 1 mol H 6.08 g Pt 4.87 g Ag mol Ag 1 mol Ag 1.22 mol Ti 106 g Ti 1 mol Ti d.94 g Li 3. 2.62 g Kr 0.02 1023 atoms Pb 153 g Pb 102.22 1023 107.32 1025 g Ag d.2) Level 1 101.65 1025 atoms H 1.67 106 g U 1 mol U b.62 g Kr to mol Kr 1 mol Kr 5. l. 5.44 1023 atoms Pb 4.75 1014 mol Pt 195.33 1022 mol Sb 1 mol Sb 1. 25.02 1023 atoms Hg 1 mol Hg 7.27 1016 g Pt c. Make the following conversions.50 mol Li to g Li 6.65 1025 atoms H 8.0550 mol As 4. 113 mol O 1 mol CuSO45H2O 6.51 1023 atoms Fe 0. 0. 10. Thus.0 moles of carbon and 10.25 102 mol CuSO45H2O 9 mol O 0. 1 mol P. Inc.02 1023 atoms Fe 0. 0. 10.02 1023 atoms Zn 3.02 1023 atoms C 1 mol C 1.0 g C contains more atoms.0340 g Zn 1 mol Zn 0.02 1023 atoms 10.0 moles of calcium contain the same number or atoms.0 moles of calcium? How many does each have? One mole of any substance contains 6.0 mol O b. A mixture contains 0.20 g C.0 moles of carbon or 10.02 1024 atoms Solutions Manual Chemistry: Matter and Change • Chapter 11 155 . Which has more atoms. a division of the McGraw-Hill Companies.124 g Mg 24.9 mol CO2 91.02 1023 atoms Ar 2.CHAPTER 11 b.01 g C 1 mol C 6. Which has more atoms.250 mol Fe 1 mol Fe 1.8 mol O 1 mol CO2 c.0 g of calcium? How many atoms does each have? 1 mol C 6. 45. 1.0340 g Zn 63.50 1023 atoms Ca 10.124 g Mg 1 mol Mg 0. How many moles of oxygen atoms are contained in the following? a.601 1023 atoms C d.59 g Zn 6.50 mol KMnO4 4 mol O 2.51 1023 atoms Fe 1 mol Ar 150 g Ar 39.3 1024 atoms Ar 1 mol Ar 0. 1. SOLUTIONS MANUAL 5.01 1023 atoms C 1 mol Ca 6.0 g C 12.02 1023 representative particles.31 g Mg 6.02 1023 atoms Ca 10. 150 g Ar Level 2 106. 4 mol O 108.07 1021 atoms Mg 104.08 g Ca 1 mol Ca 1. 10.95 g Ar 6. how many moles of sodium are represented? How many moles of phosphorus? How many moles of oxygen? 3 mol Na.0 g of carbon or 10. In the formula for sodium phosphate (Na3PO4).02 1023 atoms Mg 1 mol Mg 3.9 mol CO2 2 mol O 45.02 1023 atoms C 10.50 mol KMnO4 1 mol KMnO4 10. 2. What is the total number of atoms in the mixture? 6.0 g Ca 40.3) Level 1 107. 105.13 1020 atoms Zn 1 mol Zn c.250 mol Fe and 1.11 1023 total atoms Chemical Formulas (11.25 102 mol CuSO45H2O 1.20 g C 12.0 mol 1 mol 6.01 g C 1 mol C Copyright © Glencoe/McGraw-Hill.601 1023 atoms C 2. 18 g/mol molar mass 110.3) Level 1 111.01 g N 1 mol N 16.81 1024 molecules CCl4 1 molecule CCl4 35.00 mol CCl4 1 mol CCl4 80.81 1024 molecules CCl4 58.00 g O 1 mol O 16.81 1024 molar mass 129. Atoms per formula unit Level 2 109.24 1024 atoms Cl 9.02 g/mol HNO3 b.11 CHAPTER SOLUTIONS MANUAL The Composition of a Compound 6 5 4 3 2 1 Molar Mass (11.00 mol carbon tetrachloride (CCl4)? How many carbon atoms? How many chlorine atoms? How many total atoms? 6.83 g/mol CoCl2 4 mol atoms Cl molecules CCl4 1 molecule CCl4 7. What is its formula? What is its molar mass? . Inc. cobalt chloride (CoCl2) 1.02 1023 molecules CCl4 3. a division of the McGraw-Hill Companies.05 1024 total atoms 156 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.048 g H 16.008 g H 1 mol H 1.04 g C 1.008 g H 6 mol H 1 mol H 6.01 g N 2 mol N 28.032 g H 1 mol H 1.00 g O molar mass 81.90 g Cl 1 mol Cl 1.05 g/mol NH4NO3 c. ammonium nitrate (NH4NO3) 14.00 g O 1 mol O Ca C H Atoms O CaC4H6O4 (calcium acetate) 40.00 g O molar mass 158.00 g O 4 mol O 1 mol O 64.45 g Cl 2 mol Cl 70.93 g Co 1 mol Co 1 mol atoms C 1.08 g Ca 1 mol Ca 40. The graph shows the numbers of atoms of each element in a compound.08 g Ca 1 mol Ca molar mass 63. nitric acid (HNO3) 1.008 g H 4 mol H 4.00 g O 1 mol O 1 mol O 16.39 g Zn 1 mol Zn 65.39 g/mol ZnO d. How many carbon tetrachloride molecules are in 3.00 g O 3 mol O 48. zinc oxide (ZnO) 65.81 1024 atoms C 7.00 g O 3 mol O 48.93 g Co 1 mol Co 58.01 g N 1 mol N 14.24 1024 atoms Cl 1. a.39 g Zn 1 mol Zn 16.008 g H 1 mol H 14.02 g N 1 mol N 12.01 g C 4 mol C 1 mol C 48.81 1024 atoms C 1. Determine the molar mass of each of the following. 07 g S 1 mol S molar mass Solutions Manual 183.008 g H 5 mol H 5.01 g N 1 mol N 1 mol N 14. The chemical formula of allyl sulfide is (C3H5)2S.09 g/mol H2SO4 molar mass 107.00 g O 6 mol O 96.01 g C 1 mol C 1.00 g O 1 mol O 16.064 g H 1 mol H 16.87 g Ag 1 mol Ag Copyright © Glencoe/McGraw-Hill.07 g S 1 mol S 16.07 g S 1 mol S 1 mol S 32.00 g O molar mass Mass-Mole Conversions (11. a. sulfuric acid (H2SO4) 1.040 g H 1 mol H molar mass 14.3) Level 1 114. Calculate the molar mass of each of the following.0 g CH3OH 32. the compound responsible for the smell of garlic.01 g N 16.00 g O 1 mol O 1. How many moles are in 100.21 g/mol (C3H5)2S b.00 g O 1 mol O 16.04 g CH3OH 3.01 g C 6 mol C 72.00 g O 1 mol O 32.06 g C 1.12 g/mol C6H8O6 molar mass 113.01 g N 1 mol N 32.080 g H 1 mol H 32.00 g O 1 mol O 176.0 g of each of the following compounds? a.07 g S 1 mol S 32.CHAPTER 11 SOLUTIONS MANUAL 112.00 g O 3 mol O 48.00 g O 1 mol O 98.06 g C 1 mol C 1. ascorbic acid (C6H8O6) 12.008 g H 4 mol H 4.00 g O 3 mol O 1 mol O 48.19 g/mol C7H5NO3S Chemistry: Matter and Change • Chapter 11 157 . 12.02 g/mol 1 mol N2O 2. 14. Inc.02 g N 1 mol N 16.01 g N 2 mol N 28.00 g O 1 mol O c.016 g H 1 mol H 32.00 g O 4 mol O 64.01 g C 7 mol C 84. dinitrogen oxide (N2O) 169.02 g N2O b.008 g H 2 mol H 2.07 g C 1 mol C 16.07 g S 1 mol S 32.01 g N 1 mol N 14. Determine the molar mass of allyl sulfide.0 g N2O 44. saccharin (C7H5NO3S) molar mass 44.88 g/mol AgNO3 d. methanol (CH3OH) 12.01 g C 1 mol C 12.04 g/mol 1 mol CH3OH 100.87 g Ag 1 mol Ag 107.008 g H 10 mol H 10. silver nitrate (AgNO3) 14.01 g C 6 mol C 1 mol C 72.27 mol N2O 100.12 mol CH3OH 16.032 g H 1 mol H 12.07 g S molar mass 114. a division of the McGraw-Hill Companies.008 g H 8 mol H 8. 50 102 mol 1 mol 1.00 g O molar mass 242.26 102 g Ca(OH)2 116.145 g H2S 34.25 102 74. Benzoyl peroxide is a substance used as an acne medicine.145 g H2S 20. 0.0 g Al2O3 101. How many moles of aluminum ions are in 45. CHAPTER .14 g C 1 mol C 1.55 g Cu 1 mol Cu 35.008 g H 2 mol H 1 mol H 8.50 102 moles of benzoyl peroxide (C14H10O4)? 12.95 1025 molecules HF 22.01 g HF 1 mol HF 1 mol HF 6.00 g F 1 mol F 1 mol F 19.00 g O 3 mol O 1 mol O 48.25 103 mol H2S 158 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. Determine the number of moles in each of the following.01 g C 14 mol C 168.10 g/mol 1.0 g of aluminum oxide? 26.96 g Al2O3 32.96 g/mol 1.00 g O molar mass 74. Hydrofluoric acid is a substance used to etch 16.96 g Al 1 mol Al 16.008 g H 2 mol H 2.22 g 3.11 SOLUTIONS MANUAL 115.60 mol Na2S b.07 g S 1 mol S 1 mol S 32.35 g CuCl2 4.45 g Cl 2 mol Cl 1 mol Cl 70.10 g Ca(OH)2 mol Ca(OH)2 1 mol Ca(OH)2 9. Determine the mass of 4.07 g S molar mass 78.22 g/mol 40.883 mol Al3 ions 1 mol Al2O3 molar mass 34. 1.25 102 g Na2S 78.98 g Na 1 mol Na 32.08 g Ca 1 mol Ca 242.09 g/mol 1 mol H2S 0. What is the mass of each of the following? a.90 g Cl molar mass 134.55 g Cu 1 mol Cu 63. a division of the McGraw-Hill Companies.05 g/mol 1 mol Na2S 1.00 g O molar mass 101. 1.07 g S 1 mol S 32.99 g Na 2 mol Na 45.25 102 mol Ca(OH)2 Level 2 117.08 g Ca 1 mol Ca 40.016 g H 1 mol H 1 mol Al2O3 45.02 1023 molecules HF 1650 g HF 119.00 g F molar mass 20.50 102 mol CuCl2 63.45 g/mol 134.09 g H2S 4.008 g H 1 mol H 1.01 g/mol 4.98 g Al 2 mol Al 53.07 g S 1 mol S 2 mol Al3 ions 0.008 g H 1 mol H 19.95 1025 HF molecules.008 g H 10 mol H 10.080 g H 1 mol H 16.00 g O 4 mol O 1 mol O 64. 1.05 g CuCl2 b. a.05 g Na2S 1. 4.50 102 mol CuCl2 1 mol CuCl2 6.25 102 g Na2S glass. What is the mass in grams of 3. Inc.48 g benzoyl peroxide 2.016 g H 118.00 g O 2 mol O 1 mol O 32. 01 g C 2 mol C 1 mol C 24.100 mol K2CrO4 3 mol ions 0.35 104 mol ions 1 mol AgNO3 Benzene C6H6 0.87 g Ag 1 mol Ag 14.008 g H 12 mol H 12.33 g Ba 1 mol Ba 137.33 g Ba 1 mol Ba 16.00 g O molar mass 166.02 g C 1.65 1021 0.88 g AgNO3 Glucose C6H12O6 1.500 g Ba(OH)2 171.008 g H 3 mol H 1 mol H 3.500 g Ba(OH)2 Copyright © Glencoe/McGraw-Hill.082 1024 2 mol ions 2.16 g/mol Chemistry: Matter and Change • Chapter 11 159 .35 g/mol 1 mol Ba(OH)2 0.0 2.87 g Ag 1 mol Ag 107. 0.4178 100.00 g O 6 mol O 96. a division of the McGraw-Hill Companies.91 g AgC2H3O2 2.50 417 1.00 109 mol Na2CO3 1 mol Na2CO3 3.300 mol ions 137.00 109 mol ions Silver acetate: 107. Mass-Particle Conversions (11. Inc.87 g Ag 1 mol Ag 107.009 39 0.91 g/mol 166.733 5. How many moles of ions are in the following? a. Calculate the values that will complete the table.00 g O 2 mol O 1 mol O 32.10 g H 1 mol H 16.024 g H 16.06 g C 1.3) Level 1 121.01 g N 1 mol N 1 mol N 14.00 g O 3 mol O 1 mol O 48.02 1023 formula units 2.0200 g AgNO3 107.516 1023 Lead(II) sulfide PbS b.00 g O 1.798 324.51 1024 formula units Glucose: 12. 1. 0.75 103 mol ions 1 mol Ba(OH)2 d.50 mol AgC2H3O2 1 mol AgC2H3O2 1. 0. Mass.35 g Ba(OH)2 3 mol ions 8.0 1.88 g/mol Table 11-2 Moles.00 g O 1 mol O molar mass Solutions Manual 180. 12.01 g N 16.00 g O 2 mol O 1 mol O 32. and Representative Particles Compound Number Mass(g) of moles Representative particles Silver acetate Ag(C2H3O2) 2.100 mol K2CrO4 1 mol K2CrO4 0.87 g Ag 1 mol Ag c.0200 g AgNO3 169.008 g H 2 mol H 1 mol H 2.50 mol AgC2H3O2 1 mol AgC2H3O2 417 g AgC2H3O2 6.016 g H molar mass 171.00 109 mol Na2CO3 3 mol ions 1.51 1024 1 mol AgNO3 0.CHAPTER 11 SOLUTIONS MANUAL 120.00 g O molar mass 169.01 g C 6 mol C 1 mol C 72. 02 1023 formula units 1.11 SOLUTIONS MANUAL 1 mol glucose 324.798 mol PbCl2 1 mol PbCl2 1.50 g Au 196.3 g/mol 1 mol ethanol 47.85 g Fe 1 mol Fe 35.798 mol glucose 1 mol glucose 6.3 g PbS 6.798 mol PbCl2 500.516 1023 formula units 122.11 g benzene mol benzene 9.1 g PbCl2 6.45 g Cl 3 mol Cl 1 mol Cl 106.16 g glucose 1.02 1023 atoms Au 1.35 1023 chloride ions 1 mol chloride ions 6.048 g H 1 mol H 3.02 1023 molecules glucose molar mass 78.00 g O 1 mol O 16.85 g Fe 1 mol Fe 55.62 1024 molecules glucose 1 mol glucose 6.45 g Cl 2 mol Cl 1 mol Cl 70. 12. Chemistry: Matter and Change • Chapter 11 46.07 1022 atoms Au 1 mol Au 1.2 g Pb 1 mol Pb 16.16 g/mol.01 g C 6 mol C 72.798 mol glucose 1 mol Au 3.4178 mol PbS 239.90 g Cl molar mass 278.65 1021 molecules benzene 1 mol benzene 6. 1.02 1023 chloride ions 1 mol FeCl3 162.07 g/mol 6.2 g Pb 1 mol Pb 207.01 g C 2 mol C 24.97 g Au 6.14 1023 molecules ethanol 126.39 103 mol benzene 78.082 1024 formula units PbCl2 160 123. Inc.07 g S 1 mol S 1 mol S 32.06 g C 1 mol C From solution #121.20 g FeCl3 3 mol chloride ions 1 mol FeCl3 21.2 g Pb 1 mol Pb 207.35 g Cl molar mass 162.1 g/mol 1 mol PbCl2 1. CHAPTER .16 g glucose 1080 g glucose 1 mol glucose 125. How many formula units are present in 500.1 g FeCl3 Solutions Manual Copyright © Glencoe/McGraw-Hill.02 1023 molecules benzene 9.50 g gold.20 g/mol 2.048 g H 1 mol H Lead(II) sulfide: 207.008 g H 6 mol H 6.11 g/mol 5. Determine the number of molecules of ethanol (C2H5OH) in 47.733 g benzene 103 180.082 1024 molecules Benzene: 124.62 1024 molecules of glucose (C6H12O6).02 1023 formula units 0.02 1023 molecules 1.00 g O 1 mol O 32.2 g Pb 1 mol Pb 35.02 g C 1 mol C 1.02 1023 molecules ethanol 1 mol ethanol 6.0 g ethanol 46. Determine the number of atoms in 3. 12. a division of the McGraw-Hill Companies.0 g lead(II) chloride? 207.0 g PbS 0.0 g PbCl2 278.008 g H 6 mol H 6. Calculate the mass of 3.39 1 mol benzene 0. What mass of iron(III) chloride contains 2. the molar mass of glucose equals 180.07 g S molar mass molar mass 239.07 g ethanol 1 mol PbS 100.4178 mol PbS 1 mol PbS 2.0 g glucose 180.35 1023 chloride ions? 55.0 g. 2 g Pb 1 cm3 1 mol Pb 18. What is the mass of a mole of electrons? 9. 24.00 g O 1 mol O 1.0 kg Fe3O4 1 kg Fe3O4 1.25 g CH3COOH 14.0 g vinegar? Copyright © Glencoe/McGraw-Hill. 1.3 g/cm3.11 1028 g.55 g/mol 4.008 g H 9 mol H 9. How many moles of iron can be recovered in 250. 55. 12.02 1023 ions MgCl2 2 mol Cl 1 mol MgCl2 1 mol Cl 1.01 g C 2 mol C 24.25 1022 molecules CH3COOH 130.85 g Fe 3 mol Fe 167. The density of lead is 11.0% acetic acid (CH3COOH).0 g aluminum oxide (Al2O3).3 cm3 11.050 25.02 1023 electrons 1 mol electrons 1 electron 5. Inc.37 1023 ions Cl 133.45 g Cl 2 mol Cl 1 mol Cl 70.01 g C 8 mol C 96.00 g O 4 mol O 1 mol O 64.90 g molar mass 94.000 105 g Fe3O4 1 mol Fe3O4 1. Calculate the volume of one mole lead.05 g/mol 0. Determine the number of molecules of acetaminophen in a 500 mg tablet.55 g Fe 1 mol Fe From solution # 119.08 g C 1 mol C 16.01 g N 1 mol N 1 mol CH3COOH 1.032 g H 1 mol H 12.02 1023 molecules CH3COOH 1 mol CH3COOH molar mass 1.02 1023 molecules C8H9NO2 1 mol C8H9NO2 2 1021 molecules C8H9NO2 Solutions Manual Chemistry: Matter and Change • Chapter 11 161 .003 mol C8H9NO2 6.31 g Mg 1 mol Mg 24.3 g Pb 1 mol Pb 151.072 g H 1 mol H molar mass 60.90 mol Al3 103 g Fe3O4 100. How many molecules of acetic acid are present in 25.11 1028 g 6. Acetaminophen.0 g vinegar 1.05 g CH3COOH 16.75 g of magnesium chloride.48 104 g/mol electrons 129. the molar mass of aluminum oxide equals 101.31 g/mol 1 mol MgCl2 10.00 g O 2 mol O 32. 207.16 g/mol 1 g C8H9NO2 500 mg C8H9NO2 103 mg C8H9NO2 1 mol C8H9NO2 0.55 g Fe3O4 3 mol Fe 1296 mol Fe 1 mol Fe3O4 128. Vinegar is 5. 131.96 g Al2O3 molar mass 231.0 g Al2O3 1 mol Al2O3 101.11 CHAPTER SOLUTIONS MANUAL 127. The mass of an electron is 9.16 g C8H9NO2 0.000 105 g Fe3O4 231. Determine the number of chloride ions in 10.008 g H 4 mol H 4.01 g N 1 mol N 14. Calculate the moles of aluminum ions present from 100.0 kg Fe3O4? Level 2 132. a division of the McGraw-Hill Companies.75 g MgCl2 94.00 g O 1 mol Al2O3 2 mol Al3 250.25 g CH3COOH 60.003 mol C8H9NO2 151.00 g O 2 mol O 32.02 g C 1 mol C has the formula C8H9NO2. a common aspirin substitute.00 g O 1 mol O 6.31 g 1 mol Mg 35. 16.96 g/mol.31 g MgCl2 6. 99 g Na 1 mol Na 22.00 g O 2 mol O 32.11 SOLUTIONS MANUAL 134.30 g 6.98 g Al 2 mol Al 53.0 g CO2 44.99 g Na 1 mol Na 35.008 g H 22 mol H 22. sucrose (C12H22O11) 231.07 g S 3 mol S 1 mol S 12.96 g percent by mass 100 342.00 g O 1 mol O molar mass 44.00 g percent by mass 100 342.00 g O 12 mol O 192.42% O b. or siderite (FeCO3)? 55. 22.17 g/mol 53. Which of the following iron compounds contain the greatest percentage of iron: pyrite (FeS2).45 g Cl molar mass 58.14 g S molar mass FeS2 119.85 g Fe 3 mol Fe 167.02 1023 ions Na 2.00 g O 1 mol O molar mass 342.77% Al 96. 12.07 g S 2 mol S 1 mol S 64.0 g NaCl 58.96 g Al 1 mol Al 32.55 g 72.44 g NaCl 6.21 g percent by mass = 100 342.10% C 22.12 g C 1 mol C 1.55 g Fe 1 mol Fe 16.55 g 27.17 g 56.55 g/mol molar mass 342.18 g percent by mass 100 342.01 g C 1 mol C 16.84 1023 atoms O 176. aluminum sulfate (Al2(SO4)3) 26.30 g 42. a.44 g/mol 1 mol NaCl 25.00 g O 4 mol O 64.85 g Fe 1 mol Fe 55. a division of the McGraw-Hill Companies.00 g O 11 mol O 176. Calculate the number of sodium ions present in 25.00 g O 1 mol O 167.568 mol CO2 25.0 g sodium chloride.21 g S 16.00 g O 1 mol O molar mass Percent Composition (11.11% O 137.99 g/mol Solutions Manual Copyright © Glencoe/McGraw-Hill.85 g Fe 1 mol Fe 32.0 g carbon dioxide.568 mol CO2 1 mol CO2 1023 6.58 1023 ions Na 1 mol NaCl 135.00 g percent by mass 100 231.45 g Cl 1 mol Cl 1 mol Cl 35.4) 136.02 atoms O) 0.64% O c. magnetite (Fe3O4) 55.17 g 28.12% S 96. Determine the number of oxygen atoms present in 25.360% Fe 64.55 g percent by mass 100 231.12 g percent by mass 100 342. CHAPTER .01 g C 12 mol C 144.30 g 51. Express the composition of each of the following as the mass percent of its elements (percent composition).30 g/mol 144. Inc.01 g CO2 2(6.01 g C 1 mol C 12. hematite (Fe2O3).21 g percent by mass 100 342.18 g H 1 mol H 16.480% H 162 Chemistry: Matter and Change • Chapter 11 96.01 g/mol 1 mol CO2 0.17 g 15. 87 g Ti 1 mol Ti Solutions Manual 16. Determine the empirical formula for each of the following compounds.85 g Fe 1 mol Fe 55. Determine the percent composition of caffeine.20 g 5.191% H 56.85 g Fe 1 mol Fe 55. Caffeine.01 g C 14. c. The empirical formula is CH2. a. a stimulant found in coffee.00 g O 2 mol O 32.70 g Fe 1 mol Fe 16.01 g C 8 mol C 1 mol C 96. The empirical formula is C5H4.86% N 32.08 g percent by mass 100 194. Calculate the molar mass of caffeine.87 g/mol 55. Inc.00 g O 3 mol O 48.87 g Ti 1 mol Ti 16.99 g 46. ethylene (C2H4) Both subscripts can be divided by two.70 g/mol chemical formula C8H10N4O2.85 g percent mass 100 152.01 g C 1 mol C 1 mol C 12.00 g O molar mass TiO2 79.55% in FeS2 111.72 g/mol Chemistry: Matter and Change • Chapter 11 163 . molar mass 194. 96.00 g O 1 mol O 16.944% Fe.85 g Fe 1 mol Fe 47.70 g percent mass 100 159.04 g N 1 mol N 16.48% O 140. b. has the Hematite.CHAPTER 11 SOLUTIONS MANUAL 55. has the greatest percentage of iron. The empirical formula is C3H4O3.00 g O 1 mol O molar mass Fe2O3 159.04 g percent by mass 100 194. 12.95 g 36.85 g Fe 1 mol Fe 1. 138.70 g 69.00 g O 3 mol O 1 mol O 48. a division of the McGraw-Hill Companies.20 g 49.95 g/mol 55. a.87 g Ti 1 mol Ti 47.08 g C 55.00 g percent by mass 100 194.51% in FeCO3 Copyright © Glencoe/McGraw-Hill.00 g O 1 mol O molar mass FeCO3 152.47% C 10. with 69.87 g Ti 1 mol Ti 47.008 g H 10 mol H 10.20 g 28.01 g N 4 mol N 56. 139.20 g/mol b.944% in Fe2O3 55. ascorbic acid (C6H8O6) All subscripts can be divided by two.00 g O 2 mol O 1 mol O 32.20 g 16.08 g percent by mass 100 194.85 g Fe 2 mol Fe 111. naphthalene (C10H8) Both subscripts can be divided by two.00 g O molar mass FeTiO3 151.00 g O 3 mol O 96.85 g percent mass 100 119. Which of the titanium-containing minerals rutile (TiO2) or ilmenite (FeTiO3) has the larger percent of titanium? 47.080 g H 1 mol H 12. 00 7. 104 g/mol n 8. 50. 8.87 g percent by mass 100 79.30 g Empirical and Molecular Formulas (11.681 mol C 1.008 g H 50 mol H 50.592 mol N molar mass 164 28.29 g percent by mass 100 80.87 g 59.30 g 1 mol N 8. 13. Monosodium glutamate (MSG) is sometimes 14.00 g percent by mass 100 27.00 g O 1 mol O 7.008 g H 4.008 g H 1 mol H 142.01 g C 1 mol C 32.4) 143. a division of the McGraw-Hill Companies. 12.77% H.00 g O 1 mol O added to food to enhance flavor. has the molar mass CH 13.00 13. 141. Inc. and 37.25% C and 7.69 g 1. The hydrocarbon used in the manufacture of foam plastics is called styrene.77 g H 1.6% Na.02 g/mol 12.69 348. Analysis determined this compound to be 35.02 g percent by mass 100 9.5 g C 12.30 g 294.9% O. Analysis of styrene indicates the compound is 92.681 mol C 12.11 SOLUTIONS MANUAL 47.87% C 430.75 g H 7.69 mol H 1.01 g C 14 mol C 168.01 g C 2.521% N 294.70% H 430. 4.75% H and has a molar mass of 104 g/mol.02 g/mol formula C14H18N2O5.29 g N 14.73 mol H 18.00 g percent by mass 100 7. Aspartame.30 g/mol Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill. The formula for vitamin E is C29H50O2.008 g H 18 mol H 18.01 g C 1 mol C 12.165% H 294. CHAPTER .14 g percent by mass 100 57.29% N.008 g H 1 mol H 1. 1.144 g H 1 mol H 144.144 g percent by mass 100 6.40 g percent by mass 100 11.72 g 31.55% Ti in FeTiO3 TiO2 has the greater percent by mass of titanium.69 g The empirical formula is CH.400 g H 1 mol H 1 mol H 7.01 g C 1.96 mol C 168.01 g N 2 mol N 28.00 g O 5 mol O 80.5% C.00 7.01 g C 29 mol C 348. Determine the molecular formula for styrene.93% Ti in TiO2 47.87 g percent by mass 100 151. is thought to retard the aging process in humans. an artificial sweetener.69 g 12.29 g C 1 mol C 1 mol C 92. Vitamin E.69 mol H 1.69 molar mass 430.69 g/mol 7.008 g H 16. Determine the percent composition of aspartame.00 g O 2 mol O 32. What is the empirical formula for MSG? 1 mol C 35.30 g 1 mol H 4. found in many plants.02 g N 1 mol N 16.14 g C 1 mol C Multiply the subscripts in the empirical formula by 8 to obtain the molecular formula C8H8.25 g C 7.13% C 294. What is the percent composition of vitamin E? 80.18% O 294.01 g N 0.430% O 430. 00 mol V 1.37 mol O 4.1792 The simplest ratio is 1. Multiply the simplest ratio by two to obtain the simplest whole-number ratio.10 mol V 50.969 0.52 g Ni 0.00 mol O 0. Inc. The empirical formula is C5H8NO4Na.94 g V 1 mol O 44.969 The simplest ratio is 6. Thus.00 mol C: 8.592 The simplest whole-number ratio is 5.0% vanadium and 44.10 147.10 g N? 1 mol Ni 10.10 mol V 1. 1.01 g C 1 mol N 5.6 g Na 0.73 mol H 1. a common headache remedy.1792 0.00 g O 1 mol O molar mass C13H18O2 206. 2.73 mol H = 9. Solutions Manual Chemistry: Matter and Change • Chapter 11 165 .592 0.00 mol N 0.5 g O 0.00 g O 6.000 mol Ni 0.936 mol C: 2. The percent composition of this oxide is 56.01 g C 13 mol C 156.3470 mol C 12.592 mol Na 1.52 g Ni. 1 mol V 56.96 mol C 5.592 mol N 1.37 mol O 2.00 g O 2 mol O 32.00 mol H 0. Copyright © Glencoe/McGraw-Hill.592 2.00 g O 1. Multiply the simplest ratio by two to obtain the simplest whole-number ratio.031 mol N 0.01 mol H 0. and 5. Vanadium oxide is used as an industrial cata- lyst.0 g O 2.75 mol O 16.80% H and 15.27 g/mol The molar mass C13H18O2 equals the molar mass of ibuprofen.13 g C 1 mol C 1 mol O 37.69 g Ni 1 mol C 4. 1 mol H 8.00 mol H: 1.75 mol O 2.73 mol H 8.CHAPTER 11 SOLUTIONS MANUAL 1 mol Na 13.01 mol H: 1.00 g O ibuprofen.00 mol V: 2.01 g N 0.9 g O 16.30 mol C = 6.592 mol Na 22.50 mol O.0% oxygen.1792 0.008 g H 18 mol H 18.00 mol O 0. The empirical formula is Ni(CN)2.00 mol N: 1.30 mol C 12.00 mol O. 4.144 g H 1 mol H 2.99 g Na 12.38 g C 0.1792 mol Ni 1.3640 mol N 2.50 mol C: 9.7 g C 6.008 g H 1 mol O 15. 145.00 mol Ni: 1. 146.3640 mol N 14.0 g V 1.00 mol O.3470 mol C 1.01 g C The simplest ratio is 1. Determine the molecular formula for 16.969 mol O = 1.936 mol C 0.50 mol C 0.5% O.50 mol O 1. The empirical formula is V2O5.592 4.00 mol Na 0. What is the empirical formula of a compound that contains 10. The empirical formula is C13H18O2. Determine the empirical formula for vanadium oxide.592 0.969 8.10 g N 0.00 mol Na: 4.38 g C.031 mol N. the molecular formula is the same as the empirical formula C13H18O2.7% C.00 mol C 0.10 1 mol C 75.1792 mol Ni 58. Analysis of ibuprofen yields a molar mass of 206 g/mol and a percent composition of 75. a division of the McGraw-Hill Companies.969 mol O 16.80 g H 8. 8. CHAPTER .01 g C 1 mol H 8.000 mol O.257 254.07 g S 1 mol S 1 mol S 32. 12.12 g C 3. multiply each number by three.01 g C 3 mol C 36. sweet liquid obtained as a byproduct of the manufacture of soap.257 mol C 12. Its percent composition is 39.2 g Pb 1 mol Cl 40.32 g/mol 349. 63.75 g/mol 190.37 g Pb 0.12 g O 3.00 g O 8 mol O 1 mol O 128.2865 1.55 g Cu 3 mol Cu 190.11 g/mol.00 g O 3 mol O 48.45 g Cl 166 Chemistry: Matter and Change • Chapter 11 The simplest whole-number ratio is 1.75% hydrogen.00 g O 1.12% oxygen.146 mol Cl 35.000 mol C: 2.00 g O 10 mol O 1 mol O 160.63% 1 mol Pb 59.681 mol H 1.257 and lead reveals that the compound is 59. What is the molecular formula for glycerol? 1 mol C 39.000 mol Cl 0.0 g/mol.00 g/mol 150.2865 Cu3(OH)4SO4 The simplest ratio is 1.63 g Cl 1.008 g H 6 mol H 1 mol H 6.665 mol H 3. To obtain the simplest whole-number ratio.2865 mol Pb 207.07 g S The molar mass based on the empirical formula equals the molar mass of the compound.74% 338.2865 mol Pb 1.32 g 8.45 g Cl 4 mol Cl 141.2 g Pb 1 mol Pb 207.03 g C 1 mol C 1.008 g H 1 mol O 52. 8.11 SOLUTIONS MANUAL 148. a division of the McGraw-Hill Companies.048 g H 32. The molecular formula is PbCl4.37% lead.55 g Cu 4 mol Cu 254. molar mass Cu3(OH)4SO4 354.12% carbon.37% 40. Inc. Determine the mass percent of copper in these compounds.094 g/mol Solutions Manual Copyright © Glencoe/McGraw-Hill.258 mol O 1.20 g percent by mass 100 56.000 mol Pb: 4.258 mol O 16.07 g S molar mass Cu4(OH)6SO4 452.00 g O 35. Cu3(OH)4SO4 and Cu4(OH)6SO4. The empirical formula is C3H8O3.75 g H 8. The molar mass is 92.75 g Cu4(OH)6SO4 63. Glycerol is a thick. What is the empirical formula for the chloride? What is the molecular formula? percent Pb percent Cl 100% percent Cl 100% percent Pb 100% 59.146 mol Cl 4.80 g Cl 1 mol Cl 1.008 g H 8 mol H 8.008 g H 4 mol H 1 mol H 4. 0.000 mol Pb 0.00 g O 3.000 mol C 3.2 g Pb 1 mol Pb 16. and 52.000 mol O 3. The molar mass of the compound is 349. The Statue of Liberty turns green in air because of the formation of two copper compounds.257 149.000 mol Cl.65 g Cu 1 mol Cu 207. The empirical formula is PbCl4.07 g S 1 mol S 1 mol S 32.064 g H 1 mol H 16.00 g O 1 mol O molar mass 92.20% 452.032 g H molar mass PbCl4 32.655 mol H: 1.681 mol H 2.257 mol C 1.65 g percent by mass 100 53. Analysis of a compound containing chlorine 3.20 g Cu 1 mol Cu 16. 23 g BaCl2 1 mol H2O 14.7 g H2O 0.0566 mol n 2.00 g O 1.00 g O 1.15 g/mol Copyright © Glencoe/McGraw-Hill.02 g H2O 0.15 g/mol 16.99 g 100 37.816 mol n 2.07 g S 1 mol S 1 mol S 32.3 g BaCl2 208.15 g 152. 3.02 g H2O 0.00 g O 1 mol O 153.02 g/mol 105.02 g H2O 22.02 g 16. barium chloride dihydrate 16. Determine the formula of this hydrate and name the compound. Determine the mass percent of anhydrous sodium carbonate (Na2CO3) and water in sodium carbonate decahydrate (Na2CO310H2O).02 g) 100 62.02 g/mol 1 mol BaCl2 0.89-g sample of this hydrate was heated.008 g H 2 mol H 1 mol H 2.03% Na2CO3 286.01 g C 1 mol C 1 mol C 12.00 g O 4 mol O 1 mol O 64. A 4. calcium sulfate dihydrate Solutions Manual Chemistry: Matter and Change • Chapter 11 167 .45 g Cl 2 mol Cl 1 mol Cl 70.00 0.5) Level 1 151.00 g O 1.016 g H molar mass H2O 18.00 g O molar mass CaSO4 136.0284 mol CaSO42H2O. Inc.90 g Cl molar mass BaCl2 208.008 g H 2 mol H 2.0284 mol CaSO4 1 mol H2O 1.99 g Na 2 mol Na 1 mol Na 45.97% H2O 286.87 g 1. 22.01 g C BaCl22H2O.008 g H 2 mol H 1 mol H 2.008 g H 20 mol H 1 mol H 20.07 g S 16.87 g anhydrous calcium sulfate remained.01 g C mass of H2O mass of hydrated CaSO4 mass of H2O 4. The molecular formula is C3H8O3.016 g H molar mass H2O 18.01 g C 1 mol C 1 mol C 12.816 mol H2O 18.00 0.98 g Na 0.08 g Ca 1 mol Ca molar mass Na2CO3 105.00 g O 40. Gypsum is hydrated calcium sulfate. and after the water was driven off. a division of the McGraw-Hill Companies.33 g Ba 1 mol Ba 35.00 g O 1 mol O 1 mol O 16.23 g/mol 32.15 g 10(18.08 g Ca 1 mol Ca 40.CHAPTER 11 SOLUTIONS MANUAL The molar mass based on the empirical formula equals the molar mass of the compound.16 g H molar mass Na2CO310H2O 286. What is the formula and name for a hydrate that is 85.00 g O 1 mol O 1 mol O 16.0566 mol H2O 18.33 g Ba 1 mol Ba 137.410 mol BaCl2 85.00 g O 3 mol O 1 mol O 48.02 g/mol 1 mol CaSO4 3.99 g Na 2 mol Na 45.15 g CaSO4 0.89 g 3. The Formula for a Hydrate (11.3% barium chloride and 14.7% water? 137.00 g O 13 mol O 1 mol O 208. 1. 16.016 g H 1 mol H molar mass H2O 18.87 g CaSO4 136.99 g/mol 16.98 g Na 1 mol Na mass of hydrated CaSO4 mass of CaSO4 mass of H2O 12.410 mol 12.00 g O 1 mol O 16. Chemical analysis indicates that this hydrate is 52.35 g Initial mass of hydrate 10.8% sodium tetraborate and 47.02 g/mol 16. A 1.45 g Cl 2 mol Cl 70. Table 11-3 Data for BaCl2xH20 155.00 g O 7 mol O 1 mol O 112.008 g H 2 mol H 1 mol H 2. barium chloride dihydrate 1 mol MgI2 1.0821 mol H2O 18.072 g 0.33 g Ba 1 mol Ba 0.30 g Mass of hydrate crucible 31.11 g/mol Mass of anhydrous solid 8.016 g H molar mass H2O 18.87 g molar mass MgI2 278.33 g B 1 mol Ba 137.11 SOLUTIONS MANUAL 154.016 g H 1 mol H molar mass H2O 168 18.00 g O 1 mol O initial mass of hydrate (mass of hydrate crucible) (mass of empty crucible) 31.00 g O 1 mol O 1 mol O 16. Inc.23 g BaCl2 0.0309 mol n 8.80 g I Mass after heating 5 min 29.30 g 8.57 g BaCl2 208. Determine the formula and name the hydrate.628-g sample of a hydrate of magnesium iodide is heated until its mass is reduced to 1.628 g 1.2% water.02 g/mol Solutions Manual Copyright © Glencoe/McGraw-Hill.90 g Cl 1 mol Cl MgI28H2O molar mass BaCl2 208.556 g H2O 0.016 g H 1 mol H molar mass H2O mass of water (initial mass of hydrate) (mass of anhydrous solid) 10. The table shows data from an experiment to determine the formulas of hydrated barium chloride.0309 mol H2O 18.02 g H2O 0.98 g Na 1 mol Na 12.48 g H2O 0.57 g Chemistry: Matter and Change • Chapter 11 18. 22.00 g O 1 mol O 16. Determine the formula for the hydrate and its name.05 g 1.35 g 21.003855 mol 35.008 g H 2 mol H 2.00 g O 1 mol O 16.0412 mol BaCl22H2O.072 g and all water has been removed. a division of the McGraw-Hill Companies.01 g B 4 mol B 1 mol B 48.02 g/mol mass of anhydrous solid (mass after heating 5 min) (mass of empty crucible) 29.072 g MgI2 278.00 g O 1.0821 mol n 2.00 g O molar mass Na2B4O7 206.04 g B 16.00 g O 1 mol O 1.48 g 0. Hydrated sodium tetraborate (Na2B4O7xH2O) is commonly called borax.02 g H2O 156.99 g Na 2 mol Na 45.05 g 126.23 g/mol 16.87 g 21.0412 mol BaCl2 1 mol H2O 1.90 g I 2 mol I 1 mol I 253.30 g 10.003855 mol MgI2 137.11 g MgI2 1 mol H2O 0.00 0.02 0. What is the formula of the hydrate? mass of water mass of hydrate mass of anhydrous solid 1.57 g 1.05 g 8.31 g Mg 1 mol Mg Mass of empty crucible 21.556 g 24.31 g Mg 1 mol Mg 24.57 g 16.008 g H 2 mol H 2. CHAPTER .02 g/mol 1 mol BaCl2 8. a division of the McGraw-Hill Companies. Which of the following has the greatest number of oxygen atoms? a.CHAPTER 11 SOLUTIONS MANUAL 1 mol Na2B4O7 52.70 g Fe 1 mol Fe 6.00 g O 1 mol O 4. sodium tetraborate decahydrate 2 mol O 0.98 g Al 2 mol Al 53.07% O in TiO2 Chemistry: Matter and Change • Chapter 11 169 .75 g Zn 65.85 g Fe 2 mol Fe 111.02 1023 atoms Ag 7.45 1022 atoms Zn 1 mol Zn b.02 1023 atoms O 0.00 g O 2 mol O 32.74 g Ag c.61 mol H2O 18.00 g O 1 mol O molar mass Al2O3 54.5 mol O 6.87 g Ag 1 mol Ag 6. the number of representative particles in 3.2 g H2O 2.256 mol Na2B4O7 1 mol H2O 47.87 g/mol 159.39 g Zn 6.02 g H2O 2. Inc.63 g CO2 12.01 g/mol 79.32 1022 atoms Ag Copyright © Glencoe/McGraw-Hill. 157.61 mol n 10.21 1022 molecules CH3OH Sharpen your problem-solving skills by answering the following.4008 mol CO2 44. Determine the following: a.87 g Ti 1 mol Ti 47.21 1022 molecules CH3OH Mixed Review 3.32 1022 atoms Ag 4.01 g C 1 mol C 16.70 g/mol 26. 3.02 g Na2B4O7 0.02 1023 atoms O 1.0 g of c. Fe2O3.00 g O percent by mass 100 79.21 1022 atoms O 1 molecule CH3OH 1 mol Zn 3.87 g Ti 1 mol Ti 16.75 g Zn 1 atom O 3.00 g O 3 mol O 48.256 mol Na2B4O710H2O.02 1023 ions Na 2 mol Na 1 mol Na2O 1 mol Na 16.250 mol C6H12O6 1 mol C6H12O6 1.00 g O 1 mol O Na2O. 0.00 g O 1 mol O molar mass Solutions Manual 44.8 g Na2B4O7 206. Al2O3? 47.01 g CO2 1 mol Ag 107.2 0.98 g Na2O 55. Which of the following compounds has the greatest percent of oxygen by mass: TiO2.87 g TiO2 40. the number of sodium ions is 25. 17.0 1023 atoms O 159.250 mol C6H12O6 6 mol O 0.02 1023 atoms Zn 3.82 1023 atoms O b.00 g O 2 mol O 32.96 g Al 1 mol Al 16. 1 mol CO2 17.64 g CO2 0.8016 mol O 1 mol O 4. the mass of 4.86 1023 ions Na molar mass Fe2O3 158.5 mol O 1 mol O 9.44 g/mol 32.01 g C 1 mol C 12.4008 mol CO2 0.00 g O 3 mol O 48. molar mass TiO2 1 mol Na2O 25.0 g Na2O 61.8016 mol O 1 mol CO2 6. 05 g C 1 mol C molar mass 1.54 1023 formula units NaCl c.07 g/mol 146.21 g/mol 161. The empirical formula is C5H4. CHAPTER .109 mol Na2SO4 b.152 mol MgCl2 1 mol MgCl2 1.09 g/mol 19. the number of Cl ions in 14.00 g F 1 mol F 146. The molar mass of napthalene is 128 g/mol. Inc.31 g Mg 1 mol Mg Molecular formula is C10H8.99 g Na 2 mol Na 45. 35. 1 mol C 93.5 g MgCl2 95. is composed of 93.07 g S 1 mol S 1 mol S 32.09 g/mol 24. the mass in grams of 0. Naphthalene.83 1023 ions Cl 170 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.11 SOLUTIONS MANUAL 48.44 g Al2O3 87.2 mol H 1.07 g SF6 113 g SF6 0.008 g H 4 mol H 4.01 g C 1 mol H 6.152 mol MgCl2 2(6.80 mol C 1.5 g Na2SO4 142. 160.0 mol H. Which of the following molecular formulas are also empirical formulas: ethyl ether (C4H10O). the number of moles in 15.21 g MgCl2 0.07 g S 1 mol S 12.3 g H 6.255 mol NaCl 0.00 g O percent by mass 100 159. 1 mol MgCl2 14.2 162.07 g S 16.05 g Na2SO4 0.255 mol NaCl 6.7 g C 7. The subscripts in C4H10O and C9H8O4 represent simplest whole-number ratios.032 g H 1 mol H molar mass 64.02 1023 ion Cl) 0.01 g C 5 mol C 60.00 g O 4 mol O 1 mol O 64.25 mol C : 1.80 mol C 12. Multiply both subscripts by four to obtain the simplest whole-number ratio.7% carbon and 6. commonly known as moth balls.98 g Na 1 mol Na 32.02 1023 formula units NaCl 1 mol NaCl 1.3% hydrogen. Determine the empirical and molecular formulas for naphthalene.05 g/mol 1 mol Na2SO4 15.70 g Fe2O3 30.45 g Cl 2 mol Cl 1 mol Cl 70.00 g F 6 mol F 114. 32.00 64. aspirin (C9H8O4). glucose (C6H12O6).0 mol H 6.2 6.5 g MgCl2. 128 g/mol n 2.07 g S 1 mol S 32.25 mol C 6.00 g O percent by mass 100 54.90 g Cl molar mass 95.2 mol H 1.008 g H 7.00 g O molar mass 142.99% O in Al2O3 Al2O3 has the greatest percent of oxygen by mass.775 mol SF6 The simplest ratio is 1.775 mol SF6 1 mol SF6 d.31 g Mg 1 mol Mg 24. Calculate each of the following: a. a division of the McGraw-Hill Companies. the number of formula units in 0.5 g Na2SO4 22.06% O in Fe2O3 48. butyl dichloride (C4H8Cl2). and nitrogen.18 g H 1 mol H The simplest ratio is 1. Multiply the subscripts in the empirical formula by two to obtain the molecular formula C2F4.80 1022 atoms of helium. 166.0% carbon and 76.01 g/mol 100.00 g F are present in 1.0 g F 4.42% 12. 45% of the helium had escaped. hydrogen.200 g C 0.01 g/mol N 22.00 mol F.01 g C 6.0 g/mol.00 mol C : 2.00 mol F 19.00 g F 1 mol F molar mass CF2 50.39 1022 atoms He Copyright © Glencoe/McGraw-Hill. Tetrafluoroethylene.00 mol F 2.08 g C 114.02 1023 atoms C 2.96% 52.00 1.22 g/mol 0. 1 mol C 24. which is used in the production of Teflon.25 carat 5 102 g C 1 carat 1 mol C 5 102 g C 12. Diamond is a naturally occurring form of carbon.01 g C The molar mass of the compound is two times greater than the molar mass of the empirical formula.5 1021 atoms C 1 mol C 168.00 mol C 12.00 mol C 1. how many carbon atoms are present? (1 carat 0.80 1022 atoms He 5. Inc.68% H 4. oxygen.00 g F 2 mol F 38.680 g/mL) 2. After 24 hours.0 g/mol n 2.0 g C 2.200 g) 0.25-carat diamond.00 mol F 2.680 g isooctane 103 mL 1.000 50.96 g H 164.02 1023 atoms Pb 207.80 102 g isooctane Solutions Manual Chemistry: Matter and Change • Chapter 11 171 . molar mass 96. How many molecules of isooctane (C8H18) 1 mol F 76.00 12.00 mol C 2. Determine the empirical and molecular formulas of this compound.00 L 1L 1 mL 6. Calculate the mass in grams of one atom of lead. The graph shows the percent composition of a compound containing carbon. a division of the McGraw-Hill Companies. is composed of 24. 22. 4. 19.01 g C 1 mol C 12.01 g C 8 mol C 1 mol C 4. 1 mol Pb 1 atom Pb 6.42 g O.95 g N. If you have a 0.00 L? (density of isooctane 0.55 9.44 1022 g Pb 1 mol Pb 167. 38. How many atoms of helium remained? percent remaining 100% percent that escaped 100% 45% 55% 0.008 g H 18 mol H 18. How many grams of each element are present in 100 g of the compound? C 19.01 g C 1 mol C 165.68 g C. A party balloon was filled with 9.0% fluorine and has a molar mass of 100.CHAPTER 11 SOLUTIONS MANUAL 163.02 g Pb 3.95% O 52. The empirical formula is CF2. Writing in Chemistry 174. which ore would yield the greater quantity of copper? Explain your answer.00 g/cm3.00 103 m3 1 g H2O (102 cm)3 4. or is there more than one way to interpret the expression? Explain.95 mol isooctane 5. Chalcocite would yield the greater quantity of copper because the ore has the greater percentage copper by mass. molar mass. the advantages of each. Concept maps will vary. Student pamphlets will vary but should discuss how octane ratings are found. Octane rating of 100% means pure octane.95 mol isooctane 6.00 1032 g H2O 18. CHAPTER .0 m 5. 172.0 m 20. Determine the ratio of moles of water to moles of anhydrous compound.00 m in depth. 172 Chemistry: Matter and Change • Chapter 11 SOLUTIONS MANUAL 171. Inc. volume length width height volume 40. Allow the dish to cool. and 5. Specify atoms or molecules. percent composition.00 109 g H2O 1 mol H2O 4.9% copper by mass.80 102 g isooctane 114. Calculate the number of moles of anhydrous compound and water. Octane ratings are used to identify certain grades of gasoline.11 1 mol isooctane 6.02 g H2O 6. number of particles. Octane rating of 90% means 90% octane and 10% heptane. Avogadro’s number.02 1023 molecules H2O 1 mol H2O 1. Determine the masses of the anhydrous solid and the water lost. Communicating If you use the expression “a mole of nitrogen.02 1023 molecules isooctane 1 mol isooctane 3. Chalcopyrite (CuFeS2) is 34. Designing an Experiment Design an experi- ment that can be used to determine the amount of water in alum (KAl(SO4)2xH2O).58 1024 molecules isooctane 169. Use the wholenumber ratio of the moles as the coefficient of H2O in the formula. Calculate the number of molecules of water in a swimming pool that is 40.26 g isooctane 5.00 103 m3 3 1m 1 cm3 4.0 m in width. and when each grade is used. Analyze and Conclude A mining company has two possible sources of copper: chalcopyrite (CuFeS2) and chalcocite (Cu2S). Solutions Manual Copyright © Glencoe/McGraw-Hill. Include moles. Heat the evaporating dish gently for 5 minutes.6% copper by mass (determined from percent composition) and chalcocite (Cu2S) is 79.0 m in length. and measure and record the mass. Assume that the density of water is 1. a division of the McGraw-Hill Companies.34 1032 molecules H2O Thinking Critically 170. Research octane rating and prepare a pamphlet for consumers identifying the different types of gasoline. Determine and record the mass of an empty evaporating dish. 20. Concept Mapping Design a concept map that illustrates the mole concept. and molecular formula. and strongly for another 5 minutes to evaporate all the water. Add about 2 g of the hydrate.00 m 4. If the mining conditions and the extraction of copper from the ore were identical for each of the ores.” is it perfectly clear what you mean. empirical formula. 173. Measure and record the mass. How could you change the expression to make it more precise? The phrase could be interpreted to mean a mole of nitrogen atoms or a mole of nitrogen molecules. 4 34.5 40. a division of the McGraw-Hill Companies. empirical formula c.5 323 Copyright © Glencoe/McGraw-Hill. Avogadro formulated his hypothesis as an explanation for earlier works by Gay-Lussac and Ritter. c.1 9. Standardized Test Practice Chapter 11 page 351 Interpreting Graphs Use the graph to answer questions 1–4.233 0. (Chapter 2) a. Cumulative Review Refresh your understanding of previous chapters by answering the following. chemical properties Acetaldehyde and butanoic acid must have the same empirical formula because they have the same number of grams and therefore the same number of moles of each element in a 100. Aqueous solutions of sulfuric acid and Amedeo Avogadro (1776–1856) and how his work led scientists to the number of particles in a mole. 18. H2SO4(aq) 2KOH(aq) 0 K2SO4(aq) 2H2O(l) 179.3 54.4 36. Inc. (82. Students should mention Avogadro’s hypothesis. Research the life of the Italian chemist c. a. Two isotopes of an element will have the same atomic number but different mass numbers.7 Ethanol 9. How can you tell if a chemical equation is balanced? (Chapter 10) There are equal numbers of each kind of atom on both sides of the equation. How do these two numbers compare for isotopes of an element? (Chapter 4) Atomic number equals the number of protons.5 54.7 potassium hydroxide undergo a double replacement reaction. 53.44 4. Magnesium metal and water combine to form solid magnesium hydroxide and hydrogen gas.125 406 Percent by mass b.92) 0.0 6. b N2O4(g) 0 2NO2(g) Solutions Manual Chemistry: Matter and Change • Chapter 11 173 . Dinitrogen tetroxide gas decomposes into nitrogen dioxide gas. Acetaldehyde and butanoic acid must have the same _____ . Percent Composition of Some Organic Compounds 438. Mass number equals the number of protons plus the number of neutrons.CHAPTER 11 SOLUTIONS MANUAL 175.2 40 number.1 Formaldehyde Acetaldehyde Butanoic acid Compound name 1. Distinguish between atomic number and mass %C %H %O 0 13. Avogadro died before seeing his ideas accepted. molar mass d.8 30 20 10 177. molecular formula b. His ideas were rejected by chemists of his day but were revived later by the Italian chemist Stanislao Cannizzaro. Mg(s) 2H2O(l) 0 Mg(OH)2(s) H2(g) b. (Chapter 10) a.23 456.0 36. Write balanced equations for the following reactions. Express the following answers with the correct number of significant figures. 4. 176. 178.0131 60 50 52.0 g sample. C6HO2 d. Molar mass of CH2O 30. The empirical formula of formaldehyde is the same as its molecular formula.33 mol C 1.9 mol H 5.35 mol C 2.54 mol C 12.33 6.00 1 mol O 2.28 9. CHAPTER .0 g Determine the empirical formula for formaldahyde. Assume a 100.54 mol C 1.28 mol O 16.06 g/mol The molecular formula for butanoic acid (C2H4O)2 C4H8O2. 200. 60.33 mol O 1.01 g C 1 mol H 6.9 mol H 1.0 g d.28 mol O 1. c. Inc.008 g H 1 mol O 34.2 g C 4.03 mol H 3.3 g O 3.00 g O 3. The mass of the empirical formula is 44. 88. then what is its molecular formula? a.00 mol O 1 mol O 3. C4H8O2 Determine the empirical formula. C4HO3 C52H13O35 C2H6O C4H13O2 1 mol C 52.11 2.65 mol H 1. b.0 g C 3.03 mol H 1.65 mol H 2. If the molar mass of butanoic acid is 88.03 g 2. 1 mol C 54.35 mol C 12. How many grams are in 2.1 g H 9.00 mol C 1 mol C 3.99 2 mol C 2.1 g/mol 2.01 g C 1 mol H 9.00 g O 174 Chemistry: Matter and Change • Chapter 11 Solutions Manual Copyright © Glencoe/McGraw-Hill.0 g sample.000 mol 60. 4.18 2.1 g/mol. a division of the McGraw-Hill Companies.8 g O 1 mol O 16.7 g H 6.06 g formaldahyde 1 mol b 1 mol H 13. d 3.33 3.18 12.92 mol H 6 mol H 2.06 g c. 1 mol C 40. What is the empirical formula of ethanol? a.000 moles of formaldehyde? a. 4.33 The empirical formula for formaldahyde is CH2O. 182. C2H4O c. 30.5 g C 4.00 g b.00 mol O 1 mol O 2.0 g H 12.33 mol O 16.18 The empirical formula for ethanol is C2H6O.28 2. d.4 g O 2.33 mol C 12.00 mol H 2 mol H 3.03 g/mol 30.008 g H 1 mol O 53.00 44.96 4 mol H 2.00 mol C 2 mol C 2. CH2O b.00 g O Determine the mole ratios.28 The empirical formula for butanoic acid is C2H4O.06 g/mol.18 mol O 1.01 g C SOLUTIONS MANUAL Determine the mole ratios. c 4.008 g H 1 mol O 36. c 7.91 g P 16.39 101 mol b.59 amu)? a.74 g/mol 1 mol Co2TiO4 7.14 102 mol b Solutions Manual Chemistry: Matter and Change • Chapter 11 175 .88 g Ti 1 mol Ti 44. Avogadro’s number of molecules of a compound c.00 g O 12 mol O 192. How many atoms are in 0. 3. 4. 6. 9. 3.22 101 mol d.13 g Co2TiO4 227 g Co2TiO4 3.CHAPTER 11 SOLUTIONS MANUAL 5. a.00 g F 1 mol F 1 mol F 19. 442 g/mol d.93 g Co 2 mol Co 117. The molar mass of fluorapatite (Ca5(PO4)3F) is _____ .00 g O 1 mol O 19.73 1025 b.76 1023 atoms Ge Copyright © Glencoe/McGraw-Hill. the number of atoms in exactly 12 g of pure 12C d.17 102 mol Find the molar mass of Co2TiO4. Inc. 314 g/mol b. 3.00 g O 1 mol O Molar mass Co2TiO4 226.00 g F molar mass 504.00 g O 4 mol O 64. 58.31 g/mol d 8.97 g P 3 mol P 1 mol P 92. 2. A mole is all of the following EXCEPT _____ .63 1023 atoms Ge 6.625 mol Ge 1 mol Ge 1023 3. a division of the McGraw-Hill Companies.13 g of the compound? a.14 102 mol c.625 moles of Ge (atomic mass 72.86 g Co 1 mol Co 44. 504 g/mol 40. 2.02 0.08 g Ca 5 mol Ca 200.99 1025 c. the atomic or molar mass of an element or compound b. 344 g/mol c. a.76 1023 d. How many moles of cobalt(III) titanate (Co2TiO4) are in 7. the SI measurement unit for the amount of a substance A mole is not a mass so a is the answer.88 g Ti 1 mol Ti 16. a 6.40 g Ca 1 mol Ca 30. MgSO4 readily absorbs water to form two different hydrates.64 1022 g/molecule BaSiF6 1 mol H2O 13.09 g Si 1 mol Si 1 mol Si 28.02 g H2O c 0.0 g sample of the hydrate. magnesium sulfate monohydrate a 176 Chemistry: Matter and Change • Chapter 11 Solutions Manual .0 g H2O 0.68 1026 g b. 2.721 mol H2O 0.0% H2O and 87.0 g MgSO4 and 13. 6.0 g MgSO4 120.16 1021 g c.33 g Ba 1 mol Ba 28.38 g MgSO4 279. magnesium sulfate monohydrate b.CHAPTER 11 9. magnesium sulfate dihydrate c. The hydrate is MgSO4H2O. 137. One of these hydrates is found to contain 13. 4. magnesium sulfate hexahydrate d.00 g F 6 mol F 1 mol F 114.723 mol MgSO4 Copyright © Glencoe/McGraw-Hill.09 g Si Molar mass MgSO4 120.723 mol MgSO4 4. a.42 g BaSiF6 1 mol BaSiF6 1 mol BaSiF6 6. What is the name of this hydrate? a. magnesium sulfate heptahydrate SOLUTIONS MANUAL 10. 0. 1.42 g/mol BaSiF6 1 mol MgSO4 87. The mass of one molecule of barium hexafluo- rosilicate (BaSiF6) is _____ .02 1023 molecules BaSiF6 Assume a 100.02 1023 g Determine the molar mass of BaSiF6.997 1 1 1 0.0 g H2O.38 g/mol 19.02 g/mol Molar mass 279. a division of the McGraw-Hill Companies. the sample contains 87.0% MgSO4. Inc.721 mol H2O 18.64 1022 g d. Magnesium sulfate (MgSO4) is often added to water-insoluble liquid products of chemical reactions to remove any unwanted water.33 g Ba 1 mol Ba 137. Then.00 g F Molar mass H2O 18.
Copyright © 2024 DOKUMEN.SITE Inc.