= 11.0 g (assuming 100 washers is exact).

CHAPTER 8 1. 100 washers 0.110 g 1 washer 100. g 1 washer 0.110 g = 11.0 g (assuming 100 washers is exact). = 909 washers 2. The empirical formula is CFH from the structure given. The empirical formula represents the smallest whole number ratio of the number and types of atoms present. 3. The atomic mass unit (amu) is defined by scientists to more simply describe relative masses on an atomic or molecular scale. One amu is equivalent to 1.66 10 24 g. 4. The average atomic mass takes into account the various isotopes of an element and the relative abundances in which those isotopes are found. 5. a. 125 C atoms b. 5 10 6 K atoms c. 1.04 10 22 Li atoms d. 1 Mg atom 12.01 amu 1 C atom = 1.50 103 amu 39.10 amu 1 K atom = 1.955 108 amu = 2 10 8 amu 24.31 amu 1 Mg atom 6.941 amu 1 Li atom = 7.22 1022 amu = 24.31 amu e. 3.011 10 I atoms 126.9 amu 1 I atom = 3.821 1025 amu 6. a. 40.08 amu Ca 1 Ca atom = 1 Ca atom 40.08 amu b. 919.5 amu W 1 W atom 183.9 amu = 5 W atoms c. 549.4 amu Mn d. 6345 amu I e. 2072 amu 1 Mn atom = 10 Mn atoms 54.94 amu 1 I atom 126.9 amu = 50 I atoms 1 Pb atom = 10 Pb atoms 207.2 amu 117

7. One average iron atom has a mass of 55.85 amu 55.85 amu 299 iron atoms would weigh: 299 atoms 1 atom = 1.67 104 amu 5529.2 amu represents: 5529.2 amu 8. One tin atom has a mass of 118.7 amu. 1 atom 55.85 amu A sample containing 35 tin atoms would weigh: 35 2967.5 amu of tin would represent: 2967.5 amu 9. Avogadro s number (6.022 10 atoms; 1.00 mol) 10. 26.98 g (1.00 mol) 11. molar masses: Na, 22.99 g; K, 39.10 g Na 11.50 g Na = 0.5002 mol Na 22.99 g 0.5002 mol Na 6.033 10 Na 0.5002 mol K 39.10 g K = 19.56 g K = 3.012 10 atoms = 99 atoms. 118.7 amu 1 atom = 4155 amu; 1 tin atom = 25 tin atoms. 118.7 amu 12. Since 10.09 g of neon represents half the molar mass of neon (20.18 g), then 10.09 g of neon contains 1 2 (6.022 10 ) = 3.011 10 neon atoms (0.500 mol). Also, 0.500 mol of helium gas would contain the same number of atoms as 0.500 mol of neon gas: this would be 1 2 4.003 g = 2.002 g of helium gas. 13. The ratio of the atomic mass of H to the atomic mass of N is (1.008 amu/14.01 amu), and the mass of hydrogen is given by 7.00 g N 1.008 amu = 0.504 g H. 14.01 amu 14. The ratio of the atomic mass of Co to the atomic mass of F is (58.93 amu/19.00 amu), and the mass of cobalt is given by 57.0 g 58.93 amu = 177 g Co. 19.00 amu 118

15. mass of a magnesium atom = 3.82 10 g 24.31 amu Mg 22.99 amu Na = 4.04 10 g 16. mass of a carbon atom = 4.48 10 g 12.01 amu C 26.98 amu Al = 1.99 10 g 17. of He atoms = 4.003 g 4 mol of H atoms 1.008 g H = 4.032 g 4 mol of H atoms has a mass slightly larger than of He atoms. 18. 0.50 mol Ne atoms 20.18 g = 10.09 g Ne = 10. g Ne (two significant figures) B atoms 10.81 g B = 10.81 g B = 11 g B (two significant figures) The neon sample weighs less. 19. a. 4.95 g Ne b. 72.5 g Ni 20.18 g 58.69 g = 0.245 mol Ne = 1.24 mol Ni c. 115 mg Ag 1 g = 1.07 10 3 mol Ag 1000 mg 107.9 g 1 g d. 6.22μg U = 2.61 10 8 mol U 6 10 μg 8.0 g e. 135 g I 126.9 g = 1.06 mol I 20. a. 66.50 g F 19.00 g = 3.500 mol of F atoms b. 401.2 mg Hg c. 84.27 g Si 1 mmol = 2.000 mmol Hg (1 mmol = 1/1000 mol) 200.6 mg = 3.000 mol Si 28.09 g d. 48.78 g Pt 195.1 g = 0.2500 mol Pt e. 2431 g Mg 24.31 g = 100.0 mol Mg 119

f. 47.97 g 95.94 g = 0.5000 mol Mo 21. a. 0.25 Li b. 1.5 Al 6.941 g Li Li 26.98 g Al Al c. 8.75 10 2 mol Pb d. 125 mol Cr 52.00 g Cr Cr e. 4.25 10 3 mol Fe f. 0.000105 mol Mg = 1.74 g Li = 40.7 g Al 207.2 g Pb Pb 55.85 g Fe 22. a. 0.00552 mol Ca 40.08 g = 18.1 g Pb = 6.50 10 3 g Cr = 2.37 10 5 g Fe 24.31 g Mg Mg = 2.55 10 3 g Mg = 0.221 g Ca 10.81 g b. 6.25 millimol B = 0.0676 g B 3 10 millmol c. 135 mol Al 26.98 g d. 1.34 10 7 mol Ba 137.3 g e. 2.79 mol P 30.97 g = 3.64 10 3 g Al = 86.4 g P f. 0.0000997 mol As 74.92 g = 1.84 10 5 g Ba = 7.47 10 3 g As. a. 1.50 g Ag 6.022 10 Ag atoms 107.9 g Ag = 8.37 10 21 Ag atoms b. 0.0015 mol Cu c. 0.0015 g Cu d. 2.00 kg = 2.00 10 3 g 6.022 10 Cu atoms 6.022 10 Cu atoms 63.55 g Cu = 9.0 10 20 Cu atoms = 1.4 10 19 Cu atoms 2.00 10 3 g Mg 6.022 10 Mg atoms 24.31 g Mg = 4.95 10 25 Mg atoms 120

e. 1.000 oz = 28.35 g 2.34 oz 28.35 g 6.022 10 Ca atoms = 9.97 10 Ca atoms 1.000 oz 40.08 g Ca f. 2.34 g Ca 6.022 10 Ca atoms 40.08 g Ca = 3.52 10 22 Ca atoms g. 2.34 mol Ca 6.022 10 Ca atoms Ca = 1.41 10 24 Ca atoms 55.85 g Fe 24. a. 125 Fe atoms 6.022 10 Fe atoms b. 125 Fe atoms c. 125 g Fe d. 125 mol Fe e. 125 g Fe 55.85 amu 1 Fe atom = 6.98 103 amu Fe 55.85 g Fe 55.85 g Fe Fe = 2.24 mol Fe = 6.98 10 3 g Fe 6.022 10 Fe atoms 55.85 g Fe = 1.16 10 20 g = 1.35 10 24 Fe atoms f. 125 mol Fe 25. molar mass 6.022 10 Fe atoms Fe = 7.53 10 25 Fe atoms 26. The molar mass is calculated by summing the individual atomic masses of the atoms in the formula. 27. a. H 3 PO 4 phosphoric acid mass of 3 mol H = 3(1.008 g) = 3.024 g mass of P = 30.97 g mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of H 3 PO 4 = (3.024 g + 30.97 g + 64.00 g) = 97.99 g b. Fe 2 O 3 ferric oxide, iron(iii) oxide mass of 2 mol Fe = 2(55.85 g) = 111.7 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of Fe 2 O 3 = (111.7 g + 48.00 g) = 159.7 g c. NaClO 4 sodium perchlorate mass of Na = 22.99 g mass of Cl = 35.45 g 121

mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of NaClO 4 = (22.99 g + 35.45 g + 64.00 g) = 122.44 g d. PbCl 2 plumbous chloride, lead(ii) chloride mass of Pb = 207.2 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g molar mass of PbCl 2 = (207.2 g + 70.90 g) = 278.1 g e. HBr hydrogen bromide mass of H = 1.008 g mass of Br = 79.90 g molar mass of HBr = (1.008 g + 79.90 g) = 80.91 g f. Al(OH) 3 aluminum hydroxide mass of Al = 26.98 g mass of 3 mol H = 3(1.008) = 3.024 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of Al(OH) 3 = (26.98 g + 3.024 g + 48.00 g) = 78.00 g 28. a. KHCO 3 potassium hydrogen carbonate, potassium bicarbonate mass of K = 39.10 g mass of H = 1.008 g mass of C = 12.01 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of KHCO 3 = (39.10 + 1.008 + 12.01 + 48.00) = 100.12 g b. Hg 2 Cl 2 mercurous chloride, mercury(i) chloride mass of 2 mol Hg = 2(200.6 g) = 401.2 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g molar mass of Hg 2 Cl 2 = (401.2 g + 70.90 g) = 472.1 g c. H 2 O 2 hydrogen peroxide mass of 2 mol H = 2(1.008 g) = 2.016 g mass of 2 mol O = 2(16.00 g) = 32.00 g molar mass of H 2 O 2 = (2.016 g + 32.00 g) = 34.02 g d. BeCl 2 beryllium chloride mass of Be = 9.012 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g molar mass of BeCl 2 = (9.012 g + 70.90 g) = 79.91 g 122

e. Al 2 (SO 4 ) 3 aluminum sulfate mass of 2 mol Al = 2(26.98 g) = 53.96 g mass of 3 mol S = 3(32.07 g) = 96.21 g mass of 12 mol O = 12(16.00 g) = 192.0 g molar mass of Al 2 (SO 4 ) 3 = (53.96 g + 96.21 g + 192.0 g) = 342.2 g f. KClO 3 potassium chlorate mass of K = 39.10 g mass of Cl = 35.45 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of KClO 3 = 122.55 g 29. a. BaCl 2 mass of Ba = 137.3 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g molar mass of BaCl 2 = (137.3 g + 70.90 g) = 208.2 g b. Al(NO 3 ) 3 mass of Al = 26.98 g mass of 3 mol N = 3(14.01 g) = 42.03 g mass of 9 mol O = 9(16.00 g) = 96.00 g molar mass of Al(NO 3 ) 3 = (26.98 g + 42.03 g + 96.00 g) = 213.0 g c. FeCl 2 mass of Fe = 55.85 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g molar mass of FeCl 2 = (55.85 g + 70.90 g) = 126.75 g d. SO 2 mass of S = 32.07 g mass of 2 mol O = 2(16.00 g) = 32.00 g molar mass of SO 2 = (32.07 g + 32.00 g) = 64.07 g e. Ca(C 2 H 3 O 2 ) 2 mass of Ca = 40.08 g mass of 4 mol C = 4(12.01 g) = 48.04 g mass of 6 mol H = 6(1.008 g) = 6.048 g mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of Ca(C 2 H 3 O 2 ) 2 = (40.08 g + 48.04 g + 6.048 g + 64.00 g) = 158.17 g 1

30. a. LiClO 4 mass of Li = 6.941 g mass of Cl = 35.45 g mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of LiClO 4 = (6.941 g + 35.45 g + 64.00 g) = 106.39 g b. NaHSO 4 mass of Na = 22.99 g mass of H = 1.008 g mass of S = 32.07 g = 32.07 g mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of NaHSO 4 = (22.99 g + 1.008 g + 32.07 g + 64.00 g) = 120.07 g c. MgCO 3 mass of Mg = 24.31 g mass of C = 12.01 g mass of 3 mol O = 3(16.00 g) = 48.00 g molar mass of MgCO 3 = (24.31 g + 12.01 g + 48.00 g) = 84.32 g d. AlBr 3 mass of Al = 26.98 g mass of 3 mol Br = 3(79.90 g) = 9.7 g molar mass of AlBr 3 = (26.98 g + 9.7 g) = 266.7 g e. Cr 2 S 3 mass of 2 mol Cr = 2(52.00 g) = 104.0 g mass of 3 mol S = 3(32.07 g) = 96.21 g molar mass of Cr 2 S 3 = (104.0 + 96.21 g) = 200.2 g 31. a. molar mass of NO 2 = 46.01 g; 21.4 mg = 0.0214 g 0.0214 g NO 2 46.01 g = 4.65 10 4 mol NO 2 b. molar mass of Cu(NO 3 ) 2 = 187.6 g 1.56 g Cu(NO 3 ) 2 187.6 g = 8.32 10 3 mol Cu(NO 3 ) 2 c. molar mass of CS 2 = 76.15 g 2.47 g CS 2 = 0.0324 mol 76.15 g 124

d. molar mass of Al 2 (SO 4 ) 3 = 342.2 g 5.04 g Al 2 (SO 4 ) 3 342.2 g = 0.0147 mol Al 2(SO 4 ) 3 e. molar mass of PbCl 2 = 278.1 g 2.99 g 278.1 g = 0.0108 mol PbCl 2 f. molar mass of CaCO 3 = 100.09 g 62.4 g CaCO 3 100.09 g = 0.6 mol CaCO 3 32. a. molar mass Al 2 O 3 = 101.96 g 47.2 g = 0.463 mol 101.96 g b. molar mass KBr = 119.00 g 1000 g 1.34 kg = 11.3 mol 1 kg 119.00 g c. molar mass Ge = 72.59 g 1 g 521 mg = 7.18 10 3 mol 1000 mg 72.59 g d. molar mass of U = 8.0 g 1 g 56.2 μg = 2.36 10 7 mol 6 10 μg 8.0 g e. molar mass of NaC 2 H 3 O 2 = 82.03 g 29.7 g = 1.69 mol = 0.362 mol 82.03 g f. molar mass of SO 3 = 80.07 g 1.03 g = 0.0129 mol 80.07 g 33. a. molar mass MgCl 2 = 95.21 g 41.5 g = 0.436 mol 95.21 g b. molar mass Li 2 O = 29.88 g; 135 mg = 0.135 g 0.135 g 29.88 g = 4.52 10 3 mol = 4.52 mmol 125

c. molar mass Cr = 52.00 g; 1.21 kg = 1210 g 1210 g =.3 mol 52.00 g d. molar mass H 2 SO 4 = 98.09 g 62.5 g = 0.637 mol 98.09 g e. molar mass C 6 H 6 = 78.11 g 42.7 g = 0.547 mol 78.11 g f. molar mass H 2 O 2 = 34.02 g 135 g = 3.97 mol 34.02 g 34. a. molar mass of Li 2 CO 3 = 73.89 g 1.95 10 3 g 73.89 g = 2.64 10 5 mol b. molar mass of CaCl 2 = 110.98 g 1000 g 4. kg = 38. 1 kg 110.98 g c. molar mass of SrCl 2 = 158.52 g 1 g 1. mg = 7.76 10 6 mol 1000 mg 158.52 g d. molar mass of CaSO 4 = 136.15 g 4.75 g 136.15 g = 3.49 10 2 mol e. molar mass of NO 2 = 46.01 g 1 g 96.2 mg = 2.09 10 3 mol 1000 mg 46.01 g f. molar mass of Hg 2 Cl 2 = 472.1 g 12.7 g = 0.0269 mol 472.1 g 35. a. molar mass AlCl 3 = 133.3 g 1.25 mol 133.3 g = 167 g 126

b. molar mass NaHCO 3 = 84.01 g 3.35 mol 84.01 g = 281 g c. molar mass HBr = 80.91 g 80.91 mg 4.25 mmol = 344 mg = 0.344 g 1 mmol d. molar mass U = 8.0 g 1.31 10 3 mol 8.0 g = 0.312 g e. molar mass CO 2 = 44.01 g 0.00104 mol 44.01 g = 0.0458 g f. molar mass Fe = 55.85 g 1.49 10 2 mol Fe 55.85 g = 8.32 10 3 g 36. a. molar mass of H 2 S = 34.09 g 1.2 34.09 g = 41.2 g b. molar mass of Li 2 S = 45.95 g 4.22 10 3 mol 45.95 g = 0.194 g c. molar mass of FeCl 3 = 162.2 g 224 mol 162.2 g = 3.63 10 4 g d. molar mass of Na 2 CO 3 = 105.99 g 105.99 g 7.29 millmol = 0.773 g 3 10 millimol e. molar mass of NaC 2 H 3 O 2 = 82.03 g 8.14 10 3 mol 82.03 g = 6.68 10 5 g f. molar mass of PH 3 = 33.99 g 0.00793 mol 33.99 g = 0.270 g 127

37. a. molar mass of C 2 H 6 O = 46.07 g 0.25 46.07 g = 11.6 g C 2 H 6 O b. molar mass of CO 2 = 44.01 g 1.26 mol 44.01 g = 55.5 g CO 2 c. molar mass of AuCl 3 = 303.4 g 9.31 10 4 mol 303.4 g = 0.282 g AuCl 3 d. molar mass of NaNO 3 = 85.00 g 7.74 mol 85.00 g = 658 g NaNO 3 e. molar mass of Fe = 55.85 g 0.000357 mol 55.85 g = 0.0199 g Fe 38. a. molar mass C 6 H 6 = 78.11 g 0.994 mol 78.11 g = 77.6 g b. molar mass CaH 2 = 42.10 g 4.2 42.10 g = 177 g c. molar mass H 2 O 2 = 34.02 g 1.79 10 4 mol 34.02 g = 6.09 10 3 g d. molar mass C 6 H 12 O 6 = 180.16 g 105.99 g 1.22 mmol = 0.220 g 3 10 mmol e. molar mass Sn = 118.7 g 10.6 mol 118.7 g = 1.26 10 3 g f. molar mass SrF 2 = 125.62 g 0.00030 125.62 g = 0.0378 g 128

39. a. 4.75 millimol = 0.00475 mol 0.00475 mol 6.022 10 molecules b. molar mass of PH 3 = 33.99 g 4.75 g 6.022 10 molecules 33.99 g c. molar mass of Pb(C 2 H 3 O 2 ) 2 = 325.3 g = 2.86 10 2ecules = 8.42 10 22 molecules 1.25 10 2 g 6.022 10 formula units 325.3 g = 2.31 10 19 formula units 6.022 10 formula units d. 1.25 10 2 mol = 7.53 10 21 formula units e. If the sample contains a total of 5.40 mol of carbon, then because each benzene contains six carbons, there must be (5.40/6) = 0.900 mol of benzene present. 6.022 10 molecules 0.900 mol = 5.42 10 molecules 6.022 10 molecules 40. a. 6.37 mol CO = 3.84 10 24 molecules CO b. molar mass of CO = 28.01 g 6.022 10 molecules 6.37 g = 1.37 10 molecules CO 28.01 g c. molar mass of H 2 O = 18.02 g 2.62 10 6 g 6.022 10 molecules 18.02 g = 8.76 10 16 molecules H 2 O 6.022 10 molecules d. 2.62 10 6 g = 1.58 10 18 molec. H 2 O e. molar mass of C 6 H 6 = 78.11 g 6.022 10 molecules 5. g = 4.03 10 22 molecules C 6 H 6 78.11 g 41. a. molar mass of C 2 H 6 O = 46.07 g 1.271 g 46.07 g = 0.02759 mol C 2H 6 O 0.02759 mol C 2 H 6 O 2 mol C = 0.05518 mol C C H O 2 6 129

b. molar mass of C 6 H 4 Cl 2 = 147.0 g 3.982 g 147.0 g = 0.027088 mol C 6H 4 Cl 2 6 mol C 0.027088 mol C 6 H 4 Cl 2 C H Cl 6 4 2 c. molar mass of C 3 O 2 = 68.03 g 0.4438 g 68.03 g = 0.00656 mol C 3O 2 3 mol C 0.00656 mol C 3 O 2 C O 3 2 d. molar mass of CH 2 Cl 2 = 84.93 g 2.910 g 84.93 g = 0.034264 mol CH 2Cl 2 = 0.1625 mol C = 0.01957 mol C 0.034264 mol CH 2 Cl 2 C CH Cl 2 2 = 0.03426 mol C 42. a. molar mass of Na 2 SO 4 = 142.1 g Na 2SO4 S 2.01 g Na 2 SO 4 = 0.014 S 142.1 g Na SO b. molar mass of Na 2 SO 3 = 126.1 g 2 4 Na 2SO3 S 2.01 g Na 2 SO 3 = 0.0159 mol S 126.1 g Na SO c. molar mass of Na 2 S = 78.05 g 2 3 Na 2S S 2.01 g Na 2 S = 0.0258 mol S 78.05 g Na S d. molar mass of Na 2 S 2 O 3 = 158.1 g Na 2S2O3 S 2.01 g Na 2 S 2 O 3 = 0.0127 mol S 158.1 g Na S 43. molar 44. less than 45. a. mass of H present = 1.008 g = 1.008 g mass of Cl present = 35.45 g = 35.45 g mass of O present = 3(16.00 g) = 48.00 g molar mass of HClO 3 = 84.46 g 2 2 130

% H = 1.008 g H 84.46 g 100 = 1.193 %H % Cl = 35.45 g Cl 84.46 g 100 = 41.97 %Cl % O = 48.00 g O 84.46 g 100 = 56.83 %O b. mass of U present = 8.0 g = 8.0 g mass of F present = 4(19.00 g) = 76.00 g molar mass of UF 4 = 314.0 g % U = 8.0 g U 100 = 75.80% U 314.0 g % F = 76.00 g F 314.0 g 100 = 24.20% F c. mass of Ca present = 40.08 g = 40.08 g mass of H present = 2(1.008 g) = 2.016 g molar mass of CaH 2 = 42.10 g % Ca = 40.08 g C 100 = 95.21% Ca 42.10 g % H = 2.016 g H 42.10 g 100 = 4.789% H d. mass of Ag present = 2(107.9 g) = 215.8 g mass of S present = 32.07 g = 32.07 g molar mass of Ag 2 S = 247.9 g 215.8 g Ag % Ag = 100 = 87.06% Ag 247.9 g % S = 32.07 g S 247.9 g 100 = 12.94% S e. mass of Na present = 22.99 g = 22.99 g mass of H present = 1.008 g = 1.008 g mass of S present = 32.07 g = 32.07 g mass of O present = 3(16.00 g) = 48.00 g molar mass of NaHSO 3 = 104.07 g 22.99 g Na % Na = 100 = 22.09% Na 104.07 g 131

% H = 1.008 g H 104.07 g 100 = 0.9686% H % S = 32.07 g S 104.07 g 100 = 30.82% S % O = 48.00 g O 104.07 g 100 = 46.12% O f. mass of Mn present = 54.94 g = 54.94 g mass of O present = 2(16.00 g) = 32.00 g molar mass of MnO 2 = 86.94 g 54.94 g Mn % Mn = 100 = 63.19% Mn 86.94 g % S = 32.00 g O 86.94 g 100 = 36.81% O 46. a. mass of Zn present = 65.38 g mass of O present = 16.00 g molar mass of ZnO = 81.38 g %Zn = 65.38 g Zn 100 = 80.34% Zn 81.38 g %O = 16.00 g O 81.38 g 100 = 19.66% O b. mass of Na present = 2(22.99 g) = 45.98 g mass of S present = 32.07 g molar mass of Na 2 S = 78.05 g 45.98 g Na %Na = 100 = 58.91% Na 78.05 g %S = 32.07 g S 78.05 g 100 = 41.09% S c. mass of Mg present = 24.31 g mass of O present = 2(16.00 g) = 32.00 g mass of H present = 2(1.008 g) = 2.016 g molar mass of Mg(OH) 2 = 58.33 g 24.31 g Mg %Mg = 100 = 41.68% Mg 58.33 g 132

%O = 32.00 g O 58.33 g %H = 2.016 g H 58.33 g 100 = 54.86% O 100 = 3.456% H d. mass of H present = 2(1.008 g) = 2.016 g mass of O present = 2(16.00 g) = 32.00 g molar mass of H 2 O 2 = 34.02 g %H = 2.016 g H 100 = 5.926% H 34.02 g %O = 32.00 g O 34.02 g 100 = 94.06% O e. mass of Ca present = 40.08 g mass of H present = 2(1.008 g) = 2.016 g molar mass of CaH 2 = 42.10 g 40.08 g Ca %Ca = 100 = 95.20% Ca 42.10 g %H = 2.016 g H 42.10 g 100 = 4.789% H f. mass of K present = 2(39.10 g) = 78.20 g mass of O present = 16.00 g molar mass of K 2 O = 94.20 g %K = 78.20 g K 100 =83.01% K 94.20 g %O = 16.00 g O 94.20 g 100 = 16.99% O 47. a. molar mass of CH 4 = 16.04 g % C = 12.01 g C 100 = 74.88% C 16.04 g b. molar mass NaNO 3 = 85.00 g 22.99 g Na % Na = 100 = 27.05% Na 85.00 g c. molar mass of CO = 28.01 g % C = 12.01 g C 100 = 42.88% C 28.01 g 133

d. molar mass of NO 2 = 46.01 g %N = 14.01 g N 100 =30.45% N 46.01 g e. molar mass of C 8 H 18 O = 130.2 g % C = 96.08 g C 100 = 73.79% C 130.2 g f. molar mass of Ca 3 (PO 4 ) 2 = 310.1 g 120.2 g Ca % Ca = 100 = 38.76% Ca 310.1 g g. molar mass of C 12 H 10 O = 170.2 g % C = 144.1 g C 100 = 84.67% C 170.2 g h. molar mass of Al(C 2 H 3 O 2 ) 3 = 204.1 g 26.98 g Al % Al = 100 = 13.22% Al 204.1 g 48. a. molar mass of BaO 2 = 169.3 g 137.3 g Ba % Ba = 100 = 81.10% Ba 169.3 g b. molar mass of BaO = 153.3 g 137.3 g Ba % Ba = 100 = 89.56% Ba 153.3 g c. molar mass of CoBr 2 = 218.73 g 58.93 g Co % Co = 100 = 26.94% Co 218.73 g d. molar mass of CoBr 3 = 298.63 g 58.93 g Co % Co = 100 = 19.73% Co 298.63 g e. molar mass of SnCl 2 = 189.60 g 118.7 g Sn % Sn = 100 = 62.61% Sn 189.60 g f. molar mass of SnCl 4 = 260.50 g 118.7 g Sn % Sn = 100 = 45.57% Sn 260.50 g 134

g. molar mass of LiH = 7.949 g 6.941 g Li % Li = 100 = 87.32% Li 7.949 g h. molar mass of AlH 3 = 30.00 g 26.98 g Al % Al = 100 = 89.93% Al 30.00 g 49. a. molar mass of C 6 H 10 O 4 = 146.1 g % C = 72.06 g C 100 = 49.32% C 146.1 g b. molar mass of NH 4 NO 3 = 80.05 g % N = 28.02 g N 100 = 35.00% N 80.05 g c. molar mass of C 8 H 10 N 4 O 2 = 194.2 g % C = 96.09 g C 100 = 49.47% C 194.2 g d. molar mass of ClO 2 = 67.45 g 35.45 g Cl % Cl = 100 = 52.56% Cl 67.45 g e. molar mass of C 6 H 11 OH = 100.2 g % C = 72.06 g C 100 = 71.92% C 100.2 g f. molar mass of C 6 H 12 O 6 = 180.2 g % C = 72.06 g C 100 = 39.99% C 180.2 g g. molar mass of C 20 H 42 = 282.5 g % C = 240.2 g C 100 = 85.03% C 282.5 g h. molar mass of C 2 H 5 OH = 46.07 g % C = 24.02 g C 100 = 52.14% C 46.07 g 50. a. molar mass of ICl = 162.35 g %I = 126.9 g I 100 = 78.16% I 162.35 g 135

b. molar mass of N 2 O = 44.02 g %N = 28.02 g N 100 = 63.65% N 44.02 g c. molar mass of NO = 30.01 g %N = 14.01 g N 100 = 46.68% N 30.01 g d. molar mass of HgCl 2 = 271.5 g 200.6 g Hg %Hg = 100 = 73.89% Hg 271.5 g e. molar mass of Hg 2 Cl 2 = 472.1 g 401.2 g Hg %Hg = 100 = 84.98% Hg 472.1 g f. molar mass of SF 6 = 146.07 g %S = 32.07 g S 100 = 21.96% S 146.07 g g. molar mass of XeF 2 = 169.3 g 131.3 g Xe %Xe = 100 = 77.55% Xe 169.3 g h. molar mass of MnO 2 = 86.94 g 54.94 g Mn %Mn = 100 = 63.19% Mn 86.94 g 51. a. molar mass of NH 4 I = 144.94 g + NH4I NH4 + 4.25 g = 0.0293 mol NH 4 144.94 g NH I molar mass of NH + 4 = 18.04 g 0.0293 mol NH + 18.04 g NH 4 NH 4 4 + + 4 = 0.529 g NH 4 + + 2 mol NH4 b. 6.3 (NH 4 ) 2 S (NH ) S = 12.6 mol NH 4 + molar mass of NH + 4 = 18.04 g 12.6 mol NH + 18.04 g NH 4 NH 4 4 + + 4 2 = 227 g NH 4 + ion 136

c. molar mass of Ba 3 P 2 = 473.8 g 2+ Ba3P2 3 mol Ba 9.71 g = 0.0615 mol Ba 2+ 473.8 g Ba P molar mass of Ba 2+ = 137.3 g 0.0615 mol Ba 2+ 137.3 g Ba Ba 2+ 2+ 3 2 2+ 3 mol Ca d. 7.63 mol Ca 3 (PO 4 ) 2 Ca (PO ) molar mass of Ca 2+ = 40.08 g 22.9 mol Ca 2+ 40.08 g Ca Ca 2+ 2+ = 8.44 g Ba 2+ 3 4 2 = 918 g Ca 2+ = 22.9 mol Ca 2+ 52. a. molar mass of (NH 4 ) 2 S = 68.15 g; molar mass of S 2 ion = 32.07 g % S 2 2 32.07 g S = 100 = 47.06% S 2 68.15 g NH S 4 b. molar mass of CaCl 2 = 110.98 g; molar mass of Cl = 35.45 g % Cl 70.90 g Cl = 110.98 g CaCl 2 100 = 63.89% Cl c. molar mass of BaO = 153.3 g; molar mass of O 2 ion = 16.00 g % O 2 2 16.00 g O = 100 = 10.44% O 2 153.3 g BaO d. molar mass of NiSO 4 = 154.76 g; molar mass of SO 2 4 ion = 96.07 g % SO 4 2 = 2 96.07 g SO4 154.76 g NiSO 4 100 = 62.08% SO 4 2 53. To determine the empirical formula of a new compound, the composition of the compound by mass must be known. To determine the molecular formula of the compound, the molar mass of the compound must also be known. 54. The empirical formula indicates the smallest whole number ratio of the number and type of atoms present in a molecule. For example, NO 2 and N 2 O 4 both have two oxygen atoms for every nitrogen atom and therefore have the same empirical formula 55. a. NaO b. C 4 H 3 O 2 c. C 12 H 12 N 2 O 3 is already the empirical formula. d. C 2 H 3 Cl 137

56. a. yes (each of these has empirical formula CH) b. no (the number of hydrogen atoms is wrong) c. yes (both have empirical formula NO 2 ) d. no (the number of hydrogen and oxygen atoms is wrong) 57. Assume we have 100.0 g of the compound so that the percentages become masses. 89.56 g Ba = 0.6253 mol Ba 137.3 g 10.44 g O 16.00 g = 0.6525 mol O Dividing both of these numbers of moles by the smaller number of moles (0.65 mol Ba) gives 0.65 mol Ba = 1.000 mol Ba 0.65 mol 0.6525 mol O = 1.000 mol O 0.65 mol The empirical formula is BaO. 58. Assume we have 100.0 g of the compound so that the percentages become masses. 11.64 g N = 0.8308 mol N 14.01 g 88.36 g Cl 35.45 g = 2.493 mol Cl Dividing both of these numbers of moles by the smaller number of moles gives 0.8308 mol N = 1.000 mol N 0.8308 2.493 mol Cl = 3.00 Cl 0.8308 mol The empirical formula is NCl 3. 59. 0.22 g C = 0.01933 mol C 12.01 g 0.05848 g H 1.008 g = 0.05802 mol H 0.3091 g O = 0.01932 mol O 16.00 g Dividing each number of moles by the smallest number of moles (0.01932 mol C) gives 138

0.01933 mol C 0.01932 mol 0.05802 mol H 0.01932 mol = 1.00 C = 3.003 mol H 0.01932 mol O = 1.000 mol O 0.01932 mol The empirical formula is CH 3 O. 60. Assume we have 100.0 g of the compound, so that the percentages become masses. 78.14 g B = 7.228 mol B 10.81 g 21.86 g H 1.008 g = 21.69 mol H Dividing each number of moles by the smaller number of moles gives 7.228 mol B = 1.000 mol B 7.228 mol 21.69 mol H = 3.000 mol H 7.228 mol The empirical formula is BH 3. 61. The mass of chlorine in the reaction is 6.280 1.271 = 5.009 g Cl. 1.271 g Al = 0.0471 Al 26.98 g 5.009 g Cl 35.45 g = 0.1413 mol Cl Dividing each of these number of moles by the smaller (0.0471 Al) gives 0.0471 Al = 1.000 mol Al 0.0471 0.1413 mol Cl = 3.000 mol Cl 0.0471 The empirical formula is AlCl 3. 62. Consider 100.0 g of the compound so that percentages become masses. 45.56 g Sn = 0.3838 mol Sn 118.7 g 54.43 g Cl 35.45 g = 1.535 mol Cl 139

Dividing each number of moles by the smaller number of moles gives 0.3838 mol Sn 0.3838 mol = 1.000 mol Sn 1.535 mol Cl = 3.999 mol Cl 0.3838 mol The empirical formula is SnCl 4. 63. 3.269 g Zn 65.38 g = 0.05000 mol Zn 0.800 g O 16.00 g = 0.0500 mol O Because the two components are present in equal amounts on a molar basis, the empirical formula must be simply ZnO. 64. Consider 100.0 g of the compound. 55.06 g Co = 0.9343 mol Co 58.93 g If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass. 44.94 g S = 1.40 S 32.07 g Dividing each number of moles by the smaller (0.9343 mol Co) gives 0.09343 mol Co = 1.000 mol Co 0.9343 1.40 S = 1.500 mol S 0.9343 mol Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for the compound as Co 2 S 3. 65. The amount of fluorine that reacted with the aluminum sample must be (3.89 g 1.25 g) = 2.64 g of fluorine 1.25 g Al = 0.04633 mol Al 26.98 g 2.64 g F 19.00 g = 0.1389 mol F Dividing each number of moles by the smaller number of moles gives 0.04633 mol Al = 1.000 mol Al 0.04633 mol 140

0.1389 mol F 0.04633 mol = 2.999 mol F The empirical formula is just AlF 3 66. 2.50 g Al 26.98 g = 0.09266 mol Al 5.28 g F 19.00 g = 0.2779 mol F Dividing each number of moles by the smaller number of moles gives 0.09266 mol Al = 1.000 mol Al 0.09266 mol 0.2779 mol F = 2.999 mol F 0.09266 mol The empirical formula is just AlF 3. Note the similarity between this problem and question 65: they differ in the way the data is given. In question 65, you were given the mass of the product, and first had to calculate how much fluorine had reacted. 67. Consider 100.0 g of the compound. 67.61 g U = 0.284 U 8.0 g 32.39 g F 19.00 g = 1.705 mol F Dividing each number of moles by the smaller number of moles (0.284 U) gives 0.284 U = 1.000 mol U 0.284 1.705 mol F = 6.000 mol F 0.284 The empirical formula is UF 6. 68. Consider 100.0 g of the compound so that percentages become masses. 46.46 g Li = 6.694 mol Li 6.941 g 53.54 g O 16.00 g = 3.346 mol O Dividing each number of moles by the smaller number of moles gives 6.694 mol Li = 2.00 Li 3.346 mol 141

3.346 mol O = 1.000 mol O 3.346 mol The empirical formula is Li 2 O 69. Consider 100.0 g of the compound. 33.88 g Cu = 0.533 Cu 63.55 g 14.94 g N 51.18 g O 14.01 g 16.00 g = 1.066 mol N = 3.199 mol O Dividing each number of moles by the smallest number of moles (0.533 Cu) gives 0.533 Cu = 1.000 mol Cu 0.533 1.066 mol N 0.533 = 2.000 mol N 3.199 mol O = 6.00 O 0.533 The empirical formula is CuN 2 O 6 [i.e., Cu(NO 3 ) 2 ]. 70. Consider 100.0 g of the compound. 59.78 g Li = 8.613 mol Li 6.941 g 40.22 g N 14.01 g = 2.87 N Dividing each number of moles by the smaller number of moles (2.87 N) gives 8.613 mol Li = 3.000 mol Li 2.87 2.87 N = 1.000 mol N 2.87 The empirical formula is Li 3 N. 71. Consider 100.0 g of the compound. 66.75 g Cu = 1.050 mol Cu 63.55 g 10.84 g P 30.97 g = 0.3500 mol P 142

22.41 g O 16.00 g = 1.40 O Dividing each number of moles by the smallest number of moles (0.3500 mol P) gives 1.050 mol Cu = 3.000 mol Cu 0.3500 mol 0.3500 mol P 0.3500 = 1.000 mol P 1.40 O = 4.003 mol O 0.3500 mol The empirical formula is thus Cu 3 PO 4. 72. Consider 100.0 g of the compound so that percentages become masses. 71.06 g Co = 1.206 mol Co 58.93 g 28.94 g O 16.00 g = 1.809 mol O Dividing each number of moles by the smaller number of moles gives 1.206 mol Co = 1.000 mol Co 1.206 mol 1.809 mol O = 1.500 mol O 1.206 mol Multiplying these relative numbers of moles by 2 to give whole numbers gives the empirical formula as Co 2 O 3 73. The compound must contain 1.00 mg of lithium and 2.73 mg of fluorine. 1 mmol 1.00 mg Li = 0.144 mmol Li 6.941 mg 2.73 mg F 1 mmol 19.00 mg = 0.144 mol F The empirical formula of the compound is LiF. 74. Compound 1: Assume 100.0 g of the compound. 22.55 g P = 0.728 P 30.97 g 77.45 g Cl 35.45 g = 2.185 mol Cl Dividing each number of moles by the smaller (0.728 P) indicates that the formula of Compound 1 is PCl 3. 143

Compound 2: Assume 100.0 g of the compound. 14.87 g P = 0.480 P 30.97 g 85.13 g Cl 35.45 g = 2.40 Cl Dividing each number of moles by the smaller (0.480 P) indicates that the formula of Compound 2 is PCl 5. 75. The empirical formula of a compound represents only the smallest whole number relationship between the number and type of atoms in a compound, whereas the molecular formula represents the actual number of atoms of each type in a true molecule of the substance. Many compounds (for example, H 2 O) may have the same empirical and molecular formulas. 76. If only the empirical formula is known, the molar mass of the substance must be determined before the molecular formula can be calculated. 77. Assume that we have 100.00 g of the compound: then 78.14 g will be boron and 21.86 g will be hydrogen. 78.14 g B = 7.228 mol B 10.81 g 21.86 g H 1.008 g = 21.69 mol H Dividing each number of moles by the smaller number (7.228 mol B) gives the empirical formula as BH 3. The empirical molar mass of BH 3 would be [10.81 g + 3(1.008 g)] = 13.83 g. This is approximately half of the indicated actual molar mass, and therefore the molecular formula must be B 2 H 6. 78. empirical formula mass of CH = 13 g molar mass n = empirical formula mass = 78 g 13 g = 6 The molecular formula is (CH) 6 or C 6 H 6. 79. empirical formula mass of CH 2 = 14 molar mass n = empirical formula mass = 84 g 14 g = 6 molecular formula is (CH 2 ) 6 = C 6 H 12. 80. empirical formula mass of C 2 H 5 O = 46 g molar mass n = empirical formula mass = 90 g ~2 46 g = molecular formula is (C 2 H 5 O) 2 = C 4 H 10 O 2 144

81. Consider 100.0 g of the compound. 42.87 g C = 3.570 mol C 12.01 g 3.598 g H 28.55 g O 25.00 g N 1.008 g 16.00 g 14.01 g = 3.569 mol H = 1.784 mol O = 1.784 mol N Dividing each number of moles by the smallest number of moles (1.784 mol O or N) gives 3.570 mol C = 2.00 C 1.784 mol 3.569 mol H 1.784 mol 1.784 mol O 1.784 mol = 2.00 H = 1.000 mol O 1.784 mol N = 1.000 mol N 1.784 mol The empirical formula of the compound is C 2 H 2 ON and the empirical formula mass of C 2 H 2 ON = 56. n = molar mass empirical formula mass = 168 g 56 g = 3 The molecular formula is (C 2 H 2 ON) 3 = C 6 H 6 O 3 N 3. 82. For NO 2 : molar mass = 14.01 + 2(16.00) = 46.01 g %N = 14.01 g N 100 = 30.45 %N 46.01 g %O = 2(16.00 g O) 100 = 69.55 %O 46.01 g For N 2 O 4 : molar mass = 2(14.01 g) + 4(16.00 g) = 92.02 g %N = 2(14.01 g N) 100 = 30.45 %N 92.02 g %O = 4(16.00 g O) 100 = 69.55 %O 92.02 g 145

83. [1] c [6] d [2] e [7] a [3] j [8] g [4] h [9] i [5] b [10] f 84. 5.00 g Al 0.185 mol 1.12 10 atoms 0.140 g Fe 0.00250 mol 1.51 10 21 atoms 2.7 10 2 g Cu 4.3 mol 2.6 10 24 atoms 0.00250 g Mg 1.03 10 4 mol 6.19 10 19 atoms 0.062 g Na 2.7 10 3 mol 1.6 10 21 atoms 3.95 10 18 g U 1.66 10 20 mol 1.00 10 4 atoms 85. 4.24 g 0.0543 mol 3.27 10 22 molec. 3.92 10 atoms 4.04 g 0.224 mol 1.35 10 molec. 4.05 10 atoms 1.98 g 0.0450 mol 2.71 10 22 molec. 8.13 10 22 atoms 45.9 g 1.26 mol 7.59 10 molec. 1.52 10 24 atoms 126 g 6.99 mol 4.21 10 24 molec. 1.26 10 25 atoms 0.297 g 0.00927 mol 5.58 10 2ec. 3.35 10 22 atoms 86. mass of 2 mol X = 2(41.2 g) = 82.4 g mass of Y = 57.7 g = 57.7 g mass of 3 mol Z = 3(63.9 g) = 191.7 g molar mass of X 2 YZ 3 = 331.8 g %X = 82.4 g 100 = 24.8 %X 331.8 g % Y = 57.7 g 100 331.8 g = 17.4% Y % Z = 191.7 g 100 331.8 g = 57.8% Z If the molecular formula were actually X 4 Y 2 Z 6, the percentage composition would be the same, and the relative mass of each element present would not change. The molecular formula is always a whole number multiple of the empirical formula. 87. magnesium/nitrogen compound: mass of nitrogen = 1.2791 g 0.9240 = 0.3551 g N 0.9240 g Mg = 0.0380 Mg 24.31 g 0.3551 g N 14.01 g = 0.02535 mol N Dividing each number of moles by the smaller number of moles gives 146

0.0380 Mg 0.02535 mol = 1.499 mol Mg 0.02535 mol N = 1.000 mol N 0.02535 mol Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg 3 N 2. magnesium/oxygen compound: Consider 100.0 g of the compound. 60.31 g Mg = 2.48 Mg 24.31 g 39.69 g O 16.00 g = 2.48 O The empirical formula is MgO. 88. For the first compound (restricted amount of oxygen) 2.118 g Cu = 0.03333 mol Cu 63.55 g 0.2666 g O 16.00 g = 0.01666 mol O Since the number of moles of Cu (0.03333 mol) is twice the number of moles of O (0.01666 mol), the empirical formula is Cu 2 O. For the second compound (stream of pure oxygen) 2.118 g Cu = 0.03333 mol Cu 63.55 g 0.5332 g O 16.00 g = 0.03333 mol O Since the numbers of moles are the same, the empirical formula is CuO. 89. Compound Molar Mass % H HF 20.01 g 5.037% HCl 36.46 g 2.765% HBr 80.91 g 1.246% HI 127.9 g 0.7881% 90. a. molar mass H 2 O = 18.02 g 4.21 g 6.022 10 molecules = 1.41 10 molecules 18.02 g The sample contains 1.41 10 oxygen atoms and 2(1.41 10 ) = 2.82 10 hydrogen atoms. 147

b. molar mass CO 2 = 44.01 g 6.81 g 6.022 10 molecules = 9.32 10 22 molecules 44.01 g The sample contains 9.32 10 22 carbon atoms and 2(9.32 10 22 ) = 1.86 10 oxygen atoms. c. molar mass C 6 H 6 = 78.11 g 6.022 10 molecules 0.000221 g = 1.70 10 18 molec. 78.11 g The sample contains 6(1.70 10 18 ) = 1.02 10 19 atoms of each element. 6.022 10 molecules d. 2.26 mol = 1.36 10 24 molecules atoms C = 12(1.36 10 24 ) = 1.63 10 25 atoms atoms H = 22(1.36 10 24 ) = 2.99 10 25 atoms atoms O = 11(1.36 10 24 ) = 1.50 10 25 atoms 91. a. Assuming 10,000,000,000 = 1.0 10 10 1.0 10 10 28.02 g N2 molecules = 4.7 10 13 g N 2 6.022 10 molecules b. 2.49 10 20 44.01 g CO2 molecules = 0.0182 g CO 2 6.022 10 molecules c. 7.0983 mol NaCl 58.44 g d. 9.012 10 6 mol C 2 H 4 Cl 2 98.95 g = 414.8 g NaCl = 8.918 10 4 g C 2 H 4 Cl 2 92. a. molar mass of C 3 O 2 = 3(12.01 g) + 2(16.00 g) = 68.03 g % C = 36.03 g C 100 = 52.96% C 68.03 g 7.819 g C 3 O 2 52.96 g C 100.0 g C O = 4.141 g C 3 2 4.141 g C 6.022 10 molecules 12.01 g C = 2.076 10 C atoms b. molar mass of CO = 12.01 g + 16.00 g = 28.01 g % C = 12.01 g C 100 = 42.88% C 28.01 g 148

1.53 10 2ecules CO 1 C atom ecule CO = 1.53 1021 C atoms 1.53 10 21 12.01 g C C atoms = 0.0305 g C 6.022 10 C atoms c. molar mass of C 6 H 6 O = 6(12.01 g) + 6(1.008 g) + 16.00 g = 94.11 g % C = 72.06 g C 100 = 76.57% C 94.11 g 0.200 mol C 6 H 6 O 6 mol C C H O 6 6 = 1.20 mol C 1.20 mol C 12.01 g = 14.4 g C 14.4 g C 6.022 10 C atoms 12.01 g C = 7.22 10 C atoms 93. [1] g [6] i [2] c [7] f [3] b [8] h [4] a [9] e [5] j [10] d 94. 2.24 g Co 55.85 g Fe 58.93 g Co = 2.12 g Fe 95. 2.24 g Fe 58.93 g Co 55.85 g Fe = 2.36 g Co 96. 5.00 g Te 200.6 g Hg 127.6 g Te = 7.86 g Hg 97. 1.00 kg = 1.00 10 3 g 1.00 10 3 6.941 g Li g Zr 91.22 g Zr = 76.1 g Li 98. 153.8 g CCl 4 = 6.022 10 molecules CCl 4 153.8 g CCl4 ecule = 2.554 10 22 g 6.022 10 molecules 149

99. a. molar mass of C 6 H 6 = 78.11 g 2.500 g 6.048 g H 78.11 g = 0.1936 g H b. molar mass of CaH 2 = 42.10 g 2.500 g 2.016 g H = 0.1197 g H 42.10 g c. molar mass of C 2 H 5 OH = 46.07 g 2.500 g 6.048 g H 46.07 g = 0.3282 g H d. molar mass of C 3 H 7 O 3 N = 105.1 g 2.500 g 7.056 g H = 0.1678 g H 105.1 g 100. a. molar mass of C 2 H 5 O 2 N = 2(12.01 g) + 5(1.008 g) + 2(16.00 g) + 14.01 g = 75.07 g 5.000 g 14.01 g N 75.07 g = 0.9331 g N b. molar mass of Mg 3 N 2 = 3(24.31 g) + 2(14.01 g) = 100.95 g 5.000 g 28.02 g N 100.95 g = 1.388 g N c. molar mass of Ca(NO 3 ) 2 = 40.08 g + 2(14.01 g) + 6(16.00 g) = 164.10 g 5.000 g 28.02 g N 164.10 g = 0.8537 g N d. molar mass of N 2 O 4 = 2(14.01 g) + 4(16.00 g) = 92.02 g 5.000 g 28.02 g N = 1.522 g N 92.02 g 101. Consider 100.0 g of the compound. 25.45 g Cu = 0.4005 mol Cu 63.55 g 12.84 g S 4.036 g H 32.07 g 1.008 g = 0.4004 mol S = 4.004 mol H 57.67 g O 16.00 g = 3.604 mol O 150

Dividing each number of moles by the smallest number of moles (0.4004 mol) gives 0.4005 mol Cu 0.4004 mol 0.4004 mol S 0.4004 mol 4.004 mol H 0.4004 mol = 1.000 mol Cu = 1.000 mol S = 10.00 mol H 3.604 mol O = 9.00 O 0.4004 mol The empirical formula is CuSH 10 O 9 (which is usually written as CuSO 4 5H 2 O). 102. Consider 100.0 g of the compound. 16.39 g Mg = 0.6742 mol Mg 24.31 g 18.89 g N 64.72 g O 14.01 g 16.00 g = 1.348 mol N = 4.045 mol O Dividing each number of moles by the smallest number of moles 0.6742 mol Mg = 1.000 mol Mg 0.6742 mol 1.348 mol N 0.6742 mol = 1.999 mol N 4.045 mol O = 5.999 mol O 0.6742 mol The empirical formula is MgN 2 O 6 [i.e., Mg(NO 3 ) 2 ]. 103. atomic mass unit (amu) 104. We use the average mass because this average is a weighted average and takes into account both the masses and the relative abundances of the various isotopes. 105. a. 160,000 amu 1 O atom 16.00 amu = 1.0 104 O atoms (assuming exact) b. 8139.81 amu 1 N atom 14.01 amu c. 13,490 amu 1 Al atom 26.98 amu = 581 N atoms = 500 Al atoms 151

d. 5040 amu 1 H atom 1.008 amu = 5.00 103 H atoms e. 367,495.15 amu 106. 1.98 10 13 amu 3.01 10 Na atoms 1 Na atom 22.99 amu = 1.599 104 Na atoms 1 Na atom 22.99 amu = 8.61 1011 Na atoms 22.99 amu 1 Na atom = 6.92 1024 amu 107. a. 1.5 mg = 0.0015 g 0.0015 g Cr 52.00 g = 2.9 10 5 mol Cr b. 2.0 10 3 g Sr c. 4.84 10 4 g B d. 3.6 10 6 μg = 3.6 10 12 g 87.62 g = 2.3 10 5 mol Sr 10.81 g = 4.48 103 mol B 3.6 10 12 g Cf 251 g = 1.4 10 14 mol Cf e. 2000 lb 453.59 g 1 lb 9.1 10 5 g Fe = 9.1 10 5 g 55.85 g = 1.6 104 mol Fe f. 20.4 g Ba g. 62.8 g Co 137.3 g 58.93 g = 0.149 mol Ba = 1.07 mol Co 108. a. 5.0 mol K 39.10 g b. 0.000305 mol Hg 200.6 g c. 2.31 10 5 mol Mn 54.94 g d. 10.5 mol P 30.97 g = 195 g = 2.0 10 2 g K = 325 g P = 0.0612 g Hg = 1.27 10 3 g Mn 152

e. 4.9 10 4 mol Fe 55.85 g f. 125 mol Li 6.941 g g. 0.01205 mol F 19.00 g = 868 g Li = 2.7 10 6 g Fe = 0.2290 g F 109. a. 2.89 g Au 6.022 10 Au atoms 197.0 g Au = 8.83 10 21 Au atoms b. 0.000259 mol Pt c. 0.000259 g Pt 6.022 10 Pt atoms 6.022 10 Pt atoms 195.1 g Pt = 1.56 10 20 Pt atoms = 7.99 10 17 Pt atoms d. 2.0 lb 453.59 g 1 lb = 908 g 908 g Mg 6.022 10 Mg atoms 24.31 g Mg = 2.2 10 25 Mg atoms e. 1.90 ml 13.6 g 1 ml = 25.8 g Hg 25.8 g Hg 6.022 10 Hg atoms 200.6 g Hg = 7.75 10 22 Hg atoms f. 4.30 mol W g. 4.30 g W 6.022 10 W atoms 6.022 10 W atoms 183.9 g W 110. a. mass of Fe = 1(55.85 g) = 55.85 g mass of S = 1(32.07 g) = 32.07 g mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of FeSO 4 = 151.92 g b. mass of Hg = 1(200.6 g) = 200.6 g mass of 2 mol I = 2(126.9 g) = 253.8 g molar mass of HgI 2 = 454.4 g c. mass of Sn = 1(118.7 g) = 118.7 g mass of 2 mol O = 2(16.00 g) = 32.00 g molar mass of SnO 2 = 150.7 g = 2.59 10 24 W atoms = 1.41 10 22 W atoms 153

d. mass of Co = 1(58.93 g) = 58.93 g mass of 2 mol Cl = 2(35.45 g) = 70.90 g molar mass of CoCl 2 = 129.83 g e. mass of Cu = 1(63.55 g) = 63.55 g mass of 2 mol N = 2(14.01 g) = 28.02 g mass of 6 mol O = 6(16.00 g) = 96.00 g molar mass of Cu(NO 3 ) 2 = 187.57 g 111. a. mass of 6 mol C = 6(12.01 g) = 72.06 g mass of 10 mol H = 10(1.008 g) = 10.08 g mass of 4 mol O = 4(16.00 g) = 64.00 g molar mass of C 6 H 10 O 4 = 146.14 g b. mass of 8 mol C = 8(12.01 g) = 96.08 g mass of 10 mol H = 10(1.008 g) = 10.08 g mass of 4 mol N = 4(14.01 g) = 56.04 g mass of 2 mol O = 1(16.00 g) = 32.00 g molar mass of C 8 H 10 N 4 O 2 = 194.20 g c. mass of 20 mol C = 20(12.01 g) = 240.2 g mass of 42 mol H = 42(1.008 g) = 42.34 g molar mass of C 20 H 42 = 282.5 g d. mass of 6 mol C = 6(12.01 g) = 72.06 g mass of 12 mol H = 12(1.008 g) = 12.10 g mass of O = 16.00 g = 16.00 g molar mass of C 6 H 11 OH = 100.16 g e. mass of 4 mol C = 4(12.01 g) = 48.04 g mass of 6 mol H = 6(1.008 g) = 6.048 g mass of 2 mol O = 2(16.00 g) = 32.00 g molar mass of C 4 H 6 O 2 = 86.09 g f. mass of 6 mol C = 6(12.01 g) = 72.06 g mass of 12 mol H = 12(1.008 g) = 12.10 g mass of 6 mol O = 6(16.00 g) = 96.00 g molar mass of C 6 H 12 O 6 = 180.16 g 112. a. molar mass of (NH 4 ) 2 S = 68.15 g 21.2 g 68.15 g = 0.31 (NH 4) 2 S 154

b. molar mass of Ca(NO 3 ) 2 = 164.1 g 44.3 g 164.1 g = 0.270 mol Ca(NO 3) 2 c. molar mass of Cl 2 O = 86.9 g 4.35 g 86.9 g = 0.050 Cl 2O d. 1.0 lb = 454 g; molar mass of FeCl 3 = 162.2 454 g 162.2 g = 2.8 mol FeCl 3 e. 1.0 kg = 1.0 10 3 g; molar mass of FeCl 3 = 162.2 g 1.0 10 3 g 162.2 g = 6.2 mol FeCl 3 113. a. molar mass of FeSO 4 = 151.92 g 1.28 g 151.92 g = 8.43 10 3 mol FeSO 4 b. 5.14 mg = 0.00514 g; molar mass of HgI 2 = 454.4 g 0.00514 g 454.4 g = 1.13 10 5 mol HgI 2 c. 9.21 μg = 9.21 10 6 g; molar mass of SnO 2 = 150.7 g 9.21 10 6 g 150.7 g = 6.11 10 8 mol SnO 2 d. 1.26 lb = 1.26(453.59 g) = 572 g; molar mass of CoCl 2 = 129.83 g 572 g 129.83 g = 4.4 CoCl 2 e. molar mass of Cu(NO 3 ) 2 = 187.57 g 4.25 g 187.57 g = 2.27 10 2 mol Cu(NO 3 ) 2 114. a. molar mass of CuSO 4 = 159.62 g 2.6 10 2 mol 159.62 g = 4.2 g CuSO 4 b. molar mass of C 2 F 4 = 100.0 g 3.05 10 3 mol 100.0 g = 3.05 10 5 g C 2 F 4 c. 7.83 mmol = 0.00783 mol; molar mass of C 5 H 8 = 68.11 g 155

0.00783 mol 68.11 g = 0.533 g C 5 H 8 d. molar mass of BiCl 3 = 315.3 g 6.30 mol 315.3 g = 1.99 10 3 g BiCl 3 e. molar mass of C 12 H 22 O 11 = 342.3 g 12.2 mol 342.3 g = 4.18 10 3 g C 12 H 22 O 11 115. a. molar mass of (NH 4 ) 2 CO 3 = 96.09 g 3.09 mol 96.09 g = 297 g (NH 4 ) 2 CO 3 b. molar mass of NaHCO 3 = 84.01 g 4.01 10 6 mol 84.01 g = 3.37 10 4 g NaHCO 3 c. molar mass of CO 2 = 44.01 g 88.02 mol 44.01 g = 3874 g CO 2 d. 1.29 mmol = 0.00129 mol; molar mass of AgNO 3 = 169.9 g 0.00129 mol 169.9 g = 0.219 g AgNO 3 e. molar mass of CrCl 3 = 158.4 g 0.0024 mol 158.4 g = 0.38 g CrCl 3 116. a. molar mass of C 6 H 12 O 6 = 180.2 g 3.45 g 6.022 10 molecules 180.2 g = 1.15 10 22 molecules C 6 H 12 O 6 b. 3.45 mol 6.022 10 molecules c. molar mass of ICl 5 = 304.2 g 25.0 g 6.022 10 molecules 304.2 g d. molar mass of B 2 H 6 = 27.67 g = 2.08 10 24 molecules C 6 H 12 O 6 = 4.95 10 22 molecules ICl 5 1.00 g 6.022 10 molecules 27.67 g = 2.18 10 22 molecules B 2 H 6 156

e. 1.05 mmol = 0.00105 mol 0.00105 mol 6.022 10 formula units = 6.32 10 20 formula units 117. a. molar mass of NH 3 = 17.03 g 2.71 g 17.03 g = 0.159 mol NH 3 mol H = 3(0.159 mol) = 0.477 mol H b. mol H = 2(0.824 mol) = 1.648 mol H = 1.65 mol H c. 6.25 mg = 0.00625 g; molar mass of H 2 SO 4 = 98.09 g 0.00625 g 98.09 g = 6.37 10 5 mol H 2 SO 4 mol H = 2(6.37 10 5 mol) = 1.27 10 4 mol H d. molar mass of (NH 4 ) 2 CO 3 = 96.09 g 451 g 96.09 g = 4.69 mol (NH 4) 2 CO 3 mol H = 8(4.69 mol) = 37.5 mol H 118. a. mass of Ca present = 3(40.08 g) = 120.24 g mass of P present = 2(30.97 g) = 61.94 g mass of O present = 8(16.00 g) = 128.00 g molar mass of Ca 3 (PO 4 ) 2 = 310.18 g 120.24 g Ca % Ca = 310.18 g 100 = 38.76% Ca % P = 61.94 g P 310.18 g 100 = 19.97% P % O = 128.00 g O 310.18 g 100 = 41.27% O b. mass of Cd present = 112.4 g = 112.4 g mass of S present = 32.07 g = 32.07 g mass of O present = 4(16.00 g) = 64.00 g molar mass of CdSO 4 = 208.5 g 112.4 g Cd % Cd = 100 = 53.91% Cd 208.5 g % S = 32.07 g S 208.5 g 100 = 15.38% S 157

% O = 64.00 g O 208.5 g 100 = 30.70% O c. mass of Fe present = 2(55.85 g) = 111.7 g mass of S present = 3(32.07 g) = 96.21 g mass of O present = 12(16.00 g) = 192.0 g molar mass of Fe 2 (SO 4 ) 3 = 399.9 g 111.7 g Fe % Fe = 100 = 27.93% Fe 399.9 g % S = 96.21 g S 399.9 g % O = 192.0 g O 399.9 g 100 = 24.06% S 100 = 48.01% O d. mass of Mn present = 54.94 g = 54.94 g mass of Cl present = 2(35.45 g) = 70.90 g molar mass of MnCl 2 = 125.84 g 54.94 g Mn % Mn = 100 = 43.66% Mn 125.84 g % Cl = 70.90 g Cl 125.84 g 100 = 56.34% Cl e. mass of N present = 2(14.01 g) = 28.02 g mass of H present = 8(1.008 g) = 8.064 g mass of C present = 12.01 g = 12.01 g mass of O present = 3(16.00 g) = 48.00 g molar mass of (NH 4 ) 2 CO 3 = 96.09 g % N = 28.02 g N 100 = 29.16% N 96.09 g % H = 8.064 g H 96.09 g % C = 12.01 g C 96.09 g 100 = 8.392% H 100 = 12.50% C % O = 48.00 g O 96.09 g 100 = 49.95% O f. mass of Na present = 22.99 g = 22.99 g mass of H present = 1.008 g = 1.008 g 158

mass of C present = 12.01 g = 12.01 g mass of O present = 3(16.00 g) = 48.00 g molar mass of NaHCO 3 = 84.01 g 22.99 g Na % Na = 100 = 27.37% Na 84.01 g % H = 1.008 g H 84.01 g % C = 12.01 g C 84.01 g % O = 48.00 g O 84.01 g 100 = 1.200% H 100 = 14.30% C 100 = 57.14% O g. mass of C present = 12.01 g = 12.01 g mass of O present = 2(16.00 g) = 32.00 g molar mass of CO 2 = 44.01 g % C = 12.01 g C 100 = 27.29% C 44.01 g % O = 32.00 g O 44.01 g 100 = 72.71% O h. mass of Ag present = 107.9 g = 107.9 g mass of N present = 14.01 g = 14.01 g mass of O present = 3(16.00 g) = 48.00 g molar mass of AgNO 3 = 169.9 g 107.9 g Ag % Ag = 100 = 63.51% Ag 169.9 g % N = 14.01 g N 169.9 g % O = 48.00 g O 169.9 g 100 = 8.246% N 100 = 28.25% O 119. a. molar mass of NaN 3 = 65.02 g 22.99 g Na % Na = 100 = 35.36% Na 65.02 g b. molar mass of CuSO 4 = 159.62 g 63.55 g Cu % Cu = 100 = 39.81% Cu 159.62 g 159

c. molar mass of AuCl 3 = 303.4 g 197.0 g Au % Au = 100 = 64.93% Au 303.4 g d. molar mass of AgNO 3 = 169.9 g 107.9 g Ag % Ag = 100 = 63.51% Ag 169.9 g e. molar mass of Rb 2 SO 4 = 267.0 g 170.9 g Rb % Rb = 100 = 64.01% Rb 267.0 g f. molar mass of NaClO 3 = 106.44 g 22.99 g Na % Na = 100 = 21.60% Na 106.44 g g. molar mass of NI 3 = 394.7 g % N = 14.01 g N 100 = 3.550% N 394.7 g h. molar mass of CsBr = 212.8 g 132.9 g Cs % Cs = 100 = 62.45% Cs 212.8 g 120. a. % Fe = b. % Ag = 55.85 g Fe 151.92 g 215.8 g Ag 1.8 g 100 = 36.76% Fe 100 = 93.10% Ag c. % Sr = 87.62 g Sr 158.5 g 100 = 55.28% Sr d. % C = 48.04 g C 86.09 g e. % C = 12.01 g C 32.04 g 100 = 55.80% C 100 = 37.48% C f. % Al = 53.96 g Al 101.96 g 100 = 52.92% Al g. % K = 39.10 g K 106.55 g 100 = 36.70% K h. % K = 39.10 g K 74.55 g 100 = 52.45% K 160

121. 0.78 g C 12.01 g = 0.06027 mol C 0.07088 g H 1.008 g = 0.07032 mol H 0.1407 g N 0.3214 g O 14.01 g 16.00 g = 0.01004 mol N = 0.02009 mol O Dividing each number of moles by the smallest number of moles (0.01004 mol) gives 0.06027 mol C 0.01004 = 6.003 mol C 0.01004 mol N = 1.000 mol N 0.01004 The empirical formula is C 6 H 7 NO 2. 0.07032 mol H 0.01004 0.02009 mol O 0.01004 = 7.004 mol H = 2.00 O 122. 0.2990 g C 12.01 g = 0.02490 mol C 0.05849 g H 1.008 g = 0.05803 mol H 0.18 g N 0.1328 g O 14.01 g 16.00 g = 0.01655 mol N = 0.008300 mol O Dividing each number of moles by the smallest number of moles (0.008300 mol) gives 0.02490 mol C 0.008300 = 3.000 mol C 0.01655 mol N = 1.994 mol N 0.008300 The empirical formula is C 3 H 7 N 2 O. 0.05803 mol H 0.008300 0.008300 mol O 0.008300 = 6.992 mol H = 1.000 mol O 1. 2.004 g Ca 0.4670 g N 40.08 g 14.01 g = 0.05000 mol Ca = 0.03333 mol N Dividing each number of moles by the smaller number of moles gives 161

0.05000 mol Ca 0.03333 mol = 1.500 mol Ca 0.03333 mol N = 1.000 mol N 0.03333 mol Multiplying these relative numbers of moles by 2 to give whole numbers gives the empirical formula as Ca 3 N 2. 124. Mass of oxygen in compound = 4.33 g 4.01 g = 0.32 g O 4.01 g Hg 200.6 g = 0.0200 mol Hg 0.32 g O 16.00 g = 0.020 mol O Since the numbers of moles are equal, the empirical formula is HgO. 125. Mass of chlorine in compound = 3.045 g 1.00 g = 2.045 g Cl 1.00 g Cr 52.00 g = 0.0192 mol Cr 2.045 g Cl 35.45 g = 0.0577 mol Cl Dividing each number of moles by the smaller number of moles 0.0192 mol Cr 0.0192 = 1.00 mol Cr 0.0577 mol Cl = 3.0 Cl 0.0192 mol The empirical formula is CrCl 3. 126. Assume we have 100.0 g of the compound. 65.95 g Ba 34.05 g Cl 137.3 g 35.45 g = 0.4803 mol Ba = 0.9605 mol Cl Dividing each of these number of moles by the smaller number gives 0.4803 mol Ba 0.4803 mol = 1.000 mol Ba 0.9605 mol Cl 0.4803 mol = 2.000 mol Cl The empirical formula is then BaCl 2. 162