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Chapter 6 Chemical Composition

0.110 g 1. 100 washers  = 11.0 g (assuming 100 washers is exact) 1 washer

1 washer 100. g  = 909 washers 0.110 g

1 cork 2. 500. g  = 306.7 = 307 corks 1.63 g

1 stopper 500. g  = 116 stoppers 4.31 g

1 cork 1 kg (1000 g) of corks contains (1000 g  ) = 613.49 = 613 corks 1.63 g

4.31 g 613 stoppers would weigh (613 stoppers  ) = 2644 g = 2640 g 1 stopper The ratio of the mass of a stopper to the mass of a cork is (4.31 g/1.63 g). So the mass of stoppers that contains the same number of stoppers as there are corks in 1000 g of corks is 4.31 g 1000 g  = 2644 g = 2640 g 1.63 g

3. The atomic mass unit (amu) is a unit of mass 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. We use the average atomic mass of an element when performing calculations because the average mass takes into account the individual masses and relative abundance of all the isotopes of an element.

1.008 amu 5. a. 635 H  = 640. amu 1 H

183.9 amu b. 1.261  104 W atoms  = 2.319  106 amu 1 W atom

39.10 amu c. 42 K atoms  = 1642 amu 1 K atom

14.01 amu d. 7.213  1023 N atoms  = 1.011  1025 amu 1 N atom

55.85 amu e. 891 Fe atoms  = 4.976  104 amu 1 Fe atom

1 B atom 6. a. 10.81 amu  = 1.000 atom = 1 B atom 10.81 amu

1 S atom b. 320.7 amu  = 10 S atoms 32.07 amu

1 Au atom c. 19,691 amu  = 100.00 Au atoms = 100 Au atoms 197.97 amu

1 Xe atom d. 19,695 amu  = 150.0 Xe atoms = 150 Xe atoms 131.3 amu

1 Al atom e. 3588.3 amu  = 133.0 Al atoms = 133 Al atoms 26.98 amu

1 S atom 7. 8274 amu  = 258 S atoms 32.07 amu

32.07 amu 5.213  1024 S atoms  = 1.672  1026 amu 1 S atom

8. 1.204  1024

9. Avogadro’s number (6.022  1023)

10. The ratio of the atomic mass of Ca to the atomic mass of Mg is (40.08 amu/24.31 amu), and the masses of calcium are given by 40.08 amu 12.16 g Mg  = 20.05 g Ca 24.31 amu

40.08 amu 24.31 g Mg  = 40.08 g Ca 24.31 amu

11. 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 58.93 amu 57.0 g F  = 177 g Co 19.00 amu

12. 1 mol O = 16.00 g O = 6.02  1023 O atoms 16.00 g O 1 O atom  = 2.66  10–23 g O 6.022 x 1023 O atoms

16.00 g O 13. 0.50 mol O atoms  = 8.0 g O 1 mol

1.008 g 4 mol H atoms  = 4 g H 1 mol Half a mole of O atoms weighs more than 4 moles of H atoms.

1 mol Au 14. a. 26.2 g Au  = 0.133 mol Au 197.0 g

1 mol Ca b. 41.5 g Ca  = 1.04 mol Ca 40.08 g

1 g 1 mol Ba c. 335 mg Ba   = 2.44  10–3 mol Ba 103 mg 137.3 g

1 mol Pd d. 1.42  10–3 g Pd  = 1.33  10–5 mol Pd 106.4 g

1 g 1 mol Ni e. 3.05  10–5 g Ni   = 5.20  10–13 mol Ni 106 g 58.70 g

453.59 g 1 mol Fe f. 1.00 lb Fe   = 8.12 mol Fe 1 lb 55.85 g 1 mol C g. 12.01 g C  = 1.000 mol C 12.01 g

55.85 g 15. a. 2.00 mol Fe  = 112 g Fe 1 mol 58.70 g b. 0.521 mol Ni  = 30.6 g Ni 1 mol 195.1 g c. 1.23  10–3 mol Pt  = 0.240 g Pt 1 mol 207.2 g d. 72.5 mol Pb  = 1.50  104 g Pb 1 mol 24.31 g e. 0.00102 mol Mg  = 0.0248 g Mg 1 mol 26.98 g f. 4.87  103 mol Al  = 1.31  105 g Al 1 mol 6.941 g g. 211.5 mol Li  = 1468 g Li 1 mol 22.99 g h. 1.72  10–6 mol Na  = 3.95  10–5 g Na 1 mol

6.022 x 1023 Co atoms 16. a. 0.00103 g Co  = 1.05  1019 Co atoms 58.93 g Co

6.022 x 1023 Co atoms b. 0.00103 mol Co  = 6.20  1020 Co atoms 1 mol

1 mol c. 2.75 g Co  = 0.0467 mol Co 58.93 g Co

1 mol d. 5.99  1021 Co atoms  = 0.00995 mol Co 6.022 x 1023 Co atoms

58.93 g Co e. 4.23 mol Co  = 249 g Co 1 mol Co

6.022 x 1023 Co atoms f. 4.23 mol Co  = 2.55  1024 Co atoms 1 mol Co

6.022 x 1023 Co atoms g. 4.23 g Co  = 4.32  1022 Co atoms 58.93 g Co

17. molar mass

19. a. mass of 3 mol Na = 3 (22.99 g) = 68.97 g mass of 1 mol N = 1 (14.01 g) = 14.01 g molar mass of Na3N = 82.98 g

b. mass of 1 mol C = 12.01 g = 12.01 g mass of 2 mol S = 2 (32.07 g) = 64.14 g molar mass of CS2 = 76.15 g

c. mass of 1 mol N = 14.01 g = 14.01 g mass of 4 mol H = 4 (1.008 g) = 4.032 g mass of 1 mol Br = 79.90 g = 79.90 g molar mass of NH4Br = 97.942 g = 97.94 g

d. mass of 2 mol C = 2 (12.01 g) = 24.02 g mass of 6 mol H = 6 (1.008 g) = 6.048 g mass of 1 mol O = 16.00 g = 16.00 g molar mass of C2H6O = 46.068 g = 46.07 g

e. mass of 2 mol H = 2 (1.008 g) = 2.016 g mass of 1 mol S = 32.07 g = 32.07 g mass of 3 mol O = 3 (16.00 g) = 48.00 g molar mass of H2SO3 = 82.086 g = 82.09 g

f. mass of 2 mol H = 2 (1.008 g) = 2.016 g mass of 1 mol S = 32.07 g = 32.07 g mass of 4 mol O = 4 (16.00 g) = 64.00 g molar mass of H2SO4 = 98.086 g = 98.09 g

20. a. mass of 1 mol Ba = 137.3 g = 137.3 g mass of 2 mol Cl = 2 (35.45 g) = 70.90 g mass of 8 mol O = 8 (16.00 g) = 128.0 g molar mass of Ba(ClO4)2 = 336.2 g

b. mass of 1 mol Mg = 24.31 g = 24.31 g mass of 1 mol S = 32.07 g = 32.07 g mass of 4 mol O = 4 (16.00 g) = 64.00 g molar mass of MgSO4 = 120.38 g

c. mass of 1 mol Pb = 207.2 g = 207.2 g mass of 2 mol Cl = 2 (35.45 g) = 79.90 g molar mass of PbCl2 = 278.1 g

d. mass of 1 mol Cu = 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(NO3)2 = 187.57 g

e. mass of 1 mol Sn = 118.7 g = 118.7 g mass of 4 mol Cl = 4 (35.45 g) = 141.80 g molar mass of SnCl4 = 260.5 g

f. mass of 6 mol C = 6 (12.01 g) = 72.06 g mass of 6 mol H = 6 (1.008 g) = 6.048 g mass of 1 mol O = 16.00 g = 16.00 g molar mass of C6H6O = 94.11 g

21. a. molar mass of SO3 = 80.07 g 1 g 1 mol 49.2 mg SO   = 6.14  10–4 mol SO 3 1000 mg 80.07 g 3

b. molar mass of PbO2 = 239.2 g 1000 g 1 mol 7.44 x 104 kg PbO   = 3.11  105 mol PbO 2 1 kg 239.2 g 2

c. molar mass of CHCl3 = 119.37 g 1 mol 59.1 g CHCl  = 0.495 mol CHCl 3 119.37 g 3

d. molar mass of C2H3Cl3 = 133.39 g

1 g 1 mol –8 3.27 g C2H3Cl3   = 2.45  10 mol C2H3Cl3 106 g 133.39 g

e. molar mass of LiOH = 23.95 g 1 mol 4.01 g LiOH  = 0.167 mol LiOH 23.95 g

22. a. molar mass of NaH2PO4 = 120.0 g 1 mol 4.26  10–3 g NaH PO  = 3.55  10–5 mol NaH PO 2 4 120.0 g 2 4

b. molar mass of CuCl = 99.00 g 1 mol 521 g CuCl  = 5.26 mol CuCl 99.00 g

c. molar mass of Fe = 55.85 g 1000 g 1 mol 151 kg Fe   = 2.70  103 mol Fe 1 kg 55.85 g

d. molar mass of SrF2 = 125.6 g 1 mol 8.76 g SrF  = 0.0697 mol SrF 2 125.6 g 2

e. molar mass of Al = 26.98 g 1 mol 1.26  104 g Al  = 467 mol Al 26.98 g

23. a. molar mass of AlI3 = 407.7 g 407.7 g 1.50 mol AlI3  = 612 g AlI3 1 mol

b. molar mass of C6H6 = 78.11 g

–3 78.11 g 1.91  10 mol C6H6  = 0.149 g C6H6 1 mol

c. molar mass of C6H12O6 = 180.2 g 180.2 g 4.00 mol C6H12O6  = 721 g C6H12O6 1 mol

d. molar mass of C2H5OH = 46.07 g

5 46.07 g 7 4.56  10 mol C2H5OH  = 2.10  10 g C2H5OH 1 mol

e. molar mass of Ca(NO3)2 = 164.1 g 164.1 g 2.27 mol Ca(NO3)2  = 373 g Ca(NO3)2 1 mol

24. a. molar mass of CO2 = 44.01 g 1 mol 44.01 g 1.27 mmol   = 0.0559 g CO2 103 mmol 1 mol

b. molar mass of NCl3 = 120.4 g

3 120.4 g 5 4.12  10 mol NCl3  = 4.96  10 g NCl3 1 mol

c. molar mass of NH4NO3 = 80.05 g 80.05 g 0.00451 mol NH4NO3  = 0.361 g NH4NO3 1 mol

d. molar mass of H2O = 18.02 g 18.02 g 18.0 mol H2O  = 324 g H2O 1 mol

e. molar mass of CuSO4 = 159.6 g

159.6 g 4 62.7 mol CuSO4  = 1.00  10 g CuSO4 1 mol

6.022 x 1023 25. a. 6.37 mol CO  = 3.84  1024 molecules CO 1 mol

b. molar mass of CO = 28.01 g 1 mol 6.022 x 1023 molecules 6.37 g   = 1.37  1023 molecules CO 28.01 g 1 mol

c. molar mass of H2O = 18.02 g 6.022 x 1023 molecules 2.62  10–6 g  = 8.76  1016 molecules H O 18.02 g 2

23 –6 6.022 x 10 molecules 18 d. 2.62  10 mol  = 1.58  10 molecules H2O 1 mol

e. molar mass of C6H6 = 78.11 g

23 6.022 x 10 molecules 22 5.23 g  = 4.03  10 molecules C6H6 78.11 g

26. a. molar mass of Na2SO4 = 142.1 g

1 mol Na 2SO4 1 mol S 2.01 g Na2SO4   = 0.0141 mol S 142.1 g 1 mol Na 2SO4

b. molar mass of Na2SO3 = 126.1 g

1 mol Na 2SO3 1 mol S 2.01 g Na2SO3   = 0.0159 mol S 126.1 g 1 mol Na 2SO3

c. molar mass of Na2S = 78.05 g

1 mol Na 2S 1 mol S 2.01 g Na2S   = 0.0258 mol S 78.05 g 1 mol Na 2S

d. molar mass of Na2S2O3 = 158.1 g

1 mol Na 2S2O3 2 mol S 2.01 g Na2S2O3   = 0.0254 mol S 158.1 g 1 mol Na 2S2O3

27. a. mass of Na present = 2 (22.99 g) = 45.98 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 Na2SO4 = 142.05 g

45.98 g Na % Na =  100 = 32.37% Na 142.05 g

32.07 g S % S =  100 = 22.58% S 142.05 g

64.00 g O % O =  100 = 45.05% O 142.05 g

b. mass of Na present = 2 (22.99 g) = 45.98 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 Na2SO3 = 126.05 g

45.98 g Na % Na =  100 = 36.48% Na 126.05 g

32.07 g S % S =  100 = 25.44% S 126.05 g

48.00 g O % O =  100 = 38.08% O 126.05 g

c. mass of Na present = 2 (22.99 g) = 45.98 g mass of S present = 32.07 g = 32.07 g molar mass of Na2S = 78.05 g

45.98 g Na % Na =  100 = 58.91% Na 78.05 g

32.07 g S % S =  100 = 41.09% S 78.05 g

d. mass of Na present = 2 (22.99 g) = 45.98 g mass of S present = 2 (32.07 g) = 64.14 g mass of O present = 3 (16.00 g) = 48.00 g molar mass of Na2S2O3 = 158.12 g

45.98 g Na % Na =  100 = 29.08% Na 158.12 g

64.14 g S % S =  100 = 40.56% S 158.12 g

48.00 g O % O =  100 = 30.36% O 158.12 g

e. mass of K present = 3 (39.10 g) = 117.3 g mass of P present = 30.97 g = 30.97 g mass of O present = 4 (16.00 g) = 64.00 g molar mass of K3PO4 = 212.3 g

117.3 g K % K =  100 = 55.25% K 212.3 g

30.97 g P % P =  100 = 14.59% P 212.3 g

64.00 g O % O =  100 = 30.15% O 212.3 g

f. mass of K present = 2 (39.10 g) = 78.20 g mass of H present = 1.008 g = 1.008 g mass of P present = 30.97 g = 30.97 g mass of O present = 4 (16.00 g) = 64.00 g molar mass of K2HPO4 = 174.178 g = 174.18 g

78.20 g K % K =  100 = 44.90% K 174.18 g

1.008 g H % H =  100 = 0.5787% H 174.18 g

30.97 g P % P =  100 = 17.78% P 174.18 g

64.00 g O % O =  100 = 36.74% O 174.18 g

g. mass of K present = 39.10 g = 39.10 g mass of H present = 2 (1.008 g) = 2.016 g mass of P present = 30.97 g = 30.97 g mass of O present = 4 (16.00 g) = 64.00 g molar mass of KH2PO4 = 136.09 g

39.10 g K % K =  100 = 28.73% K 136.09 g

2.016 g H % H =  100 = 1.481% H 136.09 g

30.97 g P % P =  100 = 22.76% P 136.09 g

64.00 g O % O =  100 = 47.03% O 136.09 g

h. mass of K present = 3 (39.10) g = 117.3 g mass of P present = 30.97 g = 30.97 g molar mass of K3P = 148.27 g = 148.3 g

117.3 g K % K =  100 = 79.10% K 148.3 g

30.97 g P % P =  100 = 20.88% P 148.3 g

28. a. molar mass of CuBr2 = 223.4 g 63.55 g Cu % Cu =  100 = 28.45% Cu 223.4 g

b. molar mass of CuBr = 143.5 g 63.55 g Cu % Cu =  100 = 44.29% Cu 143.5 g

c. molar mass of FeCl2 = 126.75 g 55.85 g Fe % Fe =  100 = 44.06% Fe 126.75 g

d. molar mass of FeCl3 = 162.2 g 55.85 g Fe % Fe =  100 = 34.43% Fe 162.2 g

e. molar mass of CoI2 = 312.7 g 58.93 g Co % Co =  100 = 18.85% Co 312.7 g

f. molar mass of CoI3 = 439.6 g 58.93 g Co % Co =  100 = 13.41% Co 439.6 g

g. molar mass of SnO = 134.7 g 118.7 g Sn % Sn =  100 = 88.12% Sn 134.7 g

h. molar mass of SnO2 = 150.7 g 118.7 g Sn % Sn =  100 = 78.77% Sn 150.7 g

29. a. molar mass of C6H10O4 = 146.1 g 72.06 g C % C =  100 = 49.32% C 146.1 g

b. molar mass of NH4NO3 = 80.05 g 28.02 g N % N =  100 = 35.00% N 80.05 g

c. molar mass of C8H10N4O2 = 194.2 g 96.08 g C % C =  100 = 49.47% C 194.2 g

d. molar mass of ClO2 = 67.45 g 35.45 g Cl % Cl =  100 = 52.56% Cl 67.45 g

e. molar mass of C6H11OH = 100.2 g 72.06 g C % C =  100 = 71.92% C 100.2 g

f. molar mass of C6H12O6 = 180.2 g 72.06 g C % C =  100 = 39.99% C 180.2 g

g. molar mass of C20H42 = 282.5 g 240.2 g C % C =  100 = 85.03% C 282.5 g

h. molar mass of C2H5OH = 46.07 g 24.02 g C % C =  100 = 52.14% C 46.07 g

30. a. molar mass of NH4Cl = 53.49 g + molar mass of NH4 ion = 18.04 g 18.04 g NH+ % NH+ = 4  100 = 33.73% NH + 4 53.49 g 4

b. molar mass of CuSO4 = 159.62 molar mass of Cu2+ ion = 63.55 g 63.55 g Cu2+ % Cu2+ =  100 = 39.81% Cu2+ 159.62 g

c. molar mass of AuCl3 = 303.4 g molar mass of Au3+ ion = 197.0 g 197.0 g Au3+ % Au3+ =  100 = 64.93% Au3+ 303.4 g

d. molar mass of AgNO3 = 169.9 g molar mass of Ag+ ion = 107.9 g 107.9 g Ag+ % Ag+ =  100 = 63.51% Ag+ 169.9 g

31. 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.

32. The empirical formula represents the smallest whole number ratio of the elements present in a compound. The molecular formula indicates the actual number of atoms of each element found in a of the substance.

33. a. NaO

b. C4H3O2

c. C12H12N2O3 is already the empirical formula

d. C2H3Cl

34. a. yes (each of these has empirical formula CH) b. no (the number of atoms is wrong)

c. yes (both have empirical formula NO2) d. no (the number of hydrogen and oxygen atoms is wrong)

1 mol C 35. 0.1929 g C  = 0.01606 mol C 12.01 g C

1 mol H 0.01079 g H  = 0.01070 mol H 1.008 g H

1 mol O 0.08566 g O  = 0.005354 mol O 16.00 g O

1 mol Cl 0.1898 g Cl  = 0.005354 mol Cl 35.45 g Cl

Dividing each number of moles by the smallest number of moles gives:

0.01606 mol C = 3.000 mol C 0.005354 0.01070 mol H = 1.999 mol H 0.005354 0.005354 mol O = 1.000 mol O 0.005354 0.005354 Cl = 1.000 mol Cl 0.005354

The empirical formula is C3H2OCl

1 mol 36. 2.514 g Ca  = 0.06272 mol Ca 40.08 g Ca The increase in mass represents the oxygen with which the calcium reacted: 1 mol O 1.004 g O  = 0.06275 mol O 16.00 g O Since we have effectively the same number of moles of Ca and O, the empirical formula must be CaO.

37. Consider having 100.0 g of the compound. Then the percentages of the elements present are numerically equal to their masses in grams.

1 mol 58.84 g Ba  = 0.4286 mol Ba 137.3 g Ba 1 mol 13.74 g S  = 0.4284 mol S 32.07 g S

1 mol O 27.43 g O  = 1.714 mol O 16.00 g O Dividing each number of moles by the smallest number of moles (0.4284 mol S) gives

0.4286 mol Ba = 1.000 mol Ba 0.4284 0.4284 mol S = 1.000 mol S 0.4284 1.714 mol O = 4.001 mol O 0.4284

The empirical formula is BaSO4.

38. The mass of chlorine involved in the reaction is 6.280 – 1.271 = 5.009 g Cl

1 mol Al 1.271 g Al  = 0.04711 mol Al 26.98 g Al 1 mol Cl 5.009 g Cl  = 0.1413 mol Cl 34.45 g Cl Dividing each of these numbers of moles by the smaller (0.04711 mol Al) shows that the empirical formula is AlCl3.

39. Consider 100.0 g of the compound. 1 mol 55.06 g Co  = 0.9343 mol Co 58.93 g Co If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass. 1 mol 44.94 g S  = 1.401 mol S 32.07 g S Dividing each number of moles by the smaller (0.9343 mol Co) gives 0.9343 mol Co = 1.000 mol Co 0.9343 1.401 mol S = 1.500 mol S 0.9343 Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for the compound as Co2S3.

1 mol Ca 40. 2.461 g Ca  = 0.06140 mol Ca 40.08 g Ca

1 mol Cl 4.353 g Cl  = 0.1228 mol Cl 35.45 g Cl Dividing each of the number of moles by the smaller (0.06140 mol Ca) shows that the empirical formula is CaCl2.

1 mol 41. 10.00 g Cu  = 0.1574 mol Cu 63.55 g Cu

1 mol O 2.52 g O  = 0.158 mol O 16.00 g O The numbers are almost equal: the empirical formula is CuO.

42. Consider 100.0 g of the compound. 1 mol Cu 33.88 g Cu  = 0.5331 mol Cu 63.55 g Cu

1 mol N 14.94 g N  = 1.066 mol N 14.01 g N

1 mol O 51.18 g O  = 3.199 mol O 16.00 g O

Dividing each number of moles by the smaller number of moles (0.5331 mol Cu) gives 0.5331 mol Cu = 1.000 mol Cu 0.5331 1.066 mol N = 2.000 mol N 0.5331 3.199 mol O = 6.001 mol O 0.5331

The empirical formula is CuN2O6 [i.e., Cu(NO3)2]

43. Compound 1: Assume 100.0 g of the compound. 1 mol Na 83.12 g Na  = 3.615 mol Na 22.99 g Na

1 mol N 16.88 g N  = 1.205 mol Na 14.01 g N Dividing each number of moles by the smaller (1.205 mol Na) indicates that the formula of Compound 1 is Na3N.

Compound 2: Assume 100.0 g of the compound. 1 mol Na 35.36 g Na  = 1.538 mol Na 22.99 g Na

1 mol N 64.64 g N  = 4.614 mol N 14.01 g N Dividing each number of moles by the smaller (1.538 mol Na) indicates that the formula of Compound 2 is NaN3.

44. 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, H2O) may have the same empirical and molecular formulas.

45. If only the empirical formula is known, the molar mass of the substance must be determined before the molecular formula can be calculated.

46. empirical formula mass of CH2O = 30 g molar mass 90 g n = = = 3 empirical formula mass 30 g

molecular formula is (CH2O)3 = C3H6O3.

47. empirical formula mass of CH2 = 14 molar mass 84 g n = = = 6 empirical formula mass 14 g

molecular formula is (CH2)6 = C6H12.

48. empirical formula mass of CH4O = 32.04 g molar mass 192 g n = = = 6 empirical formula mass 32.04 g

molecular formula is (CH4O)6 = C6H24O6.

49. Consider 100.0 g of the compound. 1 mol C 42.87 g C  = 3.570 mol C 12.01 g C

1 mol H 3.598 g H  = 3.569 mol H 1.008 g H

1 mol O 28.55 g O  = 1.784 mol O 16.00 g O

1 mol N 25.00 g N  = 1.784 mol N 14.01 g N

Dividing each number of moles by the smallest number of moles (1.784 mol O) gives 3.570 mol C = 2.001 mol C 1.784 3.569 mol H = 2.001 mol H 1.784 1.784 mol O = 1.000 mol O 1.784

1.784 mol N = 1.000 mol N 1.784

The empirical formula of the compound is C2H2ON, empirical formula mass of

C2H2ON = 56 molar mass 168 g n = = = 3 empirical formula mass 56 g

The molecular formula is (C2H2ON)3 = C6H6O3N3.

50. Consider 100.0 g of the compound. 1 mol C 65.45 g C  = 5.450 mol C 12.01 g C 1 mol H 5.492 g H  = 5.448 mol H 1.008 g H 1 mol O 29.06 g O  = 1.816 mol O 16.00 g O

Dividing each number of moles by the smallest number of moles (1.816 mol O) gives 5.450 mol C = 3.001 mol C 1.816 5.448 mol H = 3.000 mol H 1.816 1.816 mol O = 1.000 mol O 1.816

The empirical formula is C3H3O, and the empirical formula mass is approximately 55 g. molar mass 110 g n = = = 2 empirical formula mass 55 g

The molecular formula is (C3H3O)2 = C6H6O2.

51.  c  d  e  a  j  g  h  i  b  f

52. 5.00 g Al 0.185 mol 1.12  1023 atoms 0.140 g Fe 0.00250 mol 1.51  1021 atoms 2.7  102 g Cu 4.3 mol 2.6  1024 atoms 0.00250 g Mg 1.03  10–4 mol 6.19  1019 atoms 0.062 g Na 2.7  10–3 mol 1.6  1021 atoms 3.95  10–18 g U 1.66  10–20 mol 1.00  104 atoms

53. 4.24 g 0.0543 mol 3.27  1022 molec. 3.92  1023 atoms 4.04 g 0.224 mol 1.35  1023 molec. 4.05  1023 atoms 1.98 g 0.0450 mol 2.71  1022 molec. 8.13  1022 atoms 45.9 g 1.26 mol 7.59  1023 molec. 1.52  1024 atoms 126 g 6.99 mol 4.21  1024 molec. 1.26  1025 atoms 0.267 g 0.00927 mol 5.58 x 1021 molec. 3.35 x 1022 atoms

54. magnesium/nitrogen compound: mass of nitrogen contained = 1.2791 g – 0.9240 g = 0.3551 g N 1 mol Mg 0.9240 g Mg  = 0.03801 mol Mg 24.31 g Mg

1 mol N 0.3551 g N  = 0.02535 mol N 14.01 g N

Dividing each number of moles by the smaller number of moles gives 0.03801 mol Mg = 1.499 mol Mg 0.02535 0.02535 mol N = 1.000 mol N 0.02535

Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg3N2.

magnesium–oxygen compound: Consider 100.0 g of this compound. 1 mol Mg 60.31 g Mg  = 2.481 mol Mg 24.31 g Mg

1 mol O 39.69 g O  = 2.481 mol O 16.00 g O Since the numbers of moles are the same, the compound contains the same relative number of Mg and O atoms: the empirical formula is MgO.

55. For the first compound (restricted amount of oxygen): 1 mol Cu 2.118 g Cu  = 0.03333 mol Cu 63.54 g Cu

1 mol O 0.2666 g O  = 0.01666 mol O 16.00 g 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 Cu2O.

For the second compound (stream of pure oxygen): 1 mol Cu 2.118 g Cu  = 0.03333 mol Cu 63.54 g Cu

1 mol O 0.5332 g O  = 0.03333 mol O 16.00 g O Since the numbers of moles are the same, the empirical formula is CuO.

56. We need to find the subscripts for CaHbOcSd, where a:b:c:d is the mole ratio of C:H:O:S atoms. We are given the following relationships: b = 2a a = c b = 8d We can see that d will be the smallest subscript. Thus, let d = x and we get b = 8x 2a = 8x or a = 4x since a = c, c = 4x

Thus, we have C4xH8xO4xSx. The empirical formula is C4H8O4S, which has a molar mass of about 152 g/mol [4(12.01) + 8(1.008) + 4(16.00) + 1(32.07)]. We are given that the molar mass of the compound is 152 g/mol. Thus, the molecular formula is C4H8O4S.

55.85 g Fe 57. 2.24 g Co  = 2.12 g Fe 58.93 g Co

58.93 g Co 58. 2.24 g Fe  = 2.36 g Co 55.85 g Fe

59. Consider 100.0 g of the compound. 1 mol Cu 25.45 g Cu  = 0.4005 mol Cu 63.55 g Cu

1 mol S 12.84 g S  = 0.4004 mol S 32.07 g S

1 mol H 4.036 g H  = 4.004 mol H 1.008 g H

1 mol O 57.67 g O  = 3.604 mol O 16.00 g O

Dividing each number of moles by the smallest number of moles gives 0.4005 mol Cu = 1.000 mol Cu 0.4004 0.4004 mol S = 1.000 mol S 0.4004 4.004 mol H = 10.00 mol H 0.4004 3.604 mol O = 9.001 mol O 0.4004

The empirical formula is CuSH10O9 (which is usually written as CuSO45H2O).

1 mol C 60. 0.2990 g C  = 0.02490 mol C 12.01 g C

1 mol H 0.05849 g H  = 0.05803 mol H 1.008 g H

1 mol N 0. 2318 g N  = 0.01655 mol N 14.01 g N

1 mol O 0.1328 g O  = 0.008300 mol O 16.00 g O Dividing each number of moles by the smallest number of moles (0.008300 mol O) gives 0.02490 mol C = 3.000 mol C 0.008300 0.05803 mol H = 6.992 mol H 0.008300 0.01655 mol N = 1.994 mol N 0.008300 0.008300 mol O = 1.000 mol O 0.008300

The empirical formula is C3H7N2O.

61. Mass of oxygen in compound = 4.33 g – 4.01 g = 0.32 g O 1 mol Hg 4.01 g Hg  = 0.0200 mol Hg 200.6 g Hg

1 mol O 0.32 g O  = 0.020 mol O 16.00 g O Since the numbers of moles are equal, the empirical formula is HgO.

62. Assume we have 100.0 g of the compound. 1 mol Ba 65.95 g Ba  = 0.4803 mol Ba 137.3 g Ba

1 mol Cl 34.05 g Cl  = 0.9605 mol Cl 35.45 g Cl Dividing each of these numbers of moles by the smaller number gives 0.4803 mol Ba = 1.000 mol Ba 0.4803 0.9605 mol Cl = 2.000 mol Cl 0.4803

The empirical formula is then BaCl2.

63. We need to find the subscripts for HxNyOz, where x:y:z is the mole ratio of H:N:O atoms. We are given that x = 4.0 moles H. To solve for y, we use 1 mol N 56.0 g N  = 4.00 moles N 14.01 g N To solve for z, we use 1 mol O 7.2 x 1024 atoms O  = 12 moles O 6.022 x 1023 atoms

The mole ratio is 4:4:12 or 1:1:3. The empirical formula is HNO3.

64. For every 100.0 g of A2O, we have 63.7 g A and 36.3 g O. 1 mol O 36.3 g O  = 2.27 mol O 16.00 g O The ratio between A and O is 2:1; with 2.27 mol O we must have 4.54 mol A. 63.7 g A = 14.0 g/mol 4.54 mol A

Thus, A must be nitrogen. The compound is N2O.

65.  g  i  c  f  b  h  a  e  j  d