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3 Stoichiometry Solutions to Exercises 1 mol F 68.7 g Fx = 3.62 mol F; ≈ 6 The empirical formula is C₃OF₆. (c) The mass of F is [100 g total (32.79 g Na + 13.02 Al)] 54.19 g F 32.79 g Na 1 mol Na = 1.426 mol Na; ≈ 3 13.02 0.4826 mol Al; 1 26.98 54.19 g Fx 1 mol F 2.852 mol F; ≈ The empirical formula is 3.46 See Solution 3.45 for stepwise problem-solving approach. (a) 1 mol K = 1.414 mol K; 1.414/0.4714 ≈ 3 14.6 g Px 1molP = 0.4714 mol P; 0.4714/0.4714 = 1 30.1 g Ox = 1.881 mol O; 1.881/0.4714 ≈ 4 The empirical formula is (b) 24.5 g Nax 1 mol Na = 1.066 mol Na; 1.066/0.5304 ≈ 2 14.9 g Six 1 28.09 Si mol Si 0.5304 mol si; 0.5304/0.5304 = 1 1 mol F 60.6 g Fx = 3.189 mol F; 3.189/0.5304 ≈ 6 The empirical formula is Na₂SiF₆. (c) The mass of is [100 g total (62.1 + 5.21 g H + 12.1 g N)] = 20.59 = 20.6 g 62.1 g Cx = 5.17 mol C; ≈ 6 5.21 g H = 5.17 mol O; ≈ =0.864 mol N; 20.6gOx = 1.29 mol O; ≈ 1.5 Multiplying by two, the empirical formula is C₁₂H₁₂N₂O₃. 51