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3 Stoichiometry Solutions to Exercises (a) 1(1.0) + 1(14.0) + 3(16.0) = 63.0 amu (b) 1(39.1) + 1(54.9) + 4(16.0) = 158.0 amu (c) 3(40.1) + 2(31.0) + 8(16.0) = 310.3 amu (d) 1(28.1) + 2(16.0) = 60.1 amu (e) Ga₂S₃: 2(69.7) + 3(32.1) = 235.7 amu (f) Cr₂(SO₄)₃: 2(52.0) + 3(32.1) + 12(16.0) = 392.3 amu (g) PCl₃: 1(31.0) + 3(35.5) = 137.5 amu 3.22 Formula weight in amu to 1 decimal place. (a) FW = 2(14.0) + 1(16.0) = 44.0 amu (b) HC₇H₅O₂: 7(12.0) + 6(1.0) + 2(16.0) = 122.0 amu (c) Mg(OH)₂: 1(24.3) + 2(16.0) + 2(1.0) = 58.3 amu (d) + 4(1.0) + 1(12.0) + 1(16.0) = 60.0 amu (e) 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu 3.23 Plan. Calculate the formula weight (FW), then the mass % oxygen in the compound. Solve. (a) C₁₇H₁₉NO₃: FW = 17(12.0) + 19(1.0) + 1(14.0) + 3(16.0) = 285.0 amu 3(16.0) = 285.0 amu (b) C₁₈H₂₁NO₃: FW = 18(12.0) + 21(1.0) + 1(14.0) + 3(16.0) = 299.0 amu = 3(16.0) amu x100=16.054=16.1% 299.0 amu (c) C₁₇H₂₁NO₄: FW = 17(12.0) + 21(1.0) + 1(14.0) + 4(16.0) = 303.0 amu = 4(16.0)amu = = 21.1% 303.0 amu (d) C₂₂H₂₄N₂O₈: FW = 22(12.0) + 24(1.0) + 2(14.0) + 8(16.0) = 444.0 amu = 8(16.0) (e) FW = 4(12.0) + 64(1.0) + 13(16.0) = 764.0 amu = 764 amu (f) FW = = 1448.0 amu = 24(16.0) amu 1448.0 amu 3.24 (a) %C = 2(12.0)amu 26.0 amu (b) HC₆H₇O₆: FW = 6(12.0) + 8(1.0) + 6(16.0) = 176.0 amu 43