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3 Stoichiometry Solutions to Exercises (c) (NH₄)₂SO₄: FW = 2(14.0) + 8(1.0) + 1(32.1) + 4(16.0) = 132.1 amu %H = 8(1.0) = 6.1% (d) PtCl₂(NH₃)₂: FW = 1(195.1) + 2(35.5) + 2(14.0) + 6(1.0) = 300.1 amu % Pt = 1(195.1) amu 300.1amu (e) C₁₈H₂₄O₂: FW = 18(12.0) + 24(1.0) + 2(16.0) = 272.0 amu = 2(16.0) 272.0 amu 100 = 11.8% (f) C₁₈H₂₇NO₃: FW = 18(12.0) + 27(1.0) + 1(14.0) + 3(16.0) = 305.0 amu %C = 18(12.0) amu 100 = 70.8% 305.0 amu 3.25 Plan. Follow the logic for calculating mass %C given in Sample Exercise 3.6. Solve. (a) C₇H₆O: FW = 7(12.0) + 6(1.0) + 1(16.0) = 106.0 amu %C = 7(12.0) amu (b) C₈H₈O₃: FW = 8(12.0) + 8(1.0) + 3(16.0) = 152.0 amu %C = 8(12.0) = 63.2% (c) C₇H₁₄O₂: FW = 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu %C= 7(12.0) 3.26 (a) FW = 1(12.0) + 2(16.0) = 44.0 amu %C = 12.0 amu x100 = 27.3% 44.0 amu (b) FW = 1(12.0) + 4(1.0) + 1(16.0) = 32.0 amu %C = 12.0 amu x100 = 37.5% 32.0 amu (c) C₂H₆: FW = 2(12.0) + 6(1.0) = 30.0 amu %C = 2(12.0) 100 = 80.0% 30.0 amu (d) CS(NH₂)₂: FW = 1(12.0) + 1(32.1) + 2(14.0) + 4(1.0) = 76.1 amu %C = 12.0 amu = 15.8% Avogadro's Number and the Mole (section 3.4) 3.27 (a) 6.022 X 10²³. This is the number of objects in a mole of anything. (b) The formula weight of a substance in amu has the same numerical value as the molar mass expressed in grams. 3.28 (a) exactly 12 g (b) 6.0221421 X 10²³, Avogadro's number 44