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3 Stoichiometry Solutions to Exercises 3.36 (a) Fe₂(SO₄)₃ molar mass = 2(55.845) + 3(32.07) + 12(16.00) = 399.900 = 399.9 g/mol 1.223 mol Fe₂(SO₄)₃ = (b) (NH₄)₂CO₃ molar mass = 2(14.007) + 8(1.008) + 12.011 + 3(15.9994) = 96.0872 = 96.087 g/mol X 1 mol X 2 mol = 0.1448 mol (c) C₉H₈O₄ molar mass = 9(12.01) + 8(1.008) + 4(16.00) = 180.154 = 180.2 g/mol 1.50 X 10²¹ molecules X 1 mol molecules X 1 mol aspirin = (d) 15.86 g Valium = 284.7 Valium/mol 0.05570 mol 3.37 (a) C₆H₁₀OS₂ molar mass = 6(12.01) + 10(1.008) + 1(16.00) + 2(32.07) = 162.28 = (b) Plan. mg g mol Solve. 5.00 mg allicin X 1mg X 162.3g 1 mol = 3.081 X 10⁻⁵ = 3.08 10⁻⁵ mol allicin Check. 5.00 mg is a small mass, so the small answer is reasonable. (c) Plan. Use mol from part (b) and Avogadro's number to calculate molecules. Solve. 3.081x10⁻⁵ mol allicin 6.022 molecules = mol allicin molecules Check. (3 X = 18 = 1.8 10¹⁹ (d) Plan. Use molecules from part (c) and molecular formula to calculate S atoms. Solve. allicin molecules 2S atoms S atoms 1 allicin molecule Check. Obvious. 3.38 (a) C₁₄H₁₈N₂O₅ molar mass = 14(12.01) + 18(1.008) + 2(14.01) + 5(16.00) = 294.30 g/mol (b) 1.00 mg aspartame 1 mg 294.3 = = mol aspartame (c) 3.398 10⁻⁶ mol aspartame X molecules = 2.05 10¹⁸ aspartame molecules 47