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3 Stoichiometry Solutions to Exercises NH₄NO₃: FW = 2(14.0) + 4(1.0) + 3(16.0) = 80.0 amu = [2(14.0)/80.0] X 100 = 35.0% (NH₂)₂CO: FW = 2(14.0) + 4(1.0) = 1(12.0) + 1(16.0) = 60.0 amu % N = [2(14.0)/60.0] X 100 = 46.7% N NH₃: FW = 1(14.0) + 3(1.0) = 17.0 % N = [14.0/17.0] 100 = 82.4 % N 3.88 (a) 180.2 1 mol C₉H₈O₄ = 2.7747 10⁻³ = 2.77 10⁻³ mol C₉H₈O₄ (b) 0.0027747 mol C₉H₈O₄ 6.022 molecules = 1.67 C₉H₈O₄ molecules 1 mol 9C atoms (c) 1.67 X C₉H₈O₄ molecules molecule = Catoms 3.89 (a) Analyze. Given: diameter of Si sphere (dot), density of Si. Find: mass of dot. Plan. Calculate volume of sphere in cm³, use density to calculate mass of the sphere (dot). Solve. V = 4/3π r = d/2 radius of dot = 4 nm 2 1 nm m 1 10⁻² cm m = 2x10⁻⁷ cm volume of dot = (4/3) (2 = 3.35 10⁻²⁰ = X 10⁻²⁰ cm³ 3.35 X 10⁻²⁰ cm³ 2.3 cm³ g Si = 7.707 10⁻²⁰ = 8x 10⁻²⁰ g Si in dot (b) Plan. Change g Si to mol Si using molar mass, then mol Si to atoms Si using Avogadro's number. Solve. 7.707 10⁻²⁰ g Si 1 mol Si 6.022 X mol 10²³ Si Si atoms = 1.653 X 10³ = 2 10³ Si atoms (c) Plan. A 4 nm quantum dot of Ge also has a volume of 3 X 10⁻²⁰ cm³. Use density of Ge and Avogadro's number to calculate the number of Ge atoms in a 4 nm spherical quantum dot. 3.35 10⁻²⁰ cm³ 5.325 cm³ g Ge X 72.64 1 mol g Ge Ge 6.022 mol Ge Ge atoms = 1.479 10³ = 1x 10³ Ge atoms Strictly speaking, the result has 1 sig fig (from 4 nm). A more meaningful comparison might be 1700 Si atoms 1500 Ge atoms. Although Ge has greater molar mass, it is also more than twice as dense as Si, so the numbers of atoms in the Si and Ge dots are similar. 68