1 (61)

Prévia do material em texto

3 Stoichiometry Solutions to Exercises 0.1159 g 18.02 1 mol g H₂O 1 mol 2 mol H = 0.012863 = 0.01286 mol H Calculate g C,g H and get g by subtraction. 0.0064281 mol 1 mol C C 0.012863 mol Hx H mass = 0.1005 g sample (0.07720 g C + 0.01297 g H) = 0.01033 g Calculate mol and find integer ratio of mol C: mol H: mol O. 0.01033 1 mol = X 10⁻⁴ mol Divide moles by 6.456 C: ≈ The empirical formula is C₁₀H₂₀O. FW = FW = 156 The molecular formula is the same as the empirical formula, C₁₀H₂₀O. Check. The mass of wasn't negative or greater than the sample mass; empirical and molecular formulas are reasonable. 3.54 (a) Plan. Calculate mol C and mol H, then g C and g H; get g by subtraction. Solve. 6.32 1 mol 1 mol C = X = 1.44 10⁻⁴ mol C 2.58x 10⁻³ H₂ 1 mol = = X mol H mass of = 2.78 mg sample (1.725 mg C + 0.289 mg H) = 0.77 mg 0.77 = mol O. Divide moles by 4.81x10⁻⁵. C: 1.44x10⁻⁴ H: O: The empirical formula is C₃H₆O. (b) Plan. Calculate mol C and mol H, then g C and g H. In this case, get N by subtraction. Solve. 55