CHEM 151 1st Edition Lecture 10 Outline of Last Lecture I. Avogadro’s numberII. Molar MassIII. Compounds and Molar MassOutline of Current Lecture IV. Empirical FormulaV. Molecular FormulaCurrent LectureEmpirical FormulaExample:Given40.0% C 6.71% H 53.3% OCalculate moles of each component.40.0 g C (1 mol/12.01 g C) = 3.33 mol C 6.71 g H (1 mol/1.008 g H) = 6.66 mol H53.3 g O (1 mol/16.00 g O) = 3.33 mol ODivide by lowest amount of moles.3.33 mol C / 3.33 = 1 6.66 mol H / 3.33 = 23.33 mol O / 3.33 = 1Answer: CH2OMolecular Formula (from empirical formula)Continued example:Given 90.08 g/mol CxHyOzFind molar mass of empirical formula.1 C = 1(12.01 g/mol)2 H = 2(1.008 g/mol)1 O = 1(16.00 g/mol)Total = 30.026 g/molDivide the given molar mass by the empirical formulas molar mass.(90.08 g/mol) / (30.026 g/mol) = 3Multiply the subscripts of the empirical formula by
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