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Determining a Compound’s Percentage Composition From its Formula

3.5 Composition Analysis - Determining Formulas. Determining a Compound’s Percentage Composition From its Formula Empirical, Molecular, & Structural Formulas Determining a Compound’s Empirical Formula From its Percentage Composition

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Determining a Compound’s Percentage Composition From its Formula

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  1. 3.5 Composition Analysis - Determining Formulas • Determining a Compound’s Percentage Composition From its Formula • Empirical, Molecular, & Structural Formulas • Determining a Compound’s Empirical Formula From its Percentage Composition • Determining a Compound’s Molecular Formula From its Empirical Formula and its Molar Mass

  2. 3.5 Composition Analysis - Determining Formulas Determining a Compound’s Percentage Composition From its Formula Percentage composition is the percent of a compound’s mass contributed by each type of atom in the compound. e.g. For NO2 1 mol N = (1 x 14.0 g N) = 14.0 g N / mol NO2 2 mol O = (2 x 16.0 g O) = 32.0 g O / mol NO2 Molar Mass= 46.0 g / mol NO2 % N = (14.0 g N / 46.0 g NO2) x 100 = 30.4% N % O = (32.0 g O / 46.0 g NO2) x 100 = 69.6% O text pages 142-143

  3. 3.5 Composition Analysis - Determining Formulas Empirical, Molecular, and Structural Formulas A compound’s structural formula shows how the atoms in a molecule are arranged, e.g. a compound named ethane has the structural formula: H H H – C – C – H H H Propane’s molecular formula is C2H6 and its empirical formula is CH3 text page 144

  4. e.g. For a compound that is 80.0% C and 20.0% H 80.0 g C x 1 mol C = 6.67 mol C 12.0 g C 20.0 g H x 1 mol H = 20.0 mol H 1.0 g H C6.67H20.0 = CH3 6.67 3.5 Composition Analysis - Determining Formulas Determining a Compound’s Empirical Formula From its Percentage Composition text pages 144-145

  5. 3.5 Composition Analysis - Determining Formulas Determining a Compound’s Molecular Formula From its Empirical Formula and its Molar Mass Molecular Formula = Empirical Formula x Compound’s Molar Mass Molar Mass of Empirical Formula e.g. Empirical Formula CH3 with a Molar Mass of 30.0 g/mol 1 mol C = (1 x 12.0 g C) = 12.0 g C / mol CH3 3 mol H = (3 x 1.0 g H) = 3.0 g H / mol CH3 Molar Mass= 15.0 g / mol CH3 30.0 g/mol / 15.0 g/mol = 2 2 (CH3) = C2H6 text page 146

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