1. 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 3.5 Composition Analysis - Determining Formulas

2. Determining a Compound’s Percentage Composition From its Formula text pages 142-143 Percentage composition is the percent of a compound’s mass contributed by each type of atom in the compound. 3.5 Composition Analysis - Determining Formulas e.g. For NO 2 1 mol N = (1 x 14.0 g N) = 14.0 g N / mol NO 2 2 mol O = (2 x 16.0 g O) = 32.0 g O / mol NO 2 Molar Mass = 46.0 g / mol NO 2 % N = (14.0 g N / 46.0 g NO 2 ) x 100 = 30.4% N % O = (32.0 g O / 46.0 g NO 2 ) x 100 = 69.6% O

3. Empirical, Molecular, and Structural Formulas text page 144 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 – C – C – H H H Propane’s molecular formula is C 2 H 6 and its empirical formula is CH 3 3.5 Composition Analysis - Determining Formulas

4. text pages 144-145 3.5 Composition Analysis - Determining Formulas Determining a Compound’s Empirical Formula From its Percentage Composition 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 C 6.67 H 20.0 = CH 3 6.67

5. text page 146 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 CH 3 with a Molar Mass of 30.0 g/mol 1 mol C = (1 x 12.0 g C) = 12.0 g C / mol CH 3 3 mol H = (3 x 1.0 g H) = 3.0 g H / mol CH 3 Molar Mass = 15.0 g / mol CH 3 30.0 g/mol / 15.0 g/mol = 22 (CH 3 ) = C 2 H 6