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Empirical Formulas

Empirical Formulas. Chapter 6, Section 3. What is an Empirical Formula?. Up to this point, we have been using molecular formulas to describe the chemical composition of compounds. A molecular formula is the actual formula of a compound

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Empirical Formulas

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  1. Empirical Formulas Chapter 6, Section 3

  2. What is an Empirical Formula? • Up to this point, we have been using molecular formulas to describe the chemical composition of compounds. • A molecular formula is the actual formula of a compound • By contrast, an empirical formula is a formula of a compound expressing the smallest whole-number ration of atoms in a compound

  3. Examples of Empirical Formulas • Molecular Formula: C6H12O6 • Empirical Formula: CH2O • Molecular Formula: C6H6 • Empirical Formula: CH • Molecular Formula: C12H4Cl4O2 • Empirical Formula: C6H2Cl2O

  4. Calculations of Empirical Formulas • Remember, chemical formulas are a ratio of atoms in a molecule, but it is also a ratio of moles of atoms. • Therefore, you can calculate the empirical formula for a compound given the percent composition in the compound or mass in grams

  5. Steps for Calculating Empirical Formulas • If you are given percentages change the percentages into grams. If you start in grams, go to step 2 • Convert grams of each element to moles • Divide each mole value by the smallest number. If close enough, round to the nearest whole number. If you get decimals, see step 4 • Multiply your fractions by the smallest whole number that will make all you numbers whole numbers

  6. Example 1 • Calculate the empirical formula of a compound composed of 38.67% C, 16.22% H, and 45.11% N • C = 38.67 g, H = 45.11 g, N = 45.11 g • Mole Calculations • 38.67 g C x 1 mol C = 3.220 mole C 12.01 g C • 16.22 g H x 1 mol H = 16.09 mole H 1.01 g H • 45.11 g N x 1 mol N = 3.219 mole N 14.01 g N

  7. Example 1 cont • C is the smallest. Divide all number by 3.220 mol C • 3.220 mol C / 3.220 mol C = 1C • 3.219 mol N / 3.220 mol C = 1N • 16.09 mol H/ 3.220 mol C = 5 H Our empirical formal is CH5N

  8. Example 2 • You have a sample that contains 4.151 g of Al and 3.692 g of oxygen. Calculate the empirical formula. • Already done • Mole Calculations • 4.151 g Al x 1 mol Al = 0.1539 mol Al 26.98 g • 3.692 g O x 1 mol O = 0.2308 mol O 16.00 g

  9. Example 2 cont • Al is the smallest. Divide all by 0.1539 mol Al. • 0.1539 mol Al/ 0.1529 mol Al = 1 Al • 0.2308 mol O/ 0.1529 mol Al = 1.5 O • Multiplying by 2 will get you to whole numbers • 1 Al x 2 = 2 Al • 1.5 O x 2 = 3 O Empirical formula is Al2O3

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