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Empirical Formulas

Empirical Formulas. Unit 6 - Bonding. Review…. Percent composition: The relative amount of each element in a compound by mass. % composition Strategy: Calculate GFM = Mass of Compound Determine the mass of each element Calculate % comp for each element. Review….

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Empirical Formulas

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  1. Empirical Formulas Unit 6 - Bonding

  2. Review… Percent composition: • The relative amount of each element in a compound by mass. % composition Strategy: • Calculate GFM = Mass of Compound • Determine the mass of each element • Calculate % comp for each element

  3. Review… Example: Find the % composition for the compound Mg(OH)2

  4. Empirical Formula • Empirical formula – lowest whole # ratio of atoms of elements in a compound • Example: • C6H12O6 CH2O • In lab we can determine % composition of new compounds and can then calculate the empirical formula.

  5. Empirical Formula Steps to Solving for the E.F. • Assume 100 g of compound (make the %  g) 2. Convert g of each element to moles 3. Divide each # of moles by the lowest # of moles 4. Change to whole #’s may have to multiply by – if ends in 0.5 5. Write the formula

  6. Empirical Formula • 26.8% F and 73.2% Pb 26.8 g F 1 mol F / .353 = 3.99  4 = 1.41 mol 19.00 g F 73.2 g Pb 1 mol Pb = .353 mol / .353 = 1.00  1 207.2 g Pb PbF4

  7. Empirical Formula • 70.0% Mn and 30.0% O 70.0 g Mn 1 mol Mn / 1.27 = 1.00 * 2 = 2 = 1.27 mol 54.94 g Mn 30.0 g O 1 mol O = 1.88 mol / 1.27 = 1.48 * 2 = 3 16.00 g O Mn2O3

  8. Empirical Formula • 39.7% K, 27.9% Mn, and 32.4% O 39.7 g K 1 mol K / .508 = 2.01  2 = 1.02 mol 39.10 g K 27.9 g Mn 1 mol Mn = .508 mol / .508 = 1.00  1 54.94 g Mn 32.4 g O 1 mol O = 2.03 mol / .508 = 3.996  4 16.00 g O K2MnO4

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