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Empirical Formulas

Empirical Formulas. Empirical Formulas. The lowest whole number ratio of the atoms in a compound. Examples. Molecular Formula = H 2 O 2 Empirical Formulas = HO Molecular Formula = C 2 H 6 Empirical Formulas = CH 3 Molecular Formula = C 6 H 6 Empirical Formulas = CH

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Empirical Formulas

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  1. Empirical Formulas

  2. Empirical Formulas • The lowest whole number ratio of the atoms in a compound. • Examples. • Molecular Formula = H2O2 • Empirical Formulas = HO • Molecular Formula = C2H6 • Empirical Formulas = CH3 • Molecular Formula = C6H6 • Empirical Formulas = CH • Molecular Formula = C6H12O6 • Empirical Formulas = CH2O

  3. Empirical Formulas • Most of the time Empirical Formulas and Molecular Formulas are different. • There are instances when the Empirical Formula will be the same as the Molecular Formula. • CO2 • H2O • OH • NO2

  4. Determining the Empirical Formula of a Compound • A compound with 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula? • Steps • Convert to Moles • Divide the smallest mole number by each of the other mole numbers. To get the ratios.

  5. Determining the Empirical Formula of a Compound • A compound with 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula? • Assume 100g sample that means, 25.9g of N and 74.1g O • Convert to moles. • 25.9gN 1 mol N 1.85molN 14.01g N • 74.1gO 1 mol O 4.63molO 15.999gO

  6. Determining the Empirical Formula of a Compound • 25.9gN 1 mol N 1.85molN 14.01g N • 74.1gO 1 mol O 4.63molO 15.999gO • Divide by the smallest mole • 25.9gN 1 mol N 1.85molN 14.01g N 1.85 • 74.1gO 1 mol O 4.63molO 15.999gO 1.85 • 25.9gN 1 mol N 1.85molN 1molN 14.01g N 1.85 • 74.1gO 1 mol O 4.63molO 2.5molO 15.999gO 1.85

  7. Determining the Empirical Formula of a Compound • 1 mol N x 2 = 2 mol N • 2.5 mol O x 2 = 5 mol O • Empirical Formula = N2O5

  8. Determining the Empirical Formula of a Compound • 94.1 grams of Oxygen, 5.9 grams of Hydrogen • Assume 100g sample, convert to moles. • 94.1g O 1 mol O 5.88mol O 15.999g O • 5.9g H 1 mol H 5.9 mol H 1.0079g H

  9. Determining the Empirical Formula of a Compound • Divide by the smallest mole value • 94.1g O 1 mol O 5.88mol O 1mol O 15.999g O 5.88 • 5.9g H 1 mol H 5.9 mol H 1.003mol H 1.0079g H 5.88 • Empirical Formula = OH

  10. Determining the Empirical Formula of a Compound

  11. Determining the Empirical Formula of a Compound

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