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Determining Empirical and Molecular Formulas

May 7, 2024

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Summary of the Lecture

In this lecture, we discussed how to determine the empirical and molecular formula of a compound when given its percent composition or the mass of each element in the compound. The emphasis was placed on the step-by-step conversion of these compositions to moles, obtaining the simplest whole number ratios for the empirical formula, and calculating the molecular formula using the molar mass.

Notes on Empirical Formula Calculation from Percent Composition

Example 1: Finding the Empirical Formula

Given Data:

  • Compound composition: 52.14% Carbon, 13.13% Hydrogen, 34.73% Oxygen

Steps to Calculate Empirical Formula:

  1. Convert Percent to Grams assuming 100g of the compound:

    • Carbon: 52.14 g
    • Hydrogen: 13.13 g
    • Oxygen: 34.73 g

  2. Convert Grams to Moles:

    • Carbon: 52.14 g / 12.01 g/mol = 4.34 moles
    • Hydrogen: 13.13 g / 1.008 g/mol = 13.026 moles
    • Oxygen: 34.73 g / 16 g/mol = 2.171 moles

  3. Calculate the Simplest Whole Number Ratio by dividing by the smallest mole value (oxygen moles in this case):

    • Carbon: 4.34 / 2.171 = approximately 2
    • Hydrogen: 13.026 / 2.171 = 6
    • Oxygen: 2.171 / 2.171 = 1

  4. Empirical Formula: C₂H₆O

Example 2: Finding the Molecular Formula from Empirical Formula

Given Data: Empirical Formula C₂H₆O with molar mass of the empirical formula as 46.068 g/mol and compound molar mass as 138.204 g/mol.

Steps to Find Molecular Formula:

  1. Find the Ratio of Molar Masses:
    • 138.204 g/mol / 46.068 g/mol = approximately 3
  2. Multiply the Subscripts in Empirical Formula by 3:
    • Resulting Molecular Formula: C₆H₁₈O₃

Notes on Empirical and Molecular Formula Calculation from Element Masses

Example 3: Finding the Empirical Formula from Element Masses

Given Data:

  • Mass of elements: Carbon 20.32 g, Hydrogen 5.12 g, Nitrogen 7.9 g

Steps to Calculate Empirical Formula:

  1. Convert Masses to Moles:

    • Carbon: 20.32 g / 12.01 g/mol = 1.692 moles
    • Hydrogen: 5.12 g / 1.008 g/mol = 5.079 moles
    • Nitrogen: 7.9 g / 14.01 g/mol = 0.5639 moles

  2. Calculate the Simplest Whole Number Ratio:

    • Divide each mole value by the smallest mole value (nitrogen moles):
    • Carbon: 1.692 / 0.5639 = approximately 3
    • Hydrogen: 5.079 / 0.5639 = 9
    • Nitrogen: 0.5639 / 0.5639 = 1

  3. Empirical Formula: C₃H₉N

Example 4: Finding the Molecular Formula from Empirical Formula

Given Data: Empirical Formula C₃H₉N with molar mass 59.112 g/mol, compound molar mass 236.448 g/mol.

Steps to Find Molecular Formula:

  1. Find the Ratio of Molar Masses:
    • 236.448 g/mol / 59.112 g/mol = approximately 4
  2. Multiply the Subscripts in Empirical Formula by 4:
    • Resulting Molecular Formula: C₁₂H₃₆N₄

These problem-solving steps provide a robust framework for determining empirical and molecular formulas in different scenarios.

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