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Empirical Formulas from Percent Composition Guide

Jun 5, 2025

Determining Empirical Formulas from Percent Composition

Concept Overview

  • Empirical Formula: Represents the lowest whole number ratio of atoms of each element in a compound.
  • Percent Composition: Used to determine the ratio of elements when mass data is unavailable.

Steps to Determine Empirical Formula from Percent Composition

  1. Assume Hypothetical Mass:

    • Assume a total mass of 100 grams for convenience (makes calculations straightforward as percentages directly convert to grams).

  2. Calculate Masses from Percentages:

    • Convert percentage of each element to grams based on the assumption of 100 grams total mass.

  3. Convert Mass to Moles:

    • Use the molar mass of each element to convert grams to moles.
      • Example calculation for Carbon (C):
        • 27.29% of C becomes 27.29 grams.
        • Convert to moles: ( \text{27.29 grams} \times \frac{1 \text{ mole}}{12.01 \text{ grams}} = 2.272 \text{ moles} )
      • Example calculation for Oxygen (O):
        • 72.71% of O becomes 72.71 grams.
        • Convert to moles: ( \text{72.71 grams} \times \frac{1 \text{ mole}}{16 \text{ grams}} = 4.544 \text{ moles} )

  4. Calculate Ratio of Moles:

    • Divide each mole value by the smallest number of moles calculated from the steps above.
      • For Carbon (C): ( \frac{2.272}{2.272} = 1 )
      • For Oxygen (O): ( \frac{4.544}{2.272} = 2 )

  5. Determine Empirical Formula:

    • The ratio provides the empirical formula of the compound.
    • Example: CO2 (1:2 ratio of C to O), can represent carbon dioxide or another compound with the same ratio.

Example Problem

  • Given Percentages:
    • 40% Carbon
    • 6.71% Hydrogen
    • 53.28% Oxygen
  • Assume 100g Total:
    • Carbon: 40 grams
    • Hydrogen: 6.71 grams
    • Oxygen: 53.28 grams
  • Convert to Moles:
    • Carbon: ( 40 \text{ grams} \times \frac{1 \text{ mole}}{12.01 \text{ grams}} )
    • Hydrogen: ( 6.71 \text{ grams} \times \frac{1 \text{ mole}}{1.008 \text{ grams}} )
    • Oxygen: ( 53.28 \text{ grams} \times \frac{1 \text{ mole}}{16 \text{ grams}} )
  • Calculate Molar Ratios:
    • Use smallest number of moles for division.
  • Resulting Empirical Formula:
    • CH2O

This method is robust and can be applied to any compound where percent composition is provided.

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