How to Calculate Percent Composition
Learn how to calculate the percent composition by mass of each element in a compound. Includes step-by-step examples and how to use percent composition to find empirical formulas.
What Is Percent Composition?
Percent composition is the percentage by mass that each element contributes to the total molar mass of a compound. It provides a way to describe the elemental makeup of a substance and is used to verify purity, determine empirical formulas, and compare compounds. Percent composition is calculated from the chemical formula using molar masses from the periodic table.
The Percent Composition Formula
For each element in a compound: % composition = (mass of element in one mole of compound / molar mass of compound) x 100. The mass of an element in one mole equals the element's atomic mass multiplied by the number of atoms of that element in the formula. The sum of all percent compositions in a compound must equal 100% (small deviations due to rounding are acceptable).
Step-by-Step Example: Water (H2O)
Molar mass of H2O = 2(1.008) + 15.999 = 18.015 g/mol. Percent H = (2 x 1.008 / 18.015) x 100 = 11.19%. Percent O = (15.999 / 18.015) x 100 = 88.81%. Check: 11.19 + 88.81 = 100.00%. This tells us that water is approximately 11.2% hydrogen and 88.8% oxygen by mass.
Step-by-Step Example: Ammonium Nitrate (NH4NO3)
Molar mass of NH4NO3 = 14.007 + 4(1.008) + 14.007 + 3(15.999) = 80.043 g/mol. Percent N = (2 x 14.007 / 80.043) x 100 = 35.00%. Percent H = (4 x 1.008 / 80.043) x 100 = 5.04%. Percent O = (3 x 15.999 / 80.043) x 100 = 59.96%. Ammonium nitrate is widely used as a fertilizer precisely because of its high nitrogen content.
Using Percent Composition to Find Empirical Formulas
Percent composition data from elemental analysis can be used to determine a compound's empirical formula. Assume you have 100 g of the compound so that percentages convert directly to grams. Divide each mass by the element's atomic mass to find moles. Divide all mole values by the smallest to get the mole ratio. If ratios are not whole numbers, multiply by an integer (2, 3, etc.) to clear fractions. The result is the empirical formula.
Empirical Formula Example
A compound is 40.0% C, 6.7% H, and 53.3% O by mass. In 100 g: C = 40.0/12.011 = 3.33 mol, H = 6.7/1.008 = 6.65 mol, O = 53.3/15.999 = 3.33 mol. Divide by smallest (3.33): C:H:O = 1:2:1. The empirical formula is CH2O, which matches formaldehyde, acetic acid, and glucose (all share this ratio but differ in molecular formula).
Molecular Formula from Empirical Formula
The molecular formula is a whole-number multiple of the empirical formula: molecular formula = (empirical formula) x n, where n = molar mass of compound / molar mass of empirical formula. For CH2O with empirical molar mass 30.026 g/mol, if the actual molar mass is 180.16 g/mol (glucose), n = 180.16 / 30.026 = 6. The molecular formula is C6H12O6.
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