Percent Composition

9 min read Updated Mar 26, 2026

Percent composition tells you what fraction of a compound’s mass comes from each element. It is the chemical equivalent of reading a nutrition label - instead of “20% protein, 50% carbohydrate,” you get “40% carbon, 6.7% hydrogen, 53.3% oxygen.”

Calculating Percent Composition

The formula is simple:

Example: Find the percent composition of water (H₂O).

Molar mass of H₂O = 2(1.01) + 16.00 = 18.02 g/mol
% H = [2(1.01) / 18.02] x 100 = 11.2%
% O = [16.00 / 18.02] x 100 = 88.8%
Check: 11.2 + 88.8 = 100.0% ✓

Working Backward - From Percent Composition to Empirical Formula

This is the reverse problem, and the MCAT loves it. Given percent composition data, determine the empirical formula:

Assume 100 g (percentages become grams)
Convert each to moles (divide by atomic mass)
Divide by the smallest mole value
Multiply to clear fractions if needed

Example: A compound is 26.7% C, 2.2% H, and 71.1% O.

C: 26.7 g / 12.01 g/mol = 2.22 mol
H: 2.2 g / 1.008 g/mol = 2.18 mol
O: 71.1 g / 16.00 g/mol = 4.44 mol

Divide by smallest (2.18): C = 1.02, H = 1.00, O = 2.04

Round: C₁H₁O₂ → Empirical formula: CHO₂

Combustion Analysis - The MCAT’s Favorite Method

Combustion analysis is an experimental technique for determining the molecular formula of an unknown organic compound. You burn a sample in excess oxygen and measure the masses of CO₂ and H₂O produced.

The logic is beautifully simple:

All the carbon in the original compound ends up in CO₂
All the hydrogen in the original compound ends up in H₂O
Any remaining mass is oxygen (if the compound contains oxygen)

Combustion Analysis Step-by-Step

Example: 0.255 g of an unknown compound containing only C, H, and O is burned. The combustion produces 0.561 g CO₂ and 0.306 g H₂O. The molecular weight is 180 g/mol. Find the molecular formula.

Step 1: Find moles of C from CO₂.

Moles of CO₂ = 0.561 g / 44.01 g/mol = 0.01275 mol
Moles of C = 0.01275 mol (1:1 ratio in CO₂)
Mass of C = 0.01275 mol x 12.01 g/mol = 0.153 g

Step 2: Find moles of H from H₂O.

Moles of H₂O = 0.306 g / 18.02 g/mol = 0.01698 mol
Moles of H = 2 x 0.01698 = 0.03396 mol (2 H per H₂O)
Mass of H = 0.03396 mol x 1.008 g/mol = 0.0342 g

Step 3: Find mass and moles of O by difference.

Mass of O = 0.255 - 0.153 - 0.0342 = 0.0678 g
Moles of O = 0.0678 g / 16.00 g/mol = 0.00424 mol

Step 4: Find the mole ratio.

Divide by smallest (0.00424): C = 3.01, H = 8.01, O = 1.00
Empirical formula: C₃H₈O

Step 5: Find n.

Empirical formula weight = 3(12) + 8(1) + 16 = 60 g/mol
n = 180 / 60 = 3
Molecular formula: C₉H₂₄O₃

In a combustion analysis, 0.500 g of a compound produced 1.10 g CO₂ and 0.450 g H₂O. Does the compound contain oxygen?

Click to reveal answer

Yes. Mass of C = (

\frac{1.10}{44}

) x 12 = 0.300 g. Mass of H = (

\frac{0.450}{18}

) x 2 = 0.050 g. Total C + H = 0.350 g. Sample mass = 0.500 g. Remaining 0.150 g must be oxygen (or another element). Since the problem says it is a combustion analysis of a C/H/O compound, the remainder is oxygen.

What is the percent by mass of nitrogen in urea, CO(NH₂)₂? (C = 12, N = 14, O = 16, H = 1)

Click to reveal answer

46.7%. Molar mass of CO(NH₂)₂ = 12 + 16 + 2(14) + 4(1) = 60 g/mol. Mass of N = 2(14) = 28 g/mol. % N = (

\frac{28}{60}

) x 100 = 46.7%. Nearly half of urea's mass is nitrogen, which is why it is used as a nitrogen-rich fertilizer.