Percent Composition from a Formula — and Working Backwards
Percent composition tells you what fraction of a compound's mass comes from each element. It looks like a one-line calculation, and in the forward direction it is. The marks in board and entrance papers are usually in the backward direction: how much nitrogen is in a bag of fertiliser, how pure a sample is, how much water of crystallisation a hydrate holds. This guide covers both directions with the arithmetic shown in full.
The formula
Every symbol here is a mass, so the answer is a mass percentage. The numerator is the mass that element contributes to one mole of the compound; the denominator is the molar mass of the whole compound. The percentages of all elements must add up to 100 — that is a free check on your working, and you should use it every single time.
Atomic masses used below (g/mol): H = 1.008, C = 12.011, N = 14.007, O = 15.999, S = 32.06, Cu = 63.546.
Worked example 1 — Urea, CO(NH₂)₂
Urea is the most-used nitrogen fertiliser in India, so its %N is worth knowing. Expand the formula first: 1 C, 1 O, 2 N, 4 H.
Step 1 — molar mass.
C: 1 × 12.011 = 12.011
O: 1 × 15.999 = 15.999
N: 2 × 14.007 = 28.014
H: 4 × 1.008 = 4.032
M = 12.011 + 15.999 + 28.014 + 4.032 = 60.056 g/mol
Step 2 — divide and multiply by 100.
%C = (12.011 ÷ 60.056) × 100 = 20.00%
%O = (15.999 ÷ 60.056) × 100 = 26.64%
%N = (28.014 ÷ 60.056) × 100 = 46.65%
%H = (4.032 ÷ 60.056) × 100 = 6.71%
Check: 20.00 + 26.64 + 46.65 + 6.71 = 100.00 ✓
Worked example 2 — Ammonium sulphate, (NH₄)₂SO₄
The subscript 2 outside the bracket multiplies everything inside it, so this formula holds 2 N and 8 H, not 2 N and 4 H.
N: 2 × 14.007 = 28.014
H: 8 × 1.008 = 8.064
S: 1 × 32.06 = 32.06
O: 4 × 15.999 = 63.996
M = 28.014 + 8.064 + 32.06 + 63.996 = 132.134 g/mol
%N = (28.014 ÷ 132.134) × 100 = 21.20%
Working backwards 1 — mass of an element in a sample
Once you have the percentage, the mass of that element in any quantity of the pure compound is a simple multiplication.
Question. How much nitrogen is present in 25.0 kg of pure ammonium sulphate?
mass of N = 25.0 kg × 0.2120 = 5.30 kg
Notice the unit: because a percentage is a ratio of masses, whatever unit you put in (kg here) comes straight back out. You do not have to convert to grams first.
Working backwards 2 — purity of a sample
Laboratory and industrial samples are rarely pure. If you measure a lower percentage than the pure compound should give, the shortfall is the impurity — provided the impurity contains none of the element you measured.
Question. A sample of ammonium sulphate is analysed and found to contain 19.8% nitrogen by mass. What is its purity?
The pure salt gives 21.20% N (worked example 2).
% purity = (19.8 ÷ 21.20) × 100 = 93.40%
So roughly 93.4 g in every 100 g is ammonium sulphate and about 6.6 g is something else. The assumption matters: if the impurity were, say, ammonium chloride, it would carry nitrogen of its own and this calculation would overstate the purity.
Working backwards 3 — water of crystallisation
A hydrate such as blue vitriol, CuSO₄·5H₂O, loses its water on heating. The mass lost is a direct measurement of the %H₂O, and this is a standard Class 11–12 practical.
Step 1 — molar mass of the hydrate.
CuSO₄: 63.546 + 32.06 + (4 × 15.999 = 63.996) = 159.602
5H₂O: 5 × 18.015 = 90.075
M = 159.602 + 90.075 = 249.677 g/mol
Step 2 — percentage of water.
%H₂O = (90.075 ÷ 249.677) × 100 = 36.08%
Step 3 — predict the experiment. Heat 5.00 g of the blue crystals to
constant mass:
water driven off = 5.00 × 0.36076 = 1.80 g
white anhydrous CuSO₄ left = 5.00 − 1.80 = 3.20 g
If your experimental loss is much less than 1.80 g the sample was not heated to constant mass; if it is much more, some sulphate has decomposed.
From percent composition to a formula
The reverse route — given the percentages, find the formula — is the empirical formula calculation: divide each percentage by the element's atomic mass, divide all the answers by the smallest, then multiply up to whole numbers. That is a separate calculation with its own pitfalls and it is covered in empirical vs molecular formula.
Common mistakes that cost marks
- Not expanding brackets. (NH₄)₂SO₄ has 8 hydrogens. Writing 4 gives a molar mass of 128.1 and every percentage is then wrong.
- Forgetting the water in a hydrate. The molar mass of CuSO₄·5H₂O is 249.68, not 159.60. The dot is not a decimal point and it is not a multiplication sign either — it means five water molecules per formula unit.
- Confusing mass % with mole %. Urea is 46.65% N by mass but its atoms are 2 N out of 8 atoms. Percent composition always means by mass unless a question says otherwise.
- Rounding too early. Keep three decimals through the working; round only the final answer. Rounding the molar mass to 60 in example 1 shifts %N to 46.69%.
- Not adding up to 100. If your percentages sum to 99.2% or 101%, you have an arithmetic error — go back rather than submitting.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE / ICSE Class 11 | Mole concept: %composition, then empirical formula |
| CBSE Class 12 practicals | Water of crystallisation by loss on heating |
| JEE / NEET | Purity and assay problems, combustion analysis |
| IIT-JAM / CUET-PG | Gravimetric analysis, hydrate stoichiometry |
| GATE / CSIR-NET | Assay of reagents, elemental analysis (CHN) data |
Check every percentage in one step. The Molar Mass & Composition calculator takes any formula — including brackets and hydrates such as CuSO4.5H2O — and returns the molar mass together with the element-wise mass percentages, so you can verify all of the numbers above.
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