Elemental composition of pure substances

A pure substance has a single, fixed elemental makeup. Every sample of water, anywhere, is the same 11.2% hydrogen and 88.8% oxygen by mass. That fixed makeup is its percent composition: the fraction of the total mass contributed by each element.

You can read percent composition straight off a formula. For each element, take the number of its atoms times its atomic mass, divide by the compound's molar mass, and multiply by 100:

mass % of X = (atoms of X × atomic mass of X) ÷ (molar mass) × 100

Run in reverse, the same idea recovers a formula from lab data. If you know how many grams of each element a compound contains, you can find the empirical formula — the simplest whole-number ratio of atoms — and, with the molar mass, the molecular formula.

Given percent composition (or masses) and asked for a formula, this is the whole method. The engine, as always, is moles — ratios of atoms only make sense once you count particles, not grams.

1. Assume 100 g of the compound. Then every percent becomes grams: 40.0% carbon is simply 40.0 g of carbon. This is a free move — ratios do not care about sample size.
2. Convert each element's grams to moles. Divide each mass by that element's molar mass. Now you have a mole ratio, which is a ratio of atoms.
3. Divide every mole amount by the smallest one. This scales the ratio so the least-abundant element becomes 1. If the results are not close to whole numbers, multiply them all by a small integer (2, 3…) until they are.
4. Read off the empirical formula — then scale to molecular if asked. The whole-number ratio is the empirical formula. To get the molecular formula, divide the compound's molar mass by the empirical-formula mass and multiply every subscript by that whole number.

The empirical formula is only a ratio; the molecular formula is the real count of atoms in one molecule. You cannot tell them apart from composition alone — that final subscript-scaling step needs the molar mass.

Quick reference card. Keep straight what is a ratio and what is a real count.

Question. A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, with a molar mass of 180 g/mol. Find its empirical and molecular formulas. (C = 12, H = 1, O = 16 g/mol.)

Step 1 — assume 100 g. That gives 40.0 g C, 6.7 g H, and 53.3 g O.

Step 2 — grams to moles. 40.0 ÷ 12 = 3.33 mol C; 6.7 ÷ 1 = 6.6 mol H; 53.3 ÷ 16 = 3.33 mol O.

Step 3 — divide by the smallest (3.33). C: 3.33 ÷ 3.33 = 1; H: 6.6 ÷ 3.33 ≈ 2; O: 3.33 ÷ 3.33 = 1. The ratio is 1 : 2 : 1, so the empirical formula is CH₂O.

Step 4 — scale to molecular. The empirical-formula mass of CH₂O is 12 + 2 + 16 = 30 g/mol. Divide the molar mass: 180 ÷ 30 = 6. Multiply every subscript by 6: the molecular formula is C₆H₁₂O₆ — glucose.

Sanity check. Oxygen owns the largest mass share (53.3%) even though there are equal numbers of C and O atoms, because each O atom is heavier than each C atom. Mass percent weighs atoms; the formula counts them.