Empirical Formula Mole Ratio from Percent Composition

A combustion analysis hands you percentages by mass, and you want subscripts. The bridge is that subscripts count atoms, so the percentages have to be converted to moles before they mean anything. Assume a 100 g sample, so each percent becomes a mass in grams; divide each by that element's molar mass; and the ratio of those two numbers is the ratio of the atom counts.

Worked: a compound analyses at 40.00% carbon and 6.71% hydrogen. Carbon, 40.00/12.011 = 3.3303 mol. Hydrogen, 6.71/1.008 = 6.6567 mol. The ratio H:C is 6.6567/3.3303 = 1.9989, which is 2 to four figures. So the empirical formula has twice as many hydrogens as carbons — with the remaining 53.29% being oxygen, and the same arithmetic giving one oxygen per carbon, the answer is CH₂O. Which is glucose's empirical formula, and also formaldehyde's, and also acetic acid's.

That last sentence is the honest limit of the method: an empirical formula is a ratio, not a molecule. CH₂O is C₆H₁₂O₆ divided by six. To get the molecular formula you need one more measurement — the molar mass, from mass spectrometry or a colligative property — and then you divide it by the empirical formula mass to find the multiplier.

The step this page cannot do for you is the rounding, because it is a judgement call rather than arithmetic. A ratio of 1.9989 is plainly 2. A ratio of 1.50 is not 2 and is not 1 — it means the whole set doubles, giving 3:2. Ratios near 1.33 triple to 4:3, near 1.25 quadruple to 5:4. Rounding 1.50 to the nearest integer is the classic way Fe₂O₃ becomes FeO₂ and a whole analysis is thrown away at the last step.

Two practical notes. Divide every element by the smallest mole figure in the set, so the smallest count becomes 1 and the rest are read against it. And percentages that do not add to 100 usually mean an element was determined by difference — commonly oxygen, which combustion analysis cannot measure directly.