Empirical Formula Calculator
Find the formula of a compound from its composition: the mass percentages, the grams of each element, or a combustion analysis. Every step is shown — grams to moles, dividing by the smallest, making whole numbers — and with the molar mass you also get the molecular formula.
Empirical and Molecular Formula
The empirical formula gives the simplest whole-number ratio of the atoms in a compound: glucose, C6H12O6, has the empirical formula CH2O. The molecular formula is a whole multiple of it, and the molar mass tells you which: 180 g/mol is 6 × the 30 g/mol of CH2O.
A chemical analysis gives masses, but formulas count atoms. So the masses are turned into moles (divide by the atomic mass), and the moles are compared: dividing by the smallest gives the ratio. Measurement errors make the ratio slightly off (1.996 instead of 2); a value close to x.5 or x.33 means the ratio still has to be multiplied by 2 or 3.
From Grams to a Formula
formulaStep-by-Step Solution
Worked Techniques
Take 100 g, so the percentages are grams. Divide by the atomic masses and by the smallest number of moles: CH2O.
Never round 1.5 to 2: multiply every ratio by the same small whole number until all are whole.
Burn a known mass of the compound: the carbon is found back in CO2, the hydrogen in H2O, and the rest of the mass is oxygen.
Sources & Verification
- T. L. Brown et al., Chemistry: The Central Science, 14th ed., Pearson, 2018 — Sec. 3.5 (empirical formulas from analyses, molecular formulas, combustion analysis).
- IUPAC Commission on Isotopic Abundances and Atomic Weights, Standard Atomic Weights, 2021 — the atomic masses used.
Frequently Asked Questions
How do you find the empirical formula from mass percentages?
Turn the masses into moles and compare them. Example: a compound with 40.0% C, 6.7% H and 53.3% O.
1. Take 100 g: that is 40.0 g C, 6.7 g H and 53.3 g O.
2. Divide by the atomic masses: 40.0 / 12.011 = 3.33 mol C, 6.7 / 1.008 = 6.65 mol H, 53.3 / 15.999 = 3.33 mol O.
3. Divide by the smallest (3.33): C 1, H 2.00, O 1.00.
4. The empirical formula is CH₂O.
How do you get the molecular formula from the empirical formula?
Divide the molar mass of the compound by the mass of the empirical formula; the answer is a whole number, and every subscript is multiplied by it. CH₂O weighs 30.03 g/mol. If the compound has a molar mass of 180.16 g/mol, it is 180.16 / 30.03 = 6 times as heavy, so the molecular formula is C₆H₁₂O₆ (glucose).
What do I do when the ratio is not a whole number?
Do not round a ratio such as 1.5 or 1.33: multiply all the ratios by the same small number until they are all whole. Iron oxide with 69.94% Fe and 30.06% O gives Fe : O = 1 : 1.5, and × 2 gives Fe₂O₃. Propane with 81.8% C and 18.2% H gives 1 : 2.65 (close to 2.67), and × 3 gives C₃H₈. Only ratios within about 0.1 of a whole number are rounded directly.
How does a combustion analysis give the empirical formula?
Burn a weighed sample in oxygen and weigh the CO₂ and H₂O that form. All the carbon is in the CO₂ (12.011 of every 44.009 g) and all the hydrogen in the H₂O (2.016 of every 18.015 g); the rest of the sample is oxygen. A 0.2000 g sample giving 0.2933 g CO₂ and 0.1200 g H₂O contains 0.0800 g C, 0.0134 g H and 0.1065 g O, which works out to CH₂O.
What is the difference between the empirical and the molecular formula?
The empirical formula only gives the simplest ratio of the atoms; the molecular formula gives the real number of atoms in one molecule. Glucose (C₆H₁₂O₆), acetic acid (C₂H₄O₂) and formaldehyde (CH₂O) all have the empirical formula CH₂O. For many compounds, such as water (H₂O) or ammonia (NH₃), the two are the same.