Empirical Formula CalculatorPercent composition, grams or combustion data to the simplest formula
Get the empirical formula with every step, then scale it to the molecular formula when you know the molar mass.
By Manzoor Ahmad, web development expert · Updated How do you find an empirical formula?
Convert each element’s mass to moles, divide by the smallest mole value, and multiply by a whole number until every ratio is whole. A compound that is 40.0% C, 6.71% H and 53.3% O gives 3.33 : 6.66 : 3.33 mol, a ratio of 1 : 2 : 1, so the empirical formula is CH₂O. With a molar mass of 180 g/mol it becomes C₆H₁₂O₆.
| Element | Mass (g) | Moles | ÷ smallest | × mult | Subscript |
|---|
The four-step empirical formula method
For the molecular formula, divide the measured molar mass by the empirical formula mass: n = M ÷ MEF. Multiply each subscript by n.
Worked examples
1. Vitamin C: a ratio that needs ×3
Ascorbic acid is 40.92% C, 4.58% H and 54.50% O. Moles: C 3.407, H 4.544, O 3.406. Dividing by 3.406 gives 1 : 1.334 : 1. The 1.33 tells you to multiply by 3: C₃H₄O₃. With a molar mass of 176.12 g/mol, n = 2, so the molecular formula is C₆H₈O₆.
2. Iron oxide: a ratio of 1.5
69.94% Fe and 30.06% O give 1.252 mol Fe and 1.879 mol O, a ratio of 1 : 1.50. Multiply by 2: Fe₂O₃.
3. Combustion of an alcohol
Burning 1.000 g of a C, H, O compound gives 2.197 g CO₂ and 1.199 g H₂O. Carbon = 2.197 × 12.011 ÷ 44.009 = 0.5996 g; hydrogen = 1.199 × 2.016 ÷ 18.015 = 0.1342 g; oxygen = 1.000 − 0.5996 − 0.1342 = 0.2662 g. The mole ratio is 3 : 8 : 1, giving C₃H₈O (propanol).
Decimal endings and the multiplier to use
| Ratio ends in about | Multiply all by | Example |
|---|---|---|
| .50 | 2 | 1 : 1.5 → 2 : 3 |
| .33 or .67 | 3 | 1 : 1.33 → 3 : 4 |
| .25 or .75 | 4 | 1 : 2.25 → 4 : 9 |
| .20, .40, .60, .80 | 5 | 1 : 1.2 → 5 : 6 |
| .17 or .83 | 6 | 1 : 1.17 → 6 : 7 |
Only round when a value is within about 0.1 of a whole number. Rounding 1.5 down to 1 or up to 2 is the most common empirical formula mistake.
Empirical formula FAQs
An empirical formula shows the simplest whole-number ratio of atoms in a compound. Glucose's molecular formula is C6H12O6, but its empirical formula is CH2O.
Assume a 100 g sample so percentages become grams, convert each to moles by dividing by atomic weight, divide every mole value by the smallest, then multiply by a small whole number if needed to clear fractions.
Multiply all ratios by 2. Common fractional endings tell you the multiplier: .5 → ×2, .33 or .67 → ×3, .25 or .75 → ×4, .2 → ×5. The calculator finds this automatically.
Divide the compound's molar mass by the empirical formula mass, round to a whole number n, then multiply every subscript by n. CH2O has a formula mass of 30.03; with a molar mass of 180.16, n = 6, giving C6H12O6.
Yes. Water (H2O), carbon dioxide (CO2) and many other compounds already have their atoms in the simplest ratio, so the two formulas are identical.
Combustion analysis burns a sample of an organic compound in excess oxygen and weighs the CO2 and H2O produced. All the carbon ends up in CO2 and all the hydrogen in H2O, so their masses reveal how much C and H the sample contained.
Oxygen is found by difference: sample mass minus the mass of carbon and hydrogen (and any other measured elements). Oxygen from the burning gas cannot be separated from oxygen in the sample.
Experimental data contain small errors. Values within about 0.1 of a whole number are normally rounded. A ratio like 1.48 should be treated as 1.5, which means multiplying by 2, not rounding to 1.
No. Percent data are easiest with an assumed 100 g sample, but any masses work because only the ratios matter. You can enter grams directly in this calculator.
The calculator uses Hill order for carbon compounds: C first, H second, then other elements alphabetically. Without carbon, all elements are alphabetical.
No. Different compounds can share an empirical formula. Formaldehyde, acetic acid and glucose are all CH2O. You need the molar mass or structural data to tell them apart.
Elemental analysis results are usually reported to two decimal places. Small rounding in the input rarely matters, but errors over about 0.5% can lead to the wrong multiplier for complex formulas.
Sources
This empirical formula calculator is an educational tool. Structural identification of unknown compounds requires spectroscopic and analytical confirmation.