Empirical Formula Calculator
Amounts are
Gives the molecular formula as a whole multiple of the empirical one.
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Empirical formula
CH₂O
Empirical formula mass 30.026 g/mol
Molecular formula: C₆H₁₂O₆ (180 ÷ 30.03 ≈ 6)
Working
| Element | g per 100 g | Moles | ÷ smallest | × 1, rounded |
|---|---|---|---|---|
| C | 40 | 3.3303 | 1.000 | 1 |
| H | 6.71 | 6.6567 | 1.999 | 2 |
| O | 53.3 | 3.3315 | 1.000 | 1 |
Moles = mass ÷ atomic mass. Ratios like 1.5 or 1.33 are multiplied up (×2, ×3) rather than rounded down.
Find the empirical formula of a compound from the mass percent or grams of each element, and its molecular formula when you know the molar mass. The working shows the moles, the ratio to the smallest and how fractional ratios are scaled to whole numbers.
Formula
Worked Example
In 100 g there are 40.0 ÷ 12.011 = 3.33 mol C, 6.71 ÷ 1.008 = 6.66 mol H and 53.3 ÷ 15.999 = 3.33 mol O, a 1 : 2 : 1 ratio. CH₂O weighs 30.03 g/mol, and 180 ÷ 30.03 ≈ 6, so the molecule is C₆H₁₂O₆.
Frequently Asked Questions
What is the difference between an empirical and a molecular formula?
The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number in a molecule. Glucose's empirical formula is CH₂O and its molecular formula C₆H₁₂O₆, six times larger.
What if the mole ratio comes out as 1.5 or 1.33?
Multiply every ratio by the same small whole number until all are close to integers: 1.5 → ×2, 1.33 → ×3, 1.25 → ×4. Rounding 1.5 down to 1 would give the wrong formula. Iron oxide at 69.9% Fe gives Fe : O = 1 : 1.5, so Fe₂O₃.
Do I need the masses to add up to 100%?
No. Grams from a combustion analysis or any other sample size work just as well, since only the ratio of moles matters. Percentages that don't sum to 100 can mean an element is missing, often oxygen found by difference.