Empirical Formula Calculator
Turn percent composition or measured masses into the simplest whole-number formula, then scale it to the molecular formula with the molar mass.
EF
Empirical formula
Formula mass 30.03 g/mol
MF
Molecular formula
180.16 ÷ 30.03 ≈ 6
| Element | Grams | Moles | ÷ smallest | × 1 |
|---|---|---|---|---|
| C | 40 | 3.33 | 1.000 | 1 |
| H | 6.7 | 6.647 | 1.996 | 2 |
| O | 53.3 | 3.331 | 1.000 | 1 |
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Worked example: glucose
A compound is 40.0 % C, 6.7 % H and 53.3 % O. In 100 g: C 40.0 ÷ 12.011 = 3.33 mol, H 6.7 ÷ 1.008 = 6.65 mol, O 53.3 ÷ 15.999 = 3.33 mol. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is CH₂O (30.03 g/mol). With a molar mass of 180.16 g/mol, 180.16 ÷ 30.03 = 6, so the molecular formula is C₆H₁₂O₆.
Frequently asked questions
How do you find an empirical formula from percent composition?
Assume a 100 g sample so percentages become grams. Convert each to moles with atomic masses, divide every mole value by the smallest, and if the ratios are not whole numbers multiply them all by 2, 3, 4… until they are.
What is the difference between empirical and molecular formula?
The empirical formula is the simplest whole-number ratio of atoms (CH₂O for glucose). The molecular formula is the actual number of atoms in a molecule (C₆H₁₂O₆), a whole-number multiple of the empirical formula.
How do I get the molecular formula?
Divide the molar mass of the compound (from a mass spectrum or a problem statement) by the empirical formula mass. The result, rounded to a whole number, multiplies every subscript.
What if my ratio is 1.5 or 1.33?
Those are clues to multiply: x.5 → ×2, x.33 or x.67 → ×3, x.25 or x.75 → ×4. The calculator tries multipliers 1 to 6 and picks the first that makes every ratio within 0.1 of a whole number.
My percentages don’t add to 100. Is that a problem?
Small differences come from rounding and are fine. A big gap often means oxygen was not listed and should be found by difference: O % = 100 − (sum of the others).