Chemistry and laboratory
Empirical and molecular formula calculator
Empirical formula from percent composition or grams, and the molecular formula from a molar mass, with the mole-ratio working shown.
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Result
- Empirical formula
- CH₂O
- Molecular formula
- C₆H₁₂O₆
- Empirical formula mass
- 30.026 g/mol
- Multiplier n = M ÷ empirical mass
- 6.000 → 6
Working
- Moles in a 100 g sample: C 40.0 ÷ 12.011 = 3.330; H 6.7 ÷ 1.0080 = 6.647; O 53.3 ÷ 15.999 = 3.331
- Divide by the smallest (3.330 mol): C 1.000, H 1.996, O 1.000
- n = 180.16 g/mol ÷ 30.026 g/mol = 6.000 ≈ 6
| Element | Mass in 100 g (g) | Moles | Ratio | Whole number |
|---|---|---|---|---|
| carbon (C) | 40.0 | 3.330 | 1.000 | 1 |
| hydrogen (H) | 6.7 | 6.647 | 1.996 | 2 |
| oxygen (O) | 53.3 | 3.331 | 1.000 | 1 |
- Formulas are written in Hill order (C, H, then alphabetical).
Method and limits
Enter the mass percent (or grams) of each element. The tool converts them to moles, divides by the smallest, and multiplies by the smallest whole number (up to 12) that brings every ratio within 0.1 of an integer.
nᵢ = mᵢ ÷ Ar(i) ratio = nᵢ ÷ n(min) molecular formula = (empirical formula)ₖ, k = M ÷ M(empirical)
Percentages are read as grams in a 100 g sample. Common fractional ratios are 1.5 (multiply by 2), 1.33 or 1.67 (by 3) and 1.25 or 1.75 (by 4). Atomic weights are the CIAAW 2024 abridged values.
Limits
Analytical data carry errors, so a ratio such as 1.92 may mean 2. If no multiplier up to 12 gives near-whole numbers, the best one is used and a warning appears; recheck the data. The empirical formula cannot tell isomers apart.
Worked example
C 40.0 %, H 6.7 %, O 53.3 %: 3.33 mol C, 6.65 mol H, 3.33 mol O, ratio 1 : 2.00 : 1, so CH₂O (30.03 g/mol). With M = 180.16 g/mol, k = 6.00 and the molecular formula is C₆H₁₂O₆. Iron oxide with 69.94 % Fe gives Fe : O = 1 : 1.50, doubled to Fe₂O₃.
References
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