An empirical formula calculator that turns raw mass-percentage or gram data into the simplest whole-number atom ratio — and, with a molar mass, into the full molecular formula — showing every step of the working.
How it works
The process follows the standard five-step procedure taught in A-level and university chemistry courses:
- Convert to grams. If you are working from percentage composition, assume a 100 g sample so each percentage value is numerically equal to a mass in grams.
- Convert grams to moles. Divide each element’s mass by its standard atomic mass (IUPAC 2021). For carbon: n(C) = mass(C) / 12.011 g mol⁻¹.
- Normalise to the smallest mole value. Divide every mole count by the smallest value in the set to obtain the mole ratios.
- Round to whole numbers. Ratios should be close to integers. When a ratio falls near a simple fraction (e.g. 1.5 → 3/2, 1.333 → 4/3), the tool tests multipliers from 1 to 8 to find the smallest integer factor that clears all fractions simultaneously. It then simplifies by dividing by the greatest common divisor.
- Scale to the molecular formula. If you supply the experimental molar mass M, the whole-number multiplier n = round(M / empirical formula mass) is applied to every subscript.
The calculator stores atomic masses for 86 elements to IUPAC 2021 precision.
Worked example — glucose
Combustion analysis of glucose yields:
| Element | % by mass |
|---|---|
| C | 40.00 % |
| H | 6.71 % |
| O | 53.29 % |
Step 1 — assume 100 g sample: C = 40.00 g, H = 6.71 g, O = 53.29 g
Step 2 — moles:
- n(C) = 40.00 / 12.011 = 3.330 mol
- n(H) = 6.71 / 1.008 = 6.657 mol
- n(O) = 53.29 / 15.999 = 3.331 mol
Step 3 — divide by smallest (3.330): C:H:O = 1.000 : 1.998 : 1.000 ≈ 1 : 2 : 1
Step 4 — empirical formula: CH₂O (formula mass = 30.026 g mol⁻¹)
Step 5 — molecular formula with M = 180.16 g mol⁻¹: n = 180.16 / 30.026 ≈ 6, so the molecular formula is C₆H₁₂O₆ (glucose).
The tool reproduces this result automatically and expands the full working on demand.
Formula note
The key equation is:
moles of element = mass (g) / atomic mass (g mol⁻¹)
Once mole ratios are found, the empirical formula mass (EFM) is:
EFM = Σ (subscript_i × atomic_mass_i)
The molecular formula multiplier is n = round(M / EFM), where M is the experimentally
determined molar mass (from mass spectrometry or other method). If n = 1 the
empirical and molecular formulas are identical.