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Empirical Formula Calculator

Enter the mass percentage of each element to determine the empirical formula. Optionally provide the molar mass to find the molecular formula.

From mass percent to empirical formula

The empirical formula is the simplest whole-number ratio of elements in a compound. Determining one from experimental data — usually mass percentages or combustion analysis output — is a fixed procedure: convert percentages to moles, take the ratio, clean up to whole numbers.

The method

  1. Pretend you have 100 g of sample. Mass percentages now read as grams directly. (Any sample mass works; 100 g just makes the arithmetic clean.)
  2. Convert each element’s mass to moles by dividing by atomic mass. This is the only step that requires looking anything up.
  3. Divide every mole value by the smallest one. This rescales the ratio so the least-abundant element is exactly 1, and everything else is a multiple of it.
  4. Get to whole numbers. If the ratios are within about 0.05 of integers, round. If not, multiply all of them by 2, 3, or 4 — whichever clears the fractions. A ratio of 1:1.5 multiplies to 2:3; a ratio of 1:1.33 multiplies to 3:4.
  5. Write the empirical formula with those whole numbers as subscripts.
  6. (Optional) Find the molecular formula. If you know the actual molar mass, divide it by the empirical formula mass. The integer result is the multiplier — apply it to every subscript.

The non-obvious step is step 4. Ratios like 1.33 and 1.5 are not rounding errors; they’re the calculation telling you the empirical formula has more atoms than you assumed. Forcing them to 1 or 2 produces the formula of a different compound.

Worked examples

40.00% C, 6.71% H, 53.29% O. Moles per 100 g: C = 40.00/12.011 = 3.331, H = 6.71/1.008 = 6.657, O = 53.29/15.999 = 3.331. Divide by smallest (3.331): 1.00 : 2.00 : 1.00. Empirical formula: CH₂O. If molar mass = 180.16: 180.16/30.03 = 6, so molecular formula is C₆H₁₂O₆ (glucose).

26.57% K, 35.36% Cr, 38.07% O. Moles: K = 0.6795, Cr = 0.6800, O = 2.380. Divide by smallest: K = 1.00, Cr = 1.00, O = 3.50. Multiply by 2: 2 : 2 : 7. Empirical formula: K₂Cr₂O₇ (potassium dichromate).

Combustion analysis. 0.255 g of a C/H/O compound burns to give 0.561 g CO₂ and 0.306 g H₂O. Mass C = 0.561 × (12.011/44.009) = 0.1531 g. Mass H = 0.306 × (2.016/18.015) = 0.0343 g. Mass O (by difference) = 0.255 − 0.1531 − 0.0343 = 0.0677 g. Moles: C = 0.01275, H = 0.03398, O = 0.00423. Divide by smallest: 3.01 : 8.03 : 1.00. Empirical formula: C₃H₈O (could be 1-propanol, 2-propanol, or methyl ethyl ether — empirical analysis doesn’t distinguish isomers).

Frequently Asked Questions

What is an empirical formula?
The empirical formula is the simplest whole-number ratio of atoms in a compound. Glucose has the molecular formula C₆H₁₂O₆ but the empirical formula CH₂O — the 6:12:6 ratio reduces to 1:2:1. Different compounds can share an empirical formula: formaldehyde (CH₂O), acetic acid (C₂H₄O₂), and glucose (C₆H₁₂O₆) all reduce to CH₂O. The molecular formula tells you which one you actually have.
How do you find an empirical formula from mass percent data?
Assume a 100 g sample so percentages convert directly to grams. Divide each element's mass by its atomic mass to get moles. Divide all the mole values by the smallest of them — that converts moles into a ratio normalized to the least-abundant element. Round to whole numbers if the ratios are close, or multiply through by 2, 3, or 4 if you get fractions like 1.5, 1.33, or 1.25.
What is the difference between empirical and molecular formulas?
The empirical formula is the simplest atom ratio. The molecular formula is the actual count of atoms in one molecule. Find the molecular formula by dividing the experimentally measured molar mass by the empirical formula mass — the result is an integer multiplier. CH₂O has empirical mass 30.03 g/mol; if the molar mass is 180.16, the multiplier is 6, and the molecular formula is C₆H₁₂O₆.
When do I need to multiply the mole ratios?
When dividing by the smallest mole value gives ratios that aren't close to whole numbers. A ratio ending in .5 means multiply by 2; .33 or .67 means multiply by 3; .25 or .75 means multiply by 4. So 1:1.5 becomes 2:3, and 1:1.33 becomes 3:4. Don't round 1.5 to 2 or 1.33 to 1 — that's chemically wrong and gives a different compound.
What is combustion analysis?
Combustion analysis burns a sample (usually an organic compound) in excess O₂ and traps the CO₂ and H₂O produced. The mass of CO₂ tells you how much carbon was in the sample (multiply by 12.011/44.009). The mass of H₂O tells you how much hydrogen (multiply by 2.016/18.015). Subtract C and H mass from the original sample mass to get oxygen by difference. Then run the standard mass-percent-to-empirical-formula procedure.