Empirical Formula Calculator (from mass %)

Understanding the empirical formula calculator — concept and principle

The empirical formula is the simplest whole-number ratio of atoms in a compound. It is what an elemental analysis tells you directly: combustion analysis or other methods give mass percentages, and from those you work back to the atom ratio. This calculator takes the element symbols and their mass percentages (or masses in grams) and returns the empirical formula.

Treat the percentages as grams in 100 g, convert to moles, then divide by the smallest: 1 : 2 : 1 gives CH₂O.

How it is calculated

Divide each mass or percentage by the element's standard atomic weight to get relative moles.
Divide every result by the smallest one.
If the ratios are not close to whole numbers, multiply them all by 2, 3, 4, 5 or 6 until they are.
Round to whole numbers and write the formula.

Percentages and grams work equally well, because only the ratio matters; with percentages, imagine a 100 g sample.

The principle

A formula counts atoms, but analysis measures mass. Dividing mass by atomic weight converts grams of each element into moles of atoms, which are proportional to the number of atoms. Dividing by the smallest value sets one element to 1 so the others show up as simple ratios. Ratios like 1.5 or 1.33 indicate that the true counts are multiples, which is why the multipliers are tried. The molecular formula is a whole-number multiple of the empirical formula and needs the molar mass to determine.

Worked example

Defaults: C, H, O with 40.0%, 6.7%, 53.3%.

C: 40.0 ÷ 12.011 = 3.3303; H: 6.7 ÷ 1.008 = 6.6468; O: 53.3 ÷ 15.999 = 3.3315.
Divide by the smallest (3.3303): C 1.000, H 1.996, O 1.000.
All are close to whole numbers, so no multiplier is needed.
Empirical formula: CH2O (CH₂O).

How to read the result

CH₂O is the empirical formula shared by formaldehyde, acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆). To decide which compound you have, divide its measured molar mass by the empirical formula mass (30.03 g/mol): 180 ÷ 30.03 ≈ 6 gives C₆H₁₂O₆. Analytical data usually carries errors of about 0.3–0.4%, so ratios like 1.996 are rounded confidently, but values like 1.25 call for a multiplier of 4.

Common mistakes

Rounding 1.5 to 2 instead of multiplying everything by 2.
Rounding intermediate values too early, which distorts the ratios.
Percentages that do not add up to about 100%, often because oxygen was not measured and must be found by difference.
Confusing the empirical formula with the molecular formula.

Keep in mind: The method relies on accurate analysis data. Errors of even 0.3–0.5% can make the ratio ambiguous for large molecules, and the result says nothing about structure or isomers.

Frequently asked questions

What if my percentages do not total 100%?

If one element (often oxygen) was not measured, assign it the remaining percentage before calculating.

How do I get the molecular formula?

Divide the molar mass from an independent measurement by the empirical formula mass and multiply all subscripts by that whole number.

Can I enter grams instead of percentages?

Yes, the ratios are the same either way.

Sources: Atkins, P., Jones, L., Chemical Principles, W. H. Freeman · IUPAC Commission on Isotopic Abundances and Atomic Weights, Standard atomic weights

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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