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How to Find an Empirical Formula From Percentages
Heshan Fernando
Co-founder & COO
Combustion analysis gives you percentages: 40.0% carbon, 6.7% hydrogen, 53.3% oxygen. Turning that into a formula is a four-step procedure, and step two is where the chemistry happens.
The procedure
Assume 100 g. Then each percentage is a mass in grams. 40.0% carbon becomes 40.0 g.
Convert to moles. Divide each mass by the element’s atomic mass:
- C: 40.0 ÷ 12.01 = 3.33 mol
- H: 6.7 ÷ 1.008 = 6.65 mol
- O: 53.3 ÷ 16.00 = 3.33 mol
This is the step that matters. Percentages by mass aren’t comparable directly — a gram of hydrogen contains twelve times as many atoms as a gram of carbon. Dividing by atomic mass converts mass into a count of atoms, which is what a formula describes.
Divide by the smallest. 3.33 is the smallest, giving 1 : 2 : 1. Empirical formula: CH₂O.
Scale to the molecular formula if you know the molar mass. CH₂O has a mass of 30.03. Given a molar mass of 180.16: 180.16 ÷ 30.03 = 6, so the molecular formula is C₆H₁₂O₆ — glucose.
Empirical versus molecular
The empirical formula is the simplest whole-number ratio. The molecular formula is the actual atom count in one molecule, and it’s always a whole-number multiple of the empirical.
Formaldehyde is CH₂O. Acetic acid is C₂H₄O₂. Glucose is C₆H₁₂O₆. All three reduce to the same CH₂O ratio and they’re completely different substances — which is exactly why the molar mass is needed, and why percentage composition alone can’t identify a compound.
| Compound | Molecular formula | Empirical | Molar mass |
|---|---|---|---|
| Formaldehyde | CH₂O | CH₂O | 30.03 |
| Acetic acid | C₂H₄O₂ | CH₂O | 60.05 |
| Glucose | C₆H₁₂O₆ | CH₂O | 180.16 |
When ratios aren’t whole numbers
The common complication. You divide by the smallest and get something like 1 : 1.5 : 1.
Don’t round. 1.5 is not 2 — it’s three halves, and multiplying everything by 2 gives 2 : 3 : 2, which is the correct ratio.
The multipliers to try:
- Ending in .5 → multiply by 2
- Ending in .33 or .67 → multiply by 3
- Ending in .25 or .75 → multiply by 4
Rounding 2.5 to 3 is the classic error and it produces a formula for a different compound entirely. Small deviations — 1.98 or 3.02 — are rounding in the source data and can safely be treated as whole numbers.
Why totals don’t reach 100%
Rounding in the reported percentages accounts for a fraction of a percent.
A larger gap usually means an element wasn’t reported. In combustion analysis, carbon and hydrogen are measured directly from the CO₂ and H₂O produced, and oxygen is obtained by difference — subtracting everything else from 100%. If a compound contains nitrogen or sulfur that wasn’t measured, the “oxygen” figure absorbs it and the formula comes out wrong.
A total well under 100% is a signal to ask what else is in there.
Common mistakes to avoid
- Comparing mass percentages directly without converting to moles.
- Rounding 2.5 to 3 instead of doubling everything.
- Using the molecular formula’s mass where the empirical formula’s is needed when scaling.
- Assuming oxygen by difference is correct when other elements may be present.
- Using rounded atomic masses in a calculation where the ratio comes out borderline.
How to do it with Empirical Formula Calculator
The Empirical Formula Calculator shows the mole ratios before simplifying.
- Enter the percentage composition, which should total close to 100%.
- Read the mole ratios — that’s the step where mass becomes atom count.
- Check whether the ratios need multiplying rather than rounding.
- Add the molar mass to scale up to the molecular formula.
Other chemistry tools are in the tools directory.
Frequently asked questions
What’s the difference between empirical and molecular formulas?
The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual count in one molecule. Glucose, acetic acid and formaldehyde all reduce to CH₂O.
What if the ratios aren’t whole numbers?
Multiply everything by a small integer. A ratio ending in .5 needs doubling, .33 needs tripling. Rounding 2.5 to 3 gives the wrong formula.
Why doesn’t my percentage total 100?
Rounding in the source data, or an element not reported — often oxygen, obtained by difference in combustion analysis. A total well under 100% suggests a missing element.
Final thought
Write out the mole ratios before simplifying. Every wrong answer in this calculation comes from rounding a ratio that should have been multiplied.