Chemical analysis tells you how much of each element a compound contains, but not directly how the atoms are arranged. The first step toward identifying it is the empirical formula: the simplest whole-number ratio of atoms that matches the data. This calculator converts mass percentages, element masses or the CO₂ and H₂O produced by burning a sample into moles, finds the smallest whole-number ratio and, if you know the molar mass, scales it up to the molecular formula.
How to use the empirical formula calculator
- Choose the Data you have: mass percent, grams of each element, or combustion analysis.
- For percent or grams, type one element per line in the Composition box, such as
C 40.00. You can use symbols or names (carbon 40). - For combustion analysis, enter the Mass of sample burned, the Mass of CO₂ produced and the Mass of H₂O produced. Leave “Assign any remaining mass to oxygen” ticked for compounds of carbon, hydrogen and oxygen.
- Optionally enter the Molar mass of the compound to get the molecular formula.
- Read the formula, the empirical formula mass, the multiplier and a working table showing moles, ratios and the rounding.
How the empirical formula is calculated
- Treat percentages as grams of a 100 g sample.
- Convert each mass to moles: n = m ÷ A, where A is the standard atomic weight.
- Divide every mole value by the smallest one.
- If any ratio is not close to a whole number, multiply all ratios by the smallest factor (2 to 12) that makes them whole within ±0.1.
- Round, and write the subscripts.
For the molecular formula:
In combustion analysis, the masses of carbon and hydrogen are mC = mCO₂ × 12.011 ÷ 44.009 and mH = mH₂O × 2.016 ÷ 18.015.
Worked example
A sugar from its percent composition
A compound is 40.00% C, 6.71% H and 53.29% O, with a molar mass of 180.16 g/mol.
Moles in 100 g: C 40.00 ÷ 12.011 = 3.3303; H 6.71 ÷ 1.008 = 6.6567; O 53.29 ÷ 15.999 = 3.3308
Divide by 3.3303: C 1, H 1.999, O 1.000 → empirical formula CH2O (30.026 g/mol)
Enter the molar mass of 180.16 g/mol: 180.16 ÷ 30.026 = 6.00, so the molecular formula is C6H12O6, glucose.
When ratios are fractional. A phosphorus oxide is 43.64% P and 56.36% O. The mole ratio is P 1 : O 2.5, so multiply by 2 to get P2O5. With a molar mass of 283.9 g/mol, the molecular formula is P4O10.
From combustion data. Burning 0.500 g of a compound gives 0.7329 g CO2 and 0.3001 g H2O. That is 0.2000 g C and 0.0336 g H, leaving 0.2664 g O by difference. The mole ratio again works out to CH2O.
Common fractional ratios
| Ratio after dividing | Multiply by | Example |
|---|---|---|
| x.5 | 2 | Fe 1 : O 1.5 → Fe₂O₃ |
| x.33 or x.67 | 3 | 1 : 1.33 → 3 : 4 |
| x.25 or x.75 | 4 | 1 : 1.25 → 4 : 5 |
| x.2, x.4, x.6, x.8 | 5 | 1 : 2.4 → 5 : 12 |
Do not round a ratio like 1.5 or 1.33 to the nearest integer; that is the most common mistake in empirical formula problems. Values within about 0.1 of a whole number, such as 1.98 or 3.04, can safely be rounded, since they reflect small measurement errors.
Writing the formula
For carbon compounds the calculator uses Hill order: carbon first, hydrogen second, then the other elements alphabetically. For compounds without carbon it keeps the order you typed, so enter metals first to get familiar forms like Fe₂O₃ and NaCl.
To go the other way, from a formula to its composition, use the percent composition calculator. The molar mass calculator checks the mass of the formula you found, and the ratio simplifier shows the whole-number reduction on its own.
Frequently asked questions
What is the difference between an empirical and a molecular formula?
The empirical formula is the simplest whole-number ratio of atoms; the molecular formula gives the actual number in one molecule. Glucose is C₆H₁₂O₆, but its empirical formula is CH₂O, shared with formaldehyde and acetic acid.
How do I get the molecular formula?
Divide the compound's molar mass by the empirical formula mass and round to a whole number, then multiply every subscript by it. For glucose, 180.16 ÷ 30.026 = 6, so CH₂O becomes C₆H₁₂O₆. Enter the molar mass in the optional field and the calculator does this for you.
Why do I sometimes need to multiply the ratios?
Dividing by the smallest number of moles can leave fractions such as 1.5, 1.33 or 1.25. Multiply everything by 2, 3 or 4 respectively to clear them. Iron oxide with 69.94% Fe gives Fe 1 : O 1.5, which becomes Fe₂O₃.
How does combustion analysis work?
Burning an organic compound turns all its carbon into CO₂ and all its hydrogen into H₂O. Their masses give the grams of C and H in the sample; whatever mass remains is usually oxygen. Burning 0.500 g of glucose gives 0.7329 g of CO₂ and 0.3001 g of H₂O.
What if my percentages do not add up to 100%?
Small gaps come from rounding in the analysis. If they fall short by more than a percent or two, an element is probably missing, often oxygen, which is commonly found by difference. The calculator notes the gap and rejects totals above 100.5%.