Empirical Formula Calculator
Enter the percentage or mass of each element, or the CO₂ and H₂O from a combustion, and get the simplest whole-number formula with the moles, the ratios and the multiplier shown. Add the compound's molar mass to turn it into the molecular formula. Free, no sign-up.
Elements
Add it to turn the empirical formula into the molecular formula.
Results
C3H4O3
Empirical formula
88.062 g/mol
Empirical formula mass
From masses to a whole-number ratio
| Element | Percent (%) | Atomic weight (g/mol) | Moles | ÷ smallest | × 3 | Subscript |
|---|---|---|---|---|---|---|
| C | 40.92 | 12.011 | 3.407 | 1.000 | 3.000 | 3 |
| H | 4.58 | 1.008 | 4.544 | 1.334 | 4.001 | 4 |
| O | 54.5 | 15.999 | 3.406 | 1.000 | 3.000 | 3 |
The working
C: 40.92 ÷ 12.011 g/mol = 3.407 mol
H: 4.58 ÷ 1.008 g/mol = 4.544 mol
O: 54.5 ÷ 15.999 g/mol = 3.406 mol
Divide by the smallest (O): 1.000 : 1.334 : 1.000
Those are not all whole numbers, so multiply by 3: 3 : 4 : 3
How to Use the Empirical Formula Calculator
- 1. Choose your starting data. Use percent composition or masses when you are given the share or mass of each element. Use combustion analysis when a sample was burned and you know the grams of CO₂ and H₂O it produced.
- 2. Enter the elements. Type each element's symbol and its percentage or mass in grams; add or remove rows as needed. For combustion, enter the sample mass and the two product masses, and leave the oxygen option on if the rest of the sample is oxygen.
- 3. Add the molar mass if you have it. An optional molar mass in g/mol turns the empirical formula into the molecular formula.
- 4. Read the steps. The table shows moles, the ratio to the smallest amount, any multiplier, and the final whole-number subscripts.
The Method
An empirical formula gives the simplest whole-number ratio of atoms in a compound. Experiments report masses or percentages, not atoms, so the method converts to moles first — moles are proportional to atoms:
- Treat percentages as grams in a 100 g sample, then divide each mass by the element's atomic weight to get moles.
- Divide every amount by the smallest, so the smallest becomes 1.
- If any ratio is not close to a whole number, multiply all of them by the smallest integer that fixes it, then round to whole subscripts.
| Decimal part of the ratio | Multiply every ratio by |
|---|---|
| 0.5 (1.5, 2.5 …) | 2 |
| 0.33 or 0.67 (1.33, 2.67 …) | 3 |
| 0.25 or 0.75 (1.25, 2.75 …) | 4 |
| 0.2, 0.4, 0.6 or 0.8 | 5 |
Real data are rounded, so ratios are rarely exact. The calculator rounds a ratio that is within 0.1 of a whole number (you can choose a stricter or looser limit); once a multiplier is needed, the fit must be within half that limit, so a large multiplier cannot make a poor fit look good. It tries multipliers up to 8.
Worked Example: Ascorbic Acid
Ascorbic acid (vitamin C) is 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass — the calculator's default data. In a 100 g sample:
- C: 40.92 ÷ 12.011 = 3.407 mol
- H: 4.58 ÷ 1.008 = 4.544 mol
- O: 54.50 ÷ 15.999 = 3.406 mol
- Divide by the smallest (3.406): C 1.000 : H 1.334 : O 1.000
- 1.334 is close to 1⅓, so multiply by 3: C₃H₄O₃
The empirical formula mass is 3 × 12.011 + 4 × 1.008 + 3 × 15.999 = 88.062 g/mol. If the measured molar mass is 176.12 g/mol, then 176.12 ÷ 88.062 = 2.000, so every subscript doubles and the molecular formula is C₆H₈O₆.
Worked Example: Combustion Analysis
A 1.000 g sample containing only C, H and O burns to give 1.911 g of CO₂ and 1.173 g of H₂O. All the carbon is in the CO₂ and all the hydrogen in the water, so the masses follow from the molar masses (CO₂ 44.009 g/mol, H₂O 18.015 g/mol):
- Carbon: 1.911 × 12.011 ÷ 44.009 = 0.5216 g
- Hydrogen: 1.173 × 2 × 1.008 ÷ 18.015 = 0.1313 g
- Oxygen, by difference: 1.000 − 0.5216 − 0.1313 = 0.3472 g
- Moles 0.04342 : 0.1302 : 0.02170, which is a ratio of 2.001 : 6.001 : 1.000 — C₂H₆O
What the empirical formula cannot tell you
C₂H₆O could be ethanol or dimethyl ether: the empirical formula gives the ratio only, never the structure, and the molecular formula needs a molar mass from a separate measurement such as mass spectrometry. To check the formula you find, the Percent Composition Calculator works in the opposite direction.
Common Mistakes
- Rounding 1.33 or 1.5 to a whole number. Those ratios mean a multiplier is needed (×3 and ×2); rounding gives the wrong formula.
- Rounding the moles early. Keep four or more digits until the ratios are found.
- Dividing by the largest amount. Divide by the smallest, so every subscript is at least 1.
- Forgetting oxygen in a combustion problem. It is not measured directly. Subtract the carbon and hydrogen masses from the sample.
- Reporting a molecular formula that is not a whole multiple. If the molar mass ÷ empirical mass is not close to a whole number, the data or the molar mass has an error.
Frequently Asked Questions
- How do you find the empirical formula from percent composition?
- Assume 100 g of compound so each percentage becomes grams, convert each to moles by dividing by the atomic weight, divide all the mole amounts by the smallest, and multiply by a small integer if any ratio is not whole. The whole numbers are the subscripts.
- What is the difference between an empirical and a molecular formula?
- The empirical formula is the simplest whole-number ratio, and the molecular formula is the actual number of atoms in a molecule — a whole-number multiple of it. Glucose is C₆H₁₂O₆ as a molecule and CH₂O empirically; water is H₂O in both.
- What do I do when a ratio is 1.5 or 1.33?
- Multiply every ratio by the smallest integer that makes them all whole: 1.5 needs 2, and 1.33 or 1.67 needs 3. A ratio like 1.2 needs 5. The table above lists the common cases, and the calculator applies the multiplier for you.
- How does combustion analysis give an empirical formula?
- Burning a hydrocarbon or an oxygen-containing compound in excess oxygen converts all its carbon to CO₂ and all its hydrogen to H₂O. The masses of those products give the grams of carbon and hydrogen, oxygen comes from the difference, and the rest is the usual mole-ratio method.
- Can the empirical formula be the same as the molecular formula?
- Yes. When the atoms cannot be divided by a common factor, as in H₂O, CH₄ and CO₂, the two are identical. Ionic compounds are always written as empirical formulas, such as NaCl and Fe₂O₃, because they have no discrete molecules.
- What if my percentages do not add up to 100?
- A total a little off 100 is rounding. A total well below 100 usually means an element was not reported; oxygen is often found by difference. The calculator warns when the percentages are more than 1 away from 100.