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Combustion Analysis Calculator (Empirical Formula)
In short: Combustion analysis calculator: from the masses of a sample and the CO₂ and H₂O it forms, find the mass of C, H and O, the mole ratios, the empirical formula and the molecular formula. It is free, works in your web browser and needs no download.
Enter the sample mass and the masses of CO₂ and H₂O produced. Add N₂ or the mass percent of other elements if you have them, and choose whether the remainder is oxygen or nitrogen. The calculator shows the mass of each element, the mole ratios, a smart multiplier for ratios like 1.33 or 1.5, and the molecular formula if you give the molar mass. You can also type a percent composition directly.
Other elements from separate analyses (mass % of sample)
Other elements
How combustion analysis works
The sample burns in excess oxygen. All its carbon ends up in CO₂ and all its hydrogen in H₂O, which are trapped and weighed. Oxygen in the compound cannot be measured this way because O₂ was added, so it is found by difference.
- Mass of C = mass of CO₂ × 12.011/44.009 (0.2729 g of C per g of CO₂).
- Mass of H = mass of H₂O × 2.016/18.015 (0.1119 g of H per g of H₂O).
- Add N (from N₂ collected, or from a separate analysis) and any other elements.
- Mass of O = sample mass − (C + H + other elements).
- Convert each mass to moles and divide by the smallest number of moles.
- If a ratio ends in about .5, .33, .25, .2 or .17, multiply all ratios by 2, 3, 4, 5 or 6. Then round.
- Molecular formula = (empirical formula) × n, where n = molar mass ÷ empirical formula mass.
| Ratio ends in about | .5 | .33 or .67 | .25 or .75 | .2, .4, .6, .8 | .17 or .83 |
|---|---|---|---|---|---|
| Multiply by | 2 | 3 | 4 | 5 | 6 |
Atomic masses used: C 12.011, H 1.008, N 14.007, O 15.999, S 32.06, Cl 35.45, Br 79.904, P 30.974, F 18.998.
Worked examples
Glucose
0.2000 g burns to 0.2931 g CO₂ and 0.1200 g H₂O.
C = 0.0800 g, H = 0.0134 g, O = 0.2000 − 0.0934 = 0.1066 g. Moles: C 0.00666, H 0.0133, O 0.00666, ratio 1 : 2 : 1.
CH₂O; with M = 180.16, C₆H₁₂O₆
Vitamin C (multiply by 3)
1.000 g gives 1.500 g CO₂ and 0.409 g H₂O.
C 0.4094 g, H 0.0458 g, O 0.5449 g. Moles 0.0341 : 0.0454 : 0.0341 = 1 : 1.33 : 1. Multiply by 3.
C₃H₄O₃; with M = 176.1, C₆H₈O₆
Urea, with nitrogen
0.3000 g gives 0.2198 g CO₂, 0.1800 g H₂O and 0.1399 g N₂.
C 0.0600 g, H 0.0201 g, N 0.1399 g, O by difference 0.0800 g. Ratio 1 : 4 : 2 : 1.
CH₄N₂O
Hydrocarbon (JEE style)
0.4300 g of a hydrocarbon gives 1.3175 g CO₂ and 0.6292 g H₂O. M = 86.18 g/mol.
C 0.3596 g, H 0.0704 g (total 0.4300 g, so no O). Ratio C : H = 1 : 2.33. Multiply by 3.
C₃H₇, molecular formula C₆H₁₄ (hexane)
From percent composition
A compound is 54.5% C and 9.1% H, the rest O. M = 88 g/mol.
O = 36.4%. Moles in 100 g: C 4.54, H 9.03, O 2.28. Ratio 2 : 4 : 1.
C₂H₄O; molecular formula C₄H₈O₂
Checking for oxygen
If C + H add up to the sample mass, there is no oxygen. If they fall short, the difference is oxygen (or another element you have not measured).
Choose "Nothing" to see how much is unaccounted for.
Mass balance tells you whether O is present
Common mistakes
- Forgetting the 2 in H₂O. Each H₂O carries two H atoms: mol H = 2 × mol H₂O.
- Rounding too early. Keep at least four significant figures until the final ratio. 1.33 rounded to 1 gives the wrong formula.
- Rounding 1.5 to 2. A ratio near x.5 means multiply everything by 2, not round.
- Measuring O directly from the products. O in CO₂ and H₂O comes partly from the O₂ supply. Always get O by difference.
- Mixing up empirical and molecular formulas. The empirical formula is the simplest ratio. You need the molar mass to get the molecular formula.
- Percentages that do not add to 100. A large gap usually means an element is missing (often O) or a typing error.
Found your formula? Draw possible structures for it in MolDraw, a free online chemical structure editor, and check their molar mass.
Open MolDraw editorFrequently asked questions
How do I find the empirical formula from combustion analysis?
Convert the mass of CO₂ to grams of C (× 12.011/44.009) and the mass of H₂O to grams of H (× 2.016/18.015). Find O by difference. Convert each to moles, divide by the smallest, and multiply to whole numbers if needed.
Why is oxygen found by difference?
The sample burns in added oxygen, so the O atoms in CO₂ and H₂O come partly from the O₂ supply. Subtracting the masses of C, H and other elements from the sample mass gives the oxygen that was in the compound.
What do I do if a mole ratio is 1.5 or 1.33?
Multiply all ratios by the same small whole number: ×2 for .5, ×3 for .33 or .67, ×4 for .25 or .75, ×5 for .2, ×6 for .17. A ratio of 1 : 1.33 becomes 3 : 4.
How do I get the molecular formula?
Divide the molar mass by the empirical formula mass to get n, a whole number. Multiply every subscript by n. CH₂O (30.03 g/mol) with M = 180.16 g/mol gives n = 6 and C₆H₁₂O₆.
How is nitrogen measured?
In the Dumas method nitrogen is collected as N₂ gas and its mass (or volume) gives the mass of N. In the Kjeldahl method N is converted to NH₃ and titrated, giving a percentage. Enter either one; the calculator handles O by difference afterwards.
Can I enter percent composition instead?
Yes. Switch to percent composition, type the percentages and leave O blank to fill it by difference. The calculator treats the percentages as grams in a 100 g sample.
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