Combustion analysis is the classic way to find out what an unknown organic compound is made of. A carefully weighed sample is burned in a stream of pure oxygen; every carbon atom ends up in carbon dioxide and every hydrogen atom in water. The two gases are trapped in separate absorbers – water in a desiccant such as magnesium perchlorate, CO₂ in sodium hydroxide on a support – and the absorbers are weighed before and after.
Because one mole of CO₂ contains exactly one mole of carbon and one mole of water contains two moles of hydrogen, the gain in mass of each absorber tells you how much carbon and hydrogen were in the sample. Oxygen cannot be measured this way (the sample burned in oxygen), so for compounds containing C, H and O it is found by difference: whatever mass is not carbon or hydrogen is oxygen.
Turning the masses into moles and dividing by the smallest amount gives the simplest whole-number ratio of atoms – the empirical formula. A separate molar mass measurement, for example from mass spectrometry or freezing point depression, then tells you how many empirical units make up one molecule. Modern CHN analysers automate exactly this procedure and report mass percentages directly.
How to use the Combustion Analysis Calculator
- Type of compound
- Mass of sample burned
- Mass of CO₂ collected
- Mass of H₂O collected
- Molar mass of the compound (0 if unknown) — Optional – needed only for the molecular formula
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Choose whether the sample contains oxygen or is a pure hydrocarbon, enter the masses of CO₂ and H₂O (and the sample mass for C–H–O compounds), and optionally the molar mass. The calculator converts CO₂ and H₂O into grams and moles of C and H, finds oxygen by difference, divides by the smallest mole amount, tries multipliers 1 to 12 until every value is within 0.1 of a whole number, and scales the empirical formula by M ÷ M(empirical).
Worked example
A 0.900 g sample of a C–H–O compound burns to 1.319 g CO₂ and 0.540 g H₂O. Carbon: 1.319 × 12.011 / 44.009 = 0.3600 g = 0.02997 mol. Hydrogen: 0.540 × 2 / 18.015 = 0.05995 mol = 0.0604 g. Oxygen by difference: 0.900 − 0.3600 − 0.0604 = 0.4796 g = 0.02998 mol. Dividing by 0.02997 gives C : H : O = 1.000 : 2.000 : 1.000, so the empirical formula is CH₂O (30.026 g/mol). With a measured molar mass of 90.08 g/mol, n = 90.08 / 30.026 = 3.00 and the molecular formula is C₃H₆O₃ – lactic acid. The mass composition is 40.00% C, 6.71% H and 53.29% O.
Turning awkward ratios into whole numbers
| Ratio ends in about | Multiply by | Example |
|---|---|---|
| .50 | 2 | 1 : 1.5 → 2 : 3 |
| .33 or .67 | 3 | 1 : 1.33 → 3 : 4 |
| .25 or .75 | 4 | 1 : 2.25 → 4 : 9 |
| .20, .40, .60, .80 | 5 | 1 : 1.6 → 5 : 8 |
| .17 or .83 | 6 | 1 : 1.17 → 6 : 7 |
The calculator tries multipliers from 1 to 12 automatically and stops at the first one that puts every value within 0.1 of a whole number. If none does, it shows the closest ratio and warns you – usually a sign of incomplete combustion or a weighing error.
Tips
- Count halogens like hydrogens and ignore oxygen and sulfur in degree-of-unsaturation formulas.
- Balance combustion with O₂ last – it is the easiest to adjust.
- Use dry mass for combustion analysis and protein content.
- Everything is calculated in your browser.
Common mistakes to avoid
- Counting nitrogen like a hydrogen in degree of unsaturation (it subtracts one H).
- Forgetting that one mole of CO₂ contains one mole of C.
- Using the wrong Kjeldahl factor (6.25 is generic).
Formula
m(C) = m(CO₂) × 12.011 / 44.009 · m(H) = m(H₂O) × 2.016 / 18.015 · m(O) = m(sample) − m(C) − m(H) · divide each mole amount by the smallest, scale to whole numbers · n = M / M(empirical)Worked example
Inputs — Type of compound: Contains C, H and O (oxygen by difference); Mass of sample burned: 0.9 g; Mass of CO₂ collected: 1.319 g; Mass of H₂O collected: 0.54 g; Molar mass of the compound (0 if unknown): 90.08 g/mol.
Results — Empirical formula = CH₂O; Empirical formula mass = 30.026 g/mol; Multiplier n = M ÷ empirical mass = 3; Molecular formula = C₃H₆O₃ (M = 90.08 g/mol); Mass of carbon in the sample = 0.36; Mass of hydrogen in the sample = 0.0604.
Key terms
- Empirical formula
- The simplest whole-number ratio of atoms in a compound, e.g. CH₂O for glucose and lactic acid.
- Molecular formula
- The actual number of each atom in one molecule – a whole-number multiple of the empirical formula.
- Oxygen by difference
- Sample mass minus the masses of all measured elements, assigned to oxygen.
- Absorber
- A tube of reagent that traps one combustion product (water or CO₂) so its mass gain can be weighed.
Frequently asked questions
How do you find the empirical formula from combustion analysis?
Convert the CO₂ mass to moles of C (÷ 44.009 g/mol) and the H₂O mass to moles of H (÷ 18.015 × 2), find oxygen by difference, then divide all mole amounts by the smallest and round to whole numbers – multiplying by 2, 3 or 4 first if you get values like 1.5 or 1.33.
Why is oxygen found by difference?
The sample is burned in excess oxygen, so the O atoms in CO₂ and H₂O come partly from the oxygen gas. Only the sample mass minus the masses of C and H can tell you how much oxygen the compound itself contained.
What fraction of CO₂ is carbon?
12.011 ÷ 44.009 = 27.29% by mass. So 1.000 g of CO₂ holds 0.2729 g of carbon.
What fraction of water is hydrogen?
2.016 ÷ 18.015 = 11.19% by mass. So 1.000 g of H₂O holds 0.1119 g of hydrogen.
What if I get a ratio like 1 : 1.6?
1.6 is close to 8/5, so multiply every ratio by 5: a hydrocarbon giving 3.300 g CO₂ and 1.080 g H₂O has C : H = 1 : 1.599, which becomes C₅H₈.
How do I get the molecular formula?
Divide the molar mass by the empirical formula mass and multiply every subscript by that whole number. CH₂O (30.03 g/mol) with M = 90.08 g/mol gives n = 3 and C₃H₆O₃ (lactic acid).
Does nitrogen or sulfur change the method?
Yes – N₂ and SO₂ need their own measurement (CHNS analysers use gas chromatography). This calculator handles C–H–O compounds and hydrocarbons; for other elements, oxygen-by-difference would wrongly include them.
How precise do the masses need to be?
Microanalysis weighs 1–3 mg samples to ±0.001 mg, and journals usually accept results within ±0.4 percentage points of the theoretical composition.