Audited 29 Jul 2026·Last updated 28 Oct 2026·4 citations·Tier 3·0 uses

Combustion Analysis Calculator

Work out carbon, hydrogen and oxygen-by-difference masses and the empirical formula from a burned sample's weight and the CO₂ and H₂O it produced.

Combustion Analysis Calculator

The mass of the unknown compound you weighed before combustion. Every error in this number goes straight into the oxygen figure, because oxygen is found by difference.
g
Mass gained by the carbon dioxide absorber. All of its carbon came from your sample; its oxygen mostly did not.
g
Mass gained by the water absorber. All of its hydrogen came from your sample. Make sure the trap was dry before the run — a wet trap inflates hydrogen and destroys the oxygen figure.
g
Sample contains
Empirical formula
C3H8O
The simplest whole-number ratio of carbon, hydrogen and oxygen atoms, written in Hill order (C, then H, then O); a ratio, not necessarily the molecule.
Carbon in sample
0.1531 g
Hydrogen in sample
0.0342 g
Oxygen (by difference)
0.0676 g
Carbon
60.04
Hydrogen
13.43
Unexplained mass
0 g
Empirical formula mass
60.096 g/mol

Background.

A quarter of a gram of an unknown liquid burns completely in a stream of oxygen, and two absorber tubes downstream get heavier: one by 0.306 g as it soaks up water, the other by 0.561 g as it captures carbon dioxide. Those three weighings are enough to show that the liquid's empirical formula is C₃H₈O.

The reasoning is atom bookkeeping. Every carbon atom in the CO₂ tube came from the sample, and so did every hydrogen atom in the water tube, so the sample's carbon and hydrogen follow from the two mass gains by simple proportion. Oxygen cannot be traced that way. The sample burned in a flood of O₂, so most of the oxygen locked inside the collected products came from the gas stream rather than the compound. Sample oxygen is instead taken as whatever mass is left once carbon and hydrogen are subtracted, which is also how ASTM's practice for the ultimate analysis of coal (D3176) obtains it.

Two things follow. Every weighing error, in the sample and in both tubes, ends up in the oxygen figure. And any other element present — nitrogen, sulfur, chlorine, or ash left behind in the boat — would be silently booked as oxygen. Use these results for compounds of carbon, hydrogen and possibly oxygen, and nothing else.

What is combustion analysis calculator?

Combustion analysis is the classical way to find the elemental make-up of an organic compound. A weighed sample is burned completely in excess oxygen; the gases pass first through an absorber that holds water and then through one that holds carbon dioxide, and each absorber is weighed before and after the run. What comes out is an empirical formula, the smallest whole-number ratio of atoms, not the number of atoms in one molecule. Modern CHN analysers run the same chemistry with instrumental detectors — ASTM D5373 measures the CO₂, water vapour and nitrogen in the combustion gas — and oxygen is still not among the things they determine.

How to use this calculator.

  1. Enter the mass of sample you burned, in grams, weighed on the same balance as the absorbers. A milligram of error here becomes a milligram of error in the oxygen, because oxygen is the leftover.
  2. Enter the mass gained by the CO₂ absorber and by the H₂O absorber, each as the after-minus-before difference in grams. Moisture already sitting in the train before the run is counted as sample hydrogen, so dry it first.
  3. Leave 'Sample contains' on carbon, hydrogen and oxygen for most unknowns. Switch to hydrocarbon only when you want to test the claim that there is no oxygen: the leftover mass then shows up as unexplained mass instead of being called oxygen.
  4. Read the empirical formula, then compare the carbon and hydrogen percentages with those calculated for the compound you suspect. For a molecular formula, take the empirical formula mass and a measured molar mass to the empirical formula calculator.

The 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)

Carbon in the sample is the CO₂ mass multiplied by the share of CO₂ that is carbon, 12.011 ÷ 44.009 = 0.272921. Hydrogen works the same way with water, except that each water molecule carries two hydrogen atoms, so the factor is 2.016 ÷ 18.015 = 0.111907. Oxygen is m(O) = m(sample) − m(C) − m(H). The atomic weights behind those factors — C 12.011, H 1.008, O 15.999 — are the CIAAW abridged standard values.

To reach a formula, each element mass is divided by its atomic weight to give moles, and every mole figure is divided by the smallest. Ratios such as 1.5 or 1.33 are not rounded away. The ratios are multiplied by 1, then 2, and so on up to 12, and the first multiplier that brings every one of them within 0.05 of a whole number is kept; for subscripts of 4 or more the allowance widens to 1.5% of the whole number. Only then are the subscripts rounded.

Two judgement calls sit beside the chemistry. In carbon-hydrogen-oxygen mode, a residual smaller than 0.1% of the sample mass is treated as weighing noise rather than oxygen, so a genuine hydrocarbon does not pick up a phantom O. And if carbon plus hydrogen comes out more than 0.5% heavier than the sample itself, no formula is given, because the weighings contradict each other; a smaller overshoot is reported as a negative unexplained mass.

A worked example.

Example

Take the default run: 0.255 g of sample (255 mg), 0.561 g of CO₂ and 0.306 g of water, analysed as carbon, hydrogen and oxygen. Carbon: 0.561 g × 0.272921 = 153.11 mg, which is 60.0427% of the sample. Hydrogen: 0.306 g × 0.111907 = 34.24 mg, or 13.4288%. Oxygen by difference: 255 − 153.11 − 34.24 = 67.65 mg, shown as 0.0676 g. In moles, carbon is 153.11 ÷ 12.011 = 12.75 mmol, hydrogen 34.24 ÷ 1.008 = 33.97 mmol and oxygen 67.65 ÷ 15.999 = 4.228 mmol. Dividing by the smallest, oxygen's, gives C 3.015, H 8.034 and O 1. At a multiplier of 1 both are already within tolerance of whole numbers, so the empirical formula is C3H8O, with an empirical formula mass of 60.096 g/mol. That ratio happens to settle the molecule as well. A compound of carbon, hydrogen and oxygen can hold at most two hydrogens per carbon plus two, and C₃H₈O is already at that limit, so no larger multiple can exist. The liquid is propan-1-ol, propan-2-ol or methoxyethane; combustion analysis cannot tell those three apart, because their compositions are identical.

h2o Mass0.306
sample Mass0.255
co2 Mass0.561
sample Compositioncho

Frequently asked questions.

If the CO₂ contains oxygen, why can't that oxygen be credited to the sample?
Because the sample burns in a large excess of O₂ gas, and the oxygen atoms in the CO₂ and water come mostly from that stream. Nothing in the trap masses separates gas-stream oxygen from sample oxygen, so the sample's share has to be found by subtracting carbon and hydrogen from the sample mass.
Can I use this for an amine, a thiol or a chlorinated compound?
No. Nitrogen, sulfur, halogens and any inorganic ash would all fall into the residual and be reported as oxygen, producing a formula that is wrong with no warning. A full ultimate analysis measures nitrogen, sulfur and ash separately and subtracts them before taking oxygen by difference; these results do not.
I entered the products of burning butane and got C2H5. Is something wrong?
No, that is butane's correct empirical formula. Composition only fixes a ratio, and C₄H₁₀ has the same carbon-to-hydrogen ratio as C₂H₅. With a measured molar mass you find the multiplier: 58.124 ÷ 29.062 = 2, which gives C₄H₁₀. A tenth of a gram of butane would yield about 0.3029 g of CO₂ and 0.155 g of water.
What is the unexplained mass telling me in hydrocarbon mode?
It is the mass your no-oxygen assumption leaves unaccounted for. Run the default numbers as a hydrocarbon and oxygen is forced to zero, the unexplained mass becomes 0.0676 g, over a quarter of the sample, and the formula turns into C3H8. A figure that large means the sample is not a hydrocarbon.
How close should my carbon and hydrogen percentages be to the theoretical values?
Many chemistry journals accept an elemental analysis as support for purity when each found percentage is within 0.4 percentage points of the calculated one. Propan-1-ol, for example, should come out near 60.0% carbon and 13.4% hydrogen; a gap of a full point usually means a wet or impure sample.
Why did I get a formula like C16H46O5?
That pattern points to a bad weighing, often a water trap that picked up moisture. Adding just 0.030 g to the default water mass turns C3H8O into C16H46O5, which is impossible: sixteen carbons can carry at most 34 hydrogens. Incomplete combustion errs differently: soot left in the tube is carbon that never reached the CO₂ absorber.

References& sources.

  1. [1]Flowers, P., Theopold, K., Langley, R. & Robinson, W. R. (2019). Chemistry 2e, Section 3.2 'Determining Empirical and Molecular Formulas'. OpenStax, Rice University. Peer-reviewed, CC BY 4.0. PRIMARY SOURCE for the second half of this calculation: the procedure of converting element masses to moles, dividing by the smallest, and multiplying to whole-number subscripts, together with the definitions of empirical and molecular formula. Retrieved 2026-07-29. Open access.
  2. [2]ASTM D3176-15, 'Standard Practice for Ultimate Analysis of Coal and Coke', ASTM International, West Conshohocken, PA. It mandates precisely the treatment used here — there being no satisfactory direct ASTM test method for determining oxygen, oxygen shall be calculated by subtracting from 100 the sum of the other components of the ultimate analysis (carbon, hydrogen, nitrogen, sulfur and ash). It agrees with the by-difference method, and its inclusion of nitrogen, sulfur and ash in that subtraction is the reason this page is scoped to C/H/O compounds only. Retrieved 2026-07-29. Standards document, paywalled: the scope and the oxygen-by-difference clause are quoted from the publicly available scope summary.
  3. [3]ASTM D5373-21, 'Standard Test Methods for Determination of Carbon, Hydrogen and Nitrogen in Analysis Samples of Coal and Carbon in Analysis Samples of Coal and Coke', ASTM International. The standardised instrumental form of this experiment: carbon dioxide, water vapour and elemental nitrogen in the combustion gas stream are determined by instrumental detection, giving carbon, hydrogen and nitrogen. Oxygen is not among the determinands, which corroborates why it must be taken by difference. Retrieved 2026-07-29. Standards document, paywalled: scope quoted from the publicly available summary.
  4. [4]Meija, J. et al. (2021). 'Atomic weights of the elements 2021 (IUPAC Technical Report)'. Pure and Applied Chemistry 93(5), 573–600, doi:10.1515/pac-2019-0603, as maintained in the CIAAW table 'Abridged Standard Atomic Weights 2024' (named revision: 2024). The named, versioned source of the three atomic weights and therefore of both gravimetric factors: C 12.011, H 1.0080, O 15.999, giving M(CO2) = 44.009 and M(H2O) = 18.015 g/mol. Retrieved 2026-07-29. Open access.

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Method
m(C) = m(CO₂) × 12.011⁄44.009 · m(H) = m(H₂O) × 2.016⁄18.015 · m(O) = m(sample) − m(C) − m(H)
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