What this calculator does
Burning a known mass of a CHO compound and weighing the carbon dioxide and water produced gives its empirical formula. Half a gram producing 0.9553 g of CO₂ and 0.5866 g of water works out as C₂H₆O, which is ethanol.
Carbon and hydrogen are measured directly from the combustion products, but oxygen is found only by subtracting them from the sample mass. That makes the oxygen figure the least reliable of the three: it inherits the errors of both other measurements, and for a compound with little oxygen it can be swamped by them entirely.
The formula
The carbon comes from the CO₂ mass scaled by the ratio of carbon to carbon dioxide molar masses, and the hydrogen similarly from the water. Oxygen is the sample mass minus those two. Each mass is converted to moles and divided by the smallest to give the ratio, which is then rounded to whole numbers.
| Term | Meaning |
|---|---|
| Empirical formula | The simplest whole-number atom ratio. |
| By difference | Oxygen is not measured but inferred, which is why it carries the most error. |
| Combustion train | The apparatus trapping and weighing the CO₂ and water separately. |
| Molecular formula | A whole-number multiple of the empirical formula, needing a separate molar mass to determine. |
The inputs explained
| Field | What to enter |
|---|---|
| Sample mass (g) | Mass of compound burned, in grams. |
| CO2 produced (g) | Mass of carbon dioxide collected. |
| H2O produced (g) | Mass of water collected. |
When to use it
Identifying an organic unknown
The classic application, giving an empirical formula from a controlled burn.
Confirming a synthesis
Elemental analysis of a product should match the expected composition within a small tolerance.
Teaching stoichiometry
The calculation chains several conversions together and rewards careful bookkeeping.
Worked examples
Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.
What formulae do these combustion results give?
Three compounds with their combustion products.
| CO₂ produced (H₂O varies with it) | Empirical ratio C : H : O | Mass of carbon | Mass of oxygen (by difference) |
|---|---|---|---|
| 0.7329 g | 1.14 : 4.45 : 1.00 | 0.2000 g | 0.2343 g |
| 0.9553 g | 2.00 : 6.00 : 1.00 | 0.2607 g | 0.1736 g |
| 1.5145 g | 26.20 : 49.58 : 1.00 | 0.4133 g | 0.0210 g |
Questions
Why is oxygen found by difference?
Because burning a compound in oxygen means the oxygen in the products comes from both the sample and the air supply, so it cannot be measured directly. Subtracting the carbon and hydrogen from the sample mass is the only route, and it inherits both their errors.
What if the oxygen comes out as zero or negative?
The compound is a hydrocarbon with no oxygen, or the measurements are slightly inconsistent. This calculator floors the oxygen at zero and reports a two-element ratio. A genuinely negative value means a measurement error somewhere in the chain.
How do I get the molecular formula?
You need the molar mass from a separate measurement, usually mass spectrometry. Divide it by the empirical formula mass to get the multiplier. Combustion analysis alone cannot distinguish CH₂O from C₆H₁₂O₆, since both have the same composition by mass.
Does this work for nitrogen-containing compounds?
Not as set up here. Nitrogen would be included in the oxygen-by-difference figure and corrupt it. Determining nitrogen needs a separate method, typically Dumas or Kjeldahl, and the result is then subtracted before the oxygen is calculated.
For the same calculation from percent composition, see the empirical formula calculator. For balancing the combustion, see the combustion reaction balancer.