Combustion Formula Calculator

Complete combustion of a hydrocarbon converts every carbon to carbon dioxide and every hydrogen to water:

$\mathrm{C}_x\mathrm{H}_y + \left(x + \frac{y}{4}\right)\mathrm{O_2} \longrightarrow x\,\mathrm{CO_2} + \frac{y}{2}\,\mathrm{H_2O}$

• $x$ — carbons per molecule, which fixes the $\mathrm{CO_2}$ coefficient directly.
• $y$ — hydrogens per molecule; each water takes two, so the coefficient is $y/2$.
• The oxygen coefficient follows by counting: $2x$ oxygen atoms in the $\mathrm{CO_2}$ plus $y/2$ in the water, divided by the 2 atoms in each $\mathrm{O_2}$.

If the fuel already contains oxygen, $\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z$, subtract it: the coefficient becomes $x + y/4 - z/2$.

What "complete" assumes. Excess oxygen and only two products, $\mathrm{CO_2}$ and $\mathrm{H_2O}$. Incomplete combustion — oxygen-limited — yields CO or carbon as well, and no single formula covers it because the product mix depends on conditions. Any nitrogen or sulfur in the fuel leaves as separate oxides and must be handled on its own.

Fractional coefficients like $\tfrac{13}{2}$ are perfectly valid algebra; multiply the whole equation through when whole numbers are wanted.

Combustion analysis runs the formula backwards. A sample of known mass is burned completely, and the masses of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ collected are used to deduce the formula.

The logic

All the carbon in the $\mathrm{CO_2}$ came from the sample, and all the hydrogen in the water came from the sample. Oxygen cannot be traced this way, because oxygen in the products came from both the sample and the supplied $\mathrm{O_2}$ — so it is found by difference.

Step by step

1. $n(\mathrm{C}) = n(\mathrm{CO_2}) = \dfrac{m(\mathrm{CO_2})}{44.01}$ — one carbon per molecule.
2. $n(\mathrm{H}) = 2\,n(\mathrm{H_2O}) = 2 \times \dfrac{m(\mathrm{H_2O})}{18.02}$ — two hydrogens per molecule.
3. Convert both back to masses and subtract from the sample mass. What remains is oxygen: $n(\mathrm{O}) = m(\mathrm{O})/15.999$. If the remainder is essentially zero, the compound is a hydrocarbon.
4. Divide all mole counts by the smallest to get the empirical formula, scaling up if a ratio lands on a half or a third.
5. With a known molar mass, divide it by the empirical formula mass and multiply the subscripts by that integer.

Significant figures

The oxygen mass is a small difference between larger numbers, so it loses precision fastest. Carry full precision through every intermediate step and round only the final ratios.

• Forgetting the factor of 2 for hydrogen. Each $\mathrm{H_2O}$ carries two hydrogen atoms; using $n(\mathrm{H}) = n(\mathrm{H_2O})$ halves the hydrogen subscript.
• Trying to get oxygen from the products. Oxygen in the $\mathrm{CO_2}$ and $\mathrm{H_2O}$ mostly came from the air supply. It is always found by subtracting the C and H masses from the sample mass.
• Rounding the mole ratios too early. A ratio of 2.50 means a 5:2 formula, not 3:1.
• Assuming a hydrocarbon. If the C and H masses fall short of the sample mass by more than rounding error, the compound contains oxygen.
• Stopping at the empirical formula. $\mathrm{CH_3}$ and $\mathrm{C_2H_6}$ have the same ratio; only a molar mass separates them.
• Using the complete-combustion formula for a limited-oxygen case. If CO appears among the products, the equation no longer applies.