Combustion Reaction Calculator

A combustion reaction is a reaction in which a fuel combines with oxygen, $\mathrm{O_2}$, and releases energy. In a complete combustion the only products are carbon dioxide and water:

$\text{fuel} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}$

General formula for combustion of a hydrocarbon $\mathrm{C}_x\mathrm{H}_y$:

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

Here $x$ is the number of carbon atoms and $y$ the number of hydrogen atoms per molecule of fuel. If the fuel already contains oxygen — an alcohol, ester, or sugar written $\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z$ — that oxygen is subtracted:

$\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z + \left(x + \frac{y}{4} - \frac{z}{2}\right)\mathrm{O_2} \rightarrow x\,\mathrm{CO_2} + \frac{y}{2}\,\mathrm{H_2O}$

What these equations assume: oxygen in excess and complete conversion of the fuel. When oxygen is limited the reaction is incomplete and yields carbon monoxide, $\mathrm{CO}$, or solid carbon along with the water, and the coefficients above no longer apply.

How to identify one: $\mathrm{O_2}$ on the reactant side with $\mathrm{CO_2}$ and $\mathrm{H_2O}$ as the products is the signature of complete combustion of an organic fuel.

Step-by-Step

1. Write the skeleton with the fuel coefficient fixed at 1: fuel $+\ \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}$.
2. Balance carbon: the $\mathrm{CO_2}$ coefficient is $x$.
3. Balance hydrogen: the $\mathrm{H_2O}$ coefficient is $y/2$.
4. Balance oxygen last. Count the oxygen atoms now fixed in the products, $2x + y/2$, subtract the $z$ atoms the fuel supplied, and halve the remainder — giving $x + y/4 - z/2$ for $\mathrm{O_2}$.
5. Clear fractions: if that coefficient is a half-integer, multiply every coefficient by 2.
6. Check each element on both sides.

Oxygen goes last because it is the only element appearing in both products; fixing carbon and hydrogen first leaves exactly one unknown.

Why the Method Always Works

Balancing is a small linear system — one conservation equation per element, one unknown per species. With four species and three elements there is a single solution up to an overall scale factor, which is why the smallest whole-number coefficients are unique.

Coefficients Are Mole Ratios

From $\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O}$, complete combustion of 2.0 mol of methane consumes 4.0 mol of $\mathrm{O_2}$ and produces 2.0 mol of $\mathrm{CO_2}$.

• Changing subscripts instead of coefficients — turning $\mathrm{H_2O}$ into $\mathrm{H_2O_2}$ balances the count but changes the substance. Only the number in front may change.
• Balancing oxygen first — oxygen sits in both products, so balancing it early forces you to redo it after fixing carbon and hydrogen.
• Forgetting the oxygen inside the fuel — ethanol needs 3 $\mathrm{O_2}$, not the $2 + 6/4 = 3.5$ you get by ignoring its own oxygen atom.
• Leaving a fractional coefficient — $\mathrm{C_8H_{18}} + 12.5\,\mathrm{O_2}$ is arithmetically correct but conventionally doubled to whole numbers.
• Calling any reaction with oxygen a combustion — rusting consumes $\mathrm{O_2}$ but produces no $\mathrm{CO_2}$; it is oxidation, not combustion.
• Writing $\mathrm{CO}$ as a product of complete combustion — carbon monoxide is the incomplete-combustion product.