Combustion Reaction Calculator

Balance the combustion of any hydrocarbon or C, H, O fuel.

Handles both hydrocarbons (γ = 0) and oxygenated C, H, O fuels.

Updated October 18, 2026

CreatorFrank Zhao

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What this calculator does

This calculator writes the complete combustion equation of a C, H, O fuel for you. You enter how many carbon, hydrogen and oxygen atoms one molecule of the fuel contains, and it returns the smallest whole-number coefficients for the fuel, oxygen, carbon dioxide and water, together with the finished equation.

Balancing that equation by hand is not hard, but it is easy to get wrong in one specific place. The oxygen coefficient is usually a fraction — hexane needs $9.5$ molecules of oxygen — so after you fix carbon and hydrogen you have to multiply the whole row by something before the equation looks like a textbook answer. The calculator does that scaling, and it also tells you when a formula has no combustion equation at all, which is the case people tend to misread rather than miscalculate.

It is a working tool for anyone who has a molecular formula in hand and needs the reaction that goes with it: coursework and exam practice, preparing a lab calculation, sizing the oxygen demand of a fuel before an air-fuel or energy calculation, or reading a fuel spec sheet and wondering what its formula actually burns into.

Nothing about the answer is looked up. The calculator solves the three element balances for the formula you type and reduces the result to the smallest whole numbers, so the same input always gives the same equation, and the equation it prints conserves every atom on both sides. If you mistype a formula, though, it will balance that formula just as correctly — the arithmetic is never a check on the chemistry you meant.

Three counts in

Carbon, hydrogen and oxygen atoms per molecule

Four coefficients out

Fuel, oxygen, carbon dioxide and water coefficients, plus the written equation

γ = 0 for hydrocarbons

Oxygenated fuels enter their own oxygen count instead

Quick start

Every fuel is described by the same three numbers, so the whole job is reading subscripts off a formula and typing them in.

Write the molecular formula down first

You need the atom counts, not the name. Ethanol is $\mathrm{C_2H_6O}$ , so it is two carbons, six hydrogens and one oxygen; methane is $\mathrm{CH_4}$ , so one, four and none.

Enter the carbon count in Total atoms of carbon C (α)

For $\mathrm{C_6H_{14}}$ this is $6$ .

Enter the hydrogen count in Total atoms of hydrogen H (β)

For the same fuel this is $14$ .

Enter the total oxygen count in Total atoms of oxygen O (γ)

Count every oxygen in the formula, including the ones inside groups such as $\mathrm{OH}$ . A hydrocarbon contains none, so this box takes $0$ .

Read the balanced reaction and its coefficients

The reaction boxes show your substance and the finished equation; the coefficient boxes below them list the four numbers on their own.

A first calculation, step by step

Take methane, $\mathrm{CH_4}$ . Enter $\alpha = 1$ , $\beta = 4$ and $\gamma = 0$ . Carbon and hydrogen settle the products immediately, and the oxygen balance gives the oxygen coefficient:

a = \alpha + \frac{\beta}{4} - \frac{\gamma}{2}

=

1 + \frac{4}{4} - 0

=

2

\mathrm{CH_4} + 2\,\mathrm{O_2}

\longrightarrow

\mathrm{CO_2} + 2\,\mathrm{H_2O}

The equation balances: one carbon, four hydrogens and four oxygens on each side. The coefficients are mole ratios, so they scale to any batch size. When you need to carry that forward — how many moles of $\mathrm{CO_2}$ come out of a given mass of fuel, for instance — the molar ratio calculator converts between the two species you pick.

How the equation is balanced

Complete combustion means every carbon leaves as carbon dioxide and every hydrogen as water. That fixes the products, and the three element balances then fix the coefficients.

\mathrm{C_\alpha H_\beta O_\gamma} + a\,\mathrm{O_2} \longrightarrow b\,\mathrm{CO_2} + c\,\mathrm{H_2O}

The three balances

Carbon appears only in the fuel and in $\mathrm{CO_2}$ , hydrogen only in the fuel and in water, and the formulas are already fixed, so the first two coefficients are read straight off the subscripts:

b = \alpha

\qquad

c = \frac{\beta}{2}

\qquad

a = \alpha + \frac{\beta}{4} - \frac{\gamma}{2}

The last one is the oxygen balance: the fuel's own oxygen (γ) is subtracted because it already counts towards the oxygen in the products. That subtraction is why an oxygen-rich formula such as oxalic acid needs so little oxygen from the air, and why a formula can end up needing none at all.

Clearing the fractions

Balances are written over the common denominator 4, which turns the row $\left(F,\; a,\; b,\; c\right)$ into whole numbers, and the row is then divided by its greatest common divisor so no common factor is left. Hexane is the classic case:

a = 6 + \frac{14}{4} - 0

=

9.5

Written over the denominator 4 the row is $\left(4,\; 38,\; 24,\; 28\right)$ , and its greatest common divisor is $2$ , so the answer is the row $\left(2,\; 19,\; 12,\; 14\right)$ :

2\,\mathrm{C_6H_{14}} + 19\,\mathrm{O_2}

\longrightarrow

12\,\mathrm{CO_2} + 14\,\mathrm{H_2O}

Dividing by the greatest common divisor matters: the row $\left(4,\; 38,\; 24,\; 28\right)$ balances too, but it counts four hexane molecules where two will do, and the reduced row is the one every textbook and data sheet quotes.

When no combustion equation exists

An oxygen coefficient of zero is a real answer, not a rounding artefact. A fuel needs oxygen only while its own oxygen falls short of what the products hold, which is the same as saying the row has to satisfy:

4\alpha + \beta - 2\gamma > 0

\Longleftrightarrow

a > 0

When that fails, the coefficient comes out as zero or negative and the calculator reports the substance as not valid rather than printing a combustion that cannot happen. The number itself stays on screen, so you can see how far off the formula is.

From coefficients to air and energy

The oxygen coefficient is the stoichiometric oxygen demand in moles per mole of fuel, which is the starting point for an air-fuel ratio: the air-fuel ratio calculator turns it into mass of air per mass of fuel, and the heat of combustion of the fuel turns the same balanced equation into energy released.

Worked examples

Three fuels that behave differently enough to be worth reading side by side: a hydrocarbon, an oxygen-rich compound that still burns, and a formula that turns out not to be a fuel at all.

Octane, a gasoline benchmark

Situation: You want the reaction a spark-ignition fuel runs on, using octane as the usual surrogate molecule for gasoline.

Counts entered

Carbon (α)8

Hydrogen (β)18

Oxygen (γ)0

Balanced equation

2\,\mathrm{C_8H_{18}} + 25\,\mathrm{O_2}

\longrightarrow

16\,\mathrm{CO_2} + 18\,\mathrm{H_2O}

The row was doubled to clear the half from $a = 8 + \frac{18}{4} = 12.5$ , which is why the fuel coefficient is $2$ and not $1$ . The same row is the input to the heat of combustion calculator, which turns the moles of fuel into energy released.

In practice: One mole of octane consumes 12.5 mol of oxygen and releases 8 mol of carbon dioxide, so a fuel tank's worth of octane scales straight from these numbers. Pair that ratio with the fuel's heat of combustion to get from moles to energy.

Oxalic acid, where the fuel brings its own oxygen

Situation: An oxygenated compound has to be checked rather than assumed: is there enough oxygen left in it that combustion needs almost no air?

Counts entered

Carbon (α)2

Hydrogen (β)2

Oxygen (γ)4

Balanced equation

a = 2 + \frac{2}{4} - \frac{4}{2}

=

0.5

2\,\mathrm{C_2H_2O_4} + \mathrm{O_2}

\longrightarrow

4\,\mathrm{CO_2} + 2\,\mathrm{H_2O}

Doubling clears the half again: the fuel coefficient is $2$ and the oxygen coefficient is $1$ .

In practice: The γ term does the work here: two fuel molecules consume a single oxygen molecule, which is why oxidized compounds and carbohydrates need so much less air per mole than hydrocarbons of the same size.

Carbonic acid, which is not a fuel

Situation: You try a formula that is already as oxidized as the combustion products, to see what the calculator does with it.

Counts entered

Carbon (α)1

Hydrogen (β)2

Oxygen (γ)3

Balanced equation

a = 1 + \frac{2}{4} - \frac{3}{2}

=

0

$\mathrm{CH_2O_3}$ — carbonic acid, $\mathrm{H_2CO_3}$ — already holds exactly the oxygen its one carbon and two hydrogens need as $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . The oxygen coefficient box shows $0$ in red, the reaction boxes stay empty, and the note explains that the substance is not valid for this reaction. A negative coefficient means the same thing more strongly: the formula carries more oxygen than its products can hold, so no combustion equation exists.

In practice: This is the case to recognise on sight: a zero or negative oxygen coefficient means the formula is not a combustion fuel, and the red note under the coefficient says so. It is a decomposition, not a reaction with oxygen.

Getting the formula right

Almost every wrong answer from this calculator comes from the formula that went in, not from the balancing. These four checks catch most of them.

Enter the molecular formula, not the empirical one

A formula reduced to its simplest ratio describes a different molecule, and the calculator will balance it faithfully. The empirical formula of cyclohexane is $\mathrm{CH_2}$ , which gives:

2\,\mathrm{CH_2} + 3\,\mathrm{O_2}

\longrightarrow

2\,\mathrm{CO_2} + 2\,\mathrm{H_2O}

The real molecule, $\mathrm{C_6H_{12}}$ , gives $\mathrm{C_6H_{12}} + 9\,\mathrm{O_2} \longrightarrow 6\,\mathrm{CO_2} + 6\,\mathrm{H_2O}$ . The ratio between the coefficients is identical in both; the coefficients themselves count molecules, so they expose which formula you entered. If you are starting from a percent composition instead of a formula, the empirical formula calculator gets you the ratio first, and you scale it to the molecule using its molar mass.

A term with no number in front of it is one molecule

The reaction boxes drop coefficients of $1$ , exactly as chemistry writing does. A term that looks bare is one molecule of that substance, never zero of it.

If the reaction boxes are empty instead, the row has not been solved: either a count is still missing, or the substance has no combustion reaction (see the third worked example).

Count atoms on both sides as your own check

Take the octane row and multiply every subscript by its coefficient. Carbon: $2 \times 8 = 16$ on the left and $16 \times 1 = 16$ on the right. Hydrogen: $2 \times 18 = 36$ against $18 \times 2 = 36$ . Oxygen: $25 \times 2 = 50$ against $16 \times 2 + 18 = 50$ .

Remember that water carries two hydrogens

Water is the step people misread most often: the coefficient in front of $\mathrm{H_2O}$ is half the hydrogen count, not the whole of it, because each water molecule takes two hydrogens. In the same way, $\mathrm{CO_2}$ carries one carbon, so its coefficient always equals the carbon count times the fuel coefficient.

Questions

Can it balance a fuel that contains nitrogen or sulfur?

No. This calculator is written for C, H, O fuels, where complete combustion produces only $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . Fuel nitrogen and sulfur leave the products as species this equation has no room for — nitrogen as $\mathrm{N_2}$ , sulfur as $\mathrm{SO_2}$ in the usual complete-combustion treatment.

For those fuels, write the reaction out with every product you expect and use the chemical equation balancer, which takes whole formulas instead of atom counts and is not restricted to combustion.

Does the phase of the water change the coefficients?

It does not. The coefficients come from counting atoms, and liquid water, water vapour and ice all contain the same two hydrogens and one oxygen per molecule. What the phase does change is the energy side of the calculation: steam carries away the latent heat, which is why the two heating values of a fuel differ. That distinction belongs to the energy calculation, not to balancing.

Can I use it for gasoline, diesel or biomass?

Only through a surrogate molecule. Those fuels are mixtures, and a mixture has no single formula, so the honest input is a well-known representative — for example $\mathrm{C_8H_{18}}$ for gasoline or $\mathrm{C_{12}H_{26}}$ for diesel — and the answer describes that representative, not the fuel in the tank. Biomass adds a further complication: its oxygen content is high and variable, and it usually carries nitrogen and ash, so a single C, H, O formula is a rough stand-in at best.

Model limits

The arithmetic is exact for the formula you enter. The assumptions behind the equation are what you have to keep in mind.

What the balanced equation does not say

•Complete combustion only. Every carbon is assumed to leave as $\mathrm{CO_2}$ and every hydrogen as $\mathrm{H_2O}$ . A real flame short of air also makes carbon monoxide, soot, hydrogen and partly burned fragments, and those equations balance to different products.
•No air, no excess oxygen, no nitrogen. The equation shows pure oxygen as the oxidizer. Feeding air brings nitrogen along without taking part in these products, and running with excess air uses more oxygen than the stoichiometric row needs.
•Nothing about energy or ignition. A balanced equation says what the reactants and products are, not whether the mixture will light, how fast it burns or how hot it gets. Those need thermodynamic and kinetic data, not a material balance.
•The formula is taken at face value. It is balanced as written, whether or not the compound exists, and isomers are indistinguishable here: ethanol and dimethyl ether are both $\mathrm{C_2H_6O}$ and give the same coefficients. Only whole-number atom counts are accepted.
•Not a design calculation. For equipment sizing, emission work, permits or anything safety-related, treat this as the starting stoichiometry and confirm it against your own material balance and the applicable standards.

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