Combustion Reaction Calculator

Background.

Burning a hydrocarbon is bookkeeping with atoms. Every carbon atom in the fuel leaves as a molecule of carbon dioxide, every pair of hydrogen atoms leaves as a molecule of water, and the oxygen has to supply exactly the oxygen atoms those products contain. Nothing more is needed to balance the equation, so the whole calculation comes down to two short rules.

For a fuel written CₓHᵧ, the x carbon atoms become x CO₂, which hold 2x oxygen atoms, or x molecules of O₂. The y hydrogen atoms become y/2 H₂O, holding y/2 oxygen atoms, or y/4 molecules of O₂. One mole of fuel therefore needs x + y/4 moles of oxygen: 5 for propane, C₃H₈; 2 for methane, CH₄; 12.5 for octane, C₈H₁₈. Multiply by the moles of fuel and you have the oxygen demand and the amount of each product.

That figure is the stoichiometric amount, just enough oxygen for complete combustion with none left over. Real flames seldom work that way. Furnaces and engines draw their oxygen from air, which is about 21% oxygen by volume, so every mole of O₂ arrives with roughly 3.76 moles of nitrogen that must be heated and pushed through. Burners usually run with some excess air to make sure the fuel burns out, while an oxygen-starved flame makes carbon monoxide and soot instead of CO₂.

None of that is modelled here. The page treats pure hydrocarbons burned completely to CO₂ and water. Fuels that contain oxygen, sulfur or nitrogen, such as ethanol, need a different balance, and the heat released is not calculated. Nor does it check that the formula you type is a real molecule; C₃H₉ would be balanced without complaint.

What is combustion reaction calculator?

A combustion reaction, as treated here, is the complete oxidation of a hydrocarbon by oxygen to carbon dioxide and water: CₓHᵧ + (x + y/4) O₂ → x CO₂ + (y/2) H₂O. The stoichiometric oxygen is the exact amount that balanced equation calls for, with neither excess nor shortfall. All amounts on this page are in moles, which convert to grams through molar masses and to gas volumes through the ideal gas law.

How to use this calculator.

1. 01Read your fuel's molecular formula and enter the subscript on C as the carbon atoms and the subscript on H as the hydrogen atoms. Propane, C₃H₈, is 3 and 8; methane, CH₄, is 1 and 4.
2. 02Average formulas used for fuel blends often have decimal subscripts, and those can be entered as they are; the balance works the same way.
3. 03Enter the amount of fuel in moles. To convert from grams, divide by the molar mass: 88.2 g of propane, at about 44.1 g/mol, is 2 mol.
4. 04Read the oxygen required and the moles of CO₂ and water formed. If the oxygen will come from air, divide the oxygen figure by 0.21 for the approximate moles of air.

The formula.

The balanced equation is CₓHᵧ + (x + y/4) O₂ → x CO₂ + (y/2) H₂O, and each output is a coefficient multiplied by the moles of fuel, n. Carbon balance gives CO₂ = x·n. Hydrogen balance gives H₂O = (y/2)·n, because each water molecule carries two hydrogen atoms. Oxygen balance then fixes the oxygen: the products contain 2x + y/2 oxygen atoms per molecule of fuel, and halving that for diatomic O₂ gives x + y/4, so O₂ = (x + y/4)·n. Per atom, hydrogen adds to the oxygen demand only a quarter as fast as carbon does, which is why methane needs 2 mol of O₂ per mole while each extra CH₂ unit in a longer alkane adds 1.5. Mass is conserved automatically: grams of fuel plus oxygen equal grams of CO₂ plus water.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax, Rice University. Chemistry 2e, 2019. Chapters 4, 9, and 12. Retrieved 2026-08-06.
2. [2]NIST. Chemistry WebBook current 2026. Thermochemistry and species data. Retrieved 2026-08-06.
3. [3]OpenStax, Rice University. University Physics, 2016. Volume 1 mechanics and Volume 2 thermodynamics. Retrieved 2026-08-06.