The combustion reaction calculator above writes the complete combustion equation for any fuel made of carbon, hydrogen and oxygen, balances it to whole-number coefficients, and converts those coefficients into real quantities. Give it a fuel formula and an amount and it returns the oxygen required, the carbon dioxide and water produced, the theoretical air demand and the air-fuel ratio by mass, with an allowance for excess air.
Arb Digital publishes free calculators for the calculations that sit between a chemical equation and a physical quantity. Combustion is the clearest example: the equation is trivial to write once you know the pattern, and the step people actually want is the one after it, where a coefficient becomes a mass of air or a mass of carbon dioxide. This page does both and shows the equation it used.
What This Combustion Reaction Calculator Does
You supply the number of carbon, hydrogen and oxygen atoms in the fuel molecule. The page derives the oxygen coefficient from the balance, clears any fraction by scaling the whole equation, reduces it to lowest terms, and displays the result. It then applies your fuel amount to give the masses involved, and converts the oxygen requirement into an air requirement using the oxygen content of dry air.
The products of complete combustion are fixed by the elements present: all the carbon leaves as carbon dioxide, all the hydrogen as water, and any oxygen already in the fuel reduces the amount that must be supplied. There is no choice in this, which is why a combustion equation can be generated rather than solved.
Two boundaries are worth stating. The chemical equation balancer takes an equation you have already written and finds coefficients for it, whatever the reaction; this page generates the equation itself from a fuel formula, which it can only do because combustion products are predetermined. The combustion analysis calculator runs the reverse: it takes measured product masses from an unknown compound and works back to its formula. This page starts from a known fuel and predicts the products.
How to Use It
- Enter the fuel formula as atom counts, or pick a fuel from the selector to fill them in.
- Set the oxygen count if the fuel contains any. An alcohol, an ester or a carbohydrate needs less supplied oxygen than a hydrocarbon of the same carbon count.
- Enter the amount of fuel in kilograms, grams or moles. Mass units are converted using the molar mass derived from the formula.
- Set the excess air if you want the actual air demand rather than the theoretical one. Zero gives stoichiometric air.
- Read the equation first, then the quantities. If the equation is not what you expected, the atom counts are wrong and every number below it is too.
The Formula and How It Is Calculated
For a fuel CxHyOz, complete combustion gives x molecules of carbon dioxide and y/2 molecules of water. Balancing oxygen across the equation gives the coefficient x + y/4 − z/2: the x accounts for the carbon dioxide, the y/4 for the water, and the z/2 subtracts the oxygen the fuel brought with it. The whole equation is then multiplied by 2 or 4 if needed to clear the fraction, and divided by the greatest common divisor of the four coefficients.
Working the default, propane has x = 3, y = 8 and z = 0, so the oxygen coefficient is 3 + 2 = 5 and the equation is C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Octane gives 8 + 4.5 = 12.5, a fraction, so the whole equation doubles to 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O.
Converting to quantities, one kilogram of propane at 44.10 g/mol is 22.68 moles. That needs 5 × 22.68 = 113.4 moles of oxygen, which is 3.63 kg, and produces 68.0 moles of carbon dioxide at 2.99 kg and 90.7 moles of water at 1.63 kg. The masses balance: 1.00 + 3.63 in, 2.99 + 1.63 out. The molar masses come from the standard atomic weights published by the Commission on Isotopic Abundances and Atomic Weights.
Why the Oxygen Coefficient Is Often a Fraction
The y/4 term produces a fraction whenever the hydrogen count is not a multiple of four, and that is most of the time. Octane's 12.5, ethyne's 2.5 and ethanol's 3 are all consequences of the same term. The fraction is not an error and does not need to be avoided; it is a perfectly valid statement of the ratio, and in thermochemistry it is often kept deliberately so that the equation refers to exactly one mole of fuel.
Where whole numbers are wanted, the fix is to scale the entire equation rather than to round anything. Multiplying by two clears a half, and by four clears a quarter. This page finds the smallest multiplier that clears every fraction, then divides through by any common factor so the coefficients are as small as they can be. The result is the conventional form seen in textbooks.
The scaling has a consequence worth being aware of when reading heats of combustion. An enthalpy quoted per mole of fuel corresponds to the unscaled equation with its fraction intact. If the equation has been doubled to make the coefficients whole, the enthalpy for that equation as written is twice the molar value. Mixing the two conventions is a common source of a factor-of-two error, and it is why the heat of combustion calculator is careful about which basis it is on.
Air Is Not Oxygen
Burners run on air, not on oxygen, and the difference is large. Dry air is about 20.95 percent oxygen by volume, so supplying a given amount of oxygen means supplying roughly 4.77 times as much air by mole. On a mass basis, using an average molar mass of about 28.96 g/mol for air against 32.00 for oxygen, the multiplier works out at around 4.3.
This is where the stoichiometric air-fuel ratio comes from. For propane the calculation gives about 15.7 kilograms of air per kilogram of fuel, and for octane about 15.1, which is why gasoline engines are described as running near 14.7 to 1 for a typical fuel blend. Those numbers are properties of the fuel's composition, and this page derives them rather than looking them up.
The nitrogen that makes up most of the air does not take part in the combustion in this model, but it does not vanish either. It passes through, absorbing heat and leaving in the flue gas, which is why flame temperatures in air are far below those in pure oxygen. At high enough temperatures some of it does react to form nitrogen oxides, which is a real and important product of real combustion and is entirely outside the stoichiometric equation this page computes.
Excess Air and What This Page Does Not Model
Real burners supply more air than the stoichiometry requires, because mixing is imperfect and a fuel molecule that does not meet enough oxygen burns incompletely. The excess air figure on this page scales the theoretical air demand accordingly, and the surplus oxygen simply passes through with the nitrogen. Too little excess air leaves unburnt fuel and carbon monoxide; too much wastes heat in warming air that takes no part in the reaction.
What the page computes is complete combustion, and that is a deliberate limit. Complete combustion produces only carbon dioxide and water. Real combustion also produces carbon monoxide, unburnt hydrocarbons, soot and, at high temperature, nitrogen oxides, and the amounts of those depend on burner geometry, mixing, residence time and temperature rather than on stoichiometry. No composition-based calculation predicts them, and this page does not attempt to.
Two further omissions are worth naming. Fuel-bound sulfur and nitrogen are not handled here, since the inputs cover carbon, hydrogen and oxygen only; both produce their own oxides on combustion. And the water in the products is treated as a mass, without regard to whether it leaves as vapour or condenses, which is precisely the distinction between higher and lower heating values. Where product volumes rather than masses are needed, the ideal gas law calculator converts moles to a volume at a stated temperature and pressure.
Oxygen in the Fuel Changes the Answer
The z/2 term is easy to overlook and it matters. Ethanol is C₂H₆O, and its oxygen coefficient is 2 + 1.5 − 0.5 = 3, against 3.5 for ethane which has the same carbon count. Oxygen already bonded into the fuel has already done part of the oxidising, so less needs to be supplied.
The effect compounds as the oxygen content rises. Glucose is C₆H₁₂O₆, and its coefficient is 6 + 3 − 3 = 6, the same as its carbon count, so it needs exactly one oxygen molecule per carbon. This is why oxygenated fuels release less energy per kilogram than hydrocarbons and why biomass fuels need less combustion air than their carbon content alone would suggest.
Taken far enough, the term reaches zero and the calculation stops being about a fuel. A compound with as much oxygen as complete combustion requires is already fully oxidised, and this page says so rather than returning a zero coefficient without comment. For the molar masses and stoichiometric conversions that all of this depends on, the molar mass calculator, the molar ratio calculator and the moles to grams calculator cover the general cases, and measured thermochemical data for real fuels is held in the NIST Chemistry WebBook.
Arb Digital publishes hundreds of free calculators across chemistry, maths, finance and marketing — no sign-up, no limits. If something you need is missing, tell us and we will look at building it.
Browse All Free Tools Suggest a ToolCommon Mistakes to Avoid
- Forgetting the oxygen already in the fuel — the z/2 term means an alcohol needs less supplied oxygen than the hydrocarbon with the same carbon count.
- Rounding a fractional oxygen coefficient — scale the whole equation by two or four instead, because rounding 12.5 to 12 unbalances it.
- Treating the oxygen requirement as an air requirement — air is only about 21 percent oxygen, so the air demand is several times larger.
- Expecting the product mass to equal the fuel mass — the oxygen that reacted is part of the products, so they always weigh considerably more than the fuel did.
- Reading complete combustion as a description of a real burner — carbon monoxide, soot and nitrogen oxides are real products that no stoichiometric equation predicts.
Related Free Tools From Arb Digital
For an equation you have already written, use the chemical equation balancer, and to work backwards from measured products to an unknown fuel use the combustion analysis calculator. The heat of combustion calculator covers the energy released, the molar mass calculator turns a formula into g/mol, and the molar ratio calculator and moles to grams calculator handle the general stoichiometric conversions. The ideal gas law calculator converts product moles into volumes, and the percent yield calculator compares theory against a measured result. The full free online tools hub lists everything else.
Frequently Asked Questions
All the carbon becomes carbon dioxide and all the hydrogen becomes water, so the product coefficients are x and y/2 for a fuel with x carbons and y hydrogens. The oxygen coefficient is then x plus y over four, minus z over two for any oxygen in the fuel.
Because the hydrogen term divides by four, so any hydrogen count that is not a multiple of four gives a fraction. Multiplying the whole equation by two or four clears it without changing the ratio.
Far more than the oxygen alone. Dry air is about 20.95 percent oxygen by volume, so the air requirement is roughly 4.77 times the oxygen requirement by mole, or about 4.3 times by mass.
It is the mass of air needed per unit mass of fuel for complete combustion with no excess. It is a property of the fuel's composition, coming out near 15.7 for propane and around 15.1 for octane.
Yes. Every oxygen atom in the fuel removes half an oxygen molecule from the requirement, which is why ethanol needs three oxygen molecules where ethane needs three and a half for the same carbon count.
Because the oxygen that reacted is now part of them. Mass is conserved across the whole reaction, so the products weigh as much as the fuel and the oxygen combined, which is often several times the fuel mass alone.
No. This is complete combustion, which produces only carbon dioxide and water. Carbon monoxide, soot and nitrogen oxides depend on burner design, mixing and temperature rather than on stoichiometry, and no composition-based calculation predicts them.
This calculator is provided for education and general reference. It computes stoichiometric combustion quantities and is not combustion engineering, appliance, ventilation or safety guidance; fuel-burning equipment must be designed, installed and commissioned to the applicable codes by a qualified person.
