A complete combustion reaction is the reaction of a fuel with enough oxygen to turn all of its carbon into carbon dioxide and all of its hydrogen into water. It is the reaction behind gas stoves, car engines, candles and power stations, and it is the reference reaction for tabulated heats of combustion.
Balancing a combustion equation follows a fixed recipe: balance carbon with CO₂, hydrogen with H₂O, and oxygen last with O₂, because O₂ is the only species that contains just one element. If the O₂ coefficient comes out as a fraction (as for hexane, 9.5 O₂), multiply everything by two. For sulfur-containing fuels such as thiols and coal, sulfur becomes SO₂; nitrogen in the fuel is conventionally written as N₂, although real flames also make some nitrogen oxides.
The calculator does the balancing with an exact linear-algebra solver, then uses the mole ratios to scale everything to the amount of fuel you enter. It also converts the oxygen requirement into a mass of air (dry air is 20.95% O₂ by mole), which gives the stoichiometric air–fuel ratio that engine and burner designers use to set mixtures.
How to use the Combustion Reaction Calculator
- Enter the fuel as
- Fuel formula — e.g. CH4, C8H18, C2H5OH, C6H12O6, C2H5SH
- Carbon atoms (α)
- Hydrogen atoms (β)
- Oxygen atoms (γ)
- Nitrogen atoms
- Sulfur atoms
- Amount of fuel given as
- Mass of fuel burned
- Moles of fuel burned
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
Type the fuel formula (or atom counts) and the mass or moles burned. The equation is balanced element by element, the coefficients are converted to moles per mole of fuel, and each amount is multiplied by the molar masses to give grams. Gas volumes are at STP (0 °C, 1 atm, 22.414 L/mol).
Worked example
Burn 10 g of hexane, C₆H₁₄ (86.178 g/mol). Balanced equation: 2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O, i.e. 9.5 mol O₂, 6 mol CO₂ and 7 mol H₂O per mole of fuel. 10 g ÷ 86.178 g/mol = 0.11604 mol of hexane, which needs 0.11604 × 9.5 = 1.1024 mol O₂ = 35.27 g, and produces 0.6962 mol CO₂ = 30.64 g and 0.8123 mol H₂O = 14.63 g. Mass check: 10 + 35.27 = 45.27 g in, 30.64 + 14.63 = 45.27 g out. The oxygen comes from 1.1024 ÷ 0.2095 = 5.262 mol of air = 152.4 g, an air–fuel ratio of 15.24 : 1.
Oxygen demand of common fuels
| Fuel | Balanced equation | O₂ per mol fuel | Air–fuel ratio (mass) |
|---|---|---|---|
| Methane CH₄ | CH₄ + 2 O₂ → CO₂ + 2 H₂O | 2 | 17.2 |
| Propane C₃H₈ | C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O | 5 | 15.7 |
| Octane C₈H₁₈ | 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O | 12.5 | 15.1 |
| Ethanol C₂H₅OH | C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O | 3 | 9.0 |
| Hydrogen H₂ | 2 H₂ + O₂ → 2 H₂O | 0.5 | 34.3 |
| Glucose C₆H₁₂O₆ | C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O | 6 | 4.6 |
Air–fuel ratios use dry air with 20.95 mol% O₂ and a molar mass of 28.965 g/mol. Commercial gasoline is a blend, so its quoted value (about 14.7) differs a little from pure octane.
Tips
- Count halogens like hydrogens and ignore oxygen and sulfur in degree-of-unsaturation formulas.
- Balance combustion with O₂ last – it is the easiest to adjust.
- Use dry mass for combustion analysis and protein content.
- Everything is calculated in your browser.
Common mistakes to avoid
- Counting nitrogen like a hydrogen in degree of unsaturation (it subtracts one H).
- Forgetting that one mole of CO₂ contains one mole of C.
- Using the wrong Kjeldahl factor (6.25 is generic).
Formula
CₐHᵦO_γ + (α + β/4 − γ/2) O₂ → α CO₂ + (β/2) H₂O · S → SO₂ · N → N₂ · air = n(O₂) / 0.2095 · AFR = m(air) / m(fuel)
Worked example
Inputs — Enter the fuel as: Molecular formula; Fuel formula: C6H14; Amount of fuel given as: Mass; Mass of fuel burned: 10 g.
Results — Balanced combustion equation = 2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O; Molar mass of the fuel = 86.178 g/mol; O₂ needed per mole of fuel = 9.5 mol; Moles of fuel = 0.116; Oxygen needed (moles) = 1.1024; Oxygen needed (mass) = 35.2736.
Key terms
- Complete combustion
- Burning with enough oxygen to convert all carbon to CO₂ and all hydrogen to H₂O.
- Stoichiometric mixture
- A fuel–oxygen (or fuel–air) mix with exactly enough oxidiser and no excess.
- Air–fuel ratio (AFR)
- Mass of air divided by mass of fuel; a mixture above the stoichiometric value is lean, below it is rich.
- STP
- 0 °C and 1 atm, where one mole of ideal gas occupies 22.414 L.
Frequently asked questions
How do you balance a combustion reaction?
Balance C with CO₂, then H with H₂O, then count the oxygen atoms on the right, subtract any oxygen already in the fuel and divide by two to get O₂. Clear fractions by doubling: C₆H₁₄ + 9.5 O₂ → 6 CO₂ + 7 H₂O becomes 2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O.
What is the general formula for complete combustion?
For CₐHᵦO_γ: CₐHᵦO_γ + (α + β/4 − γ/2) O₂ → α CO₂ + β/2 H₂O.
What is the combustion equation for methane?
CH₄ + 2 O₂ → CO₂ + 2 H₂O. Burning 16.04 g (1 mol) of methane uses 64.00 g of O₂ and gives 44.01 g of CO₂ and 36.03 g of water.
How much CO₂ does burning 10 g of hexane make?
10 g C₆H₁₄ is 0.1160 mol; each mole gives 6 mol CO₂, so 0.6962 mol = 30.64 g of CO₂ – about three times the mass of the fuel, because most of the CO₂ mass is oxygen from the air.
What is the stoichiometric air–fuel ratio?
The mass of air that supplies exactly enough O₂. With dry air at 20.95 mol% O₂ and 28.965 g/mol, methane needs 17.2 kg of air per kg and pure octane about 15.1 kg per kg.
What is the difference between complete and incomplete combustion?
Complete combustion gives only CO₂ and H₂O. With too little oxygen, some carbon ends up as carbon monoxide or soot, and less energy is released.
Does an alcohol need less oxygen than the alkane?
Yes – the oxygen already in the fuel counts. Ethanol (C₂H₆O) needs 3 mol O₂ per mole, while ethane (C₂H₆) needs 3.5 mol.
What happens to sulfur and nitrogen in the fuel?
Sulfur burns to SO₂ (the cause of acid rain from coal), and fuel nitrogen is written as N₂ in the ideal equation; real flames also produce some NO and NO₂.