Air-Fuel Ratio Calculator
Pure octane gives 15.1; the 14.7 quoted for petrol reflects a real blend, not the chemistry.
The formula
O2 needed = C + H/4 - O/2 ; AFR = O2 * 4.76 * 28.96 / fuel mass
Where 14.7 comes from
Burning one mole of a fuel CxHyOz completely needs x + y/4 − z/2 moles of oxygen. Air is only about a fifth oxygen, so each mole of O₂ drags 3.76 moles of nitrogen along with it — 4.76 moles of air in total, weighing 28.96 g each. Divide by the fuel's molar mass and the ratio falls out. Pure octane gives 15.1; the 14.7 quoted for petrol reflects the additives and lighter fractions in a real blend.
Oxygen in the fuel changes everything
Ethanol carries its own oxygen atom, so it needs less air: its stoichiometric ratio is 9.0 against petrol's 14.7. An engine switched to ethanol must therefore inject far more fuel for the same air, which is exactly why flex-fuel vehicles need a wider-range injector and why fuel consumption rises on E85.
Lambda and phi
Lambda is the actual air-fuel ratio divided by the stoichiometric one; the equivalence ratio phi is its reciprocal. Lambda above one is lean, below one is rich. A three-way catalyst only works within about a percent of lambda equals one, which is what the oxygen sensor is there to maintain.
Empirical is not molecular
Every method on these pages returns a ratio, because that is all a composition can determine. Glucose, formaldehyde and acetic acid all reduce to CH₂O, and no amount of elemental analysis will separate them. A molar mass from mass spectrometry or a freezing-point depression is what turns the ratio into a formula, and it has to come from a second, independent measurement.