Stoichiometric Air-Fuel Ratio
Calculate the stoichiometric air-fuel ratio (AFR) by mass of a hydrocarbon C_xH_y, AFR = (x + y/4) × 137.93 ÷ M, from the number of carbon atoms x, hydrogen atoms y and the fuel molar mass M (g/mol). The result (kg air per kg fuel) is the exact amount of air needed for complete combustion, with no leftover air or fuel. Methane (CH₄) has AFR ≈ 17.2; gasoline ≈ 14.7. The 137.93 constant comes from the air mass per mole of O₂ (32 ÷ 0.232). It is the base parameter of combustion control and mixture in engines and burners. Enter x, y and the fuel molar mass.
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Stoichiometric air-fuel ratio
Burning a fuel completely takes the exact amount of air — no more, no less. That proportion is the stoichiometric air-fuel ratio (AFR). For a generic hydrocarbon C_xH_y, the complete reaction is C_xH_y + (x + y/4)·O₂ → x·CO₂ + (y/2)·H₂O, so each mole of fuel requires (x + y/4) moles of oxygen. Since air holds only 21% O₂ by volume and 23.2% by mass, the required air mass follows and gets divided by the fuel mass: AFR = (x + y/4) × 137.93 ÷ M, where M is the molar mass of the fuel and 137.93 = 32 ÷ 0.232 is the mass of air per mole of O₂. The result comes out in kg of air per kg of fuel. Familiar values: methane (CH₄): AFR ≈ 17.2; gasoline (~C₈H₁₈): ≈ 14.7; ethanol: ≈ 9.0 (less air, as it already carries oxygen); hydrogen: ≈ 34.3. The stoichiometric AFR is fundamental in combustion control: modern gasoline engines use lambda sensors to hold the mixture close to 14.7:1, ideal for catalytic converters, while industrial burners run with some excess air (a higher AFR) to guarantee complete burning. Mixtures with less air than stoichiometric (rich) produce CO and soot; with more air (lean), they waste energy. Enter x, y and the molar mass of the fuel.
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Actual Air-Fuel Ratio
Calculate the actual air-fuel ratio, AFR_actual = AFR_stoichiometric × (1 + excess air ÷ 100), from the stoichiometric air-fuel ratio and the excess air (%). The result (kg air per kg fuel) is the amount of air actually supplied in practice, always greater than stoichiometric, because real combustion needs excess air to ensure complete burning (the mixture is never perfect). This value sizes the air supply (fans), the exhaust gas flow and influences the flame temperature and efficiency. Enter the stoichiometric air-fuel ratio and the excess air.
Excess Air (from Flue Gas)
Calculate the excess air of a combustion from the oxygen in dry flue gas, EA = O₂ ÷ (20.9 − O₂) × 100%, from the measured O₂ percentage in the stack. The result, in %, shows how much air was supplied beyond stoichiometric — measured by the leftover oxygen in the exhaust gases. Some excess air (10-30%) is needed to ensure complete combustion (avoid CO and soot), but too much wastes energy heating useless air that leaves hot through the stack. Gas analyzers measure O₂ and compute the excess air to optimize combustion efficiency. Enter the O₂ percentage in the gases.
Theoretical Combustion Air Volume
Calculate the theoretical air volume needed for complete combustion of a hydrocarbon C_xH_y, V_air = (x + y/4) × 22.4 ÷ (0.21 × M), from the carbon x, hydrogen y atoms and the molar mass M. The result, in Nm³ of air per kg of fuel (at normal conditions), is the stoichiometric air — the basis for sizing fans, burners and combustion air systems. The factor 22.4 is the ideal gas molar volume (L/mol at STP) and 0.21 the volume fraction of O₂ in air. Methane needs ~13.3 Nm³/kg. Multiplied by the excess air, it gives the actual air supplied. Enter x, y and the molar mass.
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Calculate the fuel consumption per hectare of a mechanized operation, consumption = hourly consumption ÷ field capacity, dividing the tractor's hourly consumption (L/h) by the effective field capacity (ha/h). The result, in liters per hectare, is the practical indicator to budget the fuel cost of a farming operation and compare the energy efficiency of machines and settings. Heavy operations (subsoiling) consume far more L/ha than light ones (spraying). Combined with the diesel price and the total area, it gives the season's fuel cost. Enter the hourly consumption and the field capacity.
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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.