Last updated: July 3, 2026
AFR Calculator (Air-Fuel Ratio)
Creators
Dharmendra SinghReviewers
Creators
Dharmendra SinghReviewers
Quick Answer
The AFR calculator divides air mass by fuel mass for actual AFR and estimates stoichiometric AFR from CxHyOz using O2 = x + y/4 − z/2 and dry air as 23.2% oxygen by mass. It also reports lambda and equivalence ratio for rich or lean mixture interpretation.
Air-fuel ratio is the mass of air divided by the mass of fuel. Stoichiometric AFR is found from fuel formula oxygen demand, and lambda is actual AFR divided by stoichiometric AFR.
Key Takeaways
- AFR is mass of air divided by mass of fuel.
- Stoichiometric AFR for CxHyOz uses O2 = x + y/4 − z/2 mol per mole of fuel.
- Air mass is estimated from oxygen mass using 0.232 as the oxygen mass fraction of dry air.
- Lambda equals actual AFR divided by stoichiometric AFR; equivalence ratio is its reciprocal.
- Lambda below 1 or φ above 1 means rich; lambda above 1 or φ below 1 means lean.
Creators
Dharmendra SinghReviewers
Creators
Dharmendra SinghReviewers
Formula
AFR_actual = mass_air / mass_fuel; O2_stoich = x + y/4 − z/2; AFR_stoich = (O2_stoich × 32.00 / 0.232) / molar_mass_fuel; λ = AFR_actual / AFR_stoich; φ = AFR_stoich / AFR_actual
Where:
- AFR_actual=Actual air-fuel ratio by mass(kg air/kg fuel)
- AFR_stoich=Stoichiometric air-fuel ratio by mass(kg air/kg fuel)
- x=Carbon atoms in CxHyOz(atoms per molecule)
- y=Hydrogen atoms in CxHyOz(atoms per molecule)
- z=Oxygen atoms in CxHyOz(atoms per molecule)
- λ=Lambda or relative air-fuel ratio(dimensionless)
- φ=Equivalence ratio(dimensionless)
Worked Examples
Methane stoichiometric AFR
Find the theoretical air needed to burn methane, CH4.
- 1For CH4, oxygen demand is x + y/4 − z/2 = 1 + 4/4 − 0 = 2 mol O2 per mole fuel.
- 2Convert oxygen to air mass using 23.2% O2 by mass: 2 × 32.00 / 0.232 = 275.86 g air.
- 3Divide by methane molar mass: 275.86 g air ÷ 16.04 g fuel = 17.20.
Octane stoichiometric AFR
Estimate theoretical AFR for octane, C8H18, a gasoline reference fuel.
- 1Oxygen demand is 8 + 18/4 = 12.5 mol O2 per mole octane.
- 2Air mass per mole fuel is 12.5 × 32.00 / 0.232 = 1724.14 g air.
- 3Stoichiometric AFR is 1724.14 ÷ 114.23 = 15.09 kg air/kg fuel.
Rich gasoline-like mixture
Compare an actual AFR of 13.0 against a stoichiometric reference of about 14.7.
- 1Actual AFR = 13.0 ÷ 1 = 13.0 kg air/kg fuel.
- 2The chosen surrogate formula gives AFRstoich ≈ 14.68, close to the gasoline reference 14.7.
- 3Lambda = 13.0 ÷ 14.68 ≈ 0.885 and equivalence ratio φ = 14.68 ÷ 13.0 ≈ 1.13, so the mixture is rich.
Introduction
Air-fuel ratio (AFR) expresses how much air is mixed with a unit mass of fuel before combustion. This calculator reports the measured mass AFR, estimates the stoichiometric AFR from a fuel formula CxHyOz, and converts between lambda (λ) and equivalence ratio (φ). It is useful alongside combustion analysis and combustion reaction balancing. The oxygen-in-air mass fraction follows standard engineering practice described by Engineering ToolBox and combustion texts.
What is air-fuel ratio?
AFR is the mass of air divided by the mass of fuel. An AFR of 14.7 means 14.7 kg of air for each 1 kg of fuel. The actual AFR describes the mixture supplied to a burner or engine; the stoichiometric AFR is the exact theoretical ratio that supplies just enough oxygen for complete conversion of carbon to CO2 and hydrogen to H2O.
Low AFR means fuel-rich combustion with excess fuel.
High AFR means lean combustion with excess air.
The stoichiometric point depends on fuel composition, not just fuel name.
For reaction coefficients, compare with the molar ratio calculator.
Stoichiometric AFR from CxHyOz
For one mole of fuel CxHyOz, complete combustion needs x + y/4 − z/2 moles of O2. Multiplying by 32.00 g/mol gives oxygen mass. Dividing by 0.232 converts oxygen mass to dry-air mass because air is about 23.2% oxygen by mass. Finally, divide by the fuel molar mass to get kg air per kg fuel; the gram units cancel.
If the fuel already contains oxygen, z reduces external oxygen demand. Alcohols therefore have lower stoichiometric AFRs than hydrocarbons with similar carbon count.
Lambda and equivalence ratio
Lambda is actual AFR divided by stoichiometric AFR. Equivalence ratio is the reciprocal: φ = AFRstoich / AFRactual. Lambda below 1 (φ above 1) is rich; lambda above 1 (φ below 1) is lean. These dimensionless metrics let engineers compare fuels with different stoichiometric AFRs.
Engine-control literature often uses lambda, while combustion chemistry and burner studies often report equivalence ratio.
Typical stoichiometric AFR values
Common reference values are approximate because commercial fuels are mixtures. Gasoline is often modeled near 14.7:1, diesel around 14.5:1, methane about 17.2:1, and ethanol about 9.0:1. Use the calculator with the actual elemental formula or surrogate composition when accuracy matters.
| Fuel | Formula or surrogate | Stoichiometric AFR |
|---|---|---|
| Methane | CH4 | ≈ 17.2 |
| Octane | C8H18 | ≈ 15.1 |
| Ethanol | C2H6O | ≈ 9.0 |
| Gasoline | mixture | ≈ 14.7 |
| Diesel | mixture | ≈ 14.5 |
How to use this calculator
Enter air and fuel masses in matching units for the actual AFR. Then enter x, y, z, and molar mass for the fuel formula CxHyOz to calculate stoichiometric AFR. You can get fuel molar mass from formula data or use a molarity calculator workflow when converting solution fuel concentrations before combustion tests.
Use matching mass units for air and fuel; kg/kg equals g/g.
Use dry-air composition unless your engineering standard specifies humid air.
Check oxygenated fuels carefully because fuel oxygen reduces O2 demand.
Report lambda or φ with AFR to clarify rich/lean operation.
Assumptions and limitations
The stoichiometric calculation assumes complete combustion to CO2 and H2O and dry air containing 23.2% oxygen by mass. Real flames can form CO, unburned hydrocarbons, NOx, soot, or dissociation products, especially at high temperature or poor mixing. For thermodynamic modeling, validate assumptions with detailed references such as NIST Chemistry WebBook data and combustion handbooks.
AFR is a mass ratio; do not confuse it with mole fraction, volume percent oxygen, or excess-air percentage.
Quick Reference Card
Air-Fuel Ratio — Quick Reference
Quick reference • AFR Calculator (Air-Fuel Ratio)
AFR = m_air/m_fuel; AFR_stoich = [(x + y/4 − z/2) × 32.00 / 0.232] / M_fuelValid range: AFR > 0; practical engines and burners commonly operate from rich (< stoich) to excess-air lean conditions
Common Values
⚠ Watch Out
- •Use the same mass units for air and fuel.
- •Do not use volume ratio unless density conversions are included.
- •Fuel oxygen lowers the external oxygen and air requirement.
- •Humid air and exhaust dissociation can shift real combustion from the ideal calculation.
Pro Tips
- →Use lambda for sensor and engine-control comparisons.
- →Use equivalence ratio when comparing flames across different fuels.
- →For blended fuels, calculate an effective elemental formula or use component mass fractions.
- →Round final AFR to match uncertainty in fuel composition and air measurement.
FAQs
What does AFR mean in combustion?
AFR means air-fuel ratio: the mass of air supplied divided by the mass of fuel supplied. It is usually written as kg air per kg fuel or as a ratio such as 14.7:1.
How do I calculate actual AFR?
Divide the measured mass of air by the measured mass of fuel. The units must match, so 14.7 kg air per 1 kg fuel and 14.7 g air per 1 g fuel both give AFR = 14.7.
How is stoichiometric AFR calculated from a formula?
For CxHyOz, calculate moles of oxygen needed as x + y/4 − z/2. Convert that oxygen to air mass using 32.00 g/mol O2 and the oxygen mass fraction of air, then divide by fuel molar mass.
What is lambda in an AFR calculation?
Lambda is actual AFR divided by stoichiometric AFR. Lambda equals 1 at stoichiometric combustion, is below 1 for rich mixtures, and above 1 for lean mixtures.
What is equivalence ratio?
Equivalence ratio φ is stoichiometric AFR divided by actual AFR. It is the reciprocal of lambda, so φ above 1 is fuel-rich and φ below 1 is lean.
Why is gasoline often listed as 14.7:1?
Gasoline is a mixture, but common hydrocarbon surrogates average near 14.7 kg air per kg fuel for complete combustion with dry air. Actual batches vary with composition and oxygenates.