What does lambda (λ) mean?

Lambda compares actual AFR to stoichiometric AFR. λ = 1 is stoichiometric, λ > 1 is lean (excess air), λ < 1 is rich (insufficient air).

Why is 3.76 used for nitrogen?

Dry air is approximately 21% O2 and 79% N2 by volume, so N2/O2 ≈ 79/21 ≈ 3.76.

All calculators

AFR (Air–Fuel Ratio) Calculator

Q: Do oxygenated fuels change AFR?

Yes. Oxygen in the fuel reduces required external O2 and lowers the stoichiometric AFR (e.g., ethanol vs octane).

Q: Why is 3.76 used for nitrogen?

Dry air is approximately 21% O2 and 79% N2 by volume, so N2/O2 ≈ 79/21 ≈ 3.76.

Compute stoichiometric air–fuel ratio from a fuel formula (e.g., C₈H₁₈, C₂H₅OH, H₂, CO). Includes balanced reaction, required O₂, λ and φ from actual AFR, percent excess air, and an optional mini gauge.

Background

For complete combustion to CO₂ and H₂O, a generic fuel C_cH_hO_oS_s requires ν_O₂ = c + h/4 − o/2 + s moles of O₂ per mole of fuel. Air is approximated as O₂ + 3.76 N₂. Stoichiometric AFR (mass basis) is: AFR_st = ν_O₂ × (M_O₂ + 3.76 M_N₂) / M_fuel.

Enter fuel formula and optional actual AFR:

Fuel formula:

Supports parentheses and element counts (e.g., C₆H₁₂O₆, Ca(OH)₂). O, H, C, S affect O₂ demand; N in fuel is treated inert.

Actual AFR (optional):

If provided, we’ll compute λ (lambda), φ (equivalence ratio), and % excess air vs stoichiometric.

Show:

Step-by-step working

Mini gauge: λ (lean ↔ stoich ↔ rich)

Round to sensible significant figures

Quick picks:

Tip: Enter an actual AFR to get λ and % excess air. We assume dry air: O₂ + 3.76 N₂.

Result:

No results yet. Enter a fuel and click Calculate.

How to use this calculator

Fuel: Type a chemical formula (e.g., C₈H₁₈, C₂H₅OH, H₂, CO). Parentheses are supported.
Output: Stoichiometric AFR (mass basis), O₂ requirement, moles of air, and a balanced reaction.
Actual AFR (optional): Enter air/fuel to get λ (lambda), φ, and % excess air. λ = AFR / AFR_st; φ = 1/λ.
Assumptions: Complete combustion to CO₂, H₂O, SO₂; N in fuel is inert → N₂; air = O₂ + 3.76 N₂.

Formula & Equation Used

Oxygen demand: ν_O₂ = c + h/4 − o/2 + s (for fuel C_cH_hO_oS_s)

Air model: air ≈ O₂ + 3.76 N₂

Stoichiometric AFR: AFR_st = ν_O₂ × (M_O₂ + 3.76·M_N₂) / M_fuel

Equivalence & lambda: λ = AFR/AFR_st, φ = 1/λ, % excess air = (λ − 1) × 100%

Example Problems & Step-by-Step Solutions

Example 1 — Propane (C₃H₈)

ν_O₂ = 3 + 8/4 = 5. M_fuel = 44.10 g/mol. AFR_st = 5 × (32.00 + 3.76 × 28.0134) / 44.10 ≈ 15.6.

Example 2 — Ethanol (C₂H₅OH)

c=2, h=6, o=1 → ν_O₂ = 2 + 6/4 − 1/2 = 3.0. M_fuel ≈ 46.07 g/mol → AFR_st ≈ 9.0.

Frequently Asked Questions

Q: What does λ (lambda) mean?

λ = 1 is stoichiometric. λ > 1 is lean (excess air). λ < 1 is rich (insufficient air).

Q: Do oxygenated fuels change AFR?

Yes. Oxygen in the fuel reduces external O₂ demand and lowers AFR_st (e.g., ethanol vs octane).

Q: Why use 3.76 for N₂?

Dry air is ~21% O₂ and ~79% N₂ by volume → N₂/O₂ ≈ 79/21 ≈ 3.76.

Balancing Chemical Equations