Combustion Analysis Calculator

Last updated: June 2026

Use this combustion analysis calculator to balance complete combustion equations, calculate the oxygen required and the products formed, and determine an empirical formula from carbon dioxide and water data. It is designed for O-Level, IGCSE, A-Level and Singapore H2 Chemistry students studying organic chemistry and stoichiometry.

Key takeaway: During complete combustion, every carbon atom in the fuel becomes part of CO2, while every hydrogen atom becomes part of H2O. Oxygen already present in the fuel reduces the amount of O2 that must come from the air.
How to use this calculator
  1. Choose Combustion equation to balance a fuel and calculate the amounts of O2, CO2 and H2O.
  2. Enter a molecular formula containing only C, H and optional O, such as C3H8 or C2H6O. Entering a fuel amount is optional if you only need the balanced equation.
  3. Enter the available O2 only when you want the calculator to identify the limiting reactant and any excess reactant.
  4. Choose Analysis data to calculate an empirical formula from the masses of CO2 and H2O produced.
  5. Enter the original sample mass when the compound may contain oxygen. Add the compound's molar mass only when you also want its molecular formula.

Organic Combustion Equation Balancer

For an organic compound containing carbon, hydrogen and oxygen, the complete combustion equation can be represented as follows:

CxHyOz + O2 → CO2 + H2O

Carbon is balanced first because each carbon atom in the fuel produces one CO2 molecule. Hydrogen is balanced next because two hydrogen atoms are needed for each H2O molecule. Oxygen is balanced last.

CxHyOz + (x + y/4 − z/2)O2 → xCO2 + (y/2)H2O

Fractions are chemically valid during the working, but the final equation is normally written using the smallest whole-number coefficients. The calculator automatically multiplies every coefficient by the same number when this is necessary.

Combustion Stoichiometry Calculator

Once the equation is balanced, its coefficients become mole ratios. For propane:

C3H8 + 5O2 → 3CO2 + 4H2O

Therefore, 1 mol of C3H8 reacts with 5 mol of O2 to produce 3 mol of CO2 and 4 mol of H2O. If only 2 mol of O2 is available, oxygen becomes the limiting reactant and only part of the propane can burn completely.

When a fuel mass is entered, the calculator first converts it into moles using the molar mass calculated from C = 12.0, H = 1.0 and O = 16.0. It then applies the balanced mole ratio and shows each stage of the working.

Empirical Formula from Combustion Data

Combustion analysis works because all the carbon in the original compound appears in CO2, while all the hydrogen appears in H2O. The amount of carbon is therefore equal to the amount of CO2 formed:

n(C) = n(CO2) = mass of CO2 ÷ 44.0

Each mole of water contains two moles of hydrogen atoms, so:

n(H) = 2n(H2O) = 2 × mass of H2O ÷ 18.0

If the original compound contains only carbon, hydrogen and oxygen, the oxygen mass is obtained by subtraction:

mass of O = sample mass − mass of C − mass of H

The amounts of C, H and O are then divided by the smallest amount. If the ratio contains values such as 1.5 or 1.33, every ratio is multiplied by the same small whole number until whole-number subscripts are obtained.

Worked example: propane combustion

Begin with C3H8 + O2 → CO2 + H2O. Three carbon atoms require 3CO2, while eight hydrogen atoms require 4H2O. The products now contain ten oxygen atoms altogether, so 5O2 is required.

C3H8 + 5O2 → 3CO2 + 4H2O

Worked example: combustion analysis

Suppose complete combustion produces 4.40 g of CO2 and 1.80 g of H2O. The amount of carbon is 4.40 ÷ 44.0 = 0.100 mol. The amount of water is 1.80 ÷ 18.0 = 0.100 mol, so the amount of hydrogen atoms is 2 × 0.100 = 0.200 mol.

The C:H ratio is therefore 0.100:0.200 = 1:2, giving CH2 from the carbon and hydrogen data. However, this result does not prove that oxygen is absent. The original sample mass is needed before oxygen in the compound can be found by difference.

Common exam traps

Trap 1: Complete combustion produces CO2 and H2O. Incomplete combustion may also produce CO or carbon, so the same calculation cannot be used without additional information.
Trap 2: Oxygen in the fuel counts towards the oxygen atoms in the products. It therefore reduces the O2 coefficient rather than increasing it.
Trap 3: From CO2, calculate moles of C. From H2O, multiply the amount of water by two to obtain moles of H atoms.
Trap 4: If no original sample mass is given, the C:H ratio can be found, but oxygen in the compound cannot be determined by subtraction.
Trap 5: Combustion analysis concerns composition and formula. Enthalpy of combustion is an energetics calculation and belongs to Hess' law or bond-energy calculations.

Frequently Asked Questions

What does this combustion analysis calculator calculate?

It balances the complete combustion of compounds containing C, H and optional O. When a fuel amount is entered, it also calculates O2 required, CO2 produced, H2O produced and any excess reactant. Its second mode finds empirical and molecular formulae from combustion data.

Can it balance ethanol and other oxygen-containing fuels?

Yes. For ethanol, C2H6O, the oxygen atom already present in each ethanol molecule is included when the O2 coefficient is calculated.

Why must I enter a molecular formula rather than a displayed formula?

The balancer needs the total numbers of C, H and O atoms in one fuel molecule. A molecular formula such as C2H6O states these totals directly and avoids ambiguity.

How do I know whether oxygen is present in the original compound?

Calculate the masses of carbon and hydrogen represented by the CO2 and H2O data. If their total is smaller than the original sample mass, the remaining mass is oxygen, provided the compound contains only C, H and O.

Why can the empirical formula differ from the molecular formula?

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of each type of atom and is a whole-number multiple of the empirical formula. A molar mass is needed to identify that multiple.

Scroll to Top