5
mol/mol
3
mol/mol
4
mol/mol
44.097
g/mol
159.99
g/mol
132.027
g/mol
72.06
g/mol
3.628138
g/g
2.994013
g/g
1.634125
g/g
5
mol
3
mol
4
mol
159.99
g
132.027
g
72.06
g
5
mol/mol
3
mol/mol
4
mol/mol
44.097
g/mol
159.99
g/mol
132.027
g/mol
72.06
g/mol
3.628138
g/g
2.994013
g/g
1.634125
g/g
5
mol
3
mol
4
mol
159.99
g
132.027
g
72.06
g
The Combustion Reaction Calculator balances the complete combustion equation for any organic fuel with the general formula $$C_xH_yO_z$$. Complete combustion in excess oxygen converts all carbon to CO₂ and all hydrogen to H₂O. This calculator determines the exact stoichiometric coefficients for oxygen consumed, carbon dioxide produced, and water produced. Combustion reactions are the most important class of chemical reactions in energy production — from burning natural gas ($$CH_4$$) in power plants to the oxidation of octane ($$C_8H_{18}$$) in automobile engines. Understanding the stoichiometry is essential for calculating air-fuel ratios, estimating carbon dioxide emissions, designing combustion chambers, and performing environmental impact assessments. This tool also computes the mass of O₂ needed and CO₂ released per gram of fuel, enabling quick engineering estimates.
The general combustion reaction for an organic compound $$C_xH_yO_z$$ is:
$$C_xH_yO_z + \left(x + \frac{y}{4} - \frac{z}{2}\right) O_2 \rightarrow x\,CO_2 + \frac{y}{2}\,H_2O$$
The balancing follows atom conservation:
The mass ratios are calculated using molar masses:
$$\text{O}_2 \text{ per g fuel} = \frac{n_{O_2} \times 31.998}{MW_{fuel}}$$
$$\text{CO}_2 \text{ per g fuel} = \frac{n_{CO_2} \times 44.009}{MW_{fuel}}$$
The balanced equation shows the exact molar ratios. If the O₂ coefficient is fractional, multiply all coefficients by 2 for whole numbers (common practice in textbooks). The CO₂ per gram fuel is the specific carbon emission factor — a critical parameter for greenhouse gas accounting. Methane (CH₄) produces 2.74 g CO₂/g fuel, while octane (C₈H₁₈) produces 3.08 g CO₂/g. Fuels with higher carbon content produce more CO₂ per gram. Oxygenated fuels (like ethanol, C₂H₅OH) require less external oxygen and produce less CO₂ per gram because they already contain oxygen. The O₂ required determines the minimum air supply for complete combustion — in practice, 10-20% excess air is used to ensure complete burning.
Inputs
Results
C3H8 + 5 O2 → 3 CO2 + 4 H2O. Each gram of propane requires 3.63 g O2 and produces 2.99 g CO2.
Inputs
Results
C2H6O + 3 O2 → 2 CO2 + 3 H2O. Ethanol produces 1.91 g CO2/g — less than hydrocarbons due to its oxygen content.
Complete combustion occurs when a fuel reacts with sufficient oxygen to convert all carbon to CO2 and all hydrogen to H2O. In contrast, incomplete combustion (insufficient O2) produces carbon monoxide (CO), soot (elemental carbon), or other partially oxidized products.
When the hydrogen count y is not divisible by 4, the O2 coefficient contains a fraction. For example, CH4 gives O2 = 1 + 4/4 = 2 (integer), but C2H6 gives O2 = 2 + 6/4 = 3.5 (fractional). Multiplying all coefficients by 2 gives: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O.
Air is approximately 23.2% oxygen by mass. Divide the O2 mass per gram of fuel by 0.232 to get the stoichiometric air-fuel ratio. For example, propane: 3.628/0.232 = 15.6 g air per g fuel. Add 10-20% for practical excess air.
Pure carbon (coal) produces 3.66 g CO2/g. Hydrocarbons produce less because hydrogen contributes mass without adding CO2. Methane (CH4) produces 2.74 g CO2/g due to its high H/C ratio. Oxygenated fuels like ethanol produce even less (1.91 g CO2/g).
No. This calculator assumes complete combustion with excess oxygen. Incomplete combustion produces CO, C, and other products whose proportions depend on oxygen availability and temperature, requiring more complex modeling.
Yes, if you know the average molecular formula. Biodiesel (e.g., methyl oleate C19H36O2) can be entered directly. For fuel mixtures, calculate the weighted average formula or analyze each component separately.
The heat released during combustion (enthalpy of combustion, ΔHc) depends on the fuel type but is closely related to the oxygen consumed. Approximately 13.1 kJ of heat is released per gram of O2 consumed, regardless of the fuel type — this is known as Thornton's rule.
At higher altitudes, air density decreases, reducing the available oxygen per volume. The stoichiometric requirement remains the same (by mass), but engines and burners need to process more air volume to deliver the same mass of oxygen for complete combustion.
If the fuel contains sulfur, it combusts to SO2: S + O2 → SO2. This requires one additional mole of O2 per sulfur atom and produces a pollutant (sulfur dioxide) that contributes to acid rain. This calculator covers C, H, O fuels only.
Each water molecule contains 2 hydrogen atoms. Since the fuel has y hydrogen atoms, it takes y/2 water molecules to account for all the hydrogen. This simple ratio follows directly from hydrogen atom conservation.
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