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Stoichiometry is an important concept in combustion and hybrid launchers, since they use the energy from combustible fuels. A stoichiometric mixture is one where the ratio of fuel and oxygen molecules is balanced, so all the fuel and oxygen in the chamber is (in optimal conditions) consumed in the reaction. A fuel meter system is usually designed to achieve this mixture, which will yield the most power out of the combustion reaction.


The stoichiometric ratio for gaseous fuels can easily be calculated. When the hydrogen and carbon in a fuel burns, it combines with oxygen as follows:

C + O2 --> CO2
2H2 + O2 --> 2H2O
CxHx + O2 --> CO2 + H2O

If the fuel molecule contains oxygen atoms, they will replace oxygen atoms from the air in the reaction. So for each oxygen atom the fuel contains, one less atom of atmospheric oxygen is needed. To begin calculating the stoichiometric ratio of a fuel, first find the chemical formula for the fuel, and count how many hydrogen, carbon and oxygen (if any) atoms it contains.

Since hydrogen combines with oxygen to make water (H2O), and two hydrogen atoms combine with every oxygen atom, four hydrogen atoms will consume one O2 molecule. Conversely, it can be said that one hydrogen atom will consume 0.25 O2 molecules.

Carbon atoms combine with oxygen to form CO2, so each carbon atom will consume one entire O2 molecule.

Any oxygen in the fuel will lead to one less oxygen atom being consumed from the air. Two oxygen atoms will prevent one O2 molecule from being consumed, so each oxygen atom "saves" 0.5 O2 molecules.

Now we need to find how many O2 molecules are consumed by each fuel molecule, which is calculated as follows:

  • for each hydrogen atom, add 0.25 to the amount of oxygen molecules needed
  • for each carbon atom, add 1
  • for each oxygen atom, subtract 0.5

The resulting number is the oxygen to fuel ratio. To find the fuel to oxygen ratio, divide 1 by this number. The amount of oxygen in air is 21%, so to find the fractional fuel to air ratio, the fuel to oxygen ratio is multiplied with 0.21. To find the ratio in percent, multiply this value by 100.

To find the volume of fuel needed for a particular chamber, the fractional ratio is multiplied by the chamber volume.

Concrete example

Formula of propane: C3H8

  • number of carbon atoms: 3
  • number of hydrogen atoms: 8
  • number of oxygen atoms: 0

Number of oxygen molecules needed = 3*1 + 8*0.25 = 5 molecules
The stoichiometric combustion equation is:
C3H8 + 5O2 --> 3CO2 + 4H2O

fuel/oxygen ratio = 1/5 = 0.2
fuel/air ratio = 0.2*0.21 = 0.042, or 4.2%

The actual percentage of fuel for a stoichiometric mixture will depend on whether the fuel is added to the air in the chamber or the fuel displaces some of the air in the chamber. If the fuel is added to the air in the chamber then the fuel volume is 4.2% of the chamber volume. If the fuel displaces some of the air in the chamber then the fuel volume is 4.02% of the chamber volume.

fuel volume in a 3000 cc chamber = 3000*0.042 = 126 cc if no air is displaced
fuel volume in a 3000 cc chamber = 3000*0.0402 = 120.6 cc if air is displaced

Most fuel meters do not displace air in the chamber with fuel (the pressure in the chamber rises slightly when the fuel is injected), so for a meter the fuel volume is 4.2% of the chamber volume. If the fuel is injected with a syringe, and the fuel displaces some of the air in the chamber, then the fuel volume should be 4.02% of the chamber volume.

Single formula for the fractional fuel to air ratio for hydrocarbon fuels: (1/(C+(H/4)+(O/2)))*0.21