Liquid to Gas Combustion
Name: Romanov L.
Date: May 2006
I have a volume/combustion question.
If you have one cubic centimeter of liquid gasoline, what would be
the volume of the gaseous products after a combustion? How did you
formulate your answer?
First a balanced equation would have to be written, which I will get
by assuming fully saturated octane to represent what the major
component of gasoline is, also I will assume complete combustion.
C8H18 + O2 ----> CO2 + H2O
Now the equation must be balanced so that the same number of C's,
H's and O's are on both sides of the equation:
C8H18 + 12.5 O2 ----> 9 H2O + 8 CO2
This tells us that for one molecule of gasoline, we get 9 molecules
of water and 8 molecules of CO2. The volume of 1 mole of a gas at
STP (standard temperature and pressure) is 22.4 L/mole. If we had 1
mole of octane, we would get 9 + 8 = 17 moles of gaseous products
after complete combustion, which would be 17 mole x 22.4 L/mole =
380.8 L of gaseous products.
But we have 1 mL, not 1 mole, so the next thing to do is to figure
out how many moles of octane we really do have. The formula weight
of octane is 114.2 and its density is 0.703 g/mL. If we have 1 mL,
then we have 0.703 g, which is 0.703 mole/114.2 g/mole = 0.00616
moles. This means that 1 mL of octane = 0.00616 moles x 380.8
L/mole = 2.34 L of gaseous products would result from the combustion
The correlation to gasoline is not directly applicable,
however. Gasoline is a mixture of many many different products and
depending on whether you get 87 octane or 93 octane, the ratio of
products changes. It also matters how old the gasoline is because
it has a lot of volatile components that will evaporate and change
the mixture yet again. The density of gasoline, from a quick search
is listed as 0.72 g/mL, but without an average molecular weight, it
gives you only part of the information you are searching for. If
you can find the average molecular weight of gasoline, then you can
simply work through the problem above, just replace the FW and
density figures. But again, you are also assuming that the balanced
equation is applicable as well. I think that to get a "real" answer
one would need to actually have a gasoline sample and quantitate the
component ratios, then write a balanced equation for each and weight
the volume of each component by the component ratio. Everything
else will just be a close estimate.
This question can have several answers depending upon how you
formulate the question, but the principles involved are these:
Assume gasoline is octane (C8H18). This is an oversimplification
because gasoline is a mixture of various hydrocarbons. The general
approach is to make the calculation based on 1 mol of octane, then
multiply the result by the number of mols of octane. Save this for last.
A good approximation is that the volume of any liquid reactant
or product is essentially zero compared to the volume of gas of the
same reactant or product. It is also a good approximation to assume
the ideal gas law, that is: P*V = N*R*T, where P is the pressure in
atmospheres, V is the volume in liters, N is the number of moles of
the particular GAS, T is in kelvins, and R = 0.08205 liter atm / mol kelvin.
The balanced chemical reaction of octane with oxygen is:
(25/2) O2 ---> 8 CO2 + 9 H2O
Other parameters need to be specified: Do you want to consider H2O
to be a gas or a liquid? It is evident that how you specify that
physical state greatly affects the change in the number of mols of
gaseous products and reactants. What temperature(s) do you want to
set for the reactant O2 and the products CO2 and H2O. If all are at
298 kelvins the change in volume will be very different from the
value you get if you consider O2 to be at 298 kelvins and the
products to be at 1000 kelvins. The customary way is to assume the
temperature to be a constant 298 kelvins and the water to be a
liquid. Then, on a per mole C8H18 basis the change in the number of
moles of gases is: (8 - 25/2) = - 4.5. In practice, of course, that
is not how an internal combustion engine works. You are free to do
the calculation considering the product H2O to be a gas, as well as
the CO2, and you are free to choose the temperature to be whatever
you desire. You could also choose the volume of O2 to be 298 kelvins
and the products to be at some high temperature, say 1000 kelvins.
Various selections cause major changes in the change in volume. The
key principles are to remember that it is the change in the number
of MOLS of GAS (under whatever conditions you choose for each) that
determines the change in volume. The volume of liquids are
negligible by comparison in all cases.
Finally, octane has a molecular weight of 114.2 gm/mol and a
liquid density at 25 C = 298 kelvins of about 0.7 gm / cm^3. So the
1 cm^3 of octane is about 0.006 mols of octane in 1 cm^3. Save this
"scale" factor until the end, so you do not have to keep converting
values to that amount of octane. You only need to do it once.
You would first apply some stoichiometry and then gas laws.
1) write the balanced equation (this will be a problem since
gasoline is a complex combination of compounds - you may have to
choose the most abundant liquid in gasoline) you will have to assume
perfect combustion (to make it easier)
2) convert the volume of gasoline to mass
3) calculate the mass of gaseous products (be sure you know what
temperature you are working in so that you can decide if the water
produce is gaseous or not).
1) Assume that all the gases in the system is an ideal gas
2) Calculate the volume of gas produced given a certain pressure and
temperature (use V=nRT/P)
Greg (Roberto Gregorius)
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Update: June 2012