What is the stoichiometry of gasoline and how is it calculated?

[GPracer2500]

I'm looking for some help with the stoichiometry of gasoline. 14.7:1 is commonly cited as the stoich ratio for gasoline and air. I believe this ratio is calculated using a primary reference fuel consisting of a mixture of n-heptane and iso-octane. Precisely what mixture is unknown to me.

Could someone walk me through the calculation for determining the stoich of gasoline? Is there a way to work backwards from 14.7 to determine what mixture of n-heptane and iso-octane is used for the reference gasoline? Or would the results be similar or the same no matter what mixture of those two hydrocarbons was used?

Thank you for your time!

 

[GPracer2500]

This is what I've got so far:

iso-octane = C8H18
n-heptane = C7H16
air = 21% O2, 79% N2 (not exactly, but good enough for my purposes)
combustion products = CO2 + H2O

Where do I go from here?

 

[Gokul43201]

I'm looking for some help with the stoichiometry of gasoline. 14.7:1 is commonly cited as the stoich ratio for gasoline and air. I believe this ratio is calculated using a primary reference fuel consisting of a mixture of n-heptane and iso-octane. Precisely what mixture is unknown to me.

Could someone walk me through the calculation for determining the stoich of gasoline? Is there a way to work backwards from 14.7 to determine what mixture of n-heptane and iso-octane is used for the reference gasoline?

Yes, there is.

Or would the results be similar or the same no matter what mixture of those two hydrocarbons was used?

They might be similar (can't tell until we do the calculation), but they will not be the same.

This is what I've got so far:

iso-octane = C8H18
n-heptane = C7H16
air = 21% O2, 79% N2 (not exactly, but good enough for my purposes)
combustion products = CO2 + H2O

Where do I go from here?

The next step is to write and balance the equations describing the combustion of each of these compounds, assuming complete combustion.

For instance, if you were burning ethanol (C2H6O), the balanced equation would look like this:

C_2H_6O + 3O_2 \longrightarrow 2CO2 + 3H_2O

This tells you that 1 mole of ethanol needs 3 moles of oxygen for complete combustion. Using the molar masses, convert 1 mole (ethanol) and 3 moles (O2) into the respective masses for ethanol and oxygen. Finally, figure out how much air it takes to get that mass of oxygen. The ratio of ethanol mass to air mass is your required fuel-air ratio.

Do this for iso-octane and n-heptane - what do you get for each case?

 

[GPracer2500]

Ok...

iso-octane: 2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O

and

n-heptane: C7H16 + 11 O2 ----> 7 CO2 + 8 H2O


So then I need to account for the mass of these equations and factor in the non-reactive nitrogen component somehow? Hopefully I'm not getting ahead of myself...

Molecular weights:
H = 1
O = 16
C = 12
N = 14

C8H18 = 114
C7H16 = 100
CO2 = 44
H2O = 18
O2 = 32

iso-octane:
2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O
2*114 + 25*32 ----> 16*44 + 18*18
228 + 800 ----------> 704 + 324

n-heptane:
C7H16 + 11 O2 ----> 7 CO2 + 8 H2O
100 + 11*32 ------> 7*44 + 8*18
100 + 352 ---------> 308 + 144

If air = 21% oxygen and 79% nitrogen there's 3.76x as much nitrogen as oxygen. And the molecular weight of N = 14. So for n-heptane, to get the required 11 moles oxygen we'd need 3.76*11 moles of nitrogen? 41.36 N2 (?) = 1158 units of mass?

1158 + 352 = 1510 mass units of air for every 100 mass units of n-heptane? 15.1:1 a/f ratio for stoich?

For iso-octane, 25 moles O2. 3.76*25 = 94 moles N2. 2632 mass units of nitrogen? 2632 + 800 = 3432. 3432/228 = 15.05:1 a/f ratio?

Hopefully I screwed something up because those weren't the results I was expecting. If my work is correct (a big IF), where the heck do they get 14.7:1?
Thanks a lot for the help.

 

[GPracer2500]

I worked out stoich for ethanol also and came up with 8.95:1. 9:1 is the figure I've seen for ethanol so I think I'm doing the calculations correctly. But if that's so and my iso-octane and n-heptane calculations are without error, that leads me to believe that the 14.7:1 figure is NOT based off of a primary reference fuel consisting of some mixture of iso-octane and n-heptane.

I'm aware there are hundreds of different hydrocarbon species found in typical gasoline (not to mention the possibility of oxygenates, deposit modifiers, and maybe other things I don't even know about) and I'm aware that the exact make-up of gasoline can be quite variable (seasonal, regional, batch-to-batch, etc). Nonetheless, there is an ASTM specification that defines gasoline (D4814) but I've never been able to find a copy without buying it. I wonder what it says about stoich?

I'd really like to know where the 14.7 figure is coming from.

 

[Gokul43201]

I don't see an error in your calculation (didn't actually do the arithmetic, but I trust you got that right). Also, I believe 14.7 is supposed to be the theoretical air-fuel ratio for a pure binary mixture of iso-octane and n-heptane. So, like you, I'm not sure where this number comes from.

 

[GPracer2500]

I did the math again with less rounding and I'm getting:

iso-octane: 15.03:1
n-heptane: 15.08:1

I'm a little concerned about using 3.76x as much nitrogen in the air as oxygen (21% oxygen, 79% nitrogen--I think by volume?). Could that simplistic representation of the air account for the discrepancy? The a/f ratios I'm trying to figure are by weight, not volume. I haven't yet found a good breakdown of a "standard" atmosphere by weight. I did find that Wikipedia has nitrogen listed as "78.082% by volume of dry air, 75.3% by weight in dry air."

If I know how many moles of O2 are consumed in each calculation, what does that mean for the total quantity of air needed (by weight).

thanks again

 

[Bystander]

Gasoline from the pump ain't totally saturated --- throw a little toluene, few xylenes, mesitylene into your calculations --- the "14.7" is a linear combination of combustions of various degrees of saturation.

 

[sicjeff]

Stoichiometry is useless for gasoline unless you are prepared to handle over 100 reactions simultaneously.