Why Does Calculating Ethanol's AFR Yield Different Results from Standard Values?

• D19A99G
D19A99G
TL;DR Summary
Calculation of the theoretical value of AFR of ethanol
First I have seen a few forums dotted about covering this subject already, however I couldn't find exactly what I was looking for so I apologise if its somewhere on here already.

I am trying to calculate the AFR value for pure ethanol.
I am only including oxygen in the category of "air" due to other elements being inert.

The balanced equation is C2H5OH + 3O2 = 3H2O + 2CO2.

When I then calculate the AFR while also considering that oxygen only makes up 21% of air, the value I get is approximately 9.9. Which is different to the widely used value of 9. Is there something I'm doing wrong or has the value of 9 only being achieved through practical tests which cannot be replicated via theory?

Ethanol mass = 46, Oxygen mass = 96
Air mass = Oxygen*4.76 = 456.96
AFR = 456.96/46 = 9.93
When; C=12amu, H=1amu, O=16amu

You are assuming that oxygen makes up 21% of a gas of uniform molecular mass 32 amu. You have to account for the fact that nitrogen is lighter.

D19A99G
Yeah ok, I can see what you mean. If I neglect the other gases which constitute the make up of "air" and assume oxygen = 21% and nitrogen = 79%. Then 3*(32+(3.76*28)) = 411.84. This would give me a AFR of 8.95 which is much closer.
Then considering Argon as well would give me a value of 9.01.

That makes sense thanks

berkeman
Sadly, you can't generally assume whether a generic 'percentage' is volume, mass, or whatever. Units matter. I use '%wt' and %vol' to prevent confusion (and collisions with Mars).

Bystander

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