Need check on calculation for average number of atoms in a m^3 of atmosphere at STP

mesa

So I need a check on my calculations for the average number of atoms in a 1 meter cube of atmosphere at STP.

I came up with 88.14 mols, here is how I did it:

First I calculated the average amu using percentage in air by
Nitorgen 78.08% @ 14.007 amu
Oxygen 20.95% @ 15.999 amu
Argon .93% @ 39.948 amu

I took these numbers and multiplied them by their relative percentages to get average amu for atmosphere:
10.94 + 3.35 + .37 = 14.67 amu average

We know the average density of air at STP is 1.293x10^3g/m^3
so I divided by average amu and got 88.14 mols/m^3

Is this right?

Borek

No, that's not correct - you forgot nitrogen and oxygen are present as diatomics. But approach per se is OK.

Note that there is much simpler way. Avogadro's principle says that no matter what the gas identity is, number of molecules/atoms in a given volume is always the same (as long as gas can be treated as ideal). That means you don't need to know the mixture composition, just use PV=nRT to calculate number of moles of gas in 1 cubic meter.

mesa

I'm aware that oxygen and nitrogen exist as diatomic molecules in our atmosphere, I happen to need the number of atoms as stated in the question and title of the post but I could see how that could be missed Borek :)

That's a good approach, however I would need to recalculate for number of atoms since (as you know) the IGL counts our diatomic molecules and Ar the same so the result would be slightly less than double of n.

Either way a fairly simple check and good suggestion.

Borek

Yep, missed that, my mistake.

http://www.youtube.com/watch?v=yZymBti7700

mesa

Ha ha! Since when were chemists fans of Wilder?
Pretty risque stuff for 59' but that's how he was :)