Carbons in a propane sample

Today in class we looked at a problem that some of you didn't quite get to the end of by the time class ended. Here it is. If you haven't already worked it through, give it a good try before you jump ahead to the answer...

How many carbon atoms are in a 37.43L sample of propane gas at 17.52°C and 1.472atm?

This starts out as an Ideal Gas Law problem with a single set of conditions.

PV = nRT

Plugging in the values from the problem:

(1.472atm)(37.43L) = n(0.08206^L.atm/_mol.K)((17.52+273.15)K)

n = 2.3099mols of C₃H₈(g)

{NOTE: I'm in the middle of the problem, so I'm not rounding for significant figures yet, 

but it looks like 4 sig figs would be good at this point…}

Each propane molecule contains 3 carbon atoms, so each mole of propane molecules contains 3 moles of carbon atoms.

(2.3099mols C₃H₈) (3mols C/1mol C₃H₈) = 6.92974mols C

(6.92974mols C) (6.022x10^{23 C atoms}/_{mol C}) = 4.173x10²⁴ C atoms in the sample.

DISCLAIMER: This problem assumes that propane is behaving as an ideal gas under these conditions. There's a pretty good chance that it would not be ideal under these conditions, but for the purposes of this problem let's assume it is.