Posted by Sarah on .
An unknown gaseous hydrocarbon consists of 85.63% carbon by mass. A 0.335 g sample of the gas occupies a volume of 0.107 L at STP. What is the identity of the gas?

Chemistry 
DrBob222,
Take a 100 g sample which gives you
85.63 g C
14.14.37 g H.
Convert to mols.
85.63/12 = about 7
14.37/1 = about 14
Empirical formula is CH2.
At STP a mole of gas occupies 22.4L; therefore, 0.107/22.4 = about 0.05 mols but you need to be more exact than that.
mols = grams/molar mass. You know mols and you know grams, solve for molar mass. I get approximately 72.
Since this is a hydrocarbon we know the formula will be
CH3(CH2)nCH3
7230(for the two CH3 groups) = 40 left for the CH2. Then 40/14 = about 2.8 or so, round to 3.0 so the formula must be
CH3(CH2)3CH3 or C5H12. That is a gas. Do you know the name? 
Chemistry 
Sarah,
Thanks so much, I would of never figured out that out myself past the empirical formula since I forgot about the 22.4L factor.
The molar mass turns out to be 70, and since the only answer I was given that adds up to 70 is C5H10, it's an easy pick. Pentane! 
Chemistry 
DrBob222,
Nope. Pentane is C5H12. If C5H10 is the choice you are given, that is pentene.