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?
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
72-30(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?
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!
Nope. Pentane is C5H12. If C5H10 is the choice you are given, that is pentene.