[coffee]

The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard conditions. What is the empirical formula of the hydrocarbon?
How do I solve this problem? This is what I know so far:
C?H? + O2 --> CO2 = H2O
CO2= 7.2L H2O=7.2g
Do  get CO2 to gC and H2O to gH? If so, how do I convert? 

[Mitch]

First, convert grams water to moles water and liters CO2 to moles CO2

[coffee]

ok i did that and got .4 mol H2O
I don't know how to do conversions for liters.

7.2 g L x 1000gCO2/1L CO2 x 1 mol CO2/44gCO2?   

[Mitch]

Use Pv=nRT to calculate moles CO2. They tell you its at standard conditions so you can assume 1atm pressure and 25C temperature and they give you the volume so solve for the amout of moles.

[coffee]

isnt standard temperature 273K?
anyways i got 0.00317

(1atm)(7.2L) divided by (8.31)(273K)

[Will]

isnt standard temperature 273K?

Depends! http://www.chemicalforums.com/index.php?topic=8737.0

[coffee]

ok. i usually use 273K
um anyways did i convert it correctly >,<
and what do i do after i converted both into moles

[sdekivit]

to avoid calculations, it may be useful to know the molar volumes at T = 273 K and T = 298 K at p = 10⁵ Pa:

T = 273 K --> V_m = 22,4 L / mol (for an ideal gas)
T = 298 K --> V_m = 24,5 L / mol (for any gaseous compund)

Of course, they can be calculated with the ideal gas law.