Chemistry empirical formula problem

A) A hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard conditions (S.T.P.). What is the empirical formula of the Hydrocarbon? (Recall that the molar volume of an ideal gas is 22.4 liters at S.T.P.)

B) Calculate the mass in grams of O2 required for the complete combustion of the sample of the Hydrocarbon from part A.

C) If the molecular weight of the hydrocarbon is approximately 58 g mol^-1 what is the molecular formula of the Hydrocarbon?

A) we have 7.2g of water of 7.2 / 18 moles of water = .4 moles of water
there are 2 moles of H in each moles of water so moles of H = .8
since all the H from the compound was converted to water
moles of H in the compound = .8 moles

1 mole of CO2 occupies 22.4 liters at STP so we have 7.2 / 22.4 moles of CO2 = 0.3214 moles
using similar logic to the H ... moles of C = 0.3214

Molar ratio of C : H = .3214 : .8 or after dividing bu the smallest 1 : 2.5
we multiply by 2 to get the final ratio C : H = 2 : 5

empirical formula is C2H5

B) we don't have the molar mass so if we assume the molecular formula of the hydrocarbon is 2 x the empirical formula or

C4H10 ..and

2 C4H10 + 13 O2 = > 8 CO2 + 10 H2O

we have .3214 moles of C so we also have .3214 / 4 moles of C4H10 = 0.080 moles
these react with .08 x 13/2 moles of O2 = .52 moles of O2
mass of O2 = .52 x32 = 16.64g

c) we can now check the molecular formula of C4H10 = 4x12+10x1= 58

so we assumed right ... the molecular formula is C4H10