We have #"C"_2"H"_5# as the empirical formula. We must find its molar mass. To do so, we need to know how many moles of it evaporated. Use the Ideal Gas Law:

#PV=nRT#, and here:

#P=1"atm"#

#V=9.6"L"#

#R=0.0821"L atm K"^-1"mol"^-1#

#T=293.15"K"#.

To solve for #n#, moles, we rearrange:

#n=(PV)/(RT)#, and inputting:

#n=(1*9.6)/(293.15*0.0821)#

#n=0.399"mol"~~0.4"mol"#

Another formula for moles is #n=m/M#, where #m# is the mass and #M# the molar mass.

Rearranging to solve for #M#, we get:

#M=m/n#

We have #m=23.2"g"#. Inputting:

#M=23.2/0.399#

#M=58.16"g/mol"#

We must calculate the molar mass of the empirical formula:

#M_e=(12*2)+(1*5)#

#M_e=29"g/mol"#

We see a #2:1# ratio.

Therefore, the molecular formula is #"C"_4"H"_10#, which is probably butane.