m/M=V/Vm => M=m*Vm/V=2,5*22,4/1=56g/mol
44g CO2..................12g C
35,2g CO2...............x=9,6g C
n=m/M=9,6/12=0,8 moli C
0,2 moli CxHy.................0,8 moli C
1 mol..............................x=4 atomi C
Fiind o hidrocarbura nesaturata ar putea fi o alchena => CnH2n
Formula moleculara ar fi -> C4H8 -> butena
CH2=CH-CH2-CH3