stim ca n = m/miu = V/Vm, Vm = 22,4
=> miu = mxVm/V = 44 g/mol
din formula procent. deducem ca este o hidrocarbura
=> hidrocarbura = CaHb
100% .................... 81,81% C .................. 18,18% H
44g ...................... 12a g c ...................... b g H
=> a = 3, b = 8
=> C3H8