100% - 61%C = 39% H si N
notam compusul CaHbNc
n = m/miu = V/Vm, Vm = 22,4 L
=> miu = mxVm/V = 59 g/mol
=> 12a+b+14c = 59
100% ................ 61% C
59g ................... 12a => a = 3
=> C3HbNc => b+14c = 59-3x12 = 23
din problema avem = 14c/b = 1,55
=> 14c = 1,55b => b = 9c
=> 9c+14c = 23 => c = 1
=> b = 9
=> C3H9N