notam amina alifatica saturata = CnH2n+1NH2
si avem 2a moli CH3NH2 + 3a moli CnH2n+1NH2
--------------- ---------------------
miu= 31 g/mol miu= 14n+15 g/mol
100% ................................................ 24,9% N
31x2a+(14n+15)3a amestec ........... (28a+42a) g N
=> 281,12a = 107a + 42na => n = 4
=> amina este C4H9NH2
C-C-C-C-NH2
C-C(CH3)-C-NH2
C-NH-C-C-C
C-C-NH-C-C
C-C(CH3)-NH-C
=> c.