c. ureei
d. organogene
4.
C2H2 ; C4H10 ; C6H6
5.
100% - 54,53% - 9,09% = 36,38% O
avem compusul CaHbOc
din problema a = 4
=> 100% ......................... 54,53% C
miu g/mol ................... 4x12 g C
= 88 g/mol
=> 100% .......... 9,09% H ............. 36,38% O
88 g ............... b g H ................ 16c g O
b= 8 ; c = 2
=> Fm = C4H8O2
