1.
verificam procentele sa dea 100%
100%-53,33% C - 15,55% H - 31,11% n = 0%
notam compusul organic CaHbNc
120g compus ........ 12a g C ....... b g H ....... 14c g N
100% .................. 53,33% C ....... 15,55% H .... 31,11% N
=> a = 5, b = 18, c = 3
=> C5H18N3
2.
verificam procentele sa dea 100%, diferenta va fi oxigenul
=> 100%-42,85-2,38-16,66 = 38,11% O
C = 42,85/12 = 3,571
H = 2,38/1 = 2,38
N = 16,66/14 = 1,19
O = 38,11/16 = 2,38 : 1,19
=> (C3H2NO2)n
n = 2 deoarece in problema mentioneaza ca avem 2 at N
=> C6H4N2O4
=> M = 6x12+4x1+2x14+4x16 = 168 g/mol
