Question

Ajutor va rog!!!!!!!!!!!!!!!!!!​

Answer 1

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