Question

La analiza elementală a 1,8250 g compus organic A, din clasa aminelor, cu masa molară 73 g/mol, s-au obţinut 4,40 g CO2 şi 3,08 L H2O (vapori). Se cere formula bruta si cea moleculara

Answer 1

s = 1,825g
M = 73g/mol
a = 4,4g CO2
b = 3,08L H2O

m H2O = M*V/Vm = 18*3,08/22,4 = 2,475g

%C = 300a/11s = 300*4,4/11*1,825 = 1320/20,075 = 65,75 
%H = 100b/9s = 100*2,475/9*1,825 = 247,5/16,425 = 15,06 

%N=100-(%C+%H) = 19,18

C = 65,75/12 = 5,479   |:1,37 => 4 atomi C
H = 15,06   |:1,37 => 11 atomi H
N = 19,18/14 = 1,37 => 1 atom N

Formula bruta => (C4H11N)n

nC = %C*M/AC*100 = 65,75*73/1200 = 4 atomi C
nH = %H*M/AH*100 = 15,06*73/100 = 11 atomi H
nN = %N*M/AN*100 = 19,18*73/1400 = 1 atom N

Formula moleculara => C4H11N