Question

determina formula bruta si formula moleculara a unui compus organic care contine in procente masice 40% C 6,67% H si are masa molara M=180g/mol. Se dau masele atomice Ac=12 An=1 Ao=16​

Answer 1

Răspuns:

%C=40: 12 (A C) = 3,33:3,33= 1 atg

%H=6,67: 1 (A H)= 6,67:3,33= 2 atg

%O= 100-(40+6,67)=53,33: 16=3,33:3,33= 1 atg

=> F. chimica bruta e (CH₂O)ₙ

M= 12n+2n+16n=30n

M= 180 g/mol => n=180/30=6

=> f.chim Moleculara: C₆H₁₂O₆

NE= 2*6+2-12/2=1 ∈ Z₊

nr.covalente= 4*6+12*1+6*2=48 - nr par