Question

Analiza elementara cantitativa a unui compus organic indica C:52,17% , H:13,04% restul fiind oxigen , Daca densitatea in raport cu $H_{2}$ este 23 se cere formula moleculara a compusului.

Answer 1

Diferenta pana la 100 e oxigenul
....................................|÷2,1743
C: 52,17/12=4,3475......2
H: 13,04/1=13,04..........6
O: 34,79/16=2,1743......1
(C2H6O)n
M(C2H6O)n=(24+6+16)n=46n
dH2=Msubst/MH2 => Msubst=23*2=46g/mol
46n=46 => n=1
formula moleculara: C2H6O