Question

Prin arderea a 3.12g de substanta organica avand masa molara de 104 g/mol se obtin 3.36 dm cubi de CO2 si 3240 mg de H2O.Determina formula moleculara a substantei organice.
Dau funda

Answer 1

22,4l CO2..........12g C
3.36l CO2............x
x=1,8gC

18 g H2O............2gH
3,24gH2O............y
y=0.36gH


3.12g subst....1,8gC.....0,36gH
100gsubst........a.........b

a=57,69%C
b=11,53%H
%O= 30,78

C:57,69/12=4,8
H:11,53
O: 30,78/16=1,92
acum se imparte fiecare la 1,92 si vine

C2H6O
masa molara a subst e 104 si in acelasi timp (C2H6O)n
Deci (12*2+6+16)n
46n=104=> n=2
C4H12O2