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