Calculam masele de C si H, si daca avem diferenta pana 3,12g, aceea diferenta este masa de O
1mol CO2.......22,4l.........12g C
.......................3,36l.........1,8g C
mC=1,8g
1mol H2O........18g.........2g H
......................3,24g....0,36g H
mH=0,36g
mO=msubst-mC-mH=3,12-1,8-0,36=0,96g
impartim masele elementare la masele atomice, apoi raportul cel mai mic la toate 3, obtinand formula bruta
………………………………|÷0,06
C: 1,8/12=0,15.........2,5
H: 0,36/1=0,36.........6
O: 0,96/16=0,06.......1
(C2,5H6O)n
Mbruta=(2,5*12+6+16)n=(30+6+16)n=52n
52n=104 => n=2
C5H12O2
