Răspuns:
aflu cant de CO21,792/22,4=0,08 moli
8,08x44=3,52g
notez s=1,44g subst
a=3,52g CO2
b=1,44gH2O
f procentuala%C=300a/11s=(300x3,52)/11x1,44=66,66
%H=100b/9s =(100x1,44)9x1,44=11,11
dif 100-(66,6+11,1)=22 ,3 sunt %de O
impart %la masa atomica a elem
C66,6/12=5,5
H 11,1/1=11,1 form procentuala
O 22,23/16=1,38
se imparte la cel mai mic si obt nr de atomi5,5/1,38=4C
11,1/1,38=8H
1,38/1,38=1 O
(C4H8O)n form bruta
Msubst=2,49x28,9=72
(12x4+8x1+1x16)n=72
72n=72 n=1
C4H8O form molec
Explicație:
