Question

Substanta organica A are raportul de masa C:H:O = 18:3:8 . Stiind ca , in conditii normale , 2,9g din aceasta substanta ocupa un volum de 1,12L ,
a) determina formula moleculara a substantei A
b) propune a structura pentru formula gasita .

Answer 1

m/M=V/Vm => M=m*Vm/V=2,9*22,4/1,12=58g/mol

C=18/12=1,5 moli  ||:0,5 => 3 atomi C

H=3/1= 3 moli H ||:0,5 => 6 atomi H

O=8/16=0,5 moli O => 1 atom O

F.B => (C3H6O)n 

F.M => 12*3n+6n+16n=58 => 58n=58 => n=1 => C3H6O

b) NE=1

CH3-C(=O)-CH3 -> propanona -> acetona

CH3-CH2-CH=O -> propanal