Question

Se da o substanta A. Care contine 40% carbon , 6,66 % hidrogen si restul pana la 100 oxigen, masa moleculara este 90.
Aflati formul moleculara a substanteti
Urgent!!!!

Answer 1

C:H:O = 40/12 : 6, 66 /1 : 53,34 /16 = 3,33 : 6,66 : 3,33 |÷ 3,33 = 1:2:1 raport atomic => (CH2O)n = 90 => 12•n + 2n+ 16•n = 90 => 30n = 90 => n = 3 => C3H6O3

Answer 2

%O=100-(%C+%H) = 100-(40+6,66) = 53,333%

nC=%C*M/AC*100 = 40*90/1200 = 3 atomi C

nH=%H*M/AH*100 = 6,66*90/100  = 6 atomi H

nO=%O*M/AO*100 = 53,33*90/1600 = 3 atomi O

F.M => C3H6O3