Question

Determina formula procentuala, bruta si moleculara a substantei organice care are masa moleculara egala cu 90 , iar la analiza elementală , prin combustie a condus la urmatoarele date : 0,2455g substanta s a transformat in 0,3600g dioxid de carbon si 0,1473g apa

Answer 1

%C=300a/11s*100=>300*0,36/11*0,2455 = 40%

%H=100b/9s=0,1473*100/9*0,2455 = 6,666%

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

C=40/12=3,333  ||:3,333 => 1 atom
 
H=6,666/1=6,666 => 2 atomi 

O=53,333/16=3,333 =>  1 atom

F.B => (CH2O)n 

F.M = 12n+2n+16n=90 => 30n=90 =>n=3 => C3H6O3

sau

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

nH=%H*M/AH*100=6,666*90/100=6 atomi H

nO=%O*M/AO*100=53,333*90/1600=3 atomi O

F.M => C3H6O3