Question

La arderea a 9 g de substanta (densitatea dupa aer egala cu 2,069) rezulta 13,2g CO2, 5,4g H2O. Sa se determine formula moleculra a substantei.

Answer 1

s = 9g

d aer (gaz) = 2,069

a = 13,2g CO2
b = 5,4g H2O

d gaz = M/Mgaz => M = dgaz*Mgaz = 2,069*28,9 = 60g/mol

%C = 300a/11s = 300*13,2/11*9 = 40
%H = 100b/9s = 100*5,4/9*9 = 6,666
%O = 100-(%C+%H) = 53,33 

n C = %C*M/AC*100 = 40*60/1200 = 2 atomi C
n H = %H*M/AH*100 = 6,66*60/100 = 4 atomi H
n O = %O*M/AO*100 = 53,33*60/1600 = 2 atomi O

F.M => C2H4O2 - acid acetic