Question

O subst organica are raportul de masa C:H:S:O=36:3:32:48.
a) care este formula moleculara a subst., daca prin arderea a 0,05 moli se obtin 6,72 l CO2?

Answer 1

CaHbScOd
mC/mH = 12a/b = 36/3 => 36b = 36a => a = b => CaHaScOd
mC/mS = 12a/32c = 36/32 => 1152c = 384a => c = a/3 => CaHaS(a/3)Od
mC/mO = 12a/16c = 36/48 => 576d = 576a => d = a => CaHaS(a/3)Oa
CaHaS(a/3)Oa + a/2 O2 => a CO2 + a/2 H2O
1 mol substanta............22,4 a l CO2
0.05 moli substanta.............6.72 l CO2
1.12a = 6,72 => a = 6 => FM: C6H6S3O6