Question

In conditii normale 1 dm3 dintr-o hidrocarbura gazoasa cântărește 2,5g. In urma arderii a 0.2 moli din aceeasi hidrocarbura se degaja 35,2 g de dioxide de carbon (CO2). Determina formula moleculara a hidrocarburii.

Answer 1

m/M = V/Vm => M=2,5*22,4/1 = 56g/mol

n=m/M => m=n*M=0,2*56 =11,2 g (s)

a=35,2g CO2

%C=300a/11s = (300*35,2)/(11*11,2) = 85,714 %

%H=100-%C = 14,285 %

nC = %C*M/AC*100 = (85,714*56)/(12*100) = 4 atomi C

nH=%H*M/AH*100 = 14,285*56/100 = 8 atomi H

F.M => C4H8 -> 1-butena, 2-butena, ..etc