Question

La arderea a 2,24g hidrocarbura rezulta 7.04 g CO2 si 2.88 g H2O.Stiind ca densitatea fata de H este 28 sa se determine formula moleculara.

Answer 1

dgaz=M/Mgaz => M=dgaz*Mgaz=2*28=56g/mol           MH2(gaz)=2g/mol

n=m/M=2,24/56=0,04 moli hidrocarbura

MCO2 = AC+2AO=12+2*16=44g/mol

In 44g CO2 se gasesc.........................12g C
Atunci in 7,04g CO2...........................x=1,92g C

MH2O=18g/mol

In 18g H2O.........................2gH
2,88g H2O..........................x=0,32g H

mO =2,24-1,92-0,32=0 => nu contine oxigen

n=m/M=1,92/12=0,16 moli C

n=m/M=0,32/1=0,32 moli H

0,04 moli hidrocarbura.........0,16 moli C.........0,32 moli H
1 mol....................................x=4 atomi C.........y=8 atomi H

F.M => C4H8 -> butena

Metoda II

%C=300a/11s=300*7,04/11*2,24=85,714

%H=100b/9s=14,28571

nC=%C*M/AC*100=85,714*56/1200=4 atomi C

nH=%H*M/AH*100=14,28571*56/100=8 atomi H

F.M => C4H8

a= cantitatea de CO2,g

b=cantitatea de H2O,g

s=cantitatea de hidrocarbura/substanta organica,g