Question

O substanta organica cu masa de 2,96g si densitatea vaporilor sai fata de aer de 2,55 formeaza la ardere 7,04 g de CO2 si 3,6g de apa.Stabiliti formula moleculara a substantei.Scrieti si numiti izomerii posibili (șase izomeri).

Answer 1

Metoda I:

s.o = substanta organica
a = cantitatea de CO2, g
b = cantitatea de H2O, g

s.o = 2,96g
a=7,04g CO2
b=3,6g H2O
daer=2,55

%C=300a/11s=300*7,04/11*2,96=2112/32,56=64,864%

%H=100b/9s=100*3,6/9*2,96=360/26,64=13,5135%

%O=100-(%C+%H) = 100-(64,864+13,5135)=21,622%

dgaz=M/Mgaz => M=dgaz*Mgaz=2,55*28,9=74g/mol

nC=%C*M/AC*100=64,864*74/1200=4 atomi C

nH=%H*M/AH*100=13,5135*74/100=10 atomi H

nO=%O*M/AO*100=21,622*74/1600=1 atom O

Formula moleculara este C4H10O

Metoda II:

Stim masa moleculara a substantei organice de la metoda I => 74g/mol

n=m/M=2,96/74=0.04 moli s.o

44g CO2......................12g C
7,04g............................x=1,92g C

n=m/M=1,92/12=0.16 moli C sau n=m/M=7,04/44=0,16 moli CO2

18g H2O.........................2gH2
3,6g.................................x=0,4g H2   

n=m/M=0,4/2=0,2 moli H2     sau n=m/M=3,6/18=0,2 moli H2

mO=2,96-1,92-0,4=0,64 g O  => n=m/M=0,64/16=0.04 moli O2

In 0,04 moli s.o.........0,16 moli C......0,2 moli H2.......0.04 moli O2
1 mol s.o..................x=4 atomi C......y=10 atomi H........z=1 atom O

F.m => C4H10O

Izomeri posibili:

NE=[(2a+2)-(b-d+e)]/2 = (2*4+2)-10/2 = 0 => este o hidrocarbura saturata.

CH3-CH(OH)-CH2-CH3 -> 2-butanol

C(CH3)3-OH -> tert butanol sau 2-metil-2-propanol

CH3-CO-CH2-CH3 -> etilmetil eter

atati imi vin prin minte.. mai cauta tu.