Salut!Cine ma poate ajuta cu aceste probleme?
1)Sa se determine formula bruta si formula moleculara a substantei organice care contine 54,54% C; 9,1% H si are densitatea in raport cu oxigenul 4,125.
2)Determina formulara moleculara a unei substante organice care nu are oxigen,dar contine: 53,33% C;15,55 % H si un atom de azot.

1

Răspunsuri

Cel mai inteligent răspuns!
2014-10-05T19:44:08+03:00
1)CxHyOz-subst organica
p₁=54,54%C
p₂=9,1%H2
densitatea=M supra 32
M supra 32=4,125
M=4,125*32
M=132g/mol
100g CxHyOz......54,54gC
132g CxHyOz.......12xg
x=54,54*132 supra 12*100=7299,28 supra 1200=6
x=6

100g CxHyOz.......9,1gH2
132gCxHyOz........ygH2
y=9,1*132 supra 100=1201,2 supra 100=12
y=12

6*12+12+16z=132
72+12+16z=132
84+16z=132
16z=132-84
16z=48
z=48/16
z=3
C6H12O3(formula moleculara)
(C2H4O)3    formula bruta


2)CxHyN
p1=53,33%C
p2=15,55%H2

100gCxHyN.....53,33gC
14+12x+y.......12x
12x*100=53,33(14+12x+y)
1200x=746,62+639,96x+53,33y
1200x-639,96x-53,33y=746,62
560,04x-53,33y=746,62

100g CxHyN.........15,55gH2
14+12x+y.........y
100y=15,55(14+12x+y)
100y=217,7+186,6x+15,55y
100y-15,55y-186,6x=217,7
84,45y-186,6x=217,7