Question

DAU FUNDA
La analiza a 1.18 g de substanta orga
nica s-au obtinut : 1.344 dm3 CO2 , 2.016 L H2O si 224 cm3 N2 (toate volumele au fost masurate in conditii normale). Determina formula moleculara a substantei organice, stiind ca densitatea vaporilor sai fata de azot este 2.1 .

FUNDA VA ROOOG.HELP

Answer 1

1mol CO2........22,4dm³..........12gC
......................1,344dm³............xgC
x=1,344*12/22,4=0,72g C
1mol H2O......22,4L......2gH
.....................2,016L......ygH
y=2,016*2/22,4=0,18gH
1mol N2......22,4L........28gN
..................0,224L...........zgN
z=0,224*28/22,4=0,28g N
melemente=1,18g=1,18 s.o deci nu contine O
impartim masele la masele atomice, apoi numarul cel mai mic de moli la toti molii
............................ :0,02
C: 0,72/12=0,06. 3
H: 0,18/1=0,18 9
N: 0,28/14=0,02 1
(C3H9N)n
dN2=Msubst/MN2=> Msubst=2,1*28=59g/mol
M(C3H9N)n=59n=59 => n=1
formula moleculara: C3H9N