Question

La arderea a 7.2 g de substanta organica cu densitatea relativa a vaporilor din aer egala cu 2.485 s-au obtinut 11.2 l (c.n.) dioxid de carbon IV si 10.8 g de h2o, care este denumirea substantei, daca formalu sa contine 1 atom de carbn cuaternar?

Answer 1

s.o=7.2g 

daer=2.485
(a)VCO2=11.2l
(b)mH2O=10.8g

a) Denumire=?
b) F.m=?

daer=M/Maer==>M=daer*Maer=2.485*28.9=72g/mol

m/M=V/Vmolar==>m=M*V/Vmolar=44*11.2/22.4=22g CO2

%C=300a/11s=300*22/11*7.2=6600/79.2=83.33
%H=100b/9s=100*10.8/9*7.2=1080/64.8=16.66

nC=%C*M/AC*100=83.33*72/1200=5 atomi C
nH=%H*M/AH*100=16.66*72/100=12 atomi H

Formula moleculara este C5H12..->n-pentan
Daca contine un atom de C cuaternar ==> CH3-C(CH3)-CH3 --> 2,3 dimetil propan sau izopentan

NE=(2a+2)-(b-d+e)/2=2*5+2-12/2=0/2==>formula reala, 0 leg. duble

Metoda mai usoara !

-deduc nr moli s.o: 

n=m/M=7.2/72=0.1 moli s.o

-deduc nCO2:

n=m/M=22/44=0.5 moli CO2

-deduc nH2O:

n=m/M=10.8/18=0.6 moli H2O =12 atomi H2

-deduc nr atomi de C si H:

Din 0.1mol s.o........0.5molCO2......0.6mol H2O=12 atomi H2
1mol..........x=5 atomi C.......y=12 atomi H2