Question

O hidrocarbura gazoasa contine 85,71% C si are densitatea in conditii normale ρ₀=1,25 g \ L .Formula moleculara a acestei hidrocarburi este: C₂H₄ . b) C₄H₄. c) CH₄

Answer 1

Notam f.moleculara: CxHy
Mhidrocarbura=12x+y g/mol
12x+y..............100%
12x..................85,71%
12x+y=1200/85,71=14x => 2x=y (I)
in c.n o hidrocarbura gazoasa, are densitatea=Masa moleculara/Volumul molar
d=M/Vm => M=d*Vm=1,25*22,4=28g/mol
stim ca M=12x+y=28 si y=2x
=> 14x=28 => x=2 => y=4
f.m: C2H4