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O hidrocarbura gazoasa contine 85,71% C si are densitatea in conditii normale=1,25g/l. Formula moleculara a acestei hidrocarburi este: A) C2H4: B)
C4H4: C)CH4


Răspuns :

M=1,25*22,4=28g/mol

%H=100-85,71=14,29

nC=%C*M/AC*100=85,71*28/1200=2 atomi C

nH=%H*M/AH*100=14,29*28/100=4 atomi H

F.M => C2H4-etena