Question

O cantitate de 2,8g hidrocarbura cu 85,71% C ocupa V=1 l  la 27ºC si 2,46 atm .Determinati Formula moleculara  - Raspuns : C4H8

Answer 1

V=1L 27°C(300K) =>nr. moli=(V*p):(T*R)=(1*2,46):(300*0.082)=0,1moli 2,46atm m=2.8g=>M=m:nr. moli=2.8: 0.1=28g/mol C:85.71:12=7.14 H:(100-85.71):1=14.29 Impartim ultimele 2 relatii cu cel mai mic adica 7.14 si rezulta FB=CH2 FM=(FB)n ; n=M: MCH2=28:14=2=>FM=C2H4