Alchenele au coimpozitie fixa: 85,71%C si 14,29%H
Malchena=MCnH2n=12n+2n=14n g/mol
14n.............100%
12n................x%
x=12*100/14=85,71% C => 14,29%H (aceleasi procente pentru oricare termen din seria omoloaga)
atunci %C alcool = %C alchena - 25,71% = 85,71 - 25,71 = 60% C
Fiind aciclic, saturat si monohidroxilic are formula generala: CnH2n+2O
Malcool=12n+2n+2+16=14n+18
14n+18...................100%
12n..........................60%
1200n=840n+1080 => 360n=1080 => n=3 deci avem alcool cu 3 atomi de C si aciclic => 1 sau 2 propanol
alchena: propena
22,4l 160g
CH₂=CH-CH₃ + Br₂ -> Br-CH₂-CH(Br)-CH₃
3,36l x=?
x=24g Br₂
16/100=24/ms => ms=24*100/16=150g solutie Br₂
