CH3-CH2-CH2-CH3--⇄CH3-C(CH3)-CH3
20X200/100 80X200/100
butan ramas= 40g
butan transformat= izobutan= 160g
M,butan=M,Izobutan= 58g /mol si contine un C tertiar, adica 12g
58g izobutan..........12gCtertiar
160g...................................x..................................calculeaza !!!!!
