ms = 1600 g, c% = 4%
c% = mdx100/ms
=> md = msxc%/100
= 1600x4/100 = 64 g Br2
16,4g 64g
CnH2n-2 + 2Br2 --CCl4--> CnH2n-2Br2
miu 320
=> miu = 16,4x320/64 = 82 g/mol
=> 14n-2 = 82 => n = 6
A = C6H10
din problema se intelege o alchina marginala
=> 1-hexina