Question

A compound contains 2.1% hydrogen, 12.8 % carbon and 85.1 % bromine. 1 g of compound occupies 119 cc. calculate the molecular formula of the compound

Answer 1

C=12.8% at.wt. of C=12H=2.1%. at.wt. of H=1Be=85.1%. at.wt. of Br=80Now, 12.8/12 =1.067 2.1/1 = 2.1 85.1/80=1.067Dividing with 1.067 to get simple atomic ratio,1.067/1.067=12.1/1.067=21.067/1.067=1E.F= CH2BrE.F weight=12+(2×1)+80=94Molecular weight=187.9 (given)n=187.9/94= 2M.F= (E.F)×2 =(CH2Br)×2 =(C2H4Br2)