As always, we assume a 100⋅g mass of compound, and interrogate its elemental composition in terms of moles.....
Moles of C=12.36⋅g12.01⋅g⋅mol−1=1.029⋅mol
Moles of Br=85⋅g79.90⋅g⋅mol−1=1.064⋅mol
Moles of H=2.13⋅g1.00794⋅g⋅mol−1=2.11⋅mol
We divide thru by the smallest molar quantity, to give an empirical formula of CH2Br
Now vapour density is an uncommon measurement, and reports the density of a vapour in relation to that of dihydrogen gas. And thus the molecular mass of the gas is 2×94⋅g⋅mol−1=188⋅g⋅mol−1.
As always, molecular formula is a MULTIPLE of the empirical formula; so {12.011+2×1.00794+79.9}⋅g⋅mol−1×n=188⋅g⋅mol−1
So, clearly, n=2, and our MOLECULAR formula is C2H4Br2, i.e. ethylene dibromide. Now in fact this gives rise to TWO possible isomers, i.e. Br2HC−CH3 or isomeric BrH2C−CH2Br.
