As always, we assume a 100*g mass of compound, and interrogate its elemental composition in terms of moles.....
"Moles of C"=(12.36*g)/(12.01*g*mol^-1)=1.029*mol
"Moles of Br"=(85*g)/(79.90*g*mol^-1)=1.064*mol
"Moles of H"=(2.13*g)/(1.00794*g*mol^-1)=2.11*mol
We divide thru by the smallest molar quantity, to give an empirical formula of CH_2Br
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 2xx94*g*mol^-1=188*g*mol^-1.
As always, "molecular formula" is a MULTIPLE of the "empirical formula"; so {12.011+2xx1.00794+79.9}*g*mol^-1xxn=188*g*mol^-1
So, clearly, n=2, and our MOLECULAR formula is C_2H_4Br_2, i.e. ethylene dibromide. Now in fact this gives rise to TWO possible isomers, i.e. Br_2HC-CH_3 or isomeric BrH_2C-CH_2Br.
