An organic compound has the following percentage composition C = 12.36%, H = 2.13%, Br = 85%. Its vapour density is 94. what is its molecular formula?

1 Answer
Aug 19, 2017

Answer:

We get a molecular formula of #C_2H_4Br_2#........

Explanation:

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#.