Question

A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. What is the molecular formula of the compound? (Atomic weight: H = 1.008, C = 12.0, Br = 79.9)

Answer 1

`"Molecular formula"` `-=` `C_2H_4Br_2`

Explanation:

As always with these problems, it is useful to assume `100*g` of unknown compound....and thus find an empirical formula...

`"Moles of carbon"=(12.8*g)/(12.011*g*mol^-1)=1.066*mol.`

`"Moles of hydrogen"=(2.1*g)/(1.00794*g*mol^-1)=2.084*mol.`

`"Moles of bromine"=(85.1*g)/(79.90*g*mol^-1)=1.066*mol.`

And we divide thru by the LOWEST molar quantity to give an empirical formula of.....

`CH_2Br`

But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.

And so `187.9*g*mol^-1=nxx(12.011+2xx1.00794+79.9)*g*mol^-1`

And thus `n=2`, and the molecular formula is `C_2H_4Br_2`...

I prefer questions that quote actual microanalytical data. `C_2H_4Br_2` is a liquid, and you would rarely be able to get combustion data on a liquid (the analyst would probably laugh at you!).