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)

Md Ariful Islam

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Md Ariful Islam

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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!).

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