# 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)

Sep 11, 2017

$\text{Molecular formula}$ $\equiv$ ${C}_{2} {H}_{4} B {r}_{2}$

#### Explanation:

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

$\text{Moles of carbon} = \frac{12.8 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 1.066 \cdot m o l .$

$\text{Moles of hydrogen} = \frac{2.1 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 2.084 \cdot m o l .$

$\text{Moles of bromine} = \frac{85.1 \cdot g}{79.90 \cdot g \cdot m o {l}^{-} 1} = 1.066 \cdot m o l .$

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

$C {H}_{2} B r$

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

And so $187.9 \cdot g \cdot m o {l}^{-} 1 = n \times \left(12.011 + 2 \times 1.00794 + 79.9\right) \cdot g \cdot m o {l}^{-} 1$

And thus $n = 2$, and the molecular formula is ${C}_{2} {H}_{4} B {r}_{2}$...

I prefer questions that quote actual microanalytical data. ${C}_{2} {H}_{4} B {r}_{2}$ is a liquid, and you would rarely be able to get combustion data on a liquid (the analyst would probably laugh at you!).