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?

Aug 19, 2017

We get a molecular formula of ${C}_{2} {H}_{4} B {r}_{2}$........

Explanation:

As always, we assume a $100 \cdot g$ mass of compound, and interrogate its elemental composition in terms of moles.....

$\text{Moles of C} = \frac{12.36 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1} = 1.029 \cdot m o l$

$\text{Moles of Br} = \frac{85 \cdot g}{79.90 \cdot g \cdot m o {l}^{-} 1} = 1.064 \cdot m o l$

$\text{Moles of H} = \frac{2.13 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 2.11 \cdot m o l$

We divide thru by the smallest molar quantity, to give an empirical formula of $C {H}_{2} B r$

Now $\text{vapour density}$ 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 $2 \times 94 \cdot g \cdot m o {l}^{-} 1 = 188 \cdot g \cdot m o {l}^{-} 1$.

As always, $\text{molecular formula}$ is a MULTIPLE of the $\text{empirical formula}$; so $\left\{12.011 + 2 \times 1.00794 + 79.9\right\} \cdot g \cdot m o {l}^{-} 1 \times n = 188 \cdot g \cdot m o {l}^{-} 1$

So, clearly, $n = 2$, and our MOLECULAR formula is ${C}_{2} {H}_{4} B {r}_{2}$, i.e. ethylene dibromide. Now in fact this gives rise to TWO possible isomers, i.e. $B {r}_{2} H C - C {H}_{3}$ or isomeric $B r {H}_{2} C - C {H}_{2} B r$.