# A 10.5*g mass of a hydrocarbon contains 1.5*g of hydrogen. If the molecular mass of the hydrocarbon in 210*g*mol^-1, what are the empirical and molecular formulae of the hydrocarbon?

Apr 23, 2017

$1.5 g$ of H would be $1.5 \div 1 = 1.5 m o l$
$9 g$ of C would be $9 \div 12 = 0.75 m o l$

#### Explanation:

So the mol-ratio of $C \div H = 0.75 \div 1.5 = 1 \div 2$
And the empirical formula is ${\left(C {H}_{2}\right)}_{n}$

One of those units has a mass of $12 + 2 \times 1 = 14 u$

So there are $210 \div 14 = 15$ of those units, and the molecular formula will be:

${C}_{15} {H}_{30}$

Apr 23, 2017

We find $\text{(i) the empirical formula}$, and then $\text{(ii) the molecular formula}$.

#### Explanation:

$\text{Moles of hydrogen}$ $=$ $\frac{1.5 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 1.49 \cdot m o l$

$\text{Moles of carbon}$ $=$ $\frac{9.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 0.75 \cdot m o l$.

We divide thru by the smallest molar quantity, and (clearly) we get an empirical formula of $C {H}_{2}$.

But we know that the $\text{molecular formula}$ is whole number multiple of the $\text{empirical formula}$;

i.e. $\text{molecular formula"=nxx"empirical formula}$

And thus, $210 \cdot g \cdot m o l = n \times \left(12.011 + 1.00794\right) \cdot g \cdot m o {l}^{-} 1$, and we solve for $n$, to get $n = 16.15$. This value should be closer to an integer, however, this gives a molecular formula of ${C}_{16} {H}_{32}$ based on the figures you quoted.