# A hydrocarbon is 83.33% by mass with respect to carbon. What is (i) its empirical formula, and (ii) its molecular formula if the hydrocarbon has a mass of 72*g*mol^-1?

Aug 13, 2017

I gets $\text{pentane}$...................

#### Explanation:

As always with these problems it is useful to ASSUME a mass of $100 \cdot g$ of sample, and then interrogate its ATOMIC composition.

$\text{Moles of carbon} = \frac{83.33 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 6.94 \cdot m o l \cdot C$

The balance $100 \cdot g - 83.33 \cdot g$ MUST represent the percentage of hydrogen, because we were told that we analyzed an HYDROCARBON.

$\text{Moles of hydrogen} = \frac{100.0 \cdot g - 83.33 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 16.54 \cdot m o l \cdot H$

We divide thru by the lowest molar quantity ($C$) to get a trial formula of:

$C {H}_{2.38}$; but we require the simplest WHOLE number ratio, and let us try $5 \times C {H}_{2.38} \cong {C}_{5} {H}_{12}$.

Now the MOLECULAR FORMULA is always a WHOLE number multiple of the EMPIRICAL FORMULA, and since we know the molecular mass:

$n \times \left(5 \times 12.011 + 12 \times 1.00794\right) \cdot g \cdot m o {l}^{-} 1 = 72.11 \cdot g \cdot m o {l}^{-} 1$.

Clearly $n = 1$, and we can quote the molecular formula as isomeric ${C}_{5} {H}_{12}$.