# Question #bf587

Mar 22, 2017

$\text{Empirical formula}$ $\equiv$ ${C}_{3} {H}_{6} O$

$\text{Molecular formula}$ $\equiv$ ${C}_{6} {H}_{12} {O}_{2}$

#### Explanation:

AS with all these problems, it is useful to ASSUME that there are $100 \cdot g$ of unknown compound. And we divide thru by the ATOMIC masses of the constituent elements to find the $\text{empirical formula}$.

There are thus $\frac{62.07 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 5.17 \cdot m o l \cdot \text{carbon}$

And $\frac{10.34 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 10.26 \cdot m o l \cdot \text{hydrogen}$

Now, we still have a missing percentage, and this is assumed to be oxygen (note that this is a standard assumption; there are few ways to analyse oxygen gas by combustion in that we have to burn the organic compound in oxygen to oxidize it, and so typically the oxygen percentage is calculated by difference, i.e. it is the missing percentage).

And thus$\frac{100 \cdot g - 62.07 \cdot g - 10.34 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 1.72 \cdot m o l \cdot \text{oxygen}$

So to get the empirical formula, we divide thru by the SMALLEST molar quantity (that of oxygen) to get:

$C : \frac{5.17 \cdot m o l}{1.72 \cdot m o l} = 3.0$

$H : \frac{10.26 \cdot m o l}{1.72 \cdot m o l} = 5.96$

$O : \frac{1.72 \cdot m o l}{1.72 \cdot m o l} = 1.0$

And thus we get the $\text{empirical formula}$, the simplest whole number ratio defining constituent atoms in a species as:

${C}_{3} {H}_{6} O$

Now we further know that the $\text{molecular formula}$ is a WHOLE number mulitple of the $\text{empirical formula}$.

And so $116 \cdot g \cdot m o {l}^{-} 1 = n \times \left(3 \times 12.011 + 6 \times 1.00794 + 15.999\right) \cdot g \cdot m o {l}^{-} 1 = n \times 58 \cdot g \cdot m o {l}^{-} 1$.

Clearly, $n = 2$, and the $\text{molecular formula} \equiv {C}_{6} {H}_{12} {O}_{2}$.