# Combustion of adipic acid yields percentage abundances of C:49.3%, H:6.9%, and O:43.8%. What are the empirical and molecular formulae of adipic acid?

Sep 5, 2016

Adipic acid has a molecular formula of ${C}_{6} {H}_{10} {O}_{4}$.

#### Explanation:

As with all problems of this type, we assume a quantity of $100 \cdot g$ of stuff, divide thru by the atomic weight of the component elements, and work out the elemental composition with respect to moles:

$\text{Moles of C}$ $=$ $\frac{49.3 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $4.10 \cdot m o l$.

$\text{Moles of H}$ $=$ $\frac{6.9 \cdot g}{1.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $6.83 \cdot m o l$.

$\text{Moles of O}$ $=$ $\frac{43.8 \cdot g}{15.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.74 \cdot m o l$.

We divide thru by the smallest molar quantity, that of oxygen, to get the empirical formula of ${C}_{1.5} {H}_{2.5} O$, which we of course double to give ${C}_{3} {H}_{5} {O}_{2}$ because, by definition, whole numbers are required.

Now, the molecular formula is some whole number ratio of the empirical formula:

i.e. $\text{Molecular formula}$ $=$ $\text{(Empirical formula)} \times n$.

So we have to simply solve for $n$.

$146 \cdot g \cdot m o {l}^{-} 1 = \left(3 \times 12.011 + 5 \times 1.00794 + 2 \times 16.00\right) \times n \cdot g \cdot m o {l}^{-} 1$

Thus $n = 2$, and the molecular formula is ${C}_{6} {H}_{10} {O}_{4}$.