# A compound having an empirical formula of C_3H_4O_3 has a mass of 180 g/mol. What is the molecular formula?

Jun 4, 2016

$\text{Molecular formula}$ $=$ ${C}_{6} {H}_{8} {O}_{6}$

#### Explanation:

The molecular formula is always a whole number multiple of the empirical formula:

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

So we need to solve for $n$.

Thus $180 \cdot g \cdot m o {l}^{-} 1$ $=$ $n \times \left(3 \times 12.011 + 4 \times 1.0079 + 3 \times 15.999\right) \cdot g \cdot m o {l}^{-} 1$
$=$ $n \times \left(88.0\right) \cdot g \cdot m o {l}^{-} 1$

$n \cong 2$. Thus $\text{Molecular formula}$ $=$ ${C}_{6} {H}_{8} {O}_{6}$.

Of course, the ratio is out a little bit. Such errors are common in the determination of such formulae.