An unknown sample with a molecular mass of 180.0g is analyzed to yield 40% C, 6.7% H, and 53.3% O. What is the empirical formula and the molecular formula of this compound?

Apr 19, 2016

$\text{Empirical formula } = C {H}_{2} O$

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

Explanation:

We assume $100 \cdot g$ of compound and work out the atomic proportions:

$C : \frac{40.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.33$ $m o l$.

$H : \frac{6.7 \cdot g}{1.0794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $6.65$ $m o l$.

$O : \frac{53.3 .0 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.31$ $m o l$.

We divide thru by the lowest molar quantity to give us the empirical formula, $C {H}_{2} O$, which is the simplest whole number ration defining constituent atoms in a species.

Now it is a fact that the molecular formula is always a multiple of the empirical formula.

Thus "Molecular formula " =" (Empirical formula")xxn

$\left(12.011 + 2 \times 1.00794 + 15.999\right) \cdot g \cdot m o {l}^{-} 1 \times n = 180.0 \cdot g \cdot m o {l}^{-} 1$

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