# What is molecular formula of compound that weighs 46 g/mol and analyzes to 52.2% C, 34.8% O, and 13% H?

Dec 3, 2016

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

#### Explanation:

As with all these problems, we assume that there are $100 \cdot g$ unknown material:

First we calculate an empirical formula...........

$\text{Moles of carbon:}$ $=$ $\frac{52.2 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 4.35 \cdot m o l$.

$\text{Moles of hydrogen:}$ $=$ $\frac{13.0 \cdot g}{1.079 \cdot g \cdot m o {l}^{-} 1} = 12.04 \cdot m o l$.

$\text{Moles of oxygen:}$ $=$ $\frac{34.8 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 2.18 \cdot m o l$.

Note (i) that here we have simply divided the atomic masses thru by the $\text{atomic mass}$, and (ii), normally $\text{% oxygen content}$ would not be measured. You would be given $\text{% carbon content}$, and $\text{% hydrogen content}$, and $\text{% nitrogen content}$, and $\text{% oxygen content}$ would be assessed by the balance.

And now we normalize the formula by dividing thru by the lowest molar ratio, that of oxygen to get the empirical formula:

$O = \frac{2.18 \cdot m o l}{2.18 \cdot m o l}$ $=$ $1$

$C = \frac{4.35 \cdot m o l}{2.18 \cdot m o l}$ $=$ $2$

$H = \frac{12.04 \cdot m o l}{2.18 \cdot m o l}$ $=$ $6$

And thus our empirical formula is ${C}_{2} {H}_{6} O$.

But the molecular formula is always a mulitple of the empirical formula:

i.e. $\text{(empirical formula)"xxn="(molecular formula)}$.

We KNOW the molecular mass because it has been measured in a separate experiment, and provided for us:

So $n \times \left(2 \times 12.011 + 6 \times 1.00794 + 15.999\right) \cdot g \cdot m o {l}^{-} 1 = 46.0 \cdot g \cdot m o {l}^{-} 1$,

and we solve for $n$.

Clearly $n = 1$, and here

$\text{molecular formula "-=" empirical formula} = {C}_{2} {H}_{6} O$.