# A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol. What is its molecular formula?

Feb 7, 2016

The empirical formula is $C {H}_{2} O$, and the molecular formula is some multiple of this.

#### Explanation:

In 100 g of the unknown, there are $\frac{40.0 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $C$; $\frac{6.7 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $H$; and $\frac{53.5 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1}$ $O$.

We divide thru to get, $C : H : O$ $=$ $3.33 : 6.65 : 3.34$. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of $C {H}_{2} O$, i.e. near enough to WHOLE numbers.

Now the molecular formula is always a multiple of the empirical formula; i.e. ("EF")_n = "MF".

So $60.0 \cdot g \cdot m o {l}^{-} 1 = n \times \left(12.011 + 2 \times 1.00794 + 16.00\right) g \cdot m o {l}^{-} 1$.

Clearly $n = 2$, and the molecular formula is $2 \times \left(C {H}_{2} O\right)$ $=$ ${C}_{x} {H}_{y} {O}_{z}$.