# What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

Dec 19, 2016

$Z n {O}_{2}$

#### Explanation:

With all these problems, we assume that there are $100 \cdot g$ of unknown compound, and then we proceed to calculate the empirical formula:

$\text{Moles of zinc}$ $=$ $\frac{67.1 \cdot g}{65.4 \cdot g \cdot m o {l}^{-} 1} = 1.03 \cdot m o l$

$\text{Moles of oxygen}$ $=$ $\frac{32.9 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 2.06 \cdot m o l$

Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of $Z n {O}_{2}$. The examiner has been nasty in that the expectation would have been $Z n O$. However, zinc does form a peroxo species composed of $Z {n}^{2 +}$ and ${O}_{2}^{2 -}$ ions.