# An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. What is the empirical formula for the compound?

Jan 1, 2016

I follow the normal rigmarole: assume 100 g of unknown stuff and come up with the formula $Z n {N}_{2} {O}_{6}$.

#### Explanation:

In 100 g of unknown, there are 50.68 g $O$, 34.53 g $Z n$, and 14.79 g of $N$.

We divide thru by the atomic weights of oxygen, zinc, and nitrogen respectively:

$O : \frac{50.68 \cdot g}{15.99 \cdot g \cdot m o {l}^{-} 1} = 3.17$ $m o l$ $O$.

$Z n : \frac{34.53 \cdot g}{65.4 \cdot g \cdot m o {l}^{-} 1} = 0.53$ $m o l$ $Z n$.

$N : \frac{14.79 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 1.17$ $m o l$ $O$.

We divide thru (again) by the lowest ratio (that of Zn) to give:

$Z n {N}_{2} {O}_{6}$, which is the empirical formula of the compound, that is the simplest whole number ratio that defines constituent atoms in a species. The species is LIKELY zinc nitrate, $Z n {\left(N {O}_{3}\right)}_{2}$, but we don't really know that at present.