# A compound is found to contain 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound?

May 2, 2016

$M g S {O}_{3}$

#### Explanation:

As is standard for these sorts of questions, we assume 100 g of substance, and use the percentage elemental compositions to calculate an empirical formula.

$\text{Magnesium}$ $=$ $\frac{23.3 \cdot g}{24.31 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.958 \cdot m o l$

$\text{Sulfur}$ $=$ $\frac{30.7 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.958 \cdot m o l$

$\text{Oxygen}$ $=$ $\frac{46.0 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.86 \cdot m o l$

We divide thru by the smallest molar quantity $\left(0.958 \cdot m o l\right)$ to get and empirical formula of $M g S {O}_{3}$, magnesium sulfite.

Ordinarily you would not be given the percentage of oxygen by mass. You would be given the percentages of the metal and sulfur, and would have to assume that the balance was oxygen.