# A salt contains 31.84% potassium, 28.98% chlorine, and 39.18% oxygen by mass. What is its empirical formula?

Sep 16, 2017

$K C l {O}_{3}$

#### Explanation:

As always with these problems, we assume an $100 \cdot g$ mass of compounds, and then interrogate the molar quantities of the constituent atoms...

$\text{Moles of potassium} \equiv \frac{31.84 \cdot g}{39.10 \cdot g \cdot m o {l}^{-} 1} = 0.814 \cdot m o l$.

$\text{Moles of chlorine} \equiv \frac{28.98 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1} = 0.817 \cdot m o l$.

$\text{Moles of oxygen} \equiv \frac{39.18 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 2.45 \cdot m o l$.

We divide thru by the lowest molar quantity to give an empirical formula of $K C l {O}_{3}$ (or close enuff to!), i.e. $\text{potassium chlorate}$.

And the molecular formula is always a multiple of the empirical formula.....${\left(39.1 + 35.45 + 3 \times 16\right)}_{n} \cdot g \cdot m o {l}^{-} 1 = 122 , 5 \cdot g \cdot m o {l}^{-} 1$. Clearly, $n = 1$, and the molecular formula is $K C l {O}_{3}$, $\text{potassium chlorate}$.