# How do you find the empirical formulas for 65.2% Sc, 34.8% O?

Mar 4, 2017

As with all these problems, we ASSUME a $100 \cdot g$ mass of compound.........and come up with an empirical formula of $S {c}_{2} {O}_{3}$.

#### Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species. We assume a $100 \cdot g$ mass of $S {c}_{2} {O}_{3}$, and we come up with a molar ratio:

$\text{Moles of scandium}$ $=$ $\text{Mass of scandium"/"Molar mass of scandium}$ $=$ $\frac{65.2 \cdot g}{44.96 \cdot g \cdot m o {l}^{-} 1} = 1.45 \cdot m o l$.

$\text{Moles of oxygen}$ $=$ $\text{Mass of oxygen"/"Molar mass of scandium}$ $=$ $\frac{34.8 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 2.175 \cdot m o l$.

In each instance, I divided thru by the ATOMIC MASS of the element. And now if we divide thru by the smallest molar quantity, (that of the metal), I get a formula of $S c {O}_{1.5}$. Because, by specification, the empirical formula is a WHOLE number ratio, we quote its empirical formula as $S {c}_{2} {O}_{3}$. Capisce?