# Question #29ef3

Feb 20, 2017

The empirical formula is ${K}_{2} O$

#### Explanation:

21.7 g of potassium is 0.555 mol ($21.7 g \div 39.1 \frac{g}{\text{mol}}$)

4.44 g of oxygen is 0.2775 mol ($4.44 g \div 16.0 \frac{g}{\text{mol}}$)

This change to moles is necessary, because we need to compare quantities using a unit that refers to a number of atoms rather than a total mass.

Now, look for a whole-number ratio between these two values. If it is not obvious, try dividing both by the smaller. This will give you a ratio in which one term is a one, and should be easier to recognize.

Oxygen: $0.2775 \div 0.2775 = 1.0$

Potassium: $0.555 \div 0.2775 = 2.0$

So, the ratio in the compound is 2 atoms of K for each 1 atom of O, giving a formula ${K}_{2} O$

Feb 20, 2017

${K}_{2} O$ is the $\text{empirical formula}$.

#### Explanation:

The empirical formula is the simplest whole number ratio that represents constituent atoms in a species. Given masses of constituent atoms, we simply divide thru by the atomic mass:

$\text{Moles of potassium} = \frac{21.7 \cdot g}{39.10 \cdot g \cdot m o {l}^{-} 1} = 0.555 \cdot m o l$

$\text{Moles of oxygen} = \frac{4.44 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.278 \cdot m o l$

And if I divide thru by $0.278 \cdot m o l$, clearly I get ${K}_{2} O$. Capisce?