# What is the empirical formula of a compound containing 83% potassium and 17% oxygen?

Mar 4, 2016

${K}_{2} O$,

#### Explanation:

The following method serves for every compound:

In each 100 g of the compound you have 83 g of potassium and 17 g of oxygen.

To determine the empiriacal formula you have to pass these masses into mols

$\frac{83}{39.1} = 2.12 m o l K$ and $\frac{17}{16} = 1.06 m o l O$

Now divide all the numbers by the smallest you have obatined:

$\frac{2.12}{1.06} = 2.0 K$ and $\frac{1.06}{1.06} = 1 m o l O$

So it is ${K}_{2} O$

At this stage you can have numbers that are not integer. In this case you would have to multiply the resulsts by an integer in a way the final result is a proportion of integers:

Imagine you had obtained 1.5 K for 1 O. You just had to multiply by 2 to get the integer proportion of 3:1.