Question

A `22.35*g` mass of an unknown compound consisting of potassium, chlorine, and oxygen, contained `7.13*g` of potassium, `6.47*g` of chlorine. What is its empirical formula?

Answer 1

`KClO_3`

Explanation:

As always, we divide each constitiuent mass thru by the ATOMIC mass of each constituent atom.

`"Moles of potassium"` `=` `(7.13*g)/(39.10*g*mol^-1)` `=` `0.182*mol`.

`"Moles of chlorine"` `=` `(6.47*g)/(35.45*g*mol^-1)` `=` `0.182*mol`.

`"Moles of oxygen"` `=` `((22.35-7.13-6.47)*g)/(15.999*g*mol^-1)` `=` `0.546*mol`.

And now we divide thru by the LOWEST molar quantity, that of potassium/chlorine, to give an empirical formula of `KClO_3`.

How did I know that that there were `(22.35-7.13-6.47)*g` of `O`? The mass of oxygen was not given in the problem.