Question

What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

Answer 1

`ZnO_2`

Explanation:

With all these problems, we assume that there are `100*g` of unknown compound, and then we proceed to calculate the empirical formula:

`"Moles of zinc"` `=` `(67.1*g)/(65.4*g*mol^-1)=1.03*mol`

`"Moles of oxygen"` `=` `(32.9*g)/(15.999*g*mol^-1)=2.06*mol`

Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of `ZnO_2`. The examiner has been nasty in that the expectation would have been `ZnO`. However, zinc does form a peroxo species composed of `Zn^(2+)` and `O_2^(2-)` ions.