Question

`13.2*g` of mercury metal, and `1.0*g` oxygen combine to form mercuric oxide. What is the empirical formula?

Answer 1

`"Mercuric oxide"` is decomposed according to....

Explanation:

`HgO(s) + Delta rarr Hg(l) + 1/2O_2(g)uarr`

See here, and I hope I have not broken some rule.

As to the composition of `"mercuric oxide"`, we access the empirical formula by taking the molar quantities of each element...

`"Moles of mercury"=(13.2*g)/(200.6*g*mol^-1)=0.0658*mol.`

`"Moles of oxygen"=(1.00*g)/(16.0*g*mol^-1)=0.0625*mol.`

And this is near enuff to a 1:1 molar ratio, i.e. `HgO`.

The percentage composition by mass is simply....

`(13.2*g)/(14.2*g)xx100%=93.0%` with respect to the metal, and of course, oxygen makes up the balance....