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

1 Answer
Dec 13, 2017

#"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....