What is the empirical formula of a oxide of cobalt that contains a 1.216*g mass of metal, and a 0.495*g mass of oxygen?

2 Answers
Sep 18, 2017

Co_2O_3......"cobaltic oxide"

Explanation:

We interrogate the molar quantities of metal and oxygen.....

"Moles of cobalt"=(1.216*g)/(58.9*g*mol^-1)=0.0207*mol.

"Moles of oxygen"=(0.495*g)/(16.0*g*mol^-1)=0.0309*mol.

We divide thru by the smallest molar quantity, that of the metal, to get a trial empirical formula of.....

Co_((0.0207*mol)/(0.0207*mol))O_((0.0309*mol)/(0.0207*mol))=CoO_(1.49)....

But by specification, the empirical formula is the simplest whole number ratio defining constituent atoms in a species...and so....

Co_2O_3.....

This is not a good question inasmuch as Co_2O_3 is poorly characterized and possibly unknown. There are CoO, and Co_3O_4, a mixed cobalt oxide of CoO, and Co_2O_3....The person who set the question was not an inorganic chemist.

Sep 18, 2017

Co_2O_3

Explanation:

Assuming complete reaction of both, convert the masses into moles and normalize. I’ll use a singlet oxygen because we don’t know the ratio yet, even though the actual gas would be diatomic.

(1.216g)/(58.9(g/"mol") Co) = 0.0206 mole Co ; (0.495g)/(16(g/"mol" O)) = 0.031 mole O

Co_0.0206O_0.031 ; or "Co"/O = 0.0206/0.031 = 2/3
Co_2O_3