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

Answer:

#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

Answer:

#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#