# 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?

Sep 18, 2017

$C {o}_{2} {O}_{3}$......$\text{cobaltic oxide}$

#### Explanation:

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

$\text{Moles of cobalt} = \frac{1.216 \cdot g}{58.9 \cdot g \cdot m o {l}^{-} 1} = 0.0207 \cdot m o l .$

$\text{Moles of oxygen} = \frac{0.495 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 0.0309 \cdot m o l .$

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

$C {o}_{\frac{0.0207 \cdot m o l}{0.0207 \cdot m o l}} {O}_{\frac{0.0309 \cdot m o l}{0.0207 \cdot m o l}} = C o {O}_{1.49}$....

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

$C {o}_{2} {O}_{3}$.....

This is not a good question inasmuch as $C {o}_{2} {O}_{3}$ is poorly characterized and possibly unknown. There are $C o O$, and $C {o}_{3} {O}_{4}$, a mixed cobalt oxide of $C o O$, and $C {o}_{2} {O}_{3}$....The person who set the question was not an inorganic chemist.

Sep 18, 2017

$C {o}_{2} {O}_{3}$

#### 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.

$\frac{1.216 g}{58.9 \left(\frac{g}{\text{mol}}\right) C o} = 0.0206$ mole $C o$ ; $\frac{0.495 g}{16 \left(\frac{g}{\text{mol}} O\right)} = 0.031$ mole O

$C {o}_{0.0206} {O}_{0.031}$ ; or $\frac{\text{Co}}{O} = \frac{0.0206}{0.031} = \frac{2}{3}$
$C {o}_{2} {O}_{3}$