Question

An ionic compound forms when lithium (Z = 3) reacts with oxygen (Z = 8). If a sample of the compound contains$\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$lithium ions, how many oxide ions does it contain?

Short answer

The amount of oxide ions in the sample is$\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$

Step-by-step solution

Identify the chemical formula of the ionic compound

The atomic number of lithium (Li) is 3, so the electronic configuration ofLi is 2, 1. So, for attaining a more stable state, lithium will easily lose one electron and form a positive ion, that is,${\mathbf{Li}}^{\mathbf{+}}$

The atomic number of oxygen (O) is 8, so the electronic configuration ofO is 2, 6. Thus, for attaining stability oxygen will gain two electrons and form a negative ion, that is,${\mathbf{O}}^{\mathbf{-}\mathbf{2}}$

Hence, the chemical formula of the ionic compound is${\mathbf{Li}}_{\mathbf{2}}\mathbf{O}$

Relation between the lithium-ion and the oxide ion present in the ionic compound

In ${\mathbf{Li}}_{\mathbf{2}}\mathbf{O}$the compound, 2 moles of lithium-ion is equal to the 1 mole of oxide ions.

So, the amount of oxide ions is one-half of the number of lithium ions.

Calculate the amount of oxide present in the ionic compound

The number of lithium ions present in the compound is

$$\begin{gathered} \mathbf{Amount}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}\mathbf{ions}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{Amount}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Li}}^{\mathbf{+}}\mathbf{ions}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Therefore, the compound contains$\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$oxide ions.