Q74P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: Identify the chemical formula of the ionic compound

The atomic number of lithium (Li) is 3, so the electronic configuration of Li 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 of O 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${\mathbf{Li}}_{\mathbf{2}}\mathbf{O}$ 

Step 2: 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.

Step 3: 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.