Q75P - Chemistry: Molecular Nature Of Matter And Change

Question

An ionic compound forms when calcium (Z = 20) reacts with iodide (Z = 53). If a sample of the compound contains $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$calcium ions, how many iodide ions does it contain?

Step-by-Step Solution

Answer

The amount of iodide ions in the sample is $\mathbf{14}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$

Step 1: Identify the chemical formula of the ionic compound

The atomic number of calcium (Ca) is 20, so the electronic configuration of Ca is 2, 8, 8, 2. So, for attaining a more stable state, Ca easily loses two electrons and form a positive ion, that is, ${\mathbf{Ca}}^{\mathbf{+}\mathbf{2}}$

 

The atomic number of iodine (I) is 53, so the electronic configuration of I is 2, 8, 18, 8, and 7. Thus, for attaining stability I will gain one electron and form a negative ion, that is, ${\mathbf{I}}^{\mathbf{-}}$

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

Step 2: Relation between the calcium ion and the iodide ions present in the ionic compound

In ${\mathbf{CaI}}_{\mathbf{2}}$the compound, 1 mole of calcium ion is equal to the 2 moles of iodide ions.

So, the amount of iodide ions is twice the amount of calcium ions.

Step 3: Calculate the amount of iodide present in the ionic compound

The number of calcium ions present in the compound is $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$

$$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{I}}^{\mathbf{-}}\mathbf{ }\mathbf{ions}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{2}\mathbf{\times }\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{Ca}}^{\mathbf{+}\mathbf{2}}\mathbf{ }\mathbf{ions} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{2}\mathbf{\times }\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{14}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}} \end{gathered}$$ 

Therefore, the sample contains $\mathbf{14}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$ iodide ions.