Question

write the empirical formula for at least four ionic compounds that could be formed from the following ions: fe^{3+}, pb^{4+}, clo_{3}^{-}, co_{3}^{2-}

Explanation:

Step1: Combine \(Fe^{3 +}\) with \(ClO_{3}^{-}\)

The charge on \(Fe^{3+}\) is \(+3\) and on \(ClO_{3}^{-}\) is \(- 1\). To balance the charges, we need 3 \(ClO_{3}^{-}\) ions for each \(Fe^{3+}\) ion. So the formula is \(Fe(ClO_{3})_{3}\).

Step2: Combine \(Fe^{3 +}\) with \(CO_{3}^{2 -}\)

The charge on \(Fe^{3+}\) is \(+3\) and on \(CO_{3}^{2-}\) is \(-2\). Using the cross - multiply method (charge of \(Fe^{3+}\) becomes subscript of \(CO_{3}^{2-}\) and vice - versa), we get \(Fe_{2}(CO_{3})_{3}\).

Step3: Combine \(Pb^{4 +}\) with \(ClO_{3}^{-}\)

The charge on \(Pb^{4+}\) is \(+4\) and on \(ClO_{3}^{-}\) is \(-1\). So we need 4 \(ClO_{3}^{-}\) ions for each \(Pb^{4+}\) ion. The formula is \(Pb(ClO_{3})_{4}\).

Step4: Combine \(Pb^{4 +}\) with \(CO_{3}^{2 -}\)

The charge on \(Pb^{4+}\) is \(+4\) and on \(CO_{3}^{2-}\) is \(-2\). Using the cross - multiply method, we get \(Pb(CO_{3})_{2}\).

Answer:

\(Fe(ClO_{3})_{3}\), \(Fe_{2}(CO_{3})_{3}\), \(Pb(ClO_{3})_{4}\), \(Pb(CO_{3})_{2}\)