Question

write the empirical formula for at least four ionic compounds that could be formed from the following ions: fe²⁺, fe³⁺, po₄³⁻, io₃⁻

Explanation:

Step1: Combine \(Fe^{2 +}\) with \(PO_{4}^{3-}\)

For \(Fe^{2+}\) and \(PO_{4}^{3 -}\), using the cross - multiply method (the magnitude of the charge of one ion becomes the subscript of the other ion). The formula is \(Fe_{3}(PO_{4})_{2}\) since \(2\times3 = 3\times2\) (to balance the charges: \(3\times(+ 2)=+6\) and \(2\times(-3)=-6\)).

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

For \(Fe^{2+}\) and \(IO_{3}^{-}\), using the cross - multiply method. The formula is \(Fe(IO_{3})_{2}\) because \(1\times(+2)+2\times(-1)=0\) (the charge of \(Fe^{2+}\) is \(+2\) and the charge of each \(IO_{3}^{-}\) is \(-1\)).

Step3: Combine \(Fe^{3+}\) with \(PO_{4}^{3-}\)

For \(Fe^{3+}\) and \(PO_{4}^{3 -}\), since \(3\times(+3)+3\times(-3)=0\) (using the fact that the sum of charges in a neutral compound is zero), the formula is \(FePO_{4}\).

Step4: Combine \(Fe^{3+}\) with \(IO_{3}^{-}\)

For \(Fe^{3+}\) and \(IO_{3}^{-}\), using the cross - multiply method. The formula is \(Fe(IO_{3})_{3}\) because \(1\times(+3)+3\times(-1)=0\) (the charge of \(Fe^{3+}\) is \(+3\) and the charge of each \(IO_{3}^{-}\) is \(-1\)).

Answer:

\(Fe_{3}(PO_{4})_{2}\), \(Fe(IO_{3})_{2}\), \(FePO_{4}\), \(Fe(IO_{3})_{3}\)