Question

write the empirical formula for at least four ionic compounds that could be formed from the following ions: \\(\ce{no^-_{3}}, \ce{fe^{2+}}, \ce{po^{3-}_{4}}, \ce{fe^{3+}}\\)

Explanation:

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

For \(Fe^{2+}\) and \(NO_{3}^{-}\), using the criss - cross method (charge of cation becomes subscript of anion and vice - versa, and then simplify). The charge of \(Fe^{2+}\) is \(+ 2\) and of \(NO_{3}^{-}\) is \(-1\). So the formula is \(Fe(NO_{3})_{2}\)

Step2: Combine \(Fe^{2 +}\) with \(PO_{4}^{3-}\)

The charge of \(Fe^{2+}\) is \(+2\) and of \(PO_{4}^{3 -}\) is \(-3\). Using the criss - cross method, we get \(Fe_{3}(PO_{4})_{2}\)

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

The charge of \(Fe^{3+}\) is \(+3\) and of \(NO_{3}^{-}\) is \(-1\). Using the criss - cross method, the formula is \(Fe(NO_{3})_{3}\)

Step4: Combine \(Fe^{3 +}\) with \(PO_{4}^{3-}\)

The charge of \(Fe^{3+}\) is \(+3\) and of \(PO_{4}^{3 -}\) is \(-3\). Using the criss - cross method (simplifying the ratio \(3:3 = 1:1\)), the formula is \(FePO_{4}\)

Answer:

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