Write the empirical formula for at least four ionic compounds that could be formed from the following ions: \[ \mathrm{CrO}_{4}^{2-}, \mathrm{NH}_{4}^{+}, \mathrm{Fe}^{3+}, \mathrm{PO}_{4}^{3-} \]

To write the empirical formulas for ionic compounds formed from the given ions, we need to combine the ions in such a way that the overall charge of the compound is neutral. Here are the ions provided: 1. \(\mathrm{CrO}_{4}^{2-}\) (chromate ion) 2. \(\mathrm{NH}_{4}^{+}\) (ammonium ion) 3. \(\mathrm{Fe}^{3+}\) (iron(III) ion) 4. \(\mathrm{PO}_{4}^{3-}\) (phosphate ion) Now, let's find combinations of these ions to form neutral ionic compounds: ### 1. Combining \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{NH}_{4}^{+}\) - The chromate ion has a charge of \(-2\) and the ammonium ion has a charge of \(+1\). - To balance the charges, we need 2 ammonium ions for every chromate ion. **Formula:** \(\mathrm{(NH_{4})_{2}CrO_{4}}\) ### 2. Combining \(\mathrm{Fe}^{3+}\) and \(\mathrm{PO}_{4}^{3-}\) - The iron(III) ion has a charge of \(+3\) and the phosphate ion has a charge of \(-3\). - These charges balance each other directly. **Formula:** \(\mathrm{FePO_{4}}\) ### 3. Combining \(\mathrm{Fe}^{3+}\) and \(\mathrm{CrO}_{4}^{2-}\) - The iron(III) ion has a charge of \(+3\) and the chromate ion has a charge of \(-2\). - To balance the charges, we need 2 chromate ions for every 3 iron ions. **Formula:** \(\mathrm{Fe_{2}(CrO_{4})_{3}}\) ### 4. Combining \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{PO}_{4}^{3-}\) - The ammonium ion has a charge of \(+1\) and the phosphate ion has a charge of \(-3\). - To balance the charges, we need 3 ammonium ions for every phosphate ion. **Formula:** \(\mathrm{(NH_{4})_{3}PO_{4}}\) ### Summary of Empirical Formulas 1. \(\mathrm{(NH_{4})_{2}CrO_{4}}\) 2. \(\mathrm{FePO_{4}}\) 3. \(\mathrm{Fe_{2}(CrO_{4})_{3}}\) 4. \(\mathrm{(NH_{4})_{3}PO_{4}}\) These are the empirical formulas for the ionic compounds that can be formed from the given ions.