Question

try again
your answer for at least one of your compounds is incorrect.
write the empirical formula for at least four ionic compounds that could be formed from the following ions:
nh₄⁺, ch₃co₂⁻, fe³⁺, no₃⁻
nh₄ch₃co₂, nh₄no₃, fe₃, fe₃

Explanation:

Step1: Combine cations and anions

Ionic compounds are formed by the combination of cations (\(NH_4^+\), \(Fe^{3+}\)) and anions (\(CH_3CO_2^-\), \(NO_3^-\)).

Step2: Balance charges

• For \(NH_4^+\) and \(CH_3CO_2^-\): The charge of \(NH_4^+\) is \(+ 1\) and the charge of \(CH_3CO_2^-\) is \(-1\). So the formula is \(NH_4CH_3CO_2\).
• For \(NH_4^+\) and \(NO_3^-\): The charge of \(NH_4^+\) is \(+1\) and the charge of \(NO_3^-\) is \(-1\). So the formula is \(NH_4NO_3\).
• For \(Fe^{3+}\) and \(CH_3CO_2^-\): The charge of \(Fe^{3+}\) is \(+3\) and the charge of \(CH_3CO_2^-\) is \(-1\). To balance the charges (\(+3+3\times(- 1)=0\)), the formula is \(Fe(CH_3CO_2)_3\).
• For \(Fe^{3+}\) and \(NO_3^-\): The charge of \(Fe^{3+}\) is \(+3\) and the charge of \(NO_3^-\) is \(-1\). To balance the charges (\(+3 + 3\times(-1)=0\)), the formula is \(Fe(NO_3)_3\).

Answer:

\(NH_4CH_3CO_2\), \(NH_4NO_3\), \(Fe(CH_3CO_2)_3\), \(Fe(NO_3)_3\)