Sure! Let's work through formulating empirical formulas for four possible ionic compounds that can be formed using the ions provided: [tex]\(CH_3CO_2^{-}\)[/tex], [tex]\(Fe^{2+}\)[/tex], [tex]\(NH_4^{+}\)[/tex], and [tex]\(CN^{-}\)[/tex].
### Step-by-Step Solution
1. Identifying Possible Ionic Compounds:
- [tex]\(Fe^{2+}\)[/tex] with [tex]\(CH_3CO_2^{-}\)[/tex]
- [tex]\(NH_4^{+}\)[/tex] with [tex]\(CH_3CO_2^{-}\)[/tex]
- [tex]\(Fe^{2+}\)[/tex] with [tex]\(CN^{-}\)[/tex]
- [tex]\(NH_4^{+}\)[/tex] with [tex]\(CN^{-}\)[/tex]
2. Formation of Compounds:
We need to balance the charges to determine the empirical formulas for each compound.
#### Compound 1: [tex]\(Fe^{2+}\)[/tex] with [tex]\(CH_3CO_2^{-}\)[/tex]
- Iron (II) ion: [tex]\(Fe^{2+}\)[/tex]
- Acetate ion: [tex]\(CH_3CO_2^{-}\)[/tex]
- To balance the charges, we need two acetate ions for each iron (II) ion.
- The empirical formula is [tex]\(Fe(CH_3CO_2)_2\)[/tex].
#### Compound 2: [tex]\(NH_4^{+}\)[/tex] with [tex]\(CH_3CO_2^{-}\)[/tex]
- Ammonium ion: [tex]\(NH_4^{+}\)[/tex]
- Acetate ion: [tex]\(CH_3CO_2^{-}\)[/tex]
- The charges balance one-to-one.
- The empirical formula is [tex]\(NH_4CH_3CO_2\)[/tex].
#### Compound 3: [tex]\(Fe^{2+}\)[/tex] with [tex]\(CN^{-}\)[/tex]
- Iron (II) ion: [tex]\(Fe^{2+}\)[/tex]
- Cyanide ion: [tex]\(CN^{-}\)[/tex]
- To balance the charges, we need two cyanide ions for each iron (II) ion.
- The empirical formula is [tex]\(Fe(CN)_2\)[/tex].
#### Compound 4: [tex]\(NH_4^{+}\)[/tex] with [tex]\(CN^{-}\)[/tex]
- Ammonium ion: [tex]\(NH_4^{+}\)[/tex]
- Cyanide ion: [tex]\(CN^{-}\)[/tex]
- The charges balance one-to-one.
- The empirical formula is [tex]\(NH_4CN\)[/tex].
### Conclusion:
The empirical formulas for the four possible ionic compounds formed from the given ions are:
1. [tex]\(Fe(CH_3CO_2)_2\)[/tex]
2. [tex]\(NH_4CH_3CO_2\)[/tex]
3. [tex]\(Fe(CN)_2\)[/tex]
4. [tex]\(NH_4CN\)[/tex]
These formulas represent the balanced ionic compounds derived from combining the given ions.
