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ALEKS Chem Prep: Atoms, Ions, and Molecules

Predicting the formula of ionic compounds with common polyatomic ions

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
[tex]\[ NH_4^+, CH_3CO_2^-, Fe^{2+}, ClO_3^- \][/tex]

A. [tex]\[ NH_4^+ + CH_3CO_2^- \rightarrow \text{NH}_4\text{CH}_3\text{CO}_2 \][/tex]

B. [tex]\[ NH_4^+ + ClO_3^- \rightarrow \text{NH}_4\text{ClO}_3 \][/tex]

C. [tex]\[ Fe^{2+} + CH_3CO_2^- \rightarrow \text{Fe}(\text{CH}_3\text{CO}_2)_2 \][/tex]

D. [tex]\[ Fe^{2+} + ClO_3^- \rightarrow \text{Fe}(\text{ClO}_3)_2 \][/tex]



Answer :

GinnyAnswer
Certainly! Let's solve this step by step to determine the empirical formulas of at least four ionic compounds formed from the given ions:
- [tex]\( NH_4^+ \)[/tex]
- [tex]\( CH_3CO_2^- \)[/tex]
- [tex]\( Fe^{2+} \)[/tex]
- [tex]\( ClO_3^- \)[/tex]

### Compound 1: [tex]\( NH_4^+ \)[/tex] and [tex]\( CH_3CO_2^- \)[/tex]
To form an ionic compound, the net charge must be zero.
- [tex]\( NH_4^+ \)[/tex] has a charge of [tex]\( +1 \)[/tex].
- [tex]\( CH_3CO_2^- \)[/tex] (acetate) has a charge of [tex]\( -1 \)[/tex].

The formula for the resulting compound would be [tex]\( NH_4CH_3CO_2 \)[/tex], known as ammonium acetate.

### Compound 2: [tex]\( NH_4^+ \)[/tex] and [tex]\( ClO_3^- \)[/tex]
Similarly, we can combine:
- [tex]\( NH_4^+ \)[/tex] with a charge of [tex]\( +1 \)[/tex].
- [tex]\( ClO_3^- \)[/tex] (chlorate) has a charge of [tex]\( -1 \)[/tex].

The formula for this compound would be [tex]\( NH_4ClO_3 \)[/tex], known as ammonium chlorate.

### Compound 3: [tex]\( Fe^{2+} \)[/tex] and [tex]\( CH_3CO_2^- \)[/tex]
To balance the charges, we must consider:
- [tex]\( Fe^{2+} \)[/tex] has a charge of [tex]\( +2 \)[/tex].
- [tex]\( CH_3CO_2^- \)[/tex] has a charge of [tex]\( -1 \)[/tex].

We need two [tex]\( CH_3CO_2^- \)[/tex] ions to balance one [tex]\( Fe^{2+} \)[/tex] ion:
[tex]\[ \text{Formula}: Fe(CH_3CO_2)_2 \][/tex]
This compound is known as iron (II) acetate.

### Compound 4: [tex]\( Fe^{2+} \)[/tex] and [tex]\( ClO_3^- \)[/tex]
Combining the remaining ions:
- [tex]\( Fe^{2+} \)[/tex] with a charge of [tex]\( +2 \)[/tex].
- [tex]\( ClO_3^- \)[/tex] has a charge of [tex]\( -1 \)[/tex].

You need two [tex]\( ClO_3^- \)[/tex] ions to balance the [tex]\( +2 \)[/tex] charge of [tex]\( Fe^{2+} \)[/tex]:
[tex]\[ \text{Formula}: Fe(ClO_3)_2 \][/tex]
This compound is known as iron (II) chlorate.

### Summary
The empirical formulas for the four ionic compounds are:
1. [tex]\( NH_4CH_3CO_2 \)[/tex] (ammonium acetate)
2. [tex]\( NH_4ClO_3 \)[/tex] (ammonium chlorate)
3. [tex]\( Fe(CH_3CO_2)_2 \)[/tex] (iron (II) acetate)
4. [tex]\( Fe(ClO_3)_2 \)[/tex] (iron (II) chlorate)

These formulas ensure that the total charges balance to zero for each ionic compound.

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