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Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

[tex]\[ BrO_3^{-}, PO_4^{3-}, Fe^{3+}, NH_4^{+} \][/tex]

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Example compounds:
1. [tex]\[ \text{Fe(BrO}_3\text{)}_3 \][/tex]
2. [tex]\[ (\text{NH}_4\text{)}_3\text{PO}_4 \][/tex]
3. [tex]\[ \text{FePO}_4 \][/tex]
4. [tex]\[ \text{NH}_4\text{BrO}_3 \][/tex]

Sagot :

GinnyAnsweridn
To determine the empirical formulas for ionic compounds formed by the given ions, we must ensure the total charge of each compound is neutral. The ions provided are:

- Bromate ion: [tex]\( BrO_3^- \)[/tex] with a charge of [tex]\(-1\)[/tex]
- Phosphate ion: [tex]\( PO_4^{3-} \)[/tex] with a charge of [tex]\(-3\)[/tex]
- Iron(III) ion: [tex]\( Fe^{3+} \)[/tex] with a charge of [tex]\(+3\)[/tex]
- Ammonium ion: [tex]\( NH_4^+ \)[/tex] with a charge of [tex]\(+1\)[/tex]

Let's go step-by-step through the possible combinations to achieve neutral compounds:

1. Combination of [tex]\(Fe^{3+}\)[/tex] and [tex]\(BrO_3^-\)[/tex]:

To balance the charges, we need three [tex]\(BrO_3^-\)[/tex] ions to neutralize one [tex]\(Fe^{3+}\)[/tex] ion. This results in:
[tex]\[ Fe(BrO_3)_3 \][/tex]
Here, the total charge is [tex]\(3 \times -1 + 3 = 0\)[/tex], making [tex]\(Fe(BrO_3)_3\)[/tex] an electrically neutral compound.

2. Combination of [tex]\(Fe^{3+}\)[/tex] and [tex]\(PO_4^{3-}\)[/tex]:

In this case, one [tex]\(Fe^{3+}\)[/tex] ion can neutralize one [tex]\(PO_4^{3-}\)[/tex] ion directly. Thus, the compound formed is:
[tex]\[ FePO_4 \][/tex]
The total charge is [tex]\(-3 + 3 = 0\)[/tex], ensuring [tex]\(FePO_4\)[/tex] is neutral.

3. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(PO_4^{3-}\)[/tex]:

To balance the charges, we need three [tex]\(NH_4^+\)[/tex] ions to neutralize one [tex]\(PO_4^{3-}\)[/tex] ion. This results in:
[tex]\[ (NH_4)_3PO_4 \][/tex]
Here, the total charge is [tex]\(3 \times +1 + (-3) = 0\)[/tex], making [tex]\((NH_4)_3PO_4\)[/tex] neutral.

4. Combination of [tex]\(NH_4^+\)[/tex] and [tex]\(BrO_3^-\)[/tex]:

Here, one [tex]\(NH_4^+\)[/tex] ion can balance one [tex]\(BrO_3^-\)[/tex] ion directly. Therefore, the compound formed is:
[tex]\[ NH_4BrO_3 \][/tex]
The total charge is [tex]\(+1 + (-1) = 0\)[/tex], making [tex]\(NH_4BrO_3\)[/tex] neutral.

Thus, the empirical formulas for the ionic compounds that can be formed from the given ions are:

1. [tex]\( Fe(BrO_3)_3 \)[/tex] (Iron(III) Bromate)
2. [tex]\( FePO_4 \)[/tex] (Iron(III) Phosphate)
3. [tex]\( (NH_4)_3PO_4 \)[/tex] (Ammonium Phosphate)
4. [tex]\( NH_4BrO_3 \)[/tex] (Ammonium Bromate)
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