Sure! Let's determine the empirical formula for four ionic compounds that can be formed using the ions [tex]\( Pb^{4+} \)[/tex], [tex]\( Fe^{2+} \)[/tex], [tex]\( IO_3^{-} \)[/tex], and [tex]\( OH^{-} \)[/tex]. Here's the step-by-step process:
### Compound 1: Lead(IV) Iodate
1. Identify the ions involved:
- [tex]\( Pb^{4+} \)[/tex]
- [tex]\( IO_3^{-} \)[/tex]
2. Determine the ratio of cations to anions needed to balance the charges:
- [tex]\( Pb^{4+} \)[/tex] has a +4 charge.
- [tex]\( IO_3^{-} \)[/tex] has a -1 charge.
- To balance the +4 charge, we need 4 [tex]\( IO_3^{-} \)[/tex] ions.
3. Write the empirical formula:
[tex]\[
Pb(IO_3)_4
\][/tex]
### Compound 2: Iron(II) Hydroxide
1. Identify the ions involved:
- [tex]\( Fe^{2+} \)[/tex]
- [tex]\( OH^{-} \)[/tex]
2. Determine the ratio of cations to anions needed to balance the charges:
- [tex]\( Fe^{2+} \)[/tex] has a +2 charge.
- [tex]\( OH^{-} \)[/tex] has a -1 charge.
- To balance the +2 charge, we need 2 [tex]\( OH^{-} \)[/tex] ions.
3. Write the empirical formula:
[tex]\[
Fe(OH)_2
\][/tex]
### Compound 3: Lead(IV) Hydroxide
1. Identify the ions involved:
- [tex]\( Pb^{4+} \)[/tex]
- [tex]\( OH^{-} \)[/tex]
2. Determine the ratio of cations to anions needed to balance the charges:
- [tex]\( Pb^{4+} \)[/tex] has a +4 charge.
- [tex]\( OH^{-} \)[/tex] has a -1 charge.
- To balance the +4 charge, we need 4 [tex]\( OH^{-} \)[/tex] ions.
3. Write the empirical formula:
[tex]\[
Pb(OH)_4
\][/tex]
### Compound 4: Iron(II) Iodate
1. Identify the ions involved:
- [tex]\( Fe^{2+} \)[/tex]
- [tex]\( IO_3^{-} \)[/tex]
2. Determine the ratio of cations to anions needed to balance the charges:
- [tex]\( Fe^{2+} \)[/tex] has a +2 charge.
- [tex]\( IO_3^{-} \)[/tex] has a -1 charge.
- To balance the +2 charge, we need 2 [tex]\( IO_3^{-} \)[/tex] ions.
3. Write the empirical formula:
[tex]\[
Fe(IO_3)_2
\][/tex]
These are the empirical formulas for the possible ionic compounds:
1. [tex]\( Pb(IO_3)_4 \)[/tex]
2. [tex]\( Fe(OH)_2 \)[/tex]
3. [tex]\( Pb(OH)_4 \)[/tex]
4. [tex]\( Fe(IO_3)_2 \)[/tex]
