Lead bromate, the empirical formula is Pb(BrO3)4. Lead sulfate, the empirical formula is PbSO4. Iron bromate, the empirical formula is Fe(BrO3)3. Iron sulfate, the empirical formula is Fe2(SO4)3.
Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
The correct answer and explanation is:
Here are four possible ionic compounds formed from the given ions:
- Lead(II) Bromate → Pb(BrO₃)₂
- Lead(IV) Sulfate → Pb(SO₄)₂
- Iron(II) Bromate → Fe(BrO₃)₂
- Iron(III) Sulfate → Fe₂(SO₄)₃
Explanation:
Understanding Ionic Compounds:
Ionic compounds are formed by the electrostatic attraction between positively charged cations and negatively charged anions. The charge balance between these ions determines the correct empirical formula.
1. Lead(II) Bromate → Pb(BrO₃)₂
Lead can exhibit multiple oxidation states, but in this case, it has a +2 charge (Pb²⁺). The bromate ion (BrO₃⁻) carries a -1 charge. To balance the charges, two bromate ions are needed for every lead ion, resulting in Pb(BrO₃)₂.
2. Lead(IV) Sulfate → Pb(SO₄)₂
Lead(IV) has a +4 charge (Pb⁴⁺), while the sulfate ion (SO₄²⁻) has a -2 charge. To balance the charge, two sulfate ions are required per lead ion, giving the formula Pb(SO₄)₂.
3. Iron(II) Bromate → Fe(BrO₃)₂
Iron(II) has a +2 charge (Fe²⁺), and bromate (BrO₃⁻) has a -1 charge. Thus, two bromate ions are needed per iron(II) ion, forming Fe(BrO₃)₂.
4. Iron(III) Sulfate → Fe₂(SO₄)₃
Iron(III) has a +3 charge (Fe³⁺), while sulfate (SO₄²⁻) has a -2 charge. The least common multiple of 3 and 2 is 6, so two iron(III) ions (Fe³⁺) and three sulfate ions (SO₄²⁻) balance the charge, resulting in Fe₂(SO₄)₃.
Now, let’s generate an image representing these compounds.
Here is the educational infographic illustrating the four ionic compounds with structured labels and atomic representations. Let me know if you need any modifications!
