QUESTION IMAGE
Question
fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:
some ionic compounds
cation anion empirical formula name of compound
pb²+ no₃⁻
cr³+ clo₃⁻
k+ cro₄²-
Step1: Determine the empirical formula for \(Pb^{2 +}\) and \(NO_3^-\)
The charge on \(Pb^{2+}\) is \(+ 2\) and on \(NO_3^-\) is \(-1\). To balance the charges, we need \(2\) \(NO_3^-\) ions for each \(Pb^{2+}\) ion. So the empirical formula is \(Pb(NO_3)_2\). The name of the compound: The cation is lead (\(Pb^{2+}\)) and the anion is nitrate (\(NO_3^-\)). So the name is lead(II) nitrate.
Step2: Determine the empirical formula for \(Cr^{3+}\) and \(ClO_3^-\)
The charge on \(Cr^{3+}\) is \(+3\) and on \(ClO_3^-\) is \(-1\). To balance the charges, we need \(3\) \(ClO_3^-\) ions for each \(Cr^{3+}\) ion. So the empirical formula is \(Cr(ClO_3)_3\). The name of the compound: The cation is chromium (\(Cr^{3+}\)) and the anion is chlorate (\(ClO_3^-\)). So the name is chromium(III) chlorate.
Step3: Determine the empirical formula for \(K^{+}\) and \(CrO_4^{2 -}\)
The charge on \(K^{+}\) is \(+1\) and on \(CrO_4^{2-}\) is \(-2\). To balance the charges, we need \(2\) \(K^{+}\) ions for each \(CrO_4^{2-}\) ion. So the empirical formula is \(K_2CrO_4\). The name of the compound: The cation is potassium (\(K^{+}\)) and the anion is chromate (\(CrO_4^{2-}\)). So the name is potassium chromate.
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| cation | anion | empirical formula | name of compound |
|---|---|---|---|
| \(Cr^{3+}\) | \(ClO_3^-\) | \(Cr(ClO_3)_3\) | chromium(III) chlorate |
| \(K^{+}\) | \(CrO_4^{2-}\) | \(K_2CrO_4\) | potassium chromate |