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 | |||
|---|---|---|---|
| $pb^{2 + }$ | $po_{4}^{3 - }$ | ||
| $k^{+}$ | $bro_{4}^{-}$ | ||
| $ca^{2 + }$ | $io^{-}$ | ||
| $pb^{4 + }$ | $so_{4}^{2 - }$ |
Step1: Determine formula for $Pb^{2 +}$ and $PO_{4}^{3 -}$
To form a neutral compound, we need to balance the charges. The least - common multiple of 2 and 3 is 6. So we need 3 $Pb^{2+}$ ions and 2 $PO_{4}^{3 -}$ ions. The empirical formula is $Pb_{3}(PO_{4})_{2}$, and the name is lead(II) phosphate.
Step2: Determine formula for $K^{+}$ and $BrO_{4}^{-}$
Since the charges are + 1 and - 1 respectively, the empirical formula is $KBrO_{4}$, and the name is potassium perbromate.
Step3: Determine formula for $Ca^{2+}$ and $IO^{-}$
To balance the + 2 charge of $Ca^{2+}$, we need 2 $IO^{-}$ ions. The empirical formula is $Ca(IO)_{2}$, and the name is calcium hypoiodite.
Step4: Determine formula for $Pb^{4+}$ and $SO_{4}^{2 -}$
The least - common multiple of 4 and 2 is 4. We need 1 $Pb^{4+}$ ion and 2 $SO_{4}^{2 -}$ ions. The empirical formula is $Pb(SO_{4})_{2}$, and the name is lead(IV) sulfate.
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| cation | anion | empirical formula | name of compound |
|---|---|---|---|
| $K^{+}$ | $BrO_{4}^{-}$ | $KBrO_{4}$ | potassium perbromate |
| $Ca^{2+}$ | $IO^{-}$ | $Ca(IO)_{2}$ | calcium hypoiodite |
| $Pb^{4+}$ | $SO_{4}^{2 -}$ | $Pb(SO_{4})_{2}$ | lead(IV) sulfate |