Name the following compounds, which contain one or more polyatomic ions.
a. Ca_{3}(PO_{4})_{2} b. Fe_{2}(SO_{4})_{3} c. (NH_{4})_{2}CO_{3}
a. The positive ion present is the calcium ion (Ca^{2+}). We will not need a Roman numeral to specify the charge on a Ca^{2+} ion because it is always +2. The negative ion is the polyatomic phosphate ion (PO_{4} ^{3-}). The name of the compound is calcium phosphate. As in naming binary ionic compounds, subscripts in the formula are not incorporated into the name.
b. The positive ion present is iron(III). The negative ion is the polyatomic sulfate ion (SO_{4} ^{2-}). The name of the compound is iron(III) sulfate. The determination that iron is present as iron(III) involves the following calculation dealing with charge balance:
\quad\quad\quad\quad 2(iron charge) + 3(sulfate charge) = 0
The sulfate charge is -2. (You had to memorize that.) Therefore,
\quad\quad\quad\quad 2(iron charge) + 3(-2) = 0
\quad\quad\quad\quad\quad\quad\quad 2(iron charge) = +6
\quad\quad\quad\quad\quad\quad\quad\quad Iron charge = +3
c. Both the positive and the negative ions in this compound are polyatomic—the ammonium ion (NH_{4} ^{+}) and the carbonate ion (CO_{3} ^{2-}). The name of the compound is simply the combination of the names of the two polyatomic ions: ammonium carbonate.