Skip to main content
Chemistry LibreTexts

4.4: Writing Ionic Formulas

  • Page ID
    49806
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Skills to Develop

    • Write the correct formula for an ionic compound.

    Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattice containing many ions each of the cation and anion. An ionic formula, like \(\ce{NaCl}\), is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. Sodium sulfide, another ionic compound, has the formula \(\ce{Na_2S}\). This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. This section will teach you how to find the correct ratio of ions, so that you can write a correct formula.

    Ionic Formulas

    When an ionic compound forms, the number of electrons given off by the cations must be exactly the same as the number of electrons taken on by the anions. Therefore, if calcium, which gives off two electrons, is to combine with fluorine, which takes on one electron, then one calcium atom must combine with two fluorine atoms. The formula would be \(\ce{CaF_2}\).

    To write the formula for an ionic compound:

    1. Write the symbol and charge of the cation (first word)

    a) If the element is in group 1 or 2 or is \(\ce{Al}\) with a consistent charge, you can get the charge using your periodic table.

    b) If the metal is a transition metal with a variable charge, the charge will be given to you in Roman numerals.

    2. Write the symbol and charge of the anion (second word)

    a) Look at your polyatomic ion chart first. If your anion is a polyatomic ion, write the ion in parentheses.

    b) If the anion is not on the polyatomic chart, it is a nonmetal anion from your periodic table. You can get its charge using your table.

    3. Write the correct subscripts so that the total charge of the compound will be zero.

    4. Write the final formula. Leave out all charges and all subscripts that are 1. If there is only 1 of the polyatomic ion, leave off parentheses.

    Pay close attention to how these steps are followed in the given examples.

    Example 4.4.1: aluminum chloride

    Write the formula for aluminum chloride.

    Solution

    Cation, aluminum: \(\ce{Al^{3+}}\) (you can find this charge using your periodic table)

    Anion, chloride: \(\ce{Cl^-}\) (chloride is chlorine as an ion, get its charge from your periodic table)

    To balance the charges you need \(1 \cdot \left( +3 \right)\) and \(3 \cdot \left( -1 \right)\). Giving: \(\ce{Al_1^{3+}} \ce{Cl_3^{1-}}\)

    The final formula is: \(\ce{AlCl_3}\)

    Example 4.4.2: aluminum sulfide

    Write the formula for aluminum sulfide

    Solution:

    Cation, aluminum: \(\ce{Al^{3+}}\) (you can find this charge using your periodic table)

    Anion, sulfide: \(\ce{S^{2-}}\) (sulfide is sulfur as an ion, get its charge from your periodic table)

    To balance the charges you need \(2 \cdot \left( +3 \right)\) and \(3 \cdot \left( -2 \right)\). Giving: \(\ce{Al_2^{3+}} \ce{S_3^{2-}}\)

    The final formula is: \(\ce{Al_2S_3}\)

    Example 4.4.3: lead (IV) oxide

    Write the formula for lead (IV) oxide

    Solution:

    Cation, lead (IV): \(\ce{Pb^{4+}}\) (the charge is given to you as a Roman numeral, because this is a metal with a variable charge)

    Anion, oxide: \(\ce{O^{2-}}\) (oxide is oxygen as an ion, get its charge from your periodic table)

    To balance the charges you need \(1 \cdot \left( +4 \right)\) and \(2 \cdot \left( -2 \right)\). Giving: \(\ce{Pb_1^{4+}} \ce{O_2^{2-}}\)

    The final formula is: \(\ce{PbO_2}\)

    Example 4.4.4: calcium nitrate

    Write the formula for calcium nitrate.

    Solution:

    Cation, calcium: \(\ce{Ca^{2+}}\) (you can find this charge using your periodic table)

    Anion, nitrate: \(\left( \ce{NO_3^-} \right)\) (this is a polyatomic ion)

    To balance the charges you need \(1 \cdot \left( +2 \right)\) and \(2 \cdot \left( -1 \right)\). Giving: \(\ce{Ca_1^{2+}} \left( \ce{NO_3} \right)_2^{1-}\)

    The final formula is: \(\ce{Ca(NO_3)_2}\)

    In this case you need to keep the parentheses. There are two of the group \(\left( \ce{NO_3^-} \right)\). Without the parentheses, you are merely changing the number of oxygen atoms.

    Example 4.4.5: magnesium sulfate

    Write the formula for magnesium sulfate.

    Solution:

    Cation, magnesium: \(\ce{Mg^{2+}}\) (you can find this charge using your periodic table)

    Anion, sulfate: \(\ce{(SO_4)^{2-}}\) (this is a polyatomic ion)

    To balance the charges you need \(1 \cdot \left( +2 \right)\) and \(1 \cdot \left( -2 \right)\). Giving: \(\ce{Mg_1^{2+}} \ce{(SO_4)_1^{2-}}\)

    The final formula is: \(\ce{MgSO_4}\)

    In this case you do not need parentheses. They are only required if there is more than one of the polyatomic ion.

    Example 4.4.6: copper (II) acetate

    Write the formula for copper (II) acetate.

    Solution:

    Cation, copper (II): \(\ce{Cu^{2+}}\) (the charge is given to you in a Roman numeral)

    Anion, acetate: \(\ce{(C_2H_3O_2)^-}\) (this is a polyatomic ion)

    To balance the charges you need \(1 \cdot \left( +2 \right)\) and \(2 \cdot \left( -1 \right)\). Giving: \(\ce{Cu_1^{2+}} \ce{(C_2H_3O_2)_2^{1-}}\)

    The final formula is \(\ce{Cu(C_2H_3O_2)_2}\)

    In this case you need to keep the parentheses. There are two of the group \(\ce{(C_2H_3O_2)^-}\). Without the parentheses, you are merely changing the number of oxygen atoms.

    Summary

    • Formulas for ionic compounds contain the lowest whole number ratio of subscripts such that the sum of the subscripts of the more electropositive element times its oxidation number plus the subscripts of the more electronegative element times its oxidation number equals zero.

    Vocabulary

    • Ionic formula: Includes the symbols and number of each atom present in a compound in the lowest whole number ratio.

    Contributors


    4.4: Writing Ionic Formulas is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?