4.3: Writing Formulas for Ionic Compounds
- Page ID
- 370395
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- 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.
If you know the name of a binary ionic compound, you can write its chemical formula. Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charges to cancel each other out.
Write the formulas for aluminum nitride and lithium oxide.
Solution
| Write the formula for aluminum nitride | Write the formula for lithium oxide | |
|---|---|---|
| 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. | \(\ce{Al^{3+}} \: \: \: \: \: \ce{N^{3-}}\) | \(\ce{Li^+} \: \: \: \: \: \ce{O^{2-}}\) |
| 2. Use a multiplier to make the total charge of the cations and anions equal to each other. |
total charge of cations = total charge of anions 1(3+) = 1(3-) +3 = -3 |
total charge of cations = total charge of anions 2(1+) = 1(2-) +2 = -2 |
| 3. Use the multipliers as subscript for each ion. | \(\ce{Al_1N_1}\) | \(\ce{Li_2O_1}\) |
| 4. Write the final formula. Leave out all charges and all subscripts that are 1. | \(\ce{AlN}\) | \(\ce{Li_2O}\) |
An alternative way to writing a correct formula for an ionic compound is to use the crisscross method. In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped.
Write the formula for lead (IV) oxide.
Solution
| Crisscross Method | Write the formula for lead (IV) oxide |
|---|---|
| 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. | \(\ce{Pb^{4+}} \: \: \: \: \: \ce{O^{2-}}\) |
| 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. |  |
| 3. Reduce to the lowest ratio. | \(\ce{Pb_2O_4}\) |
| 4. Write the final formula. Leave out all subscripts that are 1. | \(\ce{PbO_2}\) |
Write the chemical formula for an ionic compound composed of each pair of ions.
- the calcium ion and the oxygen ion
- the 2+ copper ion and the sulfur ion
- the 1+ copper ion and the sulfur ion
- Answer a:
- CaO
- Answer b:
- CuS
- Answer c:
- Cu2S
Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions.
Write the formula for sodium combined with sulfur.
Solution
| Crisscross Method | Write the formula for sodium combined with sulfur |
|---|---|
| 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. | \(\ce{Na^{+}} \: \: \: \: \: \ce{S^{2-}}\) |
| 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. |  |
| 3. Reduce to the lowest ratio. | This step is not necessary. |
| 4. Write the final formula. Leave out all subscripts that are 1. | \(\ce{Na_2S}\) |
Write the formula for each ionic compound.
- sodium bromide
- lithium chloride
- magnesium oxide
- Answer a:
- NaBr
- Answer b:
- LiCl
- Answer c:
- MgO