Why Ionic Compounds Use Formula Units Instead of Molecules

A grain of table salt has the chemical formula \(\ce{NaCl}\). It is tempting to picture billions of tiny particles, each containing one sodium ion joined to one chloride ion. You might even call each pair a sodium chloride molecule.

But there is a problem with that picture. In a sodium chloride crystal, every \(\mathrm{Na}^{+}\) ion is surrounded by six nearest \(\mathrm{Cl}^{-}\) ions. Each chloride ion is also surrounded by six sodium ions. There is no unique pair of ions that forms an individual \(\ce{NaCl}\) particle.

This is why chemists use the term formula unit instead of molecule for an ionic compound. An ionic solid consists of an extended three-dimensional lattice of ions, not separate molecules. Its chemical formula describes the simplest whole-number ratio of ions, rather than the composition of an individual particle.

01Why an ionic compound does not have molecules

Start with the difference between a molecule and an ionic lattice.

A molecule is a discrete group of atoms joined by covalent bonds. For example, a carbon dioxide molecule contains one carbon atom covalently bonded to two oxygen atoms.

You can identify where one \(\ce{CO2}\) molecule ends and another begins. Each molecule exists as a separate unit.

An ionic solid works differently.

Consider solid sodium chloride, \(\ce{NaCl}\). It contains positively charged sodium ions, \(\mathrm{Na}^{+}\), and negatively charged chloride ions, \(\mathrm{Cl}^{-}\). These ions are held together by electrostatic attractions throughout a repeating three-dimensional structure called an ionic lattice.

An individual sodium ion does not have just one chloride ion as its partner. It attracts several nearby chloride ions.

The same applies in reverse: each chloride ion attracts several neighbouring sodium ions.

So what does the formula NaCl actually mean?

It tells us the ratio of sodium ions to chloride ions.

For every one \(\mathrm{Na}^{+}\) ion, there is one \(\mathrm{Cl}^{-}\) ion in the crystal.

The ratio is therefore:

\[ \ce{Na+ : Cl-} = 1:1 \]

The formula \(\ce{NaCl}\) represents this simplest whole-number ratio.

Chemists call this ratio a formula unit.

A formula unit is a bookkeeping unit used to describe the composition of an ionic compound. It is not a physically separate pair of ions.

Think of it as a recipe ratio rather than a pre-packaged pair of ingredients. The formula tells you how many of each type of ion are needed to describe the composition, but it does not tell you that those ions exist as separate little packets.

The same idea applies to other ionic compounds.

Notice that every formula unit represents an electrically neutral ratio of ions.

This is important because ionic compounds must have an overall electrical charge of zero.

For example, \(\ce{CaCl2}\) does not mean that a calcium ion is physically attached to exactly two chloride ions. It means that the compound contains calcium ions and chloride ions in a 1:2 ratio.

What about compounds containing polyatomic ions?

The same rule applies.

Calcium nitrate, \(\ce{Ca(NO3)2}\), contains:

• One calcium ion, \(\mathrm{Ca}^{2+}\).
• Two nitrate ions, \(\ce{NO3-}\).

Each nitrate ion contains one nitrogen atom and three oxygen atoms, joined by covalent bonds.

The nitrate ions are identifiable polyatomic ions, but the overall calcium nitrate solid still forms an extended ionic lattice.

This distinction is useful: a polyatomic ion can be a discrete chemical entity within an ionic lattice without making the entire ionic compound a molecule.

02How X-ray diffraction helped reveal the structure

There is a historical reason this distinction became much clearer.

Knowing the chemical formula of a solid does not automatically reveal how its particles are arranged. Chemists could determine that sodium chloride contained sodium and chlorine in equal proportions, but that information alone could not establish whether the solid contained separate pairs or an extended structure.

They needed a way to investigate the arrangement inside a crystal.

X-ray diffraction provided that evidence.

The original problem

In 1912, Max von Laue and his colleagues demonstrated that X-rays could produce diffraction patterns when passed through crystals.

The important observation was that crystals produced regular patterns of diffracted X-rays.

Why?

A crystal contains particles arranged in a repeating pattern. X-rays scattered from different positions can reinforce one another in particular directions, producing measurable diffraction spots.

The pattern therefore contains information about the internal structure of the crystal.

How the apparatus worked

In the early Laue experiment, the apparatus followed this arrangement:

1. An X-ray source produced a beam of X-rays.
2. The beam was directed towards a thin crystal.
3. The X-rays passed through the crystal and were scattered by its regularly arranged particles.
4. A photographic plate behind the crystal recorded the diffracted beams as a pattern of spots.

The positions of the spots provided evidence of regular spacing within the crystal.

What was the limitation of the original design?

The original Laue method used X-rays containing a range of wavelengths.

This produced complicated diffraction patterns that could be difficult to interpret.

William Henry Bragg and his son, William Lawrence Bragg, developed improved methods for measuring X-ray reflections. Their approach used more controlled X-ray wavelengths and a spectrometer to measure the intensity of reflected X-rays at particular angles.

Instead of relying only on a complicated photographic pattern, scientists could investigate specific reflections and use their measured directions and intensities to determine crystal structures.

In 1913, William Lawrence Bragg published work determining the structure of rock salt, \(\ce{NaCl}\), using X-ray diffraction.

The results supported an extended arrangement of sodium and chloride ions. Each sodium ion had six nearest chloride neighbours, and each chloride ion had six nearest sodium neighbours.

There was no reason to identify one particular sodium ion and one particular chloride ion as an individual molecule.

The formula \(\ce{NaCl}\) therefore described the 1:1 composition of the lattice, not separate molecular particles.

The International Union of Crystallography’s historical account of early X-ray diffraction provides further detail about the development of these structural investigations.

One important limitation is worth remembering: X-ray diffraction does not produce a direct photograph of individual ions. Scientists measure diffraction patterns and use them to construct and test models of the crystal structure. The assignment of ionic charges also relies on chemical evidence.

03A unit cell is not the same as a formula unit

There is another term you might encounter when studying crystal structures: the unit cell.

A unit cell is a repeating geometric section that can be used to describe the arrangement of particles in a crystal.

It is not the same thing as a formula unit.

For example, the conventional cubic unit cell of sodium chloride contains four sodium ions and four chloride ions.

That gives:

\[ \frac{4\ \mathrm{Na}^{+}}{4\ \mathrm{Cl}^{-}}=1:1 \]

So the unit cell represents four \(\ce{NaCl}\) formula units.

It does not contain four separate sodium chloride molecules.

A quick structural scale estimate

The repeating unit in sodium chloride has a side length of approximately \(0.564\text{ nm}\) at room temperature.

Imagine a perfect cubic sodium chloride crystal measuring \(1.00\text{ mm}\) along each side.

How many unit cells fit along one edge?

Convert the lengths into metres:

\[ a=0.564\times10^{-9}\text{ m} \]

\[ L=1.00\times10^{-3}\text{ m} \]

The number of unit cells along one edge is:

\[ \begin{aligned} N_{\text{edge}} &=\frac{L}{a}\\ &=\frac{1.00\times10^{-3}} {5.64\times10^{-10}}\\ &=1.77\times10^6 \end{aligned} \]

The total number of unit cells is approximately:

\[ \begin{aligned} N_{\text{cells}} &=(1.77\times10^6)^3\\ &=5.6\times10^{18} \end{aligned} \]

Each conventional unit cell represents four formula units.

Therefore:

\[ \begin{aligned} N_{\mathrm{FU}} &=4(5.6\times10^{18})\\ &=2.2\times10^{19} \end{aligned} \]

A perfect 1 mm sodium chloride crystal therefore contains approximately \(2.2\times10^{19}\) formula-unit ratios.

This is an illustrative crystallographic estimate. It assumes a single crystal with the stated lattice spacing and ignores defects and boundary effects.

The key distinction is that a unit cell describes a repeating region of space, while a formula unit describes the simplest whole-number ratio of ions.

04Connecting formula units to mass, moles, and particles

Now we can connect the structural idea to the calculations you need in HSC Chemistry.

When calculating the amount of an ionic compound, the counted entity is the formula unit.

For example:

• One mole of \(\ce{NaCl}\) contains Avogadro’s number of \(\ce{NaCl}\) formula units.
• One mole of \(\ce{CaCl2}\) contains Avogadro’s number of \(\ce{CaCl2}\) formula units.
• One mole of \(\ce{Al2O3}\) contains Avogadro’s number of \(\ce{Al2O3}\) formula units.

The number of ions represented by those formula units depends on the subscripts in the chemical formula.

The three quantities you need to connect

The molar mass, \(M\), connects mass and amount. Its unit is \(\mathrm{g\,mol^{-1}}\).

The Avogadro constant, \(N_A\), connects amount and number:

\[ N_A=6.022\times10^{23}\text{ mol}^{-1} \]

Step 1: Convert mass into amount

Use:

\[ n=\frac{m}{M} \]

Check the units:

\[ \frac{\mathrm{g}}{\mathrm{g\,mol^{-1}}} =\mathrm{mol} \]

The result is the amount of the ionic compound, expressed in moles of formula units.

Step 2: Convert amount into number of formula units

Use:

\[ N_{\mathrm{FU}}=nN_A \]

The units cancel:

\[ \mathrm{mol}\times\mathrm{mol}^{-1} =\text{count} \]

You now have the number of formula-unit ratios.

Step 3: Convert formula units into ions

Use the subscripts in the chemical formula.

For example, consider \(\ce{CaCl2}\).

One formula-unit ratio contains:

• One \(\mathrm{Ca}^{2+}\) ion.
• Two \(\mathrm{Cl}^{-}\) ions.

Therefore:

\[ N_{\mathrm{Ca}^{2+}}=N_{\mathrm{FU}} \]

\[ N_{\mathrm{Cl}^{-}}=2N_{\mathrm{FU}} \]

The total number of ions is:

\[ N_{\text{ions}}=3N_{\mathrm{FU}} \]

This last step is where many students make mistakes. They correctly calculate the number of formula units, then forget that one formula unit can represent several ions.

07The important contrast: Ionic compounds versus molecular compounds

Let’s compare sodium chloride with solid carbon dioxide.

Both substances are solids under suitable conditions, but their structures are different.

Solid sodium chloride, \(\ce{NaCl}\):

• Contains \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions.
• Forms an extended ionic lattice.
• Has no separate \(\ce{NaCl}\) molecules.
• Uses formula units to describe the simplest ratio of ions.

Solid carbon dioxide, \(\ce{CO2}\):

• Contains discrete \(\ce{CO2}\) molecules.
• Each molecule contains one carbon atom and two oxygen atoms.
• The molecules are held together in the solid by intermolecular attractions.
• Uses molecules as the counted entity.

The difference is not simply that one substance is solid and the other is a gas or liquid. Solid carbon dioxide still contains discrete molecules, while solid sodium chloride contains an extended ionic lattice.

It is the structure of the substance that determines the appropriate counted entity.

One further qualification: not every covalent solid consists of molecules. Diamond and silicon dioxide, for example, have extended covalent structures. The important question is whether the substance actually contains discrete molecular units.

A useful decision rule

When a question asks you to calculate the number of particles in a substance, identify what the chemical formula represents before using Avogadro’s constant.

There is one more situation to recognise.

When an ionic compound dissolves in water, its ions separate from the lattice. For example:

\[ \ce{CaCl2(s) -> Ca^2+(aq) + 2Cl-(aq)} \]

The solid initially contains an ionic lattice. After complete dissolution, the dissolved species are hydrated ions rather than intact calcium chloride formula units.

You can still calculate the amount of calcium chloride added using its formula-unit stoichiometry, but if a question asks for the actual number of dissolved particles, you must count the ions.

This distinction becomes important in questions involving dissolution, concentration, and total ion numbers. See Dissolution of Ionic Compounds in Water for HSC Chemistry.

08Common mistakes to avoid

09Questions and solutions

Question 1

A student has a \(4.03\text{ g}\) sample of pure magnesium oxide, \(\ce{MgO}\).

The molar mass of magnesium oxide is \(40.3\text{ g mol}^{-1}\).

Calculate:

a. The amount of magnesium oxide in moles.

b. The number of formula units.

c. The number of magnesium ions and oxide ions in the sample.

Use \(N_A=6.02\times10^{23}\text{ mol}^{-1}\).

Solution 1

The sample contains \(0.100\text{ mol}\) of magnesium oxide, equivalent to \(6.02\times10^{22}\) formula units. It contains the same number of magnesium ions and oxide ions.

First, calculate the amount:

\[ \begin{aligned} n&=\frac{m}{M}\\ &=\frac{4.03}{40.3}\\ &=0.100\text{ mol} \end{aligned} \]

Next, calculate the number of formula units:

\[ \begin{aligned} N_{\mathrm{FU}} &=nN_A\\ &=0.100\times6.02\times10^{23}\\ &=6.02\times10^{22} \end{aligned} \]

The formula \(\ce{MgO}\) represents one \(\mathrm{Mg}^{2+}\) ion for every one \(\mathrm{O}^{2-}\) ion.

Therefore:

\[ N_{\mathrm{Mg}^{2+}}=6.02\times10^{22} \]

\[ N_{\mathrm{O}^{2-}}=6.02\times10^{22} \]

There are \(1.20\times10^{23}\) ions in total.

The common mistake would be to report \(6.02\times10^{22}\) as the total number of ions. That is the number of formula units. Each formula unit represents two ions.

Question 2

A pure aluminium oxide sample, \(\ce{Al2O3}\), contains \(9.03\times10^{22}\) oxide ions.

The molar mass of aluminium oxide is \(102.0\text{ g mol}^{-1}\).

Calculate:

a. The number of formula units.

b. The amount of aluminium oxide in moles.

c. The mass of the sample.

d. The number of aluminium ions.

Solution 2

The sample has \(3.01\times10^{22}\) formula units, an amount of \(0.0500\text{ mol}\), and a mass of \(5.10\text{ g}\). It contains \(6.02\times10^{22}\) aluminium ions.

Start by examining the formula:

\[ \ce{Al2O3} \]

Each formula unit represents two aluminium ions and three oxide ions.

Part a: Number of formula units

There are three oxide ions per formula unit.

Therefore:

\[ \begin{aligned} N_{\mathrm{FU}} &=\frac{N_{\mathrm{O}^{2-}}}{3}\\ &=\frac{9.03\times10^{22}}{3}\\ &=3.01\times10^{22} \end{aligned} \]

Part b: Amount in moles

Rearrange the Avogadro relationship:

\[ n=\frac{N_{\mathrm{FU}}}{N_A} \]

Substitute:

\[ \begin{aligned} n&=\frac{3.01\times10^{22}} {6.02\times10^{23}}\\ &=0.0500\text{ mol} \end{aligned} \]

Part c: Mass of the sample

Use:

\[ m=nM \]

Substitute:

\[ \begin{aligned} m&=0.0500\times102.0\\ &=5.10\text{ g} \end{aligned} \]

Part d: Number of aluminium ions

Each formula unit contains two aluminium ions.

\[ \begin{aligned} N_{\mathrm{Al}^{3+}} &=2N_{\mathrm{FU}}\\ &=2(3.01\times10^{22})\\ &=6.02\times10^{22} \end{aligned} \]

The tempting approach is to divide the oxide-ion count by two because aluminium has a charge of \(+3\). That is incorrect.

The subscripts determine the number of ions per formula unit. The charges determine the ratio needed for electrical neutrality. Here, the formula already tells us that there are three oxide ions for every two aluminium ions.

Question 3

Two students are given the following samples:

• Sample A: \(1.17\text{ g}\) of solid \(\ce{NaCl}\), with \(M=58.5\text{ g mol}^{-1}\).
• Sample B: \(2.22\text{ g}\) of solid \(\ce{CaCl2}\), with \(M=111.0\text{ g mol}^{-1}\).

Calculate:

a. The amount of each compound.

b. The number of formula units in each sample.

c. The total number of ions in each sample.

d. Explain why the samples can contain the same number of formula units but different total numbers of ions.

Use \(N_A=6.02\times10^{23}\text{ mol}^{-1}\).

Solution 3

Both samples contain \(0.0200\text{ mol}\) of compound and approximately \(1.20\times10^{22}\) formula units. However, the calcium chloride sample contains more ions because each formula unit represents three ions rather than two.

Part a: Amount of each compound

For sodium chloride:

\[ \begin{aligned} n(\ce{NaCl}) &=\frac{1.17}{58.5}\\ &=0.0200\text{ mol} \end{aligned} \]

For calcium chloride:

\[ \begin{aligned} n(\ce{CaCl2}) &=\frac{2.22}{111.0}\\ &=0.0200\text{ mol} \end{aligned} \]

Part b: Number of formula units

Both samples contain the same amount, so:

\[ \begin{aligned} N_{\mathrm{FU}} &=0.0200\times6.02\times10^{23}\\ &=1.20\times10^{22} \end{aligned} \]

Part c: Total number of ions

For sodium chloride:

Each formula unit represents one sodium ion and one chloride ion.

\[ \begin{aligned} N_{\text{ions}} &=2N_{\mathrm{FU}}\\ &=2(1.20\times10^{22})\\ &=2.41\times10^{22} \end{aligned} \]

For calcium chloride:

Each formula unit represents one calcium ion and two chloride ions.

\[ \begin{aligned} N_{\text{ions}} &=3N_{\mathrm{FU}}\\ &=3(1.20\times10^{22})\\ &=3.61\times10^{22} \end{aligned} \]

Part d: Explanation

The two samples contain equal numbers of formula units because they contain equal amounts in moles.

However, their formula units have different compositions.

A formula unit of \(\ce{NaCl}\) represents two ions, whereas a formula unit of \(\ce{CaCl2}\) represents three ions.

Therefore, equal numbers of formula units do not necessarily mean equal numbers of ions.

The trap is assuming that a formula unit always represents the same number of ions. It does not. You must inspect the chemical formula before counting.

Question 4

An X-ray diffraction investigation produces a structural model of sodium chloride.

The conventional cubic unit cell is identified as containing:

• Four \(\mathrm{Na}^{+}\) ion positions.
• Four \(\mathrm{Cl}^{-}\) ion positions.

A student concludes:

“There are eight ions in the unit cell, so there must be eight formula units. Alternatively, we could call them four sodium chloride molecules because there are four sodium ions and four chloride ions.”

Evaluate both claims.

Your answer must:

a. State the number of formula units represented by the unit cell.

b. Explain why the number of ions is not the number of formula units.

c. Explain why describing the four formula units as molecules is incorrect, using the local arrangement of ions in the sodium chloride lattice.

Solution 4

The unit cell represents four formula units, not eight formula units or four molecules.

The reason is that each formula unit represents one \(\mathrm{Na}^{+}\) ion for every one \(\mathrm{Cl}^{-}\) ion.

There are four sodium ions and four chloride ions in the conventional unit cell.

The number of 1:1 formula-unit ratios is therefore:

\[ N_{\mathrm{FU}}=\frac{4}{1}=4 \]

The eight ions do not represent eight formula units because each formula unit accounts for two ions in the 1:1 composition ratio.

However, this calculation is only a way of counting the composition. It does not mean that the ions form four physically separate pairs.

The second claim is also incorrect.

In the rock-salt structure, each sodium ion has six nearest chloride neighbours. Each chloride ion also has six nearest sodium neighbours.

The electrostatic attractions extend throughout the three-dimensional lattice. There is no unique way to divide the crystal into separate sodium chloride molecules.

The tempting interpretation is to group each sodium ion with one chloride ion simply because the formula is \(\ce{NaCl}\).

The constraint that changes the answer is the actual crystal structure: every ion interacts with several neighbouring ions, not just one chosen partner.

The unit cell describes a repeating geometric region, while the formula unit describes the simplest ionic ratio. Neither represents a separate sodium chloride molecule.

Question 5

A container initially holds \(0.0500\text{ mol}\) of fully dissolved calcium chloride, \(\ce{CaCl2}\).

Water is gradually evaporated until \(0.0200\text{ mol}\) of calcium chloride crystallises as a solid. The remaining solution contains \(0.0300\text{ mol}\) of dissolved calcium chloride.

Assume:

• No calcium chloride is lost from the container.
• The dissolved calcium chloride is completely dissociated.
• The solid crystal contains its normal ionic lattice.
• Water and solvent particles are not included in particle counts.

A student makes the following statement:

“After crystallisation, the solid contains \(0.0200N_A\) calcium chloride molecules and the solution contains \(0.0300N_A\) calcium chloride molecules. Therefore, the total number of solute particles remains \(0.0500N_A\).”

Answer the following:

a. Calculate the number of calcium ions and chloride ions in the solid crystal.

b. Calculate the number of calcium ions and chloride ions remaining in solution.

c. Calculate the total number of each type of ion in the container after crystallisation.

d. Determine whether the total number of ions has changed during crystallisation.

e. Evaluate the student’s statement. Explain what happens to the formula-unit ratios and the actual chemical particles as the salt changes from dissolved ions into a solid crystal.

Solution 5

The total number of calcium and chloride ions remains unchanged, even though the number of dissolved ions decreases.

The solid crystal contains \(1.20\times10^{22}\) calcium ions and \(2.41\times10^{22}\) chloride ions. The remaining solution contains \(1.81\times10^{22}\) calcium ions and \(3.61\times10^{22}\) chloride ions.

The key is to distinguish between the amount of calcium chloride, the number of ions, and the physical form in which those ions exist.

Part a: Ions in the solid crystal

The amount that crystallises is:

\[ n(\ce{CaCl2})=0.0200\text{ mol} \]

Each formula-unit ratio represents one calcium ion and two chloride ions.

Therefore, the amount of calcium ions in the solid is:

\[ n(\mathrm{Ca}^{2+})=0.0200\text{ mol} \]

The amount of chloride ions is:

\[ n(\mathrm{Cl}^{-})=2(0.0200) \]

\[ n(\mathrm{Cl}^{-})=0.0400\text{ mol} \]

Convert these amounts into numbers of ions.

Calcium ions:

\[ \begin{aligned} N_{\mathrm{Ca}^{2+}} &=0.0200\times6.02\times10^{23}\\ &=1.20\times10^{22} \end{aligned} \]

Chloride ions:

\[ \begin{aligned} N_{\mathrm{Cl}^{-}} &=0.0400\times6.02\times10^{23}\\ &=2.41\times10^{22} \end{aligned} \]

Part b: Ions remaining in solution

The amount of dissolved calcium chloride is \(0.0300\text{ mol}\).

Because it is completely dissociated:

\[ \ce{CaCl2(aq) -> Ca^2+(aq) + 2Cl-(aq)} \]

The amount of calcium ions is:

\[ n(\mathrm{Ca}^{2+})=0.0300\text{ mol} \]

The amount of chloride ions is:

\[ n(\mathrm{Cl}^{-})=2(0.0300) =0.0600\text{ mol} \]

Convert these into numbers of ions.

Calcium ions:

\[ \begin{aligned} N_{\mathrm{Ca}^{2+}} &=0.0300\times6.02\times10^{23}\\ &=1.81\times10^{22} \end{aligned} \]

Chloride ions:

\[ \begin{aligned} N_{\mathrm{Cl}^{-}} &=0.0600\times6.02\times10^{23}\\ &=3.61\times10^{22} \end{aligned} \]

Part c: Total number of each type of ion

Add the ions in the solid to those remaining in solution.

For calcium ions:

\[ \begin{aligned} N_{\mathrm{Ca}^{2+},\text{total}} &=1.20\times10^{22} +1.81\times10^{22}\\ &=3.01\times10^{22} \end{aligned} \]

For chloride ions:

\[ \begin{aligned} N_{\mathrm{Cl}^{-},\text{total}} &=2.41\times10^{22} +3.61\times10^{22}\\ &=6.02\times10^{22} \end{aligned} \]

Part d: Has the total number of ions changed?

No. The total number of ions has remained constant.

The container initially contained \(0.0500\text{ mol}\) of calcium chloride.

Therefore, its total calcium-ion amount was:

\[ n(\mathrm{Ca}^{2+})=0.0500\text{ mol} \]

Its total chloride-ion amount was:

\[ n(\mathrm{Cl}^{-})=0.100\text{ mol} \]

These amounts are unchanged after crystallisation.

The total number of ions is:

\[ \begin{aligned} N_{\text{ions}} &=(0.0500+0.100)N_A\\ &=0.150N_A\\ &=9.03\times10^{22} \end{aligned} \]

The ions have changed their arrangement, not disappeared.

Part e: Evaluating the student’s statement

The student’s conclusion is incorrect because neither phase contains separate calcium chloride molecules.

In the solid crystal, the ions form an extended ionic lattice. We can describe its composition using formula-unit ratios, but these are not physically separate \(\ce{CaCl2}\) molecules.

In the aqueous solution, calcium chloride has separated into hydrated calcium and chloride ions. There are no intact \(\ce{CaCl2}\) molecules in the assumed fully dissociated solution.

There is, however, an important quantity that remains unchanged:

\[ n_{\text{total}}(\ce{CaCl2})=0.0500\text{ mol} \]

This represents the total amount of calcium chloride, expressed through its formula-unit stoichiometry.

The corresponding number of formula-unit ratios is:

\[ \begin{aligned} N_{\mathrm{FU}} &=0.0500\times6.02\times10^{23}\\ &=3.01\times10^{22} \end{aligned} \]

This is a useful composition count, not a claim that those formula units exist as separate molecules in either phase.

The most important distinction is:

• The total amount of calcium chloride remains constant.
• The total number of calcium and chloride ions remains constant.
• The number of ions dispersed in the solution decreases.
• The ions transferred into the solid become part of an extended ionic lattice.

The tempting route is to treat formula units as permanent physical particles that survive every change of state. The actual constraint is that a formula unit describes composition, while the physical species and their arrangement depend on the phase.

10What to remember for HSC Chemistry

Use a formula unit when describing the composition of an ionic compound because its ions form an extended lattice rather than separate molecules.

The formula unit gives you the simplest whole-number ratio of ions. It also tells you how to convert between moles of an ionic compound and moles or numbers of its constituent ions.

For calculations, follow this sequence:

1. Identify the counted entity: formula units for an ionic solid, or molecules for a molecular substance.
2. Convert mass into moles using \(n=m/M\).
3. Convert moles into formula units using \(N=nN_A\).
4. Use the subscripts in the formula to calculate the number of each ion or atom.

Once you understand this distinction, you can apply the same stoichiometric reasoning to dissolution, precipitation, and calculations involving the total number of ions in a solution.