How to Identify the Correct Particle Before Using N = nNA

You dissolve 5.85 g of sodium chloride in water. You also dissolve 18.0 g of glucose in another container. Both samples contain 0.100 mol of solute. Does that mean they contain the same number of dissolved particles?

No. Sodium chloride separates into positive and negative ions, while glucose remains as individual molecules. The sodium chloride produces 0.200 mol of ions, whereas the glucose produces 0.100 mol of molecules.

The difference comes down to one decision: what exactly are you counting?

Before using Avogadro’s constant, identify the particle your question asks for. It might be an atom, molecule, ion, or formula unit. Getting this wrong can give you a perfectly calculated answer to the wrong question.

01The rule: identify the particle before using N = nNA

The equation is:

\[ N=nN_A \]

Where:

• \(N\) = number of particles (the actual entities being counted).
• \(n\) = amount of those particles, measured in moles.
• \(N_A\) = Avogadro’s constant, approximately \(6.02\times10^{23}\ \mathrm{mol^{-1}}\).

The important detail is that \(N_A\) does not decide which particle you are counting.

It simply converts an amount in moles into a number of the specified entities.

For example:

• 1 mol of water contains \(6.02\times10^{23}\) water molecules.
• 1 mol of oxygen atoms contains \(6.02\times10^{23}\) oxygen atoms.
• 1 mol of sodium ions contains \(6.02\times10^{23}\) sodium ions.

These are different particles, even though the conversion uses the same constant.

Four common particles in HSC Chemistry

There is an important distinction between molecules and formula units.

A water molecule is an actual group of three bonded atoms. An ionic solid such as sodium chloride does not contain separate \(\ce{NaCl}\) molecules. Instead, it contains an extended lattice of positive and negative ions.

We use a formula unit to represent the simplest ratio of those ions. For sodium chloride, that ratio is one sodium ion to one chloride ion.

How do you know which particle the question wants?

Look closely at the noun in the question.

Notice that the chemical formula alone does not always tell you the final particle to count.

The wording of the question matters just as much.

02What happens when you actually dissolve a substance?

Consider the sodium chloride and glucose samples from the opening problem.

We can use a simple conductivity investigation to understand why their particle counts differ.

The experimental setup

Imagine preparing two solutions:

• Sample A: 5.85 g of sodium chloride.
• Sample B: 18.0 g of glucose.

Dissolve each sample in water and make each solution up to 1.00 L in a separate volumetric flask.

Then insert a conductivity probe into each solution.

The masses are illustrative values chosen to give equal amounts in moles. The conductivity results below describe the expected qualitative behaviour, not measurements from an experiment claimed to have been performed.

Step 1: Predict what should happen

Calculate the amount of each substance.

For sodium chloride, the molar mass is \(58.5\ \mathrm{g\,mol^{-1}}\).

\[ n=\frac{m}{M} =\frac{5.85}{58.5} =0.100\ \mathrm{mol} \]

For glucose, \(\ce{C6H12O6}\), the molar mass is \(180.0\ \mathrm{g\,mol^{-1}}\).

\[ n=\frac{m}{M} =\frac{18.0}{180.0} =0.100\ \mathrm{mol} \]

Both samples contain 0.100 mol of their respective starting entities.

You might predict that they will contain the same number of dissolved particles.

But that prediction assumes that every solute remains as one particle when it dissolves.

That assumption is incorrect for sodium chloride.

Step 2: Follow what happens to each substance

Sodium chloride is an ionic compound. When it dissolves, its ions separate:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

One formula unit produces:

• One sodium ion.
• One chloride ion.

Therefore, each mole of sodium chloride produces two moles of dissolved ions, assuming complete dissociation.

Glucose behaves differently. It is a molecular substance and dissolves without separating into ions.

Its molecules remain intact:

\[ \ce{C6H12O6(s) -> C6H12O6(aq)} \]

One glucose molecule remains one dissolved glucose molecule.

Step 3: Interpret the conductivity result

The expected observations are:

• Sodium chloride solution conducts electricity because it contains mobile charged ions.
• Glucose solution has negligible electrical conductivity because its dissolved molecules are neutral.

The experiment supports the idea that sodium chloride produces mobile ions while glucose remains molecular.

However, conductivity does not directly measure the total number of particles. It also depends on factors such as ion charge, mobility, and temperature.

To calculate the actual particle count, we need the chemical formula and the amount of substance.

Step 4: Calculate the actual number of particles

For sodium chloride:

\[ n(\text{total ions})=2(0.100) =0.200\ \mathrm{mol} \]

\[ N=0.200\times6.02\times10^{23} \]

\[ \boxed{N=1.20\times10^{23}\text{ ions}} \]

For glucose:

\[ N=0.100\times6.02\times10^{23} \]

\[ \boxed{N=6.02\times10^{22}\text{ molecules}} \]

The two samples began with the same amount of solute, but they produced different amounts of dissolved particles.

This is why you must identify the particle before applying Avogadro’s constant.

03A reliable method for identifying the correct particle

When a question involves mass, moles, and particle numbers, use this sequence.

Step 1: Underline the particle being requested

Look for words such as:

• Atoms
• Molecules
• Ions
• Formula units
• Total particles

Do not start calculating until you know what the final answer must count.

Step 2: Identify the chemical substance

Determine whether the substance is:

• An element made of individual atoms, such as iron.
• A molecular substance, such as water.
• An ionic compound, such as calcium chloride.
• An elemental molecule, such as oxygen gas.

For ionic compounds, check whether the question concerns the solid compound or its dissolved ions.

Step 3: Calculate the amount of the compound

When mass is supplied, use:

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

Where:

• \(n\) is the amount in moles.
• \(m\) is the mass in grams.
• \(M\) is the molar mass in \(\mathrm{g\,mol^{-1}}\).

Check your units:

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

The amount you calculate initially refers to the substance represented by its chemical formula.

For a molecular substance, this is usually moles of molecules.

For an ionic compound, it is moles of formula units.

Step 4: Convert to the requested particle

Use the chemical formula to find how many of the requested particles correspond to one molecule or formula unit.

For example:

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

One formula unit produces:

• One calcium ion.
• Two chloride ions.
• Three ions in total.

Therefore:

\[ n(\mathrm{Cl}^{-})=2n(\ce{CaCl2}) \]

And:

\[ n(\text{total ions})=3n(\ce{CaCl2}) \]

This multiplication is essential when the requested particle differs from the entity used to calculate the original amount.

Step 5: Apply Avogadro’s constant

Once the amount is expressed in moles of the correct particle, use:

\[ N=nN_A \]

The units should cancel correctly:

\[ \mathrm{mol}\times \mathrm{entities\,mol^{-1}} =\mathrm{entities} \]

A useful combined expression is:

\[ N_{\text{target}}=r\times n_{\text{compound}}\times N_A \]

Here, \(r\) is the number of target particles represented by one molecule or formula unit.

If the target is the original molecule or formula unit, \(r=1\).

If you are counting all ions from dissolved calcium chloride, \(r=3\).

Three mistakes worth avoiding

Mistake 1: Counting molecules when the question asks for atoms.

One molecule of water contains three atoms: two hydrogen atoms and one oxygen atom.

One mole of water molecules therefore contains three moles of atoms in total.

Mistake 2: Counting formula units when the question asks for ions.

One formula unit of calcium chloride produces three ions when fully dissociated.

The total number of ions is therefore three times the number of formula units.

Mistake 3: Treating a polyatomic ion as several ions.

A sulfate ion, \(\mathrm{SO_{4}}^{2-}\), contains five atoms, but it is still one ion.

Likewise, a calcium ion, \(\mathrm{Ca}^{2+}\), is one ion, not two. Its charge does not change the number of particles.

04Worked examples

05Questions and solutions

The following questions progress from straightforward particle counting to situations where you need to distinguish between different counting units, chemical species, and physical states.

Use \(N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}\), unless otherwise stated.

Question 1

A sealed container holds 4.00 g of oxygen gas, \(\ce{O2}\).

Calculate the number of oxygen molecules in the container.

Solution 1

There are \(7.53\times10^{22}\) oxygen molecules in the container.

The question asks for molecules, not individual oxygen atoms. Each molecule of oxygen gas contains two bonded oxygen atoms, but we only need to count the molecules here.

Step 1: Calculate the molar mass of oxygen gas.

\[ M(\ce{O2})=2(16.0) =32.0\ \mathrm{g\,mol^{-1}} \]

Step 2: Calculate the amount of oxygen molecules.

\[ n=\frac{m}{M} =\frac{4.00}{32.0} =0.125\ \mathrm{mol} \]

Step 3: Apply Avogadro’s constant.

\[ N=nN_A \]

\[ =0.125\times6.02\times10^{23} \]

\[ \boxed{N=7.53\times10^{22}\text{ molecules}} \]

A common mistake would be to multiply the amount by two before using Avogadro’s constant. That would calculate oxygen atoms, not oxygen molecules.

If the question had asked for oxygen atoms, the answer would be twice as large.

Question 2

A student dissolves 11.11 g of anhydrous calcium chloride, \(\ce{CaCl2}\), in water and makes the solution up to 250.0 mL.

A 25.00 mL sample is removed using a pipette.

Assume complete dissociation and a uniform solution.

Calculate:

(a) The number of chloride ions in the 25.00 mL sample.

(b) The total number of ions remaining in the original flask.

Solution 2

The 25.00 mL sample contains \(1.20\times10^{22}\) chloride ions, and the remaining flask contains \(1.63\times10^{23}\) ions in total.

The key is to calculate the original amount first, then account for the fraction removed. We must also distinguish chloride ions from all ions.

Step 1: Calculate the molar mass of calcium chloride.

\[ M(\ce{CaCl2})=40.1+2(35.5) \]

\[ =111.1\ \mathrm{g\,mol^{-1}} \]

Step 2: Calculate the original amount of calcium chloride.

\[ n=\frac{11.11}{111.1} =0.1000\ \mathrm{mol} \]

This is the original amount of calcium chloride formula units.

Step 3: Calculate the fraction removed.

\[ \text{Fraction removed} =\frac{25.00}{250.0} =0.1000 \]

Therefore, the amount of calcium chloride in the aliquot is:

\[ n=0.1000\times0.1000 \]

\[ =0.01000\ \mathrm{mol} \]

Step 4: Calculate the amount of chloride ions in the aliquot.

The dissociation equation is:

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

Each formula unit produces two chloride ions.

\[ n(\mathrm{Cl}^{-})=2(0.01000) \]

\[ =0.02000\ \mathrm{mol} \]

Step 5: Calculate the number of chloride ions.

\[ N=0.02000\times6.02\times10^{23} \]

\[ \boxed{N=1.20\times10^{22}\text{ chloride ions}} \]

Step 6: Calculate the amount of calcium chloride remaining.

The remaining volume is:

\[ 250.0-25.00=225.0\ \mathrm{mL} \]

The fraction remaining is 0.9000.

\[ n(\ce{CaCl2})=0.1000\times0.9000 \]

\[ =0.09000\ \mathrm{mol} \]

Step 7: Calculate the amount of total ions remaining.

Each formula unit produces three ions in total.

\[ n(\text{total ions})=3(0.09000) \]

\[ =0.2700\ \mathrm{mol} \]

Step 8: Calculate the total number of remaining ions.

\[ N=0.2700\times6.02\times10^{23} \]

\[ \boxed{N=1.63\times10^{23}\text{ ions}} \]

The important detail is that the aliquot removes 10% of every dissolved species, not just 10% of the calcium ions or chloride ions. The remaining solution therefore retains 90% of the original amount of each ion.

Question 3

A 9.00 g sample of glucose, \(\ce{C6H12O6}\), is dissolved in water.

Calculate:

(a) The number of oxygen atoms present in the glucose sample.

(b) The total number of atoms represented by all the glucose molecules in the sample.

Assume glucose dissolves without breaking apart or reacting.

Solution 3

The sample contains \(1.81\times10^{23}\) oxygen atoms and \(7.22\times10^{23}\) atoms in total.

We must first count glucose molecules, then use the molecular formula to convert to atoms.

Step 1: Calculate the molar mass of glucose.

\[ M(\ce{C6H12O6}) =6(12.0)+12(1.0)+6(16.0) \]

\[ =72.0+12.0+96.0 \]

\[ =180.0\ \mathrm{g\,mol^{-1}} \]

Step 2: Calculate the amount of glucose molecules.

\[ n=\frac{9.00}{180.0} \]

\[ =0.0500\ \mathrm{mol} \]

Step 3: Calculate the amount of oxygen atoms.

Each glucose molecule contains six oxygen atoms.

Therefore:

\[ n(\text{O atoms})=6(0.0500) \]

\[ =0.300\ \mathrm{mol} \]

Step 4: Calculate the number of oxygen atoms.

\[ N=0.300\times6.02\times10^{23} \]

\[ \boxed{N=1.81\times10^{23}\text{ oxygen atoms}} \]

Step 5: Calculate the total number of atoms per glucose molecule.

Each molecule contains:

\[ 6+12+6=24\text{ atoms} \]

Therefore, the amount of total atoms is:

\[ n(\text{total atoms})=24(0.0500) \]

\[ =1.20\ \mathrm{mol} \]

Step 6: Calculate the total number of atoms.

\[ N=1.20\times6.02\times10^{23} \]

\[ \boxed{N=7.22\times10^{23}\text{ atoms}} \]

Notice that the number of molecules is not the same as the number of atoms. Each glucose molecule contains 24 atoms, so the total atom count is 24 times the molecule count.

Question 4

A solution is prepared by dissolving:

• 5.85 g of sodium chloride, \(\ce{NaCl}\).
• 9.00 g of glucose, \(\ce{C6H12O6}\).

Assume sodium chloride dissociates completely and glucose remains as intact molecules.

Calculate:

(a) The total number of dissolved solute particles.

(b) The percentage of these solute particles that are ions.

Do not include water molecules in your answer.

Solution 4

The solution contains \(1.51\times10^{23}\) dissolved solute particles, of which 80.0% are ions.

The tempting approach is to add the two original mole amounts and multiply by Avogadro’s constant. That would be incorrect because sodium chloride produces two ions per formula unit.

Step 1: Calculate the amount of sodium chloride.

\[ n(\ce{NaCl})=\frac{5.85}{58.5} \]

\[ =0.100\ \mathrm{mol} \]

Step 2: Calculate the amount of ions produced by sodium chloride.

The dissociation equation is:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

Each formula unit produces two ions.

\[ n(\text{ions})=2(0.100) \]

\[ =0.200\ \mathrm{mol} \]

Step 3: Calculate the amount of glucose molecules.

\[ n(\ce{C6H12O6})=\frac{9.00}{180.0} \]

\[ =0.0500\ \mathrm{mol} \]

Glucose remains molecular, so it produces 0.0500 mol of dissolved glucose molecules.

Step 4: Calculate the total amount of dissolved solute particles.

Add the amounts of the actual dissolved particles:

\[ n(\text{total particles}) =0.200+0.0500 \]

\[ =0.250\ \mathrm{mol} \]

Step 5: Calculate the total number of particles.

\[ N=0.250\times6.02\times10^{23} \]

\[ \boxed{N=1.51\times10^{23}\text{ particles}} \]

Step 6: Calculate the percentage that are ions.

The amount of ions is 0.200 mol, while the total amount of solute particles is 0.250 mol.

\[ \text{Percentage ions} =\frac{0.200}{0.250}\times100 \]

\[ \boxed{80.0\%} \]

The crucial step was counting the ions produced by sodium chloride rather than treating each original formula unit as one dissolved particle.

Question 5

A chemist takes a 0.100 mol sample of solid sodium chloride and dissolves it completely in water.

The solution is divided into two equal portions.

• Portion A remains as an aqueous sodium chloride solution.
• Portion B is evaporated to dryness, and the sodium chloride crystals are recovered without material loss.

A student makes the following claim:

“The aqueous portion contains twice as many particles as the solid portion because every sodium chloride formula unit separates into two ions when dissolved.”

Evaluate this claim.

Calculate the number of sodium ions, chloride ions, and total ions in each portion. Then explain why the two portions have different electrical conductivity, even if their total numbers of ions are the same.

Assume that the recovered solid is pure sodium chloride and that no chemical reactions occur.

Solution 5

The student’s claim is incorrect: both portions contain the same total number of sodium and chloride ions, namely \(6.02\times10^{22}\) ions in each portion.

The mistake is comparing formula units in the solid with ions in the solution as though the solid did not already contain ions.

This is a subtle but important distinction.

Step 1: Calculate the amount of sodium chloride in each portion.

The original sample contains 0.100 mol of sodium chloride.

It is divided equally, so each portion contains the equivalent of:

\[ n(\ce{NaCl})=\frac{0.100}{2} \]

\[ =0.0500\ \mathrm{mol} \]

Step 2: Calculate the amount of sodium ions in each portion.

The sodium chloride formula gives a 1:1 ratio of sodium ions to chloride ions.

Therefore:

\[ n(\mathrm{Na}^{+})=0.0500\ \mathrm{mol} \]

Step 3: Calculate the amount of chloride ions in each portion.

Similarly:

\[ n(\mathrm{Cl}^{-})=0.0500\ \mathrm{mol} \]

Step 4: Calculate the total amount of ions.

\[ n(\text{total ions})=0.0500+0.0500 \]

\[ =0.100\ \mathrm{mol} \]

Step 5: Calculate the number of each type of ion.

For sodium ions:

\[ N(\mathrm{Na}^{+})=0.0500\times6.02\times10^{23} \]

\[ \boxed{N(\mathrm{Na}^{+})=3.01\times10^{22}} \]

For chloride ions:

\[ N(\mathrm{Cl}^{-})=0.0500\times6.02\times10^{23} \]

\[ \boxed{N(\mathrm{Cl}^{-})=3.01\times10^{22}} \]

Step 6: Calculate the total number of ions.

\[ N(\text{total ions}) =0.100\times6.02\times10^{23} \]

\[ \boxed{N(\text{total ions})=6.02\times10^{22}} \]

These results apply to both portions.

Step 7: Explain the difference in conductivity.

Here is the conceptual trap.

In solid sodium chloride, the sodium and chloride ions are already present. They are arranged in a repeating ionic lattice.

The ions cannot move freely through the solid, so solid sodium chloride does not conduct electricity at room temperature.

In aqueous sodium chloride, the ions separate and become mobile in the water. These mobile charged particles allow the solution to conduct electricity.

The total number of ions has not increased. Their arrangement and ability to move have changed.

The formula unit is simply a way to represent the 1:1 ratio of ions in the solid. It is not an additional particle that disappears and creates two new ions.

This is why identifying the particle matters so much. A question about formula units, a question about ions, and a question about mobile charge carriers can describe the same substance but require different reasoning.

06Where to go next

Once you can identify the correct particle, the next challenge is connecting the amount of that particle to the concentration of a solution.

The relationship is:

\[ c=\frac{n}{V} \]

The same rule applies: make sure the amount in moles refers to the chemical species whose concentration the question asks for.

Continue with Molarity Calculations for HSC Chemistry: c = n/V to practise connecting moles and solution volume.

For more practice converting ionic formula units into individual ions, see How to Calculate the Total Number of Ions in a Dissolved Salt.