How to Convert Mass into Number of Molecules in HSC Chemistry

A balance tells you that a sample contains 5.50 g of carbon dioxide. But how many individual \(\ce{CO2}\) molecules are actually in that sample?

You can’t count them one by one. There are far too many. Instead, you use the sample’s mass to calculate its amount in moles, then convert moles into molecules using Avogadro’s constant.

Before calculating, make a prediction: would 5.50 g of water contain the same number of molecules as 5.50 g of carbon dioxide? No. Water molecules have a smaller mass, so the same total mass contains more of them.

The conversion method is:

\[ \boxed{N=\frac{m}{M}\times N_A} \]

In words:

1. Divide the sample’s mass by its molar mass to find the number of moles.
2. Multiply the number of moles by Avogadro’s constant to find the number of molecules.

The important part is identifying exactly what you’re counting. Let’s build the method properly.

01The three quantities you need to connect

Converting mass into molecules involves three quantities:

The missing link is the molar mass, \(M\), which tells you the mass of one mole of a substance.

For example, carbon dioxide has a molar mass of approximately \(44.01\ \mathrm{g\,mol^{-1}}\). This means that 44.01 g of carbon dioxide contains one mole of \(\ce{CO2}\) molecules.

The first calculation is therefore:

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

Where:

• \(n\) = amount of substance, in mol
• \(m\) = mass of the sample, in g
• \(M\) = molar mass, in \(\mathrm{g\,mol^{-1}}\)

Once you have the amount in moles, use Avogadro’s constant:

\[ N=nN_A \]

Where:

• \(N\) = number of molecules
• \(n\) = amount of substance, in mol
• \(N_A\) = Avogadro’s constant, \(6.02214076\times10^{23}\ \mathrm{mol^{-1}}\)

Combining these equations gives the complete conversion:

\[ \boxed{N=\frac{m}{M}N_A} \]

Why the units matter

You can check whether your calculation makes sense by following the units.

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

The grams cancel in the first fraction, leaving moles. The moles then cancel in the second multiplication.

Your final unit is molecules.

This is called dimensional analysis. It is a useful way to catch mistakes before they reach your calculator.

02Finding the molar mass

Before you can convert mass into moles, you need the correct molar mass.

For a molecular substance, calculate it by adding the relative atomic masses of every atom in its molecular formula.

Consider carbon dioxide, \(\ce{CO2}\).

Using the periodic table:

• Carbon: \(A_r=12.01\)
• Oxygen: \(A_r=16.00\)

Carbon dioxide contains one carbon atom and two oxygen atoms.

Therefore:

\[ \begin{aligned} M(\ce{CO2}) &=12.01+2(16.00)\\ &=44.01\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Notice that the oxygen subscript is 2. Forgetting to multiply the oxygen mass by two is an easy mistake.

The same method works for more complicated molecules.

For ethanol, \(\ce{C2H6O}\):

\[ \begin{aligned} M(\ce{C2H6O}) ={}&2(12.01)+6(1.008)\\ &+16.00\\ ={}&46.07\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The numerical value of the relative molecular mass, \(M_r\), is the same as the molar mass expressed in \(\mathrm{g\,mol^{-1}}\). However, they are different quantities: \(M_r\) has no units, whereas molar mass does.

A common mistake: using mass number instead of relative atomic mass

The mass number, \(A\), counts the protons and neutrons in one particular nucleus. It is not automatically the correct value to use as the molar mass.

For ordinary HSC calculations, use the relative atomic masses from the periodic table unless the question specifies particular isotopes.

For a refresher on the distinction, see Atomic Number and Mass Number Notation for HSC Chemistry.

03Why Avogadro’s constant exists

You might wonder where the enormous number \(6.02214076\times10^{23}\) comes from.

It is not an arbitrary number invented to make chemistry calculations difficult. It connects measurements we can make in a laboratory to the number of individual particles in a substance.

A mole works like a dozen. A dozen always means 12 specified objects, regardless of whether you’re counting eggs or pencils.

A mole is the same idea, but much larger:

\[ 1\ \mathrm{mol} = 6.02214076\times10^{23} \text{ specified entities} \]

Those entities could be atoms, molecules, ions, or other specified particles.

How scientists worked out the number

Counting individual atoms in a macroscopic sample is impractical. Even the most advanced optical microscopes cannot simply resolve and count every atom in a kilogram of material.

Scientists therefore developed an indirect counting method using the regular structure of crystals.

One important approach was the X-ray crystal density method, developed through international research on extremely pure silicon crystals.

The reasoning follows a particular sequence.

Step 1: Measure the sphere.

Researchers first prepared a highly purified, nearly perfect sphere of silicon-28.

The sphere was weighed using precision mass-measuring equipment. Its diameter was measured using optical interferometry, allowing its volume to be calculated.

Step 2: Measure the atomic arrangement.

Silicon atoms form a regular three-dimensional crystal lattice.

X-ray diffraction and interferometry can measure the spacing between atoms. From this spacing, scientists determine the dimensions of the repeating unit cell.

A conventional cubic unit cell of crystalline silicon contains eight atoms.

If the sphere’s volume is \(V_s\), and the volume of one unit cell is \(a^3\), where \(a\) is the unit-cell edge length, the estimated number of atoms is:

\[ N=\frac{V_s}{a^3}\times8 \]

This is the clever part. Instead of counting every atom, researchers count the repeating structural units and determine how many atoms belong to each one.

Step 3: Account for the isotopes.

The original natural-silicon approach faced an important limitation.

Natural silicon contains three stable isotopes: silicon-28, silicon-29, and silicon-30. These isotopes have different masses, so their proportions affect the sample’s molar mass.

A small uncertainty in isotope composition can affect the final calculation.

Later experiments addressed this by using highly enriched silicon-28, more accurate X-ray measurements, and improved mass spectrometry to determine isotope composition.

Step 4: Compare the atom count with the amount in moles.

Once researchers had an accurate estimate of the number of atoms, they could compare it with the amount of silicon calculated from its mass and molar mass.

The relationship is:

\[ N_A=\frac{N}{n} \]

This allowed researchers to determine the Avogadro constant.

The international silicon-sphere work contributed to establishing its modern value. Since 20 May 2019, the SI definition of the mole has fixed Avogadro’s constant at exactly:

\[ N_A=6.02214076\times10^{23}\ \mathrm{mol^{-1}} \]

This value is now exact by definition, rather than an experimentally measured value with uncertainty. You can read more about the modern definition at the Bureau International des Poids et Mesures (BIPM) and the NIST account of the silicon-sphere experiments.

A quick estimate of the scale

Imagine a 1.00 kg sphere made almost entirely of silicon-28.

For an illustrative calculation, take its molar mass as approximately \(28.0\ \mathrm{g\,mol^{-1}}\).

First, convert its mass into grams and calculate the amount:

\[ \begin{aligned} n&=\frac{1000}{28.0}\\ &=35.7\ \mathrm{mol} \end{aligned} \]

Now calculate the number of silicon atoms:

\[ \begin{aligned} N&=35.7\times6.022\times10^{23}\\ &=2.15\times10^{25}\text{ atoms} \end{aligned} \]

That’s approximately 21.5 septillion atoms in a single sphere.

The estimate uses the modern value of Avogadro’s constant to illustrate the scale; it is not the experimental procedure used to establish that constant.

04Check what you’re actually counting

There is one more important detail.

The equation \(N=nN_A\) gives you the number of whichever entities your amount in moles represents.

It does not automatically give you the number of molecules.

Consider these examples:

The important contrast: molecules versus formula units

Take solid sodium chloride, \(\ce{NaCl}\).

It is tempting to describe the solid as containing individual sodium chloride molecules. However, sodium chloride is an ionic compound.

Its structure is an extended lattice of alternating positive sodium ions and negative chloride ions.

A formula unit of \(\ce{NaCl}\) represents the simplest whole-number ratio of these ions: one \(\mathrm{Na}^{+}\) ion to one \(\mathrm{Cl}^{-}\) ion.

Suppose you have \(0.100\) mol of sodium chloride.

You have:

\[ 0.100\times6.022\times10^{23} \]

\[ =6.02\times10^{22} \text{ formula units} \]

Since each formula unit contains one sodium ion and one chloride ion, the total number of ions is:

\[ 2\times6.02\times10^{22} \]

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

The number of formula units and the total number of ions are different.

This distinction becomes important whenever a question asks for the number of atoms, ions, or molecules rather than simply the number of moles.

When should you multiply by the number of atoms?

Suppose you have one mole of carbon dioxide molecules.

Each \(\ce{CO2}\) molecule contains:

• One carbon atom.
• Two oxygen atoms.
• Three atoms altogether.

Therefore, one mole of \(\ce{CO2}\) contains:

• One mole of carbon atoms.
• Two moles of oxygen atoms.
• Three moles of atoms altogether.

If the question asks for molecules, do not multiply by three.

If it asks for total atoms, you must.

05A reliable method for HSC calculations

When you see a question asking you to convert mass into a number of molecules, use this sequence.

A few checks will help you avoid common errors:

• Always calculate molar mass from the complete formula.
• Keep mass and molar mass in consistent units.
• Use \(6.022\times10^{23}\), not \(6.022\times10^{-23}\).
• Do not confuse the number of molecules with the number of atoms inside those molecules.
• If the sample is a mixture, calculate the amount of each substance separately before adding the relevant particle counts.
• If the substance is ionic, use formula units rather than molecules.

Let’s put the method into practice.

06Worked examples

07Questions and solutions

Try each question before reading its solution. The questions move from a straightforward conversion to problems involving multiple substances and changing molecular structures.

Question 1

A student measures 3.60 g of pure water.

Calculate the number of water molecules in the sample.

Use \(M(\ce{H2O})=18.02\ \mathrm{g\,mol^{-1}}\).

Solution 1

The sample contains \(1.20\times10^{23}\) water molecules.

First, calculate the amount of water:

\[ \begin{aligned} n&=\frac{m}{M}\\ &=\frac{3.60}{18.02}\\ &=0.1998\ \mathrm{mol} \end{aligned} \]

Next, convert the amount into molecules:

\[ \begin{aligned} N&=nN_A\\ &=0.1998\times6.022\times10^{23}\\ &=1.20\times10^{23}\text{ molecules} \end{aligned} \]

The final answer is:

\[ \boxed{1.20\times10^{23}\text{ molecules}} \]

The important point is that the question asks for water molecules, not individual atoms. There is no need to multiply by three, even though each water molecule contains three atoms.

Question 2

A pure ethanol sample has a mass of 2.30 g.

Ethanol has the molecular formula \(\ce{C2H6O}\).

Calculate:

a) The number of ethanol molecules.

b) The number of hydrogen atoms in the sample.

Use \(A_r(\ce{C})=12.01\), \(A_r(\ce{H})=1.008\), and \(A_r(\ce{O})=16.00\).

Solution 2

The sample contains \(3.01\times10^{22}\) ethanol molecules and \(1.80\times10^{23}\) hydrogen atoms.

Start by calculating the molar mass:

\[ \begin{aligned} M(\ce{C2H6O}) ={}&2(12.01)+6(1.008)\\ &+16.00\\ ={}&46.07\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Part a: Number of ethanol molecules

Calculate the amount:

\[ \begin{aligned} n&=\frac{2.30}{46.07}\\ &=0.04992\ \mathrm{mol} \end{aligned} \]

Convert into molecules:

\[ \begin{aligned} N&=0.04992\times6.022\times10^{23}\\ &=3.01\times10^{22}\text{ molecules} \end{aligned} \]

Part b: Number of hydrogen atoms

Each ethanol molecule contains six hydrogen atoms.

Therefore:

\[ \begin{aligned} N_{\mathrm H} &=6\times3.01\times10^{22}\\ &=1.80\times10^{23}\text{ atoms} \end{aligned} \]

The final answers are:

• a) \(3.01\times10^{22}\) ethanol molecules.
• b) \(1.80\times10^{23}\) hydrogen atoms.

The distinction is important: the number of molecules counts complete ethanol structures, whereas the number of hydrogen atoms counts only one particular type of atom within those structures.

Question 3

A student has 5.84 g of solid sodium chloride, \(\ce{NaCl}\).

Calculate:

a) The number of formula units of sodium chloride.

b) The total number of ions in the sample.

Use \(M(\ce{NaCl})=58.44\ \mathrm{g\,mol^{-1}}\).

Solution 3

The sample contains \(6.02\times10^{22}\) formula units and \(1.20\times10^{23}\) ions in total.

The first important decision is recognising that sodium chloride is ionic. We count formula units, not molecules.

Part a: Number of formula units

Calculate the amount of sodium chloride:

\[ \begin{aligned} n&=\frac{5.84}{58.44}\\ &=0.09993\ \mathrm{mol} \end{aligned} \]

Convert into formula units:

\[ \begin{aligned} N&=0.09993\times6.022\times10^{23}\\ &=6.02\times10^{22} \end{aligned} \]

Therefore:

\[ \boxed{6.02\times10^{22}\text{ formula units}} \]

Part b: Total number of ions

Each formula unit of \(\ce{NaCl}\) contains:

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

This means each formula unit contains two ions altogether.

Therefore:

\[ \begin{aligned} N_{\text{ions}} &=2\times6.02\times10^{22}\\ &=1.20\times10^{23}\text{ ions} \end{aligned} \]

The final answer is:

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

There are \(6.02\times10^{22}\) sodium ions and the same number of chloride ions.

The trap here is calling the formula units molecules. The numerical calculation is similar, but the chemical entity must be named correctly.

Question 4

A sealed container holds a mixture of two gases:

• 2.20 g of carbon dioxide, \(\ce{CO2}\).
• 3.20 g of oxygen gas, \(\ce{O2}\).

Assume the gases do not react.

Calculate:

a) The total number of gas molecules in the container.

b) The total number of oxygen atoms in the container.

Use:

• \(M(\ce{CO2})=44.01\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{O2})=32.00\ \mathrm{g\,mol^{-1}}\)

Solution 4

The container holds \(9.03\times10^{22}\) gas molecules and \(1.81\times10^{23}\) oxygen atoms.

The key is to calculate the amount of each gas separately. You cannot use one molar mass for the entire mixture because carbon dioxide and oxygen have different molar masses.

Part a: Total number of gas molecules

First, calculate the amount of carbon dioxide:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{2.20}{44.01}\\ &=0.0500\ \mathrm{mol} \end{aligned} \]

Now calculate the amount of oxygen:

\[ \begin{aligned} n(\ce{O2}) &=\frac{3.20}{32.00}\\ &=0.100\ \mathrm{mol} \end{aligned} \]

Add the amounts because both substances contain molecules:

\[ \begin{aligned} n_{\text{total}} &=0.0500+0.100\\ &=0.150\ \mathrm{mol} \end{aligned} \]

Convert into molecules:

\[ \begin{aligned} N_{\text{total}} &=0.150\times6.022\times10^{23}\\ &=9.03\times10^{22}\text{ molecules} \end{aligned} \]

Part b: Total number of oxygen atoms

Now we need to count oxygen atoms rather than molecules.

Each carbon dioxide molecule contains two oxygen atoms.

Each oxygen molecule also contains two oxygen atoms.

Therefore, the amount of oxygen atoms is:

\[ \begin{aligned} n_{\mathrm O} &=2(0.0500)+2(0.100)\\ &=0.300\ \mathrm{mol} \end{aligned} \]

Convert into atoms:

\[ \begin{aligned} N_{\mathrm O} &=0.300\times6.022\times10^{23}\\ &=1.81\times10^{23}\text{ atoms} \end{aligned} \]

The final answers are:

• a) \(9.03\times10^{22}\) gas molecules.
• b) \(1.81\times10^{23}\) oxygen atoms.

Notice that the two answers are different even though both calculations use the same Avogadro constant.

This is because one calculation counts complete molecules, while the other counts the oxygen atoms contained within them.

Question 5

Nitrogen dioxide, \(\ce{NO2}\), can react with itself to form dinitrogen tetroxide, \(\ce{N2O4}\):

\[ \ce{2NO2 <=> N2O4} \]

A sealed vessel initially receives 9.20 g of pure nitrogen dioxide. At equilibrium, 40.0% of all the original \(\ce{NO2}\) composition units have become incorporated into \(\ce{N2O4}\) molecules. The remaining 60.0% are still present as separate \(\ce{NO2}\) molecules.

For this question, treat one \(\ce{NO2}\) composition unit as a bookkeeping unit. Each \(\ce{N2O4}\) molecule contains two of these units.

Assume that:

• No material escapes from the vessel.
• No other chemical species form.
• The stated percentages refer to the original number of \(\ce{NO2}\) composition units, not the percentage of molecules.

Calculate:

a) The total number of gas molecules at equilibrium.

b) The number of \(\ce{N2O4}\) molecules at equilibrium.

Use \(M(\ce{NO2})=46.01\ \mathrm{g\,mol^{-1}}\).

Solution 5

At equilibrium, the vessel contains \(9.63\times10^{22}\) gas molecules in total, of which \(2.41\times10^{22}\) are \(\ce{N2O4}\) molecules.

The important detail is that the mass of the sample remains constant, but the number of molecules changes when two \(\ce{NO2}\) molecules combine to form one \(\ce{N2O4}\) molecule.

A direct mass-to-molecules calculation using \(M(\ce{NO2})\) would give the number of original \(\ce{NO2}\) composition units. It would not give the final number of molecules.

Step 1: Calculate the original amount of \(\ce{NO2}\) composition units.

\[ \begin{aligned} n_{\text{units}} &=\frac{9.20}{46.01}\\ &=0.200\ \mathrm{mol} \end{aligned} \]

This represents 0.200 mol of \(\ce{NO2}\) composition units, regardless of whether those units are eventually found in monomers or dimers.

Step 2: Calculate the amount of units incorporated into \(\ce{N2O4}\).

The question states that 40.0% of the units form dimers.

\[ \begin{aligned} n_{\text{dimerised units}} &=0.400\times0.200\\ &=0.0800\ \mathrm{mol} \end{aligned} \]

However, each \(\ce{N2O4}\) molecule contains two \(\ce{NO2}\) units.

Therefore, divide by two:

\[ \begin{aligned} n(\ce{N2O4}) &=\frac{0.0800}{2}\\ &=0.0400\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the amount of remaining \(\ce{NO2}\) molecules.

The remaining 60.0% of the units stay as individual \(\ce{NO2}\) molecules.

\[ \begin{aligned} n(\ce{NO2}) &=0.600\times0.200\\ &=0.120\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the total amount of molecules.

Now add the amounts of the two different molecular species:

\[ \begin{aligned} n_{\text{total molecules}} &=0.0400+0.120\\ &=0.160\ \mathrm{mol} \end{aligned} \]

Convert into the total number of molecules:

\[ \begin{aligned} N_{\text{total}} &=0.160\times6.022\times10^{23}\\ &=9.63\times10^{22}\text{ molecules} \end{aligned} \]

Step 5: Calculate the number of \(\ce{N2O4}\) molecules.

We already know that:

\[ n(\ce{N2O4})=0.0400\ \mathrm{mol} \]

Therefore:

\[ \begin{aligned} N(\ce{N2O4}) &=0.0400\times6.022\times10^{23}\\ &=2.41\times10^{22}\text{ molecules} \end{aligned} \]

The final answers are:

• a) \(9.63\times10^{22}\) total gas molecules.
• b) \(2.41\times10^{22}\) \(\ce{N2O4}\) molecules.

Why is the tempting shortcut wrong?

You might calculate:

\[ 0.200\times6.022\times10^{23} =1.20\times10^{23} \]

That calculation is mathematically correct, but it counts the original \(\ce{NO2}\) composition units as though each one remains a separate molecule.

They don’t.

Some of those units have paired up. Two original units now belong to one \(\ce{N2O4}\) molecule.

The total mass and total number of nitrogen dioxide composition units are conserved, but the number of individual molecules decreases.

That is the crucial difference between counting atoms or composition units and counting complete molecules.

The mass-to-molecules calculation is the starting point for a much broader HSC Chemistry skill: stoichiometry.

Once you can convert between mass, moles, and particles, you can use balanced chemical equations to predict how many molecules react, how many are produced, and how much of each substance is required. The key is to identify the entity being counted at every stage, particularly when a chemical reaction changes the molecular structure.