Why Avogadro’s Constant Is the Same for Every Substance

Take 18.0 g of water and 58.5 g of sodium chloride. The samples have very different masses, and their particles are different kinds of entities. Yet each contains approximately 1.00 mol of its specified particles: the water contains about \(6.022 \times 10^{23}\) water molecules, and the sodium chloride contains the same number of formula units.

Why does the same enormous number apply to both substances?

Avogadro’s constant does not depend on the substance because it defines the relationship between the amount of a substance and the number of specified entities. It is a universal counting conversion, not a physical property of water, sodium chloride, or any other chemical.

The important distinction is that the number of particles in one mole is fixed, but the mass of one mole depends on the substance.

01The number stays the same, even when the substance changes

Before looking at the mathematics, make a prediction.

Suppose you have 0.500 mol of helium and 0.500 mol of water. Which sample contains more particles?

Neither. Both contain exactly the same number of their specified entities:

\[ N = 0.500 \times 6.02214076 \times 10^{23} \]

\[ N = 3.011 \times 10^{23} \]

The helium sample contains that many helium atoms. The water sample contains that many water molecules.

This is the central idea: one mole always represents the same number of specified entities, regardless of their chemical identity.

The current International System of Units (SI) defines one mole as exactly \(6.02214076 \times 10^{23}\) elementary entities. Consequently, the Avogadro constant is exactly:

\[ \boxed{N_A = 6.02214076 \times 10^{23}\ \text{mol}^{-1}} \]

The unit \(\text{mol}^{-1}\) means per mole. The constant tells you how many specified entities correspond to one mole.

This is not an approximate value that changes when you switch from helium to water. It is a fixed numerical value in the definition of the mole. The confirms this.

What exactly is being counted?

There is one detail you must get right: the entity being counted.

Notice that a mole of water molecules is not the same thing as a mole of water’s individual atoms.

Each water molecule contains two hydrogen atoms and one oxygen atom. Therefore, 1 mol of water molecules contains:

• 1 mol of water molecules;
• 2 mol of hydrogen atoms;
• 1 mol of oxygen atoms;
• 3 mol of atoms altogether.

The number of molecules is fixed by Avogadro’s constant. The number of atoms depends on how many atoms are contained in each molecule.

If you need a broader refresher on identifying particles and converting between moles and particle counts, see The Mole and Avogadro Constant: HSC Chemistry Guide.

02How did scientists establish such an enormous number?

It is reasonable to ask how anyone could know that one mole contains approximately \(6.022 \times 10^{23}\) entities. Nobody counted that many atoms individually.

Instead, scientists developed ways to estimate the number of atoms in a measurable sample. One particularly important approach used highly purified silicon crystals.

This work helped establish the numerical value of Avogadro’s constant before its modern definition made that value exact.

The silicon crystal experiment

The problem was to connect something measurable in a laboratory, such as the mass of a solid, to the number of atoms inside it.

A silicon crystal provided a useful solution because its atoms occupy a highly regular, repeating arrangement. If scientists could measure the spacing between atoms and the total volume of the crystal, they could calculate how many atoms it contained.

The experimental arrangement involved several measurements:

1. A nearly perfect sphere was manufactured from highly purified silicon, enriched in the isotope silicon-28.
2. Its mass was measured using a precision balance.
3. Optical interferometry was used to measure its diameter, allowing its volume to be calculated.
4. X-ray interferometry was used to measure the spacing of atoms in the crystal lattice.
5. Mass spectrometry helped determine the isotopic composition, while corrections accounted for impurities and surface layers.

The key was the crystal’s repeating structure.

A conventional cubic unit cell of crystalline silicon contains eight atoms. Once scientists knew the unit cell dimensions, they could calculate the volume occupied by one unit cell.

They could then estimate the total number of atoms by dividing the sphere’s volume by the unit cell volume and multiplying by eight.

The counting relationship was:

\[ N = \frac{8V_s}{a^3} \]

where:

• \(N\) is the number of silicon atoms;
• \(V_s\) is the volume of the silicon crystal;
• \(a\) is the edge length of the cubic unit cell;
• 8 is the number of atoms per conventional unit cell.

The equation assumes a perfect crystal. In the real experiment, scientists also had to correct for impurities and isotopic composition.

What should this experiment reveal?

If scientists could determine the number of atoms in a sample and independently establish its chemical amount in moles, they could calculate the Avogadro constant:

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

The prediction was that the resulting value would be a universal constant, not a value unique to silicon.

Why wasn’t the first approach enough?

Early measurements using natural silicon faced difficulties. Natural silicon contains different isotopes, and small amounts of contamination can affect the calculated number of atoms.

The challenge was particularly serious because the experiment depended on knowing the sample’s composition extremely accurately.

Researchers improved the approach by producing silicon crystals highly enriched in silicon-28. They also improved the surface preparation and measurements of the crystal’s shape and dimensions.

A refined 2015 measurement reported:

\[ N_A = 6.02214075(12)\times10^{23}\ \text{mol}^{-1} \]

The uncertainty was extremely small, and the result was consistent with the value eventually adopted for the modern SI definition. NIST’s account of the silicon measurement describes this work.

What changed in 2019?

Before 2019, the mole was defined through the number of atoms in 12 g of carbon-12. Avogadro’s constant therefore had to be determined experimentally.

In 2018, the General Conference on Weights and Measures adopted a revised definition, which took effect on 20 May 2019.

The new definition fixed the numerical value of Avogadro’s constant at exactly:

\[ 6.02214076\times10^{23}\ \text{mol}^{-1} \]

This reversed the relationship. Instead of defining the mole through a mass standard and measuring Avogadro’s constant, the mole was defined directly through a fixed number of entities.

The silicon experiment remains important because it helped establish the value used in the redefinition. However, the modern mole is not defined by silicon, carbon-12, or any particular substance.

That is the historical reason the constant is universal: it is now explicitly defined as the same number of specified entities per mole, whatever those entities happen to be.

The crystallographic measurement is useful historical and scientific context. You do not need to reproduce this experimental method for an ordinary HSC mole calculation.

A small scale estimate makes the method easier to appreciate.

Use the following illustrative, rounded values:

• Cubic unit cell edge length: \(a=0.5431\ \text{nm}\).
• Volume of the silicon crystal: \(1.00\ \text{cm}^3=1.00\times10^{-6}\ \text{m}^3\).
• Atoms per conventional unit cell: 8.

First, convert the edge length into metres:

\[ a=5.431\times10^{-10}\ \text{m} \]

Calculate the volume of one unit cell:

\[ a^3=(5.431\times10^{-10})^3 \]

\[ a^3\approx1.60\times10^{-28}\ \text{m}^3 \]

Now calculate the number of unit cells:

\[ \begin{aligned} N_{\text{cells}} &=\frac{1.00\times10^{-6}}{1.60\times10^{-28}}\\ &=6.25\times10^{21} \end{aligned} \]

Each cell contains eight atoms:

\[ \begin{aligned} N_{\text{atoms}} &=8(6.25\times10^{21})\\ &=5.00\times10^{22} \end{aligned} \]

The amount of silicon atoms represented by this count is:

\[ \begin{aligned} n&=\frac{N}{N_A}\\ &=\frac{5.00\times10^{22}} {6.022\times10^{23}}\\ &=0.0830\ \text{mol} \end{aligned} \]

So just \(1.00\ \text{cm}^3\) of silicon contains approximately \(5.00\times10^{22}\) atoms.

This is a scale estimate, not a reconstruction of the precision experiment. The actual research required corrections for isotopic composition, impurities, and surface effects.

The important result is that scientists could connect a measurable crystal to an enormous microscopic particle count.

03How mass, amount, and particle number fit together

Now we can connect the universal counting constant to something you can measure directly: mass.

A balance measures mass in grams. It does not measure how many atoms or molecules are present.

To convert mass into a particle count, you need two relationships.

Relationship 1: Mass to chemical amount

The first equation is:

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

where:

Molar mass tells you the mass of one mole of a particular substance.

For example:

• Aluminium has a molar mass of approximately \(26.98\ \text{g mol}^{-1}\).
• Copper has a molar mass of approximately \(63.55\ \text{g mol}^{-1}\).
• Water has a molar mass of approximately \(18.0\ \text{g mol}^{-1}\).

These values depend on the substance.

Notice the units:

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

Grams cancel, leaving moles.

Relationship 2: Chemical amount to number of entities

Once you know the amount in moles, use:

\[ N=nN_A \]

where:

• \(N\) is the number of specified entities;
• \(n\) is the chemical amount in moles;
• \(N_A\) is Avogadro’s constant, in \(\text{mol}^{-1}\).

The units cancel again:

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

The resulting number is a particle count. It is dimensionless in the SI system, but you must still identify what is being counted.

Combining both relationships gives:

\[ \boxed{N=\frac{mN_A}{M}} \]

This is the key equation connecting mass directly to particle number.

It also answers the title’s question mathematically.

The Avogadro constant stays fixed. The molar mass \(M\) changes depending on the substance.

What changes when the substance changes?

Consider what happens under three different conditions.

The second row is particularly important.

Equal masses do not necessarily contain equal numbers of particles.

A sample containing heavier particles will generally contain fewer particles than an equal-mass sample containing lighter particles.

This is why you cannot calculate a particle count from mass alone. You need the substance’s molar mass as well.

04Worked examples

05Common misconceptions to avoid

One final distinction is worth remembering.

The Avogadro constant is exact by definition, but a calculated amount of substance is not automatically exact. If your balance measures a mass to three significant figures, your calculated particle count cannot suddenly become infinitely precise.

Keep the precision of your measurements in mind.

06Questions and solutions

Use \(N_A=6.022\times10^{23}\ \text{mol}^{-1}\) unless otherwise stated. Assume all samples are pure and use the supplied molar masses.

Question 1

A sample contains \(0.125\ \text{mol}\) of argon atoms.

Calculate the number of argon atoms in the sample.

Solution 1

The sample contains \(7.53\times10^{22}\) argon atoms.

Argon is monatomic, so the specified entities are individual atoms.

Use:

\[ N=nN_A \]

Substitute:

\[ \begin{aligned} N&=0.125\times6.022\times10^{23}\\ &=7.5275\times10^{22}\\ &\approx7.53\times10^{22}\ \text{atoms} \end{aligned} \]

The result is the number of argon atoms, not the mass of the sample.

The chemical identity does not change the value of Avogadro’s constant.

Question 2

Two separate samples are prepared:

• Sample A contains \(8.80\ \text{g}\) of \(\ce{CO2}\).
• Sample B contains \(8.80\ \text{g}\) of \(\ce{O2}\).

Use:

• \(M(\ce{CO2})=44.0\ \text{g mol}^{-1}\)
• \(M(\ce{O2})=32.0\ \text{g mol}^{-1}\)

Calculate:

a. The number of molecules in each sample.

b. The number of oxygen atoms in each sample.

c. The ratio of the number of oxygen atoms in Sample B to Sample A.

Explain why the samples contain different numbers of molecules despite having equal masses.

Solution 2

Sample B contains more molecules and more oxygen atoms. It contains approximately 1.38 times as many oxygen atoms as Sample A.

Part a: Number of molecules

For Sample A, carbon dioxide:

First calculate the amount:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{8.80}{44.0}\\ &=0.200\ \text{mol} \end{aligned} \]

Then calculate the number of molecules:

\[ \begin{aligned} N(\ce{CO2}) &=0.200(6.022\times10^{23})\\ &=1.20\times10^{23}\ \text{molecules} \end{aligned} \]

For Sample B, oxygen gas:

\[ \begin{aligned} n(\ce{O2}) &=\frac{8.80}{32.0}\\ &=0.275\ \text{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} N(\ce{O2}) &=0.275(6.022\times10^{23})\\ &=1.66\times10^{23}\ \text{molecules} \end{aligned} \]

Part b: Number of oxygen atoms

Each \(\ce{CO2}\) molecule contains two oxygen atoms.

Therefore:

\[ \begin{aligned} N(\ce{O\ atoms\ in\ CO2}) &=2(1.2044\times10^{23})\\ &=2.41\times10^{23} \end{aligned} \]

Each \(\ce{O2}\) molecule also contains two oxygen atoms:

\[ \begin{aligned} N(\ce{O\ atoms\ in\ O2}) &=2(1.6561\times10^{23})\\ &=3.31\times10^{23} \end{aligned} \]

Part c: Compare the oxygen atom counts

\[ \begin{aligned} \text{Ratio} &=\frac{3.31\times10^{23}} {2.41\times10^{23}}\\ &\approx1.38 \end{aligned} \]

Sample B contains approximately 1.38 times as many oxygen atoms.

The tempting assumption is that equal masses must contain equal numbers of molecules.

However, carbon dioxide has a greater molar mass than oxygen gas. Therefore, the same mass of \(\ce{O2}\) represents a greater chemical amount and contains more molecules.

Avogadro’s constant has not changed. The molar masses have.

Question 3

A laboratory has a \(10.0\ \text{g}\) mixture containing only water, \(\ce{H2O}\), and carbon dioxide, \(\ce{CO2}\).

A particle-counting instrument reports that the mixture contains \(1.5055\times10^{23}\) molecules in total.

Use:

• \(M(\ce{H2O})=18.0\ \text{g mol}^{-1}\)
• \(M(\ce{CO2})=44.0\ \text{g mol}^{-1}\)

Calculate:

a. The total chemical amount of molecules in the mixture.

b. The chemical amount of each substance.

c. The number of molecules of each substance.

Solution 3

The mixture contains approximately \(0.0385\ \text{mol}\) of water and \(0.212\ \text{mol}\) of carbon dioxide.

The key is that Avogadro’s constant applies to the total number of specified entities, even when the sample contains more than one molecular substance.

Part a: Find the total chemical amount

Use:

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

\[ \begin{aligned} n_{\text{total}} &=\frac{1.5055\times10^{23}} {6.022\times10^{23}}\\ &=0.2500\ \text{mol} \end{aligned} \]

Part b: Determine the amount of each substance

Let:

• \(x\) = moles of water;
• \(y\) = moles of carbon dioxide.

The total amount is:

\[ x+y=0.2500 \]

The total mass is:

\[ 18.0x+44.0y=10.0 \]

Rearrange the first equation:

\[ x=0.2500-y \]

Substitute into the mass equation:

\[ 18.0(0.2500-y)+44.0y=10.0 \]

Expand:

\[ 4.50-18.0y+44.0y=10.0 \]

\[ 26.0y=5.50 \]

\[ y=0.2115\ \text{mol} \]

Now calculate \(x\):

\[ \begin{aligned} x&=0.2500-0.2115\\ &=0.0385\ \text{mol} \end{aligned} \]

Part c: Calculate the number of molecules

For water:

\[ \begin{aligned} N(\ce{H2O}) &=0.03846(6.022\times10^{23})\\ &=2.32\times10^{22}\ \text{molecules} \end{aligned} \]

For carbon dioxide:

\[ \begin{aligned} N(\ce{CO2}) &=0.21154(6.022\times10^{23})\\ &=1.27\times10^{23}\ \text{molecules} \end{aligned} \]

Check the total:

\[ 2.32\times10^{22} + 1.27\times10^{23} \approx1.50\times10^{23} \]

The result is consistent with the measured total.

The important reasoning is that each molecule contributes one entity to the total particle count, regardless of whether it is water or carbon dioxide. The different molar masses determine how the measured total mass is distributed between the two substances.

Question 4

A laboratory technician weighs out \(18.0\ \text{g}\) of pure water.

Using a molar mass of \(18.015\ \text{g mol}^{-1}\), the technician calculates the number of water molecules and reports:

“There are exactly \(6.02214076\times10^{23}\) water molecules.”

a. Calculate the estimated number of water molecules in the sample.

b. Explain why the technician’s claim of an exact particle count is not justified, even though Avogadro’s constant is exact.

Assume the measured mass is reported to three significant figures.

Solution 4

The sample contains approximately \(6.02\times10^{23}\) water molecules, but the technician cannot justify claiming that this count is exact.

Part a: Calculate the particle count

First convert mass into chemical amount:

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

\[ \begin{aligned} n&=\frac{18.0}{18.015}\\ &=0.99917\ \text{mol} \end{aligned} \]

Now calculate the number of molecules:

\[ N=nN_A \]

\[ \begin{aligned} N&=0.99917(6.02214076\times10^{23})\\ &\approx6.017\times10^{23} \end{aligned} \]

To three significant figures:

\[ N=6.02\times10^{23}\ \text{water molecules} \]

Part b: Explain the uncertainty

Avogadro’s constant is exact because its numerical value is fixed by definition.

However, the technician’s mass measurement is not exact. It is reported to only three significant figures, and the supplied molar mass also has limited precision.

Furthermore, a real sample may contain impurities or have a composition that differs slightly from the assumed pure substance.

These uncertainties affect the calculated amount of water.

The distinction is:

• \(N_A\) is an exact defined constant.
• \(n\) is calculated from measured quantities and can have uncertainty.
• \(N=nN_A\) inherits the uncertainty in \(n\).

An exact conversion factor does not make an uncertain measurement exact.

Question 5

A sealed bottle contains a pure molecular form of oxygen. The substance is known to be either ordinary oxygen gas, \(\ce{O2}\), or ozone, \(\ce{O3}\).

Two measurements are recorded:

• Mass of the sample: \(36.0\ \text{g}\).
• Number of molecules: \(4.5166\times10^{23}\).

Use \(A_r(\ce{O})=16.0\).

Determine:

a. The chemical amount of molecules in the sample.

b. Whether the substance is \(\ce{O2}\) or \(\ce{O3}\).

c. The chemical amount and number of individual oxygen atoms.

d. The mass the sample would have if it contained the same number of molecules but consisted of the alternative molecular form.

Explain why the molecular count alone is insufficient to determine the number of oxygen atoms.

Solution 5

The substance is ozone, \(\ce{O3}\). It contains \(0.7500\ \text{mol}\) of molecules and \(2.25\ \text{mol}\) of oxygen atoms.

The crucial step is to keep the molecular count separate from the atomic count. We must use both the measured mass and the molecular count to identify the substance.

Part a: Calculate the amount of molecules

Use:

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

\[ \begin{aligned} n_{\text{molecules}} &=\frac{4.5166\times10^{23}} {6.02214076\times10^{23}}\\ &\approx0.7500\ \text{mol} \end{aligned} \]

The sample contains \(0.7500\ \text{mol}\) of oxygen molecules.

Part b: Identify the molecular form

First calculate the amount of oxygen atoms represented by the measured mass.

The molar mass of an oxygen atom is:

\[ M(\ce{O})=16.0\ \text{g mol}^{-1} \]

Therefore:

\[ \begin{aligned} n(\ce{O\ atoms}) &=\frac{36.0}{16.0}\\ &=2.25\ \text{mol} \end{aligned} \]

Now compare the amount of oxygen atoms with the amount of molecules:

\[ \frac{2.25\ \text{mol atoms}} {0.7500\ \text{mol molecules}} =3.00 \]

There are three oxygen atoms per molecule.

Therefore, the substance must be ozone, \(\ce{O3}\).

The two possible models can be checked against the measured mass:

Only ozone agrees with the measured mass.

Part c: Calculate the number of oxygen atoms

Each ozone molecule contains three oxygen atoms.

Therefore:

\[ \begin{aligned} n(\ce{O\ atoms}) &=3(0.7500)\\ &=2.25\ \text{mol} \end{aligned} \]

Convert this amount into individual atoms:

\[ \begin{aligned} N(\ce{O\ atoms}) &=2.25(6.02214076\times10^{23})\\ &=1.35498\times10^{24} \end{aligned} \]

Therefore, the sample contains approximately:

\[ 1.35\times10^{24}\ \text{oxygen atoms} \]

Part d: Consider the alternative molecular form

If the sample contained the same number of molecules but consisted of \(\ce{O2}\), it would still contain \(0.7500\ \text{mol}\) of molecules.

However, the molar mass would be:

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

Its mass would therefore be:

\[ \begin{aligned} m&=nM\\ &=0.7500(32.0)\\ &=24.0\ \text{g} \end{aligned} \]

That is \(12.0\ \text{g}\) less than the measured mass.

The \(\ce{O2}\) model is therefore inconsistent with the experimental evidence.

The tempting mistake is to assume that \(0.7500\ \text{mol}\) of molecules must also mean \(0.7500\ \text{mol}\) of atoms.

It does not. The chemical formula determines how many atoms belong to each molecule.

The two independent measurements work together: the molecular count determines the amount of molecules, while the mass helps identify how many oxygen atoms each molecule contains.

Most importantly, neither result requires a different Avogadro constant.

07Where this leads next

The relationship between mass, chemical amount, and particle number is one of the foundations of quantitative chemistry.

You should now be able to distinguish three separate quantities:

• Mass depends on how much material you have and what substance it is.
• Chemical amount tells you how many moles of specified entities are present.
• Avogadro’s constant provides the universal conversion between moles and the number of those entities.

The next step is using this relationship in chemical reactions.

A balanced chemical equation gives ratios between reacting entities. Moles let us work with those ratios, while molar masses let us convert the results back into measurable grams.

That is the connection between particle counting and stoichiometry.

For further practice with the complete mass-to-particle calculation, see How to Convert Mass into Number of Molecules in HSC Chemistry.