How to Calculate the Mass of One Atom from Molar Mass

An individual atom is far too small to weigh on an ordinary laboratory balance. But you can still calculate its mass using a quantity chemists measure routinely: the molar mass of an element.

The method is to divide the molar mass by Avogadro’s constant:

\[
\boxed{m_{\text{one atom}}=\frac{M}{N_A}}
\]

Here:

• \(m_{\text{one atom}}\) is the mass of one atom, in grams (g).
• \(M\) is the molar mass of the element, in grams per mole (g mol\(^{-1}\)).
• \(N_A\) is Avogadro’s constant, \(6.02214076\times10^{23}\) entities per mole.

For example, the molar mass of carbon is approximately \(12.01\text{ g mol}^{-1}\). Divide this by Avogadro’s constant, and you obtain the mass of one carbon atom.

There is one important condition: you must know exactly what entity the molar mass counts. A mole of oxygen atoms and a mole of oxygen molecules are not the same thing. Getting that distinction right prevents a very common calculation error.

01Why dividing by Avogadro’s constant works

Imagine you have exactly one mole of carbon atoms. That sample contains \(6.02214076\times10^{23}\) carbon atoms, and its mass is approximately 12.01 g.

If you divide the total mass by the total number of atoms, you find the mass belonging to each atom:

\[
m_{\text{one atom}}
=
\frac{\text{mass of sample}}{\text{number of atoms}}
\]

Substituting the quantities for one mole:

\[
m_{\text{one atom}}
=
\frac{12.01\text{ g}}
{6.02214076\times10^{23}}
\]

\[
m_{\text{one atom}}
\approx 1.994\times10^{-23}\text{ g}
\]

That is the mass of one carbon atom.

The important idea is that molar mass describes the mass of a very large collection of particles, while Avogadro’s constant tells you how many particles are in that collection. Dividing one by the other gives the mass of a single particle.

Check the units before calculating

The units explain why the division works:

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

The mole units cancel, leaving grams per atom.

This is a useful dimensional check. If your final answer is in g mol\(^{-1}\), you have calculated a molar mass, not the mass of one atom. If you have used the correct number of entities, your answer should be an extremely small mass in grams.

02A reliable method for finding the mass of one atom

Use these steps whenever a question gives you the molar mass of an element and asks for the mass of an individual atom.

Step 1: Identify the entity.

Check whether the molar mass refers to individual atoms, molecules, or formula units. For an element such as carbon, which exists as individual atoms in its standard solid structure, the molar mass is expressed per mole of carbon atoms.

Step 2: Write down the molar mass.

Use the value supplied in the question or an appropriate periodic table. Keep its units as g mol\(^{-1}\).

Step 3: Write down Avogadro’s constant.

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

Step 4: Divide the molar mass by Avogadro’s constant.

\[
m_{\text{atom}}=\frac{M}{N_A}
\]

Step 5: State the answer with units.

For a mass in grams, use scientific notation. The result will usually be around \(10^{-22}\) to \(10^{-24}\) g for an individual atom.

03The important trap: atoms are not always the entities counted by molar mass

Now consider oxygen.

Oxygen is listed on the periodic table with a relative atomic mass of approximately 16.00. However, oxygen gas normally exists as \(\ce{O2}\), meaning each molecule contains two oxygen atoms.

This creates two different molar masses:

Suppose a question gives you the molar mass of oxygen gas, \(32.00\text{ g mol}^{-1}\), and asks for the mass of one oxygen atom.

You might predict that dividing by Avogadro’s constant gives the answer.

Not quite. It gives the mass of one oxygen molecule, because the molar mass counts oxygen molecules.

To find the mass of one oxygen atom, you must account for the two atoms in each molecule.

What if the molar mass belongs to a compound?

The same reasoning applies to compounds.

For example, one molecule of water, \(\ce{H2O}\), contains:

• Two hydrogen atoms.
• One oxygen atom.

Its molar mass is approximately \(18.016\text{ g mol}^{-1}\).

Dividing by Avogadro’s constant gives the mass of one entire water molecule:

\[
m_{\ce{H2O}}=
\frac{18.016}{N_A}
\]

But that is not the mass of one hydrogen atom or one oxygen atom.

To find the mass of one oxygen atom, use the oxygen contribution to the molar mass:

\[
m_{\ce{O}}=
\frac{16.00}{N_A}
\]

Similarly, the mass of one hydrogen atom is:

\[
m_{\ce{H}}=
\frac{1.008}{N_A}
\]

You cannot simply divide the mass of a water molecule by three and claim that each atom has that mass. Hydrogen and oxygen atoms have different masses.

For ionic compounds, the counted entity is generally a formula unit rather than a molecule. The same principle applies: establish what one mole represents before dividing by Avogadro’s constant.

If you need to calculate the molar mass from a chemical formula first, see Molar Mass of Elements and Compounds: HSC Chemistry Guide.

04Where does Avogadro’s constant come from?

There is a bigger measurement problem hiding behind this calculation.

How could scientists determine the number of atoms in a mole when individual atoms cannot be counted using an ordinary balance?

Historically, scientists developed experiments to estimate Avogadro’s constant. One particularly precise approach involved counting atoms in a nearly perfect silicon crystal.

The International Avogadro Project used highly enriched silicon-28 crystals, shaped into spheres with masses of approximately 1 kg. The aim was to establish the number of atoms in a known amount of material with extremely high precision.

How the experiment worked

1. Measure the sphere.

Researchers used an optical interferometer to measure the sphere’s diameter very precisely.

Knowing its diameter allowed them to calculate its volume. A precision balance provided its mass.

2. Measure the spacing between atoms.

Silicon forms a regular three-dimensional crystal lattice.

Researchers used X-ray diffraction and interferometry to determine the spacing within this repeating structure. A conventional cubic unit cell of silicon contains eight atoms.

The prediction was that if the crystal’s structure and dimensions were known accurately, researchers could calculate how many unit cells fitted inside the sphere, then multiply by eight to estimate the number of atoms.

3. Account for isotope composition.

Natural silicon contains several isotopes, including silicon-28, silicon-29, and silicon-30.

These isotopes have different masses. Even a small uncertainty in their proportions affects the calculated molar mass of the crystal.

The project therefore used highly enriched silicon-28. This made the average atomic mass more predictable and reduced an important source of uncertainty.

Surface contamination, crystal imperfections, and measurement precision also needed careful control. Improving the purity of the crystal and the quality of the measurements made the results more reliable.

A useful scale estimate

Consider a roughly 1.00 kg sphere of highly enriched silicon-28.

For an illustrative calculation, take its molar mass as approximately \(27.98\text{ g mol}^{-1}\).

First, calculate the amount of silicon:

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

\[
n=\frac{1000}{27.98}
=35.74\text{ mol}
\]

Using today’s Avogadro constant, the estimated number of atoms is:

\[
N=nN_A
\]

\[
N=(35.74)(6.02214076\times10^{23})
\]

\[
\boxed{N\approx2.152\times10^{25}\text{ atoms}}
\]

This is a scale estimate using the modern value of Avogadro’s constant, not the original experimental counting procedure.

The experiment used the measured crystal structure and sphere volume to establish the atom count and refine the value of the constant.

What changed in 2019?

The mole was originally defined using the number of atoms in 12 g of carbon-12.

On 20 May 2019, the definition changed. The mole is now defined as containing exactly:

\[
\boxed{6.02214076\times10^{23}\text{ specified entities}}
\]

This value is exact by definition, rather than an experimentally measured quantity with an uncertainty. The silicon experiments were among the important precision measurements that helped make the redefinition possible.

For HSC calculations, you can therefore treat Avogadro’s constant as a fixed conversion factor between moles and individual entities.

05Does every atom of an element have exactly the same mass?

There is one final detail worth understanding, particularly when questions involve isotopes.

Look at chlorine on the periodic table. Its relative atomic mass is approximately 35.45, not a whole number.

Why?

Naturally occurring chlorine contains different isotopes. Chlorine-35 and chlorine-37 have different numbers of neutrons, so they have different masses.

The periodic-table value is a weighted average based on the relative abundance of these isotopes.

This means that dividing the molar mass by Avogadro’s constant gives the average mass of one chlorine atom in a naturally occurring sample. It does not mean every individual chlorine atom has exactly that mass.

For most standard HSC calculations, you can use the molar mass supplied in the question directly.

However, if a question specifies a particular isotope, use its isotope-specific mass rather than automatically using the periodic-table average.

A quick decision table

Before dividing by Avogadro’s constant, identify what the molar mass represents.

Three checks before submitting your answer

1. Entity: Does the molar mass count atoms, molecules, or formula units?
2. Units: Does your calculation finish in grams per entity?
3. Composition: If the entity contains multiple atoms, have you accounted for the correct number and type of atoms?

If all three are correct, you have a reliable method.

06Questions and solutions

The following questions move from a standard single-atom calculation to situations where identifying the correct particle becomes more important than the arithmetic.

Question 1

A sample contains \(0.250\text{ mol}\) of copper atoms and has a mass of \(15.89\text{ g}\).

Calculate the mass of one copper atom in grams.

Solution 1

The mass of one copper atom is approximately \(1.055\times10^{-22}\text{ g}\).

First, calculate the molar mass using the given sample mass and amount:

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

\[
M=\frac{15.89}{0.250}
=63.56\text{ g mol}^{-1}
\]

The question specifies copper atoms, so this molar mass is per mole of copper atoms.

Now divide by Avogadro’s constant:

\[
m_{\text{atom}}=
\frac{63.56}
{6.02214076\times10^{23}}
\]

\[
\boxed{m_{\text{atom}}=1.055\times10^{-22}\text{ g}}
\]

The result is reasonable: one copper atom has an extremely small mass, even though one mole of copper atoms has a mass of 63.56 g.

Question 2

A \(4.00\text{ g}\) sample of oxygen gas contains only \(\ce{O2}\) molecules.

The molar mass of oxygen gas is \(32.00\text{ g mol}^{-1}\).

Calculate:

a) The number of oxygen molecules in the sample.

b) The number of oxygen atoms in the sample.

c) The mass of one oxygen atom in grams.

Solution 2

a) The sample contains approximately \(7.53\times10^{22}\) oxygen molecules.

First, calculate the amount of oxygen molecules:

\[
n(\ce{O2})=\frac{m}{M}
\]

\[
n(\ce{O2})=\frac{4.00}{32.00}
=0.1250\text{ mol}
\]

Convert moles into molecules:

\[
N=nN_A
\]

\[
N=0.1250(6.02214076\times10^{23})
\]

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

b) The sample contains approximately \(1.51\times10^{23}\) oxygen atoms.

Each oxygen molecule contains two oxygen atoms.

Therefore:

\[
N(\text{O atoms})=2N(\ce{O2})
\]

\[
N(\text{O atoms})=
2(7.53\times10^{22})
\]

\[
\boxed{N(\text{O atoms})=1.51\times10^{23}}
\]

There are twice as many oxygen atoms as oxygen molecules.

c) The mass of one oxygen atom is approximately \(2.66\times10^{-23}\text{ g}\).

Divide the total sample mass by the total number of oxygen atoms:

\[
m_{\text{atom}}=
\frac{4.00}{1.506\times10^{23}}
\]

\[
\boxed{m_{\text{atom}}=2.66\times10^{-23}\text{ g}}
\]

The important step was counting oxygen atoms rather than oxygen molecules.

Dividing the sample mass by the number of molecules would give the mass of one entire \(\ce{O2}\) molecule, which is twice the mass of one oxygen atom.

Question 3

Ammonia, \(\ce{NH3}\), has a molar mass of \(17.03\text{ g mol}^{-1}\).

Use the following relative atomic masses:

• Nitrogen: \(14.01\)
• Hydrogen: \(1.008\)

Calculate:

a) The mass of one nitrogen atom in grams.

b) The mass of one hydrogen atom in grams.

c) Explain why dividing \(17.03\) by \(4N_A\) does not give the mass of either individual atom.

Solution 3

a) The mass of one nitrogen atom is approximately \(2.326\times10^{-23}\text{ g}\).

b) The mass of one hydrogen atom is approximately \(1.674\times10^{-24}\text{ g}\).

First, examine the formula \(\ce{NH3}\).

One mole of ammonia molecules contains:

• One mole of nitrogen atoms.
• Three moles of hydrogen atoms.

The total molar mass can be separated into the contributions from each element:

\[
M(\ce{NH3})=14.01+3(1.008)
\]

\[
=14.01+3.024
=17.034\text{ g mol}^{-1}
\]

The small difference from the supplied \(17.03\) is due to rounding.

For nitrogen:

One mole of ammonia contains one mole of nitrogen atoms, contributing \(14.01\) g.

Therefore:

\[
m_{\ce{N}}=
\frac{14.01}
{6.02214076\times10^{23}}
\]

\[
\boxed{m_{\ce{N}}=2.326\times10^{-23}\text{ g}}
\]

For hydrogen:

One mole of ammonia contains three moles of hydrogen atoms, contributing \(3.024\) g.

The mass of one hydrogen atom is:

\[
m_{\ce{H}}=
\frac{1.008}
{6.02214076\times10^{23}}
\]

\[
\boxed{m_{\ce{H}}=1.674\times10^{-24}\text{ g}}
\]

c) Dividing the total molar mass by \(4N_A\) would incorrectly assume that all four atoms in an ammonia molecule have equal masses.

They do not.

The nitrogen atom is considerably heavier than each hydrogen atom. You must separate the contributions of the elements before calculating the mass of an individual atom.

Question 4

A hypothetical chlorine sample contains:

• 75.0% chlorine-35 atoms.
• 25.0% chlorine-37 atoms.

For this question, assume the isotope masses are approximately 35.0 u and 37.0 u respectively. The percentages represent the proportions of atoms by number.

Calculate:

a) The average relative atomic mass of the chlorine sample.

b) The average mass of one chlorine atom in grams.

c) Explain whether every chlorine atom in the sample has the mass calculated in part (b).

Solution 4

a) The average relative atomic mass is 35.5.

b) The average mass of one chlorine atom is approximately \(5.895\times10^{-23}\text{ g}\).

c) No. Individual chlorine-35 and chlorine-37 atoms have different masses. The calculated value is an average.

Step 1: Calculate the weighted average relative atomic mass.

Multiply each isotope’s mass by its fractional abundance:

\[
A_r(\ce{Cl})=
(0.750)(35.0)+(0.250)(37.0)
\]

\[
=26.25+9.25
\]

\[
\boxed{A_r(\ce{Cl})=35.5}
\]

Step 2: Convert this into an average molar mass.

The numerical value of the average relative atomic mass corresponds to the molar mass in grams per mole:

\[
M(\ce{Cl})=35.5\text{ g mol}^{-1}
\]

Step 3: Calculate the average mass of one chlorine atom.

\[
m_{\text{average atom}}=
\frac{35.5}
{6.02214076\times10^{23}}
\]

\[
\boxed{m_{\text{average atom}}=
5.895\times10^{-23}\text{ g}}
\]

Step 4: Interpret the result.

The calculation does not mean every chlorine atom has a mass of \(5.895\times10^{-23}\) g.

Using the approximate isotope masses:

\[
m_{\ce{^{35}Cl}}\approx
5.812\times10^{-23}\text{ g}
\]

\[
m_{\ce{^{37}Cl}}\approx
6.144\times10^{-23}\text{ g}
\]

The average lies between these two values because the sample contains both isotopes.

This is why periodic-table atomic masses are often decimals rather than whole numbers.

Question 5

A sealed gas sample contains a mixture of oxygen molecules, \(\ce{O2}\), and ozone molecules, \(\ce{O3}\).

The sample has:

• A total amount of \(0.500\text{ mol}\) of gas molecules.
• A total mass of \(18.40\text{ g}\).

Use the following molar masses:

\[
M(\ce{O2})=32.00\text{ g mol}^{-1}
\]

\[
M(\ce{O3})=48.00\text{ g mol}^{-1}
\]

Assume the sample contains no other substances.

Calculate:

a) The amount, in moles, of \(\ce{O2}\) molecules and \(\ce{O3}\) molecules.

b) The total amount of oxygen atoms in the sample, in moles.

c) The total number of oxygen atoms in the sample.

d) The mass of one oxygen atom in grams.

A student argues that the mass of one oxygen atom can be found simply by dividing \(18.40\text{ g}\) by the number of molecules in the sample.

Explain precisely why this argument gives the wrong answer.

Solution 5

a) The sample contains \(0.350\text{ mol}\) of \(\ce{O2}\) and \(0.150\text{ mol}\) of \(\ce{O3}\).

b) The sample contains \(1.150\text{ mol}\) of oxygen atoms.

c) The sample contains approximately \(6.925\times10^{23}\) oxygen atoms.

d) The mass of one oxygen atom is approximately \(2.657\times10^{-23}\text{ g}\).

This problem requires you to distinguish between the number of molecules and the number of atoms inside those molecules.

Step 1: Define the unknown amounts.

Let:

• \(x\) = amount of \(\ce{O2}\) in mol.
• \(y\) = amount of \(\ce{O3}\) in mol.

The total amount of molecules is \(0.500\) mol:

\[
x+y=0.500
\]

The total mass is \(18.40\) g:

\[
32.00x+48.00y=18.40
\]

These equations describe two different constraints on the same sample.

Step 2: Solve for the amounts of each molecule.

Rearrange the first equation:

\[
x=0.500-y
\]

Substitute into the mass equation:

\[
32.00(0.500-y)+48.00y=18.40
\]

Expand:

\[
16.00-32.00y+48.00y=18.40
\]

\[
16.00y=2.40
\]

\[
y=0.150\text{ mol}
\]

Therefore:

\[
x=0.500-0.150
\]

\[
x=0.350\text{ mol}
\]

So the mixture contains:

\[
\boxed{n(\ce{O2})=0.350\text{ mol}}
\]

\[
\boxed{n(\ce{O3})=0.150\text{ mol}}
\]

Step 3: Calculate the total amount of oxygen atoms.

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

Each \(\ce{O3}\) molecule contains three oxygen atoms.

Therefore:

\[
n(\text{O atoms})=2x+3y
\]

Substitute:

\[
n(\text{O atoms})=
2(0.350)+3(0.150)
\]

\[
=0.700+0.450
\]

\[
\boxed{n(\text{O atoms})=1.150\text{ mol}}
\]

Notice that the sample contains \(0.500\) mol of molecules but \(1.150\) mol of atoms.

The number of atoms depends on the molecular composition.

Step 4: Calculate the total number of oxygen atoms.

Use Avogadro’s constant:

\[
N=nN_A
\]

\[
N=1.150(6.02214076\times10^{23})
\]

\[
\boxed{N=6.925\times10^{23}\text{ atoms}}
\]

Step 5: Calculate the mass of one oxygen atom.

Divide the total sample mass by the total number of oxygen atoms:

\[
m_{\text{atom}}=
\frac{18.40}
{6.925\times10^{23}}
\]

\[
\boxed{m_{\text{atom}}=
2.657\times10^{-23}\text{ g}}
\]

This agrees with the result obtained using the molar mass of oxygen atoms:

\[
m_{\text{atom}}=
\frac{16.00}
{6.02214076\times10^{23}}
\]

Step 6: Explain the student’s mistake.

The student’s proposed calculation divides the total mass by the number of molecules:

\[
N_{\text{molecules}}=
0.500N_A
\]

This gives the average mass of one molecule in the mixture, not one oxygen atom.

The molecules contain different numbers of atoms. Some contain two, while others contain three.

The correct calculation must account for the actual number of oxygen atoms, which is why we first determined the composition of the mixture.

There is an interesting result here: changing the proportions of \(\ce{O2}\) and \(\ce{O3}\) changes the average mass per molecule, but it does not change the mass of an oxygen atom.

That is the distinction this question is testing.

07Where to go next

Finding the mass of one atom is the reverse of counting how many particles are present in a sample.

You now have the conversion:

\[
\text{Molar mass}
\longrightarrow
\text{Mass of one entity}
\]

The next useful step is reversing the process: starting with a known sample mass and calculating how many molecules or atoms it contains.

For that, continue with How to Convert Mass into Number of Molecules in HSC Chemistry.

For further reading on the historical measurements and modern definition of the mole, see the BIPM’s history of the mole, its current SI definition, and NIST’s explanation of the silicon-sphere experiments.