How to Calculate the Radius of a Molecule

A molecule does not usually have a sharply defined spherical edge, so there is no single radius that applies in every context. A useful estimate is its equivalent spherical radius: the radius of a sphere with the same volume as the volume assigned to one molecule in a condensed sample.

You can estimate this radius from the substance’s molar mass and density. The result is a scale estimate, not a direct measurement of the molecule’s shape or its bond lengths.

The radius formula

The molar volume is the volume occupied by one mole of a substance:

$$V_m = \frac{M}{\rho},$$

where $M$ is molar mass and $\rho$ is density. The volume assigned to one molecule is then

$$V_1 = \frac{V_m}{N_A} = \frac{M}{\rho N_A},$$

where $N_A = 6.02214076\times 10^{23}\ \mathrm{mol}^{-1}$ is Avogadro’s constant. If you approximate this volume by a sphere, $V_1 = \frac{4}{3}\pi r^3$, so

$$r = \left(\frac{3M}{4\pi\rho N_A}\right)^{1/3}.$$

Use consistent units. For example, if $M$ is in kilograms per mole and $\rho$ in kilograms per cubic metre, then $r$ comes out in metres.

Worked example: water

Estimate the equivalent spherical radius of a water molecule using approximate values $M = 18.0\ \mathrm{g/mol}$ and $\rho = 1.00\ \mathrm{g/cm^3}$. These are convenient rounded values for liquid water.

First calculate the molar volume:

$$V_m = \frac{18.0\ \mathrm{g/mol}}{1.00\ \mathrm{g/cm^3}} = 18.0\ \mathrm{cm^3/mol}.$$

Convert to cubic metres: $18.0\ \mathrm{cm^3} = 1.80\times10^{-5}\ \mathrm{m^3}$. Divide by Avogadro’s constant to get the volume assigned to one molecule:

$$V_1 = \frac{1.80\times10^{-5}\ \mathrm{m^3/mol}}{6.02214076\times10^{23}\ \mathrm{mol}^{-1}} \approx 2.99\times10^{-29}\ \mathrm{m^3}.$$

Now use the sphere formula:

$$r = \left(\frac{3(2.99\times10^{-29}\ \mathrm{m^3})}{4\pi}\right)^{1/3} \approx 1.93\times10^{-10}\ \mathrm{m}.$$

So the estimated equivalent radius is about $0.193\ \mathrm{nm}$, or $1.93\ \mathrm{\mathring{A}}$. Its corresponding sphere diameter is about $0.386\ \mathrm{nm}$. This diameter is not the distance between the hydrogen atoms in a water molecule; it represents the sphere that matches the volume-per-molecule estimate.

What the estimate means

For a liquid or solid, dividing the sample’s volume by its number of molecules gives an average volume per molecule. Converting that volume into a sphere makes it easier to describe as one length, but it hides details such as molecular shape, electron distribution, and how neighboring molecules pack.

The result also depends on the density used. Density changes with temperature and pressure, and values for different phases can differ greatly. For a solid, a crystal unit cell can give a more structure-specific estimate: divide the cell volume by the number of molecules in that cell, then convert the volume per molecule to an equivalent sphere. If you want a model of the molecule’s own volume rather than its share of a crystal’s volume, packing and the definition of molecular boundary must also be specified.

If you mean a gas molecule

Using a gas’s density in the formula gives the volume allocated to each molecule in the gas under those conditions. In a dilute gas, molecules are far apart, so the resulting radius mostly reflects their average spacing—not their physical size. For an ideal gas, the volume per molecule is approximately $k_BT/P$, which varies with temperature and pressure. To estimate a gas molecule’s physical size, use a molecular model or an experimental quantity such as a collision diameter instead.

Common mistakes

• Mixing units: Convert mass and volume units consistently before taking the cube root. A cubic-centimetre result cannot be combined directly with a density in kilograms per cubic metre.
• Using the ideal-gas volume as molecular size: The large volume per gas molecule includes empty space between molecules.
• Calling the result an exact boundary: Molecules have non-spherical shapes and diffuse electron clouds; the equivalent radius is a useful approximation, not a literal surface.
• Confusing radius with diameter: The sphere’s diameter is $2r$, not $r$.
• Ignoring the phase: Use a density appropriate to the phase and conditions you intend to describe.