# Lattice Energy Calculator

Created by Jack Bowater
Reviewed by Anna Szczepanek, PhD
Last updated: Sep 27, 2022

Atoms can come together in many different ways, and this lattice energy calculator is concerned with the energy stored when cations and anions ionically bond as a part of a larger, uniform structure. You're probably well aware of how ubiquitous ionic lattices are - you'll find them in your food, medicine, and maybe even in the walls of your house - but by learning what lattice energy is, the lattice energy formula, and the lattice energy trend, your appreciation for chemistry will surely increase. So, regardless of if you've been asked to find the lattice energy of $$\text{CaO}} for a test, or want to work out the lattice energy of \text{NaCl}}$$ to aid in dinner conversation, learning how to calculate lattice energy will aid in your understanding of the physical world.

## What is lattice energy? - The lattice energy definition

Before we get to grips with finding the lattice energy, it's important to know the lattice energy definition as it is quite peculiar. Chemists, for various reasons, like to have exact and sometimes unintuitive definitions, but they do serve a purpose, we assure you. In this case, the **lattice energy definition isn't the change in energy when any two atoms form an ionic bond that is part of an ionic lattice, but instead:

The energy required to fully dissociate a mole of an ionic lattice into its constituent ions in their gaseous state.

This can be thought of in terms of the lattice energy of $\text{NaCl}$:

$\text{NaCl}_{(\text{s})}\rightarrow \text{Na}^+_{(\text{g})}+ \text{Cl}^-_{(\text{g})}$

That the ions are in their gaseous state is important; in this form, they are thought to be infinitely far apart, i.e., there are no interactions between them. This ensures that the complete lattice energy is found, not merely the enthalpy of formation.

## How to calculate lattice energy - The lattice energy formula

Perhaps surprisingly, there are several ways of finding the lattice energy of a compound. In fact, there are five. We will discuss one briefly, and we will explain the remaining four, which are all slight variations on each other, in more detail. You can calculate the last four using this lattice energy calculator.

#### Experimental methods and the Born-Haber cycle

As one might expect, the best way of finding the energy of a lattice is to take an amount of the substance, seal it in an insulated vessel (to prevent energy exchange with the surroundings), and then heat the vessel until all of the substance is gas. After this, the amount of energy you put in should be the lattice energy, right?

Unfortunately, this is not the case. While you will end up with all of the lattice's constituent atoms in a gaseous state, they are unlikely to still be in the same form as they were in the lattice. This is because ions are generally unstable, and so when they inevitably collide as they diffuse (which will happen quite a lot considering there are over 600 sextillion atoms in just one mole of substance — as you can discover with our Avogadro's number calculator) they are going to react to form more stable products. These additional reactions change the total energy in the system, making finding what is the lattice energy directly difficult.

So, how to calculate lattice energy experimentally, then? The trick is to chart a path through the different states of the compound and its constituent elements, starting at the lattice and ending at the gaseous ions. If we then add together all of the various enthalpies (if you don't remember the concept, visit our enthalpy calculator), the result must be the energy gap between the lattice and the ions. This kind of construction is known as a Born-Haber cycle. For example, we can find the lattice energy of \text{CaO}} using the following information:

$\begin{split} &\text{CaO}_{(\text{s})}\rightarrow \text{Ca}_{(\text{s})} + \frac{1}{2}\text{O}_{2(\text{g})}\\ \\ &\text{Ca}_{(\text{s})}\rightarrow \text{Ca}_{(\text{g})}\\ \\ &\text{Ca}_{(\text{g})} \rightarrow \text{Ca}^{2+}_{(\text{g})}+2\text{e}^- \\ &\frac{1}{2}\text{O}_{2(\text{g})}\rightarrow O_{(\text{g})} \\ &\text{O}_{(\text{g})} +2\text{e}^-\rightarrow \text{O}_{(\text{g})}^{2-} \\ \end{split}$

Since we can find all of these energies experimentally, this is a surefire way of answering "What is the lattice energy of $\text{CaO}$?"

#### Hard-sphere model

There are however difficulties in getting reliable energetic readings. This has led many people to look for a theoretical way of finding the lattice energy of a compound. The first attempt was to find the sum of all of the forces, both attractive and repulsive, that contribute to the potential lattice energy.

Even though this is a type of potential energy, you can't use the standard potential energy formula here. The starting point for such a model is the potential energy between two gaseous ions:

$U=\frac{z^+z^-e^2}{4\pi\varepsilon_0r_0}$

where:

• $^+$ — Charge on the cation;
• $z^-$ — Charge on the anion;
• $e$ — Electronic charge ($e=1.602 \times 10^{-19}\ \text{C}$);
• $4\pi\varepsilon_0$ — Vacuum permittivity ($1.11 \times 10^{-10}\ \text{C}^2/(\text{J}\cdot\text{m})$); and
• $r_0$ — Interatomic distance (usually the sum of the cation's & anion's atomic radii in $\text{m}$).

Two alterations are necessary to make the above equation suitable for a mole of a lattice. First, to find the energy on a per mole basis, the equation should be multiplied by Avogadro's constant, $N_{\text{A}}$. Next, consider that this equation is for two ions acting on each other alone, while in a lattice each ion is acted on by every other ion at a strength relative to their interatomic distance. For a single atom in the lattice, the summation of all of these interactions can be found, known as the Madelung constant, $M$, which is then multiplied by the equation above. This constant varies from lattice structure to lattice structure, and the most common are present in the lattice energy calculator.

Therefore, the hard-sphere equation for lattice energy is:

$U=\frac{N_{\text{A}}z^+z^-e^2 M}{4\pi\varepsilon_0r_0}$

where:

• $N_{\text{A}}$ — Avogadro's number; and
• $M$ — Madelung constant.

#### Born-Landé equation

While the hard-sphere model is a useful approximation, it does have some issues. The truth is that atoms do not exist as single points that are either wholly positive or wholly negative, as in the hard-sphere model. They are instead surrounded by a number of electron orbitals regardless of charge (unless you have managed to remove all of the electrons, as in the case of $\text{H}^+$, of course). Because there is actually some element of repulsion between the anion and cation, the hard-sphere model tends to over-estimate the lattice energy. To correct for this, Born and Landé (yes, the same Born as in the Born-Haber cycle, prolific, we know) proposed an equation to describe this repulsive energy:

$U=\frac{N_\text{A}B}{r^n}$

where:

• $B$ — A constant that accounts for how the strength of the repulsion decreases as the distance increases, which is a constant for each lattice;
• $r$ — Interatomic distance; and
• $n$ — Born exponent, a measure of the lattice's compressibility.

By adding this correction to the hard-sphere equation, differentiating it with respect to $r$, assuming that at $r=r_0$ the potential energy is at a minimum, rearranging for $B$, and finally substituting that back into the hard-sphere equation, you end up with the Born-Landé equation:

$U=\frac{N_{\text{A}}z^+z^-e^2M}{4\pi\varepsilon_0r_0}\cdot \left(1-\frac{1}{n}\right)$

#### Born-Mayer equation

As you might expect, the Born-Landé equation gives a better prediction of the lattice energy than the hard-sphere model. It is, however, still an approximation, and improvements to the repulsion term have since been made. The first major improvement came from Mayer, who found that replacing $1/r^n$ with $e^{-\frac{r}{\rho}}$ yielded a more accurate repulsion term. In this case, $\rho$ is a factor representing the compressibility of the lattice, and letting this term equal $30\ \text{pm}$ is sufficient for most alkali metal halides. Substituting this new approximation into the Born-Landé equation gives:

$U=\frac{N_{\text{A}}z^+z^-e^2M}{4\pi\varepsilon_0r_0}\cdot \left(1-\frac{\rho}{r_0}\right)$

Since then, further improvements in our understanding of the universe have lead to a more accurate repulsion term, which in turn have given better equations for how to calculate lattice energy. The application of these new equation are, however, still quite niche and the improvements not as significant. For these reasons they have not been included in the present lattice energy calculator. Still, if you would like us to add some more, please feel free to write to us 😀

#### Kapustinskii equation

Unfortunately, some of the factors for both the Born-Landé and Born-Mayer equations require either careful computation or detailed structural knowledge of the crystal, which are not always easily available to us. Kapustinskii, a Soviet scientist, also noticed this and decided to make some improvements to the Born-Mayer equation to make it more fit for general purpose.

First, he found that **in most cases $ρ$ was equal to $0.345\ \text{pm}$, and so replaced it by $d$, equal to $3.45\times10^{−11}\ \text{m}$. Next, he replaced the measured distance between ions, $r_0$, with merely the sum of the two ionic radii, $r^++r^-$. After this, it was shown that the Madelung constant of a structure divided by the number of atoms in the structure's empirical formula was always roughly equal ($\sim0.85$), and so a constant to account for this could be used to replace the Madelung constant. Moving all of the other constants into a single factor gives the final result:

$U=K\cdot\frac{v\cdot \left|z^+\right|\cdot \left|z^-\right|}{r^++r^-}\cdot\frac{1-d}{r^++r^-}$

where:

• $K$$1.202\times10^{−4}\ \text{J}\cdot\text{m}/\text{mol}$;
• $v$ — Number of ions in the lattice's empirical formula; and
• $d$$3.45\times10^{−11}\ \text{m}$.

As you can see, the lattice energy can now be found from only the lattice's chemical formula and the ionic radii of its constituent atoms.

## Lattice energy trend

Looking at the Kapustinskii equation above, we can begin to understand some of the lattice energy trends as we move across and down the periodic table.

First, we can see that by increasing the charge of the ions, we will dramatically increase the lattice energy. It will, in fact, increase the lattice energy by a factor of four, all of things being equal, as $|z^+| \cdot |z^-|$ moves from being $1 \cdot 1$ to $2\cdot2$. This is due to the ions attracting each other much more strongly on account of their greater magnitude of charge. For example, using the Kapustinskii equation, the lattice energy of $\text{NaCl}$ is $746\ \text{kJ}/\text{mol}$, while the lattice energy of $\text{CaO}$ is $3430\ \text{kJ}/\text{mol}$.

The other trend that can be observed is that, as you move down a group in the periodic table, the lattice energy decreases. As elements further down the period table have larger atomic radii due to an increasing number of filled electronic orbitals (if you need to dust your atomic models, head to our quantum numbers calculator), the factor $r^++r^-$ increases, which lowers the overall lattice energy. Note, that while the increase in $r^++r^-$ in the electronic repulsion term actually increases the lattice energy, the other $r^++r^-$ has a much greater effect on the overall equation, and so the lattice energy decreases. The cause of this effect is less efficient stacking of ions within the lattice, resulting in more empty space. To see this trend for yourself, investigate it with our lattice energy calculator!

Find more about crystallography with our cubic cell calculator!

## FAQ

### How to calculate lattice energy?

You can either construct a Born-Haber cycle or use a lattice energy equation to find lattice energy. The Born-Haber cycle is more accurate as it is derived experimentally, but requires a larger amount of data. Lattice energy formulas, such as the Kapustinskii equation, are easy to use but are only estimates.

### What is the lattice energy of CaO?

The lattice energy of CaO is 3460 kJ/mol.

### What determines lattice energy?

Lattice energy is influenced by a number of factors:

• The number of ions in the crystal's empirical formula;
• The charge on both the anion and cation;
• The ionic radii of both the anion and cation; and
• The structure of the crystal lattice.

### What is the lattice energy of NaCl?

787.3 kJ/mol is the lattice energy of NaCl.

💡 Did you know that NaCl is actually table salt!

Jack Bowater
Method
Chosen approximation
Kapustinskii
Cation
Element
Na
Charge
pm
Anion
Element
Cl
Charge
pm
Stoichiometry
Lattice energy
Kapustinskii
kJ
/mol
People also viewed…

### Atomic mass

The atomic mass calculator takes the number of protons and neutrons in an atom and displays the atomic mass in atomic mass units and kilograms.

### Discount

Discount calculator uses a product's original price and discount percentage to find the final price and the amount you save.

### Percentage concentration to molarity

With our tool, you can easily convert percentage concentration to molarity.

### Sleep

The sleep calculator can help you determine when you should go to bed to wake up happy and refreshed.