3.1.2.2 - The Mole and the Avogadro Constant

What a mole counts

Chemistry often deals with unimaginably large numbers of particles. The mole gives us a practical way to count them by weighing.

The mole is defined using this constant, so the two ideas always go together.

The key word is specified. A mole can refer to different kinds of entities:

• 1 mol of atoms
• 1 mol of molecules
• 1 mol of ions
• 1 mol of electrons
• 1 mol of formula units in an ionic compound

So 1 mol of sodium atoms, 1 mol of chloride ions, and 1 mol of water molecules each contain the same number of particles, even though their masses are different.

For ionic substances, be careful with language. Sodium chloride does not exist as discrete molecules, so we talk about formula units of NaCl, not NaCl molecules.

The diagram below compares different kinds of entities that can each make up 1 mol while keeping the particle count the same in every case.

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Finding moles from mass

In many calculations, the first step is to find the amount of substance, $n$, in moles.

For a substance with known mass:

Here, $M$ is the molar mass in $\mathrm{g\,mol^{-1}}$. It has the same numerical value as the relative molecular/formula mass; this is why you may also see the numerical shortcut $n=m/M_r$ when mass is in grams. Here, $m$ is the mass in g. For molecules and formula units, $M_r$ is the relative molecular mass or relative formula mass. For atoms, use $A_r$ to identify the molar mass in $\mathrm{g\ mol^{-1}}$.

Use the formula that matches the particle being counted. For example:

• $A_r(\mathrm{O}) = 16.0$, so 1 mol of oxygen atoms has a molar mass of $16.0\ \mathrm{g\ mol^{-1}}$ and a mass of 16.0 g
• $M_r(\mathrm{O_2}) = 32.0$, so 1 mol of oxygen molecules has a molar mass of $32.0\ \mathrm{g\ mol^{-1}}$ and a mass of 32.0 g

That distinction matters because the particle you are counting changes the molar mass.

Worked example: How many moles are in 9.00 g of water, $\mathrm{H_2O}$?

Keep unrounded calculator values between steps. For multiplication and division, quote the final answer to the precision justified by the least precise measured value; for example, 9.00 g has three significant figures, so 0.500 mol has three too. Exact formula subscripts and equation coefficients do not limit precision. Very small particle counts are written in standard form, $a\times10^b$, with $1\leq a<10$.

Once you can find moles from mass, you can compare how many particles are present in different samples and prepare for equation work.

Using the Avogadro constant

The Avogadro constant converts between moles and number of particles.

Here, $N$ is the number of particles and $N_A$ is the Avogadro constant.

You can also rearrange this:

This works for any specified entity. For example:

• atoms in a sample of copper
• molecules in a sample of carbon dioxide
• ions in a sample of sodium chloride solution
• electrons transferred in a redox process

Worked example: How many molecules are present in 0.250 mol of carbon dioxide?

If you need atoms rather than molecules, you need an extra step. Each $\mathrm{CO_2}$ molecule contains 3 atoms, so $0.250$ mol of $\mathrm{CO_2}$ contains $3 \times 1.51 \times 10^{23}$ atoms in total.

For ionic substances, the formula tells you how many ions come from one formula unit. For example, $0.200$ mol of $\mathrm{CaCl_2}$ contains $0.200$ mol of $\mathrm{Ca^{2+}}$ ions and $0.400$ mol of $\mathrm{Cl^-}$ ions. If you wanted the number of chloride ions, you would then multiply $0.400$ mol by $N_A$.

That same idea lets you move smoothly between moles, particle counts, and ion counts.

Moles in solutions and concentration

When a substance is dissolved, we often need the amount present in a certain volume of solution.

The main relationship is:

In this equation, $c$ is in $\mathrm{mol\ dm^{-3}}$, $n$ is in mol, and $V$ must be in $\mathrm{dm^3}$.

This unit conversion is essential:

So if volume is given in $\mathrm{cm^3}$, divide by 1000 before using $c = n/V$. Rearranging gives:

or

Worked example: What is the concentration of a solution made by dissolving 0.200 mol of sodium hydroxide in $250\ \mathrm{cm^3}$ of solution?

First convert the volume:

Then calculate concentration:

Students often make two avoidable mistakes here:

• using the volume in $\mathrm{cm^3}$ directly in $c=n/V$
• forgetting that the volume is the volume of the solution, not just the solvent added

A quick unit check helps here: $250\ \mathrm{cm^3}$ must become $0.250\ \mathrm{dm^3}$ before you substitute into $c=n/V$, because the concentration formula is defined using $\mathrm{dm^3}$.

These concentration calculations are the groundwork for later titration and reacting-volume work, so getting comfortable with the units now is important.

Using the mole in equations

Balanced chemical equations do more than show which substances react. The coefficients also show the simplest whole-number ratio in which particles react, so they also give the mole ratio.

For example:

This means:

• 1 mol of $\mathrm{N_2}$ reacts with 3 mol of $\mathrm{H_2}$
• 1 mol of $\mathrm{N_2}$ forms 2 mol of $\mathrm{NH_3}$

The coefficients do not tell you a mass ratio. They tell you the ratio of moles.

A dependable route through these calculations is:

1. Write the balanced equation, or check that the one given is balanced.
2. Read off the mole ratio you need from the coefficients.
3. Convert any given mass or solution data into moles first.
4. Use the ratio to find the unknown amount in moles.

Worked example: How many moles of hydrogen are needed to react completely with $0.400$ mol of nitrogen?

From $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, the ratio $\mathrm{N_2:H_2}$ is $1:3$.

If you were given a mass of nitrogen instead, first convert that mass into moles using $n = m/M_r$, then apply the same ratio step.

Once you have found an amount in moles, you can convert it into mass, number of particles, or concentration depending on the quantity you need. That is why the mole is the bridge between grams, particles, solutions, and chemical equations.