The mole and the Avogadro constant

1. Overview

Atoms, molecules and ions are far too small to count or weigh one at a time, yet reactions happen particle by particle. The mole solves this. It is a fixed, enormous number of particles, set by the Avogadro constant, and its mass in grams is easy to work out from the Periodic Table. This section defines the mole and shows how to move between mass, amount and number of particles, the step behind almost every calculation in AS chemistry.

Key Definitions

• Amount of substance (n): a measure of how many specified particles a sample contains, counted in moles; its unit is the mole (mol).
• Mole (mol): the amount of substance that contains 6.022 × 10²³ specified particles (atoms, molecules, ions or electrons); this number of particles is given by the Avogadro constant.
• Avogadro constant (L): the number of specified particles per mole of a substance, 6.022 × 10²³ mol⁻¹.
• Molar mass (M): the mass per mole of a substance, in g mol⁻¹; it has the same numerical value as the substance's relative atomic mass, relative molecular mass or relative formula mass.
• Formula unit: the group of ions shown by the formula of an ionic compound, for example one Mg²⁺ ion and two Cl⁻ ions in MgCl₂; ionic compounds are counted in formula units because they contain no molecules.

Content

Why chemists count in moles

A balance measures mass, but reactions work by numbers of particles. In 2Mg(s) + O₂(g) → 2MgO(s), two magnesium atoms react with each oxygen molecule. To use that ratio in the lab, you need a way to turn a mass into a number of particles.

The numbers involved are huge. A single drop of water (about 0.05 g) contains roughly $2 \times 10^{21}$ molecules. So chemists count particles in a large, fixed bundle called the mole, in the same way that "a dozen" always means 12 of something.

The mole and the Avogadro constant

Amount of substance, symbol $n$, is the quantity that counts particles. Its unit is the mole (mol), one of the seven SI base units. Questions may say "amount, in mol" or "number of moles": both mean $n$.

One mole is the amount of substance that contains $6.022 \times 10^{23}$ specified particles. The number of particles per mole is the Avogadro constant, $L$ (some books write $N_{\text{A}}$):

$$L = 6.022 \times 10^{23}\ \text{mol}^{-1}$$

• "Specified particles" means you must say what you are counting: atoms, molecules, ions, electrons or formula units. One mole of anything contains $6.022 \times 10^{23}$ of that thing.
• The unit mol⁻¹ means "per mole". The bare number $6.022 \times 10^{23}$, with no unit, is sometimes called the Avogadro number.
• The number of particles, $N$, in a sample is the amount multiplied by $L$:

$$N = nL$$

The units check: mol × mol⁻¹ leaves no unit, which is right, because a number of particles is a pure count.

Older books define the mole as the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. Since 2019 the mole has been defined by fixing $L$ at exactly $6.022\,140\,76 \times 10^{23}$ mol⁻¹. The two definitions agree to better than one part in a billion, so they give the same answer in any calculation you will meet. When you are asked to define the mole, give the form based on the Avogadro constant.

Molar mass: linking moles to grams

The unified atomic mass unit, u, is one twelfth of the mass of a carbon-12 atom, and relative masses such as $A_{\text{r}}$ (relative atomic mass) and $M_{\text{r}}$ (relative molecular mass) compare particle masses with u. One gram is very nearly $6.022 \times 10^{23}$ u, the same number that appears in $L$:

$$1\ \text{u} \approx \frac{1}{L}\ \text{g} = 1.66 \times 10^{-24}\ \text{g}$$

So a particle with a relative mass of 18.0 has a mass of 18.0 u, and $L$ of those particles have a mass of 18.0 g. A carbon-12 atom has a mass of 12 u, so one mole of carbon-12 atoms has a mass of 12 g.

The mass per mole of a substance is its molar mass, $M$: numerically, the mass in grams of one mole. It has the same number as $A_{\text{r}}$, $M_{\text{r}}$ or the relative formula mass, but $M$ has the unit g mol⁻¹, while the relative masses have no unit. This gives one of the equations you will use most often:

$$n = \frac{m}{M}$$

where $m$ is the mass in g and $M$ is the molar mass in g mol⁻¹.

For an element with isotopes, $A_{\text{r}}$ is a weighted average. One mole of natural bromine atoms has a mass of 79.9 g, even though no single bromine atom has a mass of 79.9 u (its two isotopes have masses close to 79 u and 81 u).

Always say which particles

"One mole of oxygen" is ambiguous. It could mean oxygen atoms or O₂ molecules, and the two differ by a factor of 2. Name the particle every time, or use a label such as $n(\text{O}_2)$.

One mole of...	Contains	Mass
O atoms	$6.022 \times 10^{23}$ atoms	16.0 g
O₂ molecules	2 mol of O atoms	32.0 g
NH₃ molecules	1 mol of N atoms and 3 mol of H atoms (4 mol of atoms); 10 mol of electrons	17.0 g
MgCl₂ formula units	1 mol of Mg²⁺ ions and 2 mol of Cl⁻ ions (3 mol of ions)	95.3 g
Mg²⁺ ions	10 mol of electrons (12 − 2)	24.3 g
CuSO₄·5H₂O formula units	1 mol of Cu²⁺, 1 mol of SO₄²⁻ and 5 mol of H₂O	249.6 g

• Molecules: multiply by the number of each atom in the formula. One mole of NH₃ contains 3 mol of H atoms.
• Ionic compounds contain no molecules, so count formula units. A polyatomic ion such as SO₄²⁻ counts as one ion, even though it contains 5 atoms.
• Ions and electrons: adding or removing electrons barely changes the mass. Two moles of electrons have a mass of only about 0.001 g, so one mole of Mg²⁺ ions still has a mass of 24.3 g.

Moving between mass, amount and number of particles

The amount in mol is the hub. Go through it every time:

From	To	Do this
mass, $m$ (g)	amount, $n$ (mol)	divide by $M$
amount, $n$ (mol)	number of particles, $N$	multiply by $L$
number of molecules or formula units	number of atoms or ions	multiply by the number in one formula
number of particles, $N$	amount, $n$ (mol)	divide by $L$
amount, $n$ (mol)	mass, $m$ (g)	multiply by $M$

Mass of one particle. One mole has a mass of $M$ grams and contains $L$ particles, so one particle has a mass of $M/L$. For water, $18.0 \div (6.022 \times 10^{23}) = 2.99 \times 10^{-23}$ g. For an element with isotopes, this is the average mass of one atom.

Sense checks.

• A sample you can weigh contains an enormous number of particles. If you get a tiny value such as $10^{-25}$, you divided by $L$ where you should have multiplied; a value such as $10^{46}$ means the reverse.
• One atom or molecule has a tiny mass: a hydrogen atom is only about $1.7 \times 10^{-24}$ g.

Worked example 1: atoms in a molecular compound

Question: Urea, CO(NH₂)₂, is used as a fertiliser. A sample contains 4.50 g of urea. Calculate (a) the number of urea molecules and (b) the number of nitrogen atoms in the sample. ($A_{\text{r}}$: H 1.0, C 12.0, N 14.0, O 16.0; $L = 6.022 \times 10^{23}$ mol⁻¹)

Step 1: molar mass. $$M = 12.0 + 16.0 + 2 \times (14.0 + 2 \times 1.0) = 60.0\ \text{g mol}^{-1}$$

Step 2: amount of urea. $$n(\text{urea}) = \frac{m}{M} = \frac{4.50}{60.0} = 0.0750\ \text{mol}$$

Step 3: number of molecules. $$N(\text{molecules}) = nL = 0.0750 \times 6.022 \times 10^{23} = 4.5165 \times 10^{22}$$

Step 4: number of nitrogen atoms. Each molecule contains 2 N atoms. $$N(\text{N atoms}) = 2 \times 4.5165 \times 10^{22} = 9.033 \times 10^{22}$$

Answer: (a) $4.52 \times 10^{22}$ molecules; (b) $9.03 \times 10^{22}$ nitrogen atoms (3 s.f.).

You can also do Step 4 in moles: $n(\text{N atoms}) = 2 \times 0.0750 = 0.150$ mol, and $0.150 \times 6.022 \times 10^{23} = 9.03 \times 10^{22}$. Both routes give the same answer.

Worked example 2: ions in an ionic compound

Question: Iron(III) sulfate, Fe₂(SO₄)₃, is an ionic compound. For a 2.50 g sample, calculate (a) the amount, in mol, of sulfate ions, (b) the number of sulfate ions and (c) the total number of ions. ($A_{\text{r}}$: O 16.0, S 32.1, Fe 55.8)

Step 1: molar mass. $$M = 2 \times 55.8 + 3 \times (32.1 + 4 \times 16.0) = 111.6 + 288.3 = 399.9\ \text{g mol}^{-1}$$

Step 2: amount of formula units. $$n(\text{Fe}_2(\text{SO}_4)_3) = \frac{2.50}{399.9} = 6.2516 \times 10^{-3}\ \text{mol}$$

Step 3: amount of sulfate ions. One formula unit contains 2 Fe³⁺ ions and 3 SO₄²⁻ ions, 5 ions in all. $$n(\text{SO}_4^{2-}) = 3 \times 6.2516 \times 10^{-3} = 0.018755\ \text{mol}$$

Step 4: number of sulfate ions. $$N(\text{SO}_4^{2-}) = 0.018755 \times 6.022 \times 10^{23} = 1.1294 \times 10^{22}$$

Step 5: total number of ions. $$N(\text{ions}) = 5 \times 6.2516 \times 10^{-3} \times 6.022 \times 10^{23} = 1.8823 \times 10^{22}$$

Answer: (a) 0.0188 mol; (b) $1.13 \times 10^{22}$ sulfate ions; (c) $1.88 \times 10^{22}$ ions in total (3 s.f.).

Each sulfate ion contains 5 atoms, but it is one ion. Counting its atoms separately would be a different question.

Worked example 3: identifying an element from a count of atoms

Question: A sample of a metallic element contains $3.50 \times 10^{22}$ atoms and has a mass of 2.78 g. Calculate the relative atomic mass of the element and identify it.

Step 1: amount of atoms. $$n = \frac{N}{L} = \frac{3.50 \times 10^{22}}{6.022 \times 10^{23}} = 0.05812\ \text{mol}$$

Step 2: molar mass. $$M = \frac{m}{n} = \frac{2.78}{0.05812} = 47.83\ \text{g mol}^{-1}$$

Answer: $A_{\text{r}} = 47.8$ (3 s.f.). The closest value in the Periodic Table is 47.9, so the element is titanium. Its neighbours, scandium (45.0) and vanadium (50.9), are much further away.

Another route gives the same result: the average mass of one atom is $2.78 \div (3.50 \times 10^{22}) = 7.94 \times 10^{-23}$ g, and multiplying by $L$ gives 47.8 g mol⁻¹. The small gap between 47.8 and 47.9 is only rounding: both data values are given to 3 s.f., and a mass of 2.785 g, which still rounds to 2.78 g, would give 47.9 g mol⁻¹.

Comparing samples and reading equations

Equal amounts in mol contain equal numbers of particles, whatever the substance. So to compare numbers of particles, compare amounts: multiplying every amount by $L$ never changes which is larger.

For example, 3.00 g of ethane, C₂H₆ ($M$ = 30.0 g mol⁻¹), is 0.100 mol of molecules. Each molecule has 8 atoms, so the sample holds 0.800 mol of atoms. 16.0 g of argon ($M$ = 39.9 g mol⁻¹) is 0.401 mol of atoms. So the ethane sample has fewer molecules than the argon sample has atoms, but it holds about twice as many atoms in total.

The numbers in a balanced equation are ratios of particles, and so also ratios of moles. In

N₂(g) + 3H₂(g) → 2NH₃(g)

one N₂ molecule reacts with three H₂ molecules to make two NH₃ molecules. Scaling every number up by $L$ gives the same ratio in moles: 1 mol of N₂ reacts with 3 mol of H₂ to make 2 mol of NH₃. So 0.10 mol of N₂ ($6.0 \times 10^{22}$ molecules) needs 0.30 mol of H₂ ($1.8 \times 10^{23}$ molecules) and makes 0.20 mol of NH₃.

The total amount of molecules can change in a reaction (4 mol become 2 mol here), but the amount of each type of atom cannot: there are 2 mol of N atoms and 6 mol of H atoms on each side. Reacting-mass, gas-volume and solution calculations all build on this idea.

Key Equations

Number of particles ($N$ = number of particles, no unit; $n$ = amount in mol; $L$ = Avogadro constant, $6.022 \times 10^{23}$ mol⁻¹): $$N = nL \qquad n = \frac{N}{L}$$

Amount from mass ($m$ = mass in g; $M$ = molar mass in g mol⁻¹): $$n = \frac{m}{M}$$

Mass of one particle (in g; for an element with isotopes, the average mass of one atom): $$m_{\text{particle}} = \frac{M}{L}$$

Atoms or ions from molecules or formula units: $$N(\text{atoms or ions of one kind}) = N(\text{molecules or formula units}) \times \text{number of that kind in one formula}$$

Link to the unified atomic mass unit ($M$ in g mol⁻¹ has the same number as $A_{\text{r}}$, $M_{\text{r}}$ or relative formula mass): $$1\ \text{u} \approx \frac{1}{L}\ \text{g} = 1.66 \times 10^{-24}\ \text{g}$$

Common Mistakes to Avoid

• Mistake: writing "one mole of oxygen" without saying which particles. Fix: name them. One mole of O atoms has a mass of 16.0 g, but one mole of O₂ molecules has a mass of 32.0 g and contains $1.20 \times 10^{24}$ atoms.
• Mistake: stopping at the number of molecules when the question asks for atoms or ions. Fix: multiply by the number in one formula. In Worked example 1, stopping after Step 3 gives $4.52 \times 10^{22}$ instead of $9.03 \times 10^{22}$ nitrogen atoms, only half the right answer.
• Mistake: multiplying a mass by $L$. Fix: $L$ multiplies an amount in mol, never a mass in g. In Worked example 1, $4.50 \times 6.022 \times 10^{23} = 2.71 \times 10^{24}$, which is 60 times the true number of molecules.
• Mistake: multiplying by $L$ when you should divide, or the reverse. Fix: check the units: $N = nL$ (mol × mol⁻¹ leaves no unit) and $n = N/L$. In Worked example 3, $3.50 \times 10^{22} \times 6.022 \times 10^{23}$ gives $2.11 \times 10^{46}$, an absurd "amount" for 2.78 g of metal, and its unit would be mol⁻¹, not mol.
• Mistake: assuming a mole always means $6.022 \times 10^{23}$ atoms. Fix: it means $6.022 \times 10^{23}$ of whichever particle is named. One mole of NH₃ contains $6.022 \times 10^{23}$ molecules but $2.41 \times 10^{24}$ atoms, because each molecule has 4 atoms.
• Mistake: rounding the amount in mol early. Fix: keep the calculator value until the end. In Worked example 2, rounding $n$ to 0.006 mol gives $1.08 \times 10^{22}$ sulfate ions instead of $1.13 \times 10^{22}$.
• Mistake: mixing up units. Fix: urea has $M$ = 60.0 g mol⁻¹ but $M_{\text{r}}$ = 60.0 with no unit; $L$ is $6.022 \times 10^{23}$ mol⁻¹; and a count such as $4.52 \times 10^{22}$ molecules has no unit at all.

Exam Tips

• Define the mole in full: the amount of substance that contains $6.022 \times 10^{23}$ particles, the number given by the Avogadro constant. Leaving out "amount of substance" or the number makes the definition incomplete. When you use the mole, always name the particles: atoms, molecules, ions or electrons.
• Calculate the number of atoms, molecules or ions: the answer is a count, so the working must include multiplying by $L$; an amount in mol is not a number of particles. Set out the chain one step per line (mass, then amount, then particles, then atoms), and label each line, for example $n(\text{urea})$ or $N(\text{N atoms})$.
• Use the Data Booklet values: $L = 6.022 \times 10^{23}$ mol⁻¹ and $A_{\text{r}}$ from the Periodic Table. Give a count in standard form, to as many s.f. as the data justify (usually 3), and round only at the end.
• Compare samples in a multiple-choice question by finding the amount of the particle asked about (atoms, molecules or ions) for each one. There is no need to multiply by $L$, because it does not change the order.
• Identify an element from a mass and a number of atoms by working out $M$, then choosing the element with the closest $A_{\text{r}}$. Quote both values, for example "47.8, closest to titanium, 47.9".