The Mole and Avogadro Constant

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / the mole, content statements defining the mole as a unit of amount of substance; introducing the Avogadro constant $L$L; relating moles, mass and molar mass through $n = m/M$n=m/M; and converting between numbers of particles and moles via $N = nL$N=nL (refer to the official OCR H432 specification document for exact wording). The mole is the master concept of quantitative chemistry — every titration, every yield, every reaction-mass calculation in the rest of the course routes through this lesson.

The mole is A-Level chemistry's most powerful and most-misunderstood concept. It is simply a count — like "a dozen" or "a ream" — but the number it represents is so vast that it functions as the bridge between the microscopic world of individual atoms and the macroscopic world of grams and cubic decimetres. This lesson develops three connected threads. First, the definition of the mole itself, both in its pre-2019 form (the amount of substance containing as many particles as there are atoms in exactly 12 g of $^{12}\text{C}$12C) and in the post-2019 SI form (the amount containing exactly $6.022\,140\,76 \times 10^{23}$6.02214076×1023 entities, with Avogadro's constant $L$L now fixed by definition). Second, molar mass $M$M as the numerical bridge between dimensionless $A_r / M_r$Ar​/Mr​ and macroscopic grams. Third, the master equation $n = m/M$n=m/M and its companion $N = nL$N=nL, which between them solve essentially every mass↔moles↔particles question OCR will ever ask. The historical figure is Amedeo Avogadro (1776–1856), whose 1811 hypothesis that equal volumes of gases at the same temperature and pressure contain equal numbers of particles laid the conceptual groundwork; the constant bearing his name was first estimated by Johann Loschmidt in 1865 and is now an SI-defined exact number.

Key Equation: $n = \dfrac{m}{M} \qquad N = nL$n=Mm​N=nL where $n$n is amount of substance in mol, $m$m is mass in g, $M$M is molar mass in $\text{g mol}^{-1}$g mol−1, $N$N is the number of particles, and $L = 6.02 \times 10^{23}\text{ mol}^{-1}$L=6.02×1023 mol−1 is the Avogadro constant.

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Why Chemists Use Moles

Atoms and molecules are far too small to count individually. A teaspoon of water (about $5\text{ g}$5 g) contains approximately $1.7 \times 10^{23}$1.7×1023 molecules — a number so large that, even at one molecule per second, it would take the entire human population working in parallel for many trillions of years to enumerate them. Chemists count particles in bundles called moles, in the same way that a shopkeeper counts eggs in dozens or a printer counts paper in reams.

The mole is the bridge between the microscopic world (individual atoms and molecules, masses measured in $A_r$Ar​ or $M_r$Mr​) and the macroscopic world (grams and cubic decimetres measured in the lab). Every quantitative calculation in A-Level chemistry passes through this bridge.

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Definition of the Mole

Historical IUPAC definition (pre-2019)

The mole (mol) is the amount of substance that contains the same number of particles as there are atoms in exactly $12\text{ g}$12 g of $^{12}\text{C}$12C.

That number is the Avogadro constant, symbol $L$L (or $N_A$NA​):

$L = 6.02 \times 10^{23}\text{ mol}^{-1}$L=6.02×1023 mol−1

You should memorise this value; it is provided in the OCR data booklet but is needed in calculations without delay.

The 2019 SI redefinition

In May 2019 the SI definition was updated: the mole is now defined as containing exactly $6.022\,140\,76 \times 10^{23}$6.02214076×1023 elementary entities. The mass of $12\text{ g}$12 g of $^{12}\text{C}$12C is no longer the reference; instead, Avogadro's number itself is fixed by definition. This is part of the same SI redefinition that fixed Planck's constant to define the kilogram, eliminating the platinum-iridium "International Prototype Kilogram" as a physical artefact.

For A-Level purposes the practical meaning is unchanged: 1 mole $= 6.02 \times 10^{23}$=6.02×1023 particles. The new definition is more precise and philosophically cleaner.

The word "particles" is deliberately generic. A mole of anything is $6.02 \times 10^{23}$6.02×1023 of that thing:

Substance	1 mole contains
Carbon	$6.02 \times 10^{23}$6.02×1023 atoms
Water	$6.02 \times 10^{23}$6.02×1023 molecules
Sodium chloride	$6.02 \times 10^{23}$6.02×1023 formula units
Electrons	$6.02 \times 10^{23}$6.02×1023 electrons
Photons (in principle)	$6.02 \times 10^{23}$6.02×1023 photons

How big is Avogadro's number?

To appreciate the scale, a mole of golf balls would occupy a volume comparable to that of the Earth; a mole of rice grains spread evenly would coat the surface of every continent kilometres deep. And yet just $18\text{ g}$18 g of water (a tablespoon) contains this many molecules. The chasm of scale between $L$L and everyday counts is exactly what makes the mole indispensable as a chemical counting unit.

Dimensional analysis and the mole

The mole is one of the seven base SI units, alongside the second, metre, kilogram, kelvin, ampere and candela. Its dimensional symbol is N (amount of substance) — entirely distinct from mass M. The equation $n = m/M$n=m/M therefore has dimensions $[\text{N}] = [\text{M}] / [\text{M}\text{N}^{-1}]$[N]=[M]/[MN−1], which simplifies to $[\text{N}]$[N] — consistent. Many quantitative slips can be diagnosed by walking through the dimensions: if your "moles" answer carries grams in its units, the dimensional analysis has failed, and the calculation must be re-examined.

A consequence of the dimensional treatment is that molar mass has units $\text{g mol}^{-1}$g mol−1, which is dimensionally $[\text{M}][\text{N}]^{-1}$[M][N]−1. The Avogadro constant $L$L has units $\text{mol}^{-1}$mol−1, dimensionally $[\text{N}]^{-1}$[N]−1. The product $LM$LM has units $\text{g}$g — the mass of one entity (atom, molecule, or formula unit). For water, $L \cdot M(\text{H}_2\text{O}) = 6.02 \times 10^{23}\text{ mol}^{-1} \times 18\text{ g mol}^{-1} = 1.08 \times 10^{25}\text{ g}$L⋅M(H2​O)=6.02×1023 mol−1×18 g mol−1=1.08×1025 g — but this is the mass of one mole, not one molecule, because of the units cancelling. The correct calculation for the mass of one water molecule is $M / L = 18 / (6.02 \times 10^{23}) = 2.99 \times 10^{-23}\text{ g}$M/L=18/(6.02×1023)=2.99×10−23 g. Dimensional analysis catches this kind of unit-arithmetic error.

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Molar Mass $M$M

Molar mass is the mass in grams of one mole of a substance. Numerically, molar mass equals the relative atomic or molecular mass, but with units of $\text{g mol}^{-1}$g mol−1:

Species	$A_r$Ar​ or $M_r$Mr​	$M / \text{g mol}^{-1}$M/g mol−1
Na	23.0	23.0
H₂O	18.0	18.0
CaCO₃	100.1	100.1
NaOH	40.0	40.0
KMnO₄	158.0	158.0
Fe₂(SO₄)₃	399.9	399.9

The distinction is important: $A_r$Ar​ and $M_r$Mr​ are dimensionless while molar mass has units of $\text{g mol}^{-1}$g mol−1. Treating the dimensionless ratio as if it had units is a typical exam slip.

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The Master Equation $n = m/M$n=m/M

The most important equation in A-Level Chemistry quantitative work:

$n = \frac{m}{M}$n=Mm​

where $n$n is amount of substance (mol), $m$m is mass (g), $M$M is molar mass ($\text{g mol}^{-1}$g mol−1).

flowchart LR
    A[Mass m in g] -- divide by M --> B[Moles n]
    B -- multiply by M --> A
    B -- multiply by L --> C[Number of particles N]
    C -- divide by L --> B

Worked Example 1 — Mass to moles

Calculate the number of moles in $40.0\text{ g}$40.0 g of NaOH.

• $M(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0\text{ g mol}^{-1}$M(NaOH)=23.0+16.0+1.0=40.0 g mol−1
• $n = m/M = 40.0 / 40.0 =$n=m/M=40.0/40.0= $1.00\text{ mol}$1.00 mol

Worked Example 2 — Moles to mass

What mass of CaCl₂ contains $0.250\text{ mol}$0.250 mol?

• $M(\text{CaCl}_2) = 40.1 + 2(35.5) = 111.1\text{ g mol}^{-1}$M(CaCl2​)=40.1+2(35.5)=111.1 g mol−1
• $m = n \times M = 0.250 \times 111.1 =$m=n×M=0.250×111.1= $27.8\text{ g}$27.8 g

Worked Example 3 — Moles to particles

How many molecules in $0.500\text{ mol}$0.500 mol of CO₂?

• $N = nL = 0.500 \times 6.02 \times 10^{23} =$N=nL=0.500×6.02×1023= $3.01 \times 10^{23}\text{ molecules}$3.01×1023 molecules

Worked Example 4 — Particles to moles

A sample contains $1.50 \times 10^{24}$1.50×1024 atoms of iron. How many moles?

• $n = N/L = 1.50 \times 10^{24} / 6.02 \times 10^{23} =$n=N/L=1.50×1024/6.02×1023= $2.49\text{ mol}$2.49 mol

Worked Example 5 — Mass to particles (two steps)

How many molecules in $9.00\text{ g}$9.00 g of water?

• Step 1: $n = m/M = 9.00 / 18.0 = 0.500\text{ mol}$n=m/M=9.00/18.0=0.500 mol
• Step 2: $N = nL = 0.500 \times 6.02 \times 10^{23} =$N=nL=0.500×6.02×1023= $3.01 \times 10^{23}\text{ molecules}$3.01×1023 molecules

Worked Example 6 — Atoms of a specific element in a compound

How many oxygen atoms in $5.00\text{ g}$5.00 g of glucose, C₆H₁₂O₆?

• $M(\text{C}_6\text{H}_{12}\text{O}_6) = 6(12.0) + 12(1.0) + 6(16.0) = 180.0\text{ g mol}^{-1}$M(C6​H12​O6​)=6(12.0)+12(1.0)+6(16.0)=180.0 g mol−1
• $n(\text{glucose}) = 5.00 / 180.0 = 0.02778\text{ mol}$n(glucose)=5.00/180.0=0.02778 mol
• $n(\text{O atoms}) = 6 \times 0.02778 = 0.1667\text{ mol}$n(O atoms)=6×0.02778=0.1667 mol
• $N(\text{O}) = 0.1667 \times 6.02 \times 10^{23} =$N(O)=0.1667×6.02×1023= $1.00 \times 10^{23}\text{ atoms of O}$1.00×1023 atoms of O

Worked Example 7 — Mass of one atom

Calculate the mass of one $^{12}\text{C}$12C atom.

• One mole of $^{12}\text{C}$12C has mass $12.00\text{ g}$12.00 g and contains $6.02 \times 10^{23}$6.02×1023 atoms
• Mass of 1 atom $= 12.00 / (6.02 \times 10^{23}) =$=12.00/(6.02×1023)= $1.993 \times 10^{-23}\text{ g}$1.993×10−23 g

Worked Example 8 — Finding $A_r$Ar​ from a single-atom mass

An atom of an unknown element has mass $1.06 \times 10^{-22}\text{ g}$1.06×10−22 g. Estimate $A_r$Ar​.

• Mass of 1 mole $= 1.06 \times 10^{-22} \times 6.02 \times 10^{23} = 63.8\text{ g}$=1.06×10−22×6.02×1023=63.8 g
• $A_r \approx 63.8$Ar​≈63.8, matching Cu ($A_r = 63.5$Ar​=63.5) — the element is copper.

Worked Example 9 — Counting specific ions in a mixture

A chemist mixes $10.0\text{ g}$10.0 g of NaCl with $10.0\text{ g}$10.0 g of KCl. How many chloride ions in total?

• $n(\text{NaCl}) = 10.0 / 58.5 = 0.1709\text{ mol} \rightarrow 0.1709\text{ mol Cl}^-$n(NaCl)=10.0/58.5=0.1709 mol→0.1709 mol Cl−
• $n(\text{KCl}) = 10.0 / 74.6 = 0.1340\text{ mol} \rightarrow 0.1340\text{ mol Cl}^-$n(KCl)=10.0/74.6=0.1340 mol→0.1340 mol Cl−
• Total $n(\text{Cl}^-) = 0.3049\text{ mol}$n(Cl−)=0.3049 mol
• $N(\text{Cl}^-) = 0.3049 \times 6.02 \times 10^{23} =$N(Cl−)=0.3049×6.02×1023= $1.84 \times 10^{23}\text{ ions}$1.84×1023 ions

Worked Example 10 — Atoms in a polyatomic salt

How many hydrogen atoms in $12.0\text{ g}$12.0 g of ammonium sulfate, $(\text{NH}_4)_2\text{SO}_4$(NH4​)2​SO4​?

• $M = 2[14.0 + 4(1.0)] + 32.1 + 4(16.0) = 132.1\text{ g mol}^{-1}$M=2[14.0+4(1.0)]+32.1+4(16.0)=132.1 g mol−1
• $n((\text{NH}_4)_2\text{SO}_4) = 12.0 / 132.1 = 0.09084\text{ mol}$n((NH4​)2​SO4​)=12.0/132.1=0.09084 mol
• 8 H per formula unit: $n(\text{H}) = 8 \times 0.09084 = 0.7267\text{ mol}$n(H)=8×0.09084=0.7267 mol
• $N(\text{H}) = 0.7267 \times 6.02 \times 10^{23} =$N(H)=0.7267×6.02×1023= $4.37 \times 10^{23}\text{ atoms}$4.37×1023 atoms

Worked Example 10b — Reverse direction (given moles, find mass)

A reaction requires $0.0750\text{ mol}$0.0750 mol of $\text{Na}_2\text{CO}_3$Na2​CO3​. What mass should be weighed out?

• $M(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}$M(Na2​CO3​)=2(23.0)+12.0+3(16.0)=106.0 g mol−1
• $m = nM = 0.0750 \times 106.0 =$m=nM=0.0750×106.0= $7.95\text{ g}$7.95 g

Worked Example 10c — Reverse direction (given number of particles, find mass)

A sample contains $3.01 \times 10^{22}$3.01×1022 atoms of magnesium. Find its mass.

• $n = N/L = 3.01 \times 10^{22} / 6.02 \times 10^{23} = 0.0500\text{ mol}$n=N/L=3.01×1022/6.02×1023=0.0500 mol
• $m = nM = 0.0500 \times 24.3 =$m=nM=0.0500×24.3= $1.215\text{ g}$1.215 g ≈ 1.22 g

Worked Example 10d — Three-step chained calculation

How many oxygen atoms are present in $25.0\text{ g}$25.0 g of magnesium nitrate, $\text{Mg(NO}_3)_2$Mg(NO3​)2​?

• $M(\text{Mg(NO}_3)_2) = 24.3 + 2[14.0 + 3(16.0)] = 24.3 + 124.0 = 148.3\text{ g mol}^{-1}$M(Mg(NO3​)2​)=24.3+2[14.0+3(16.0)]=24.3+124.0=148.3 g mol−1
• $n(\text{Mg(NO}_3)_2) = 25.0 / 148.3 = 0.1686\text{ mol}$n(Mg(NO3​)2​)=25.0/148.3=0.1686 mol
• Each formula unit contains 6 O atoms (3 in each of 2 NO₃⁻): $n(\text{O}) = 6 \times 0.1686 = 1.0117\text{ mol}$n(O)=6×0.1686=1.0117 mol
• $N(\text{O}) = 1.0117 \times 6.02 \times 10^{23} =$N(O)=1.0117×6.02×1023= $6.09 \times 10^{23}\text{ atoms}$6.09×1023 atoms

This three-step chain (mass → moles → moles of specific atom → number of atoms) is the prototype for many OCR-style "atoms in a sample" questions. The discipline is to track which species the moles refer to at every step — moles of compound is not the same as moles of constituent atom.

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A second look at $n = m/M$n=m/M via dimensional analysis

The master mole equation $n = m/M$n=m/M is so familiar that it is rarely examined formally, but its dimensional structure is worth noting. The unit balance:

$[n] = \frac{[m]}{[M]} = \frac{\text{g}}{\text{g mol}^{-1}} = \text{g} \times \text{mol g}^{-1} = \text{mol}$[n]=[M][m]​=g mol−1g​=g×mol g−1=mol

If you ever write a calculation where the answer ends up with units of $\text{g}^2 \text{ mol}^{-1}$g2 mol−1 or some other nonsensical combination, the dimensional analysis catches the error before the marking scheme does. A useful habit at A-Level is to write the units of every quantity alongside its numerical value during the calculation, and check that the units of the final answer are those expected (mol, g, or mol dm⁻³ as appropriate).

Mass-unit Conversions

You will encounter three alternative units of mass in A-Level questions:

Unit	Conversion to g
g (gram)	Standard for molar mass
kg (kilogram)	$\times 10^3$×103
mg (milligram)	$\div 10^3$÷103
µg (microgram)	$\div 10^6$÷106
tonne	$\times 10^6$×106

Always convert to grams before using $n = m/M$n=m/M.