Introducing the Mole

Higher tier: The mole is assessed on the Higher tier. Foundation candidates should still be confident with relative formula mass from the previous lesson, but do not need the mole calculations below.

Atoms are so small that counting them one at a time is hopeless — a single carbon atom weighs only about $2 \times 10^{-23}$2×10−23 grams. So chemists count atoms in huge, fixed-size batches called moles, in the same way that a stationery shop counts paper by the ream or a farmer counts eggs by the dozen. The mole is what links the masses we can weigh on a laboratory balance to the number of particles actually taking part in a reaction. This lesson, part of Topic C3 of OCR Gateway Combined Science, defines the mole and the Avogadro constant, shows how to switch between mass and moles using $M_r$Mr​, and uses balanced equations to work out reacting masses.

By the end of this lesson you should be able to define the mole and state the Avogadro constant, use relative atomic and formula mass, convert between mass and moles, and calculate the mass of a product or reactant from a balanced equation.

This lesson builds AO1 recall of the mole and the Avogadro constant, strong AO2 application through the mass–mole conversions and reacting-mass calculations, and AO3 analysis when you check that a computed answer is sensible against the equation's ratio.

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What Is a Mole?

A mole is the unit chemists use for amount of substance. One mole of anything contains the same number of particles, and that number is the Avogadro constant:

$6.02 \times 10^{23} \text{ particles per mole}$6.02×1023 particles per mole

So one mole of carbon atoms is $6.02 \times 10^{23}$6.02×1023 carbon atoms; one mole of water molecules is $6.02 \times 10^{23}$6.02×1023 water molecules. The mole is really just an enormous counting number, and it is chosen so that the mass of one mole of a substance in grams comes out numerically equal to its relative atomic or formula mass.

Substance	$A_r$Ar​ or $M_r$Mr​	Mass of 1 mole
Carbon (C)	12	12 g
Water ($\text{H}_2\text{O}$H2​O)	18	18 g
Carbon dioxide ($\text{CO}_2$CO2​)	44	44 g
Sodium chloride ($\text{NaCl}$NaCl)	58.5	58.5 g

Exam Tip: The mass of one mole in grams equals the $A_r$Ar​ (for an element) or the $M_r$Mr​ (for a compound). So "the mass of one mole of $\text{CO}_2$CO2​" is just its $M_r$Mr​, 44, written in grams — you do not need to do anything with the Avogadro number.

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Relative Atomic and Formula Mass (recap)

The relative atomic mass ($A_r$Ar​) of an element comes straight from the periodic table — it is the larger of the two numbers given for each element. The relative formula mass ($M_r$Mr​) of a compound is the sum of the $A_r$Ar​ values of all its atoms.

Worked Example 1 — $M_r$Mr​ values you will reuse

Using $A_r$Ar​: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Ca = 40, Cu = 64.

• $\text{O}_2$O2​: $2 \times 16 = 32$2×16=32
• $\text{H}_2\text{O}$H2​O: $(2 \times 1) + 16 = 18$(2×1)+16=18
• $\text{CO}_2$CO2​: $12 + (2 \times 16) = 44$12+(2×16)=44
• $\text{CaCO}_3$CaCO3​: $40 + 12 + (3 \times 16) = 100$40+12+(3×16)=100
• $\text{H}_2\text{SO}_4$H2​SO4​: $(2 \times 1) + 32 + (4 \times 16) = 98$(2×1)+32+(4×16)=98

These few values crop up in reacting-mass calculations over and over, so they are worth committing to memory.

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Converting Between Mass and Moles

The equation at the heart of this whole topic ties together the number of moles, the mass in grams and the relative formula mass:

$\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​

It rearranges two ways:

$\text{mass} = \text{moles} \times M_r \qquad \text{and} \qquad M_r = \dfrac{\text{mass}}{\text{moles}}$mass=moles×Mr​andMr​=molesmass​

A formula triangle is a handy prompt: cover the quantity you want, and the triangle shows the calculation that gives it.

Worked Example 2 — Mass to moles

How many moles are there in $36 \text{ g}$36 g of water ($\text{H}_2\text{O}$H2​O, $M_r = 18$Mr​=18)?

Step 1 — write the equation: $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​.

Step 2 — substitute: $\text{moles} = \dfrac{36}{18}$moles=1836​.

Step 3 — calculate: $\text{moles} = 2 \text{ mol}$moles=2 mol.

Answer: 2 moles of water.

Worked Example 3 — Moles to mass

What is the mass of $0.25 \text{ mol}$0.25 mol of calcium carbonate ($\text{CaCO}_3$CaCO3​, $M_r = 100$Mr​=100)?

Step 1 — rearrange: $\text{mass} = \text{moles} \times M_r$mass=moles×Mr​.

Step 2 — substitute: $\text{mass} = 0.25 \times 100$mass=0.25×100.

Step 3 — calculate: $\text{mass} = 25 \text{ g}$mass=25 g.

Answer: 25 g.

Exam Tip: Work out and double-check your $M_r$Mr​ first, then set out three lines — equation, substitution, and answer with a unit. A wrong $M_r$Mr​ ruins the whole calculation, so count the atoms carefully and remember to multiply everything inside any bracket.

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Calculating Reacting Masses

A balanced equation gives you the ratio of moles in which substances react. Put that ratio together with $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​ and you can find the mass of any reactant or product. The method is the same four steps every time:

1. Work out the moles of the substance you are given.
2. Use the balanced equation to read off the mole ratio to the substance you want.
3. Work out the moles of the substance you want.
4. Convert those moles into a mass with $\text{mass} = \text{moles} \times M_r$mass=moles×Mr​.

Worked Example 4 — Mass of product from a metal

What mass of magnesium oxide forms when $6 \text{ g}$6 g of magnesium burns completely? Equation: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO. Using $A_r$Ar​: Mg = 24, O = 16, so $M_r(\text{MgO}) = 40$Mr​(MgO)=40.

Step 1 — moles of Mg: $\dfrac{6}{24} = 0.25 \text{ mol}$246​=0.25 mol.

Step 2 — mole ratio from the equation: $2\text{Mg} : 2\text{MgO}$2Mg:2MgO, which simplifies to $1 : 1$1:1.

Step 3 — moles of MgO $= 0.25 \text{ mol}$=0.25 mol.

Step 4 — mass of MgO $= 0.25 \times 40 = 10 \text{ g}$=0.25×40=10 g.

Answer: 10 g of magnesium oxide.

Check with conservation of mass: the oxygen added has mass $10 - 6 = 4 \text{ g}$10−6=4 g, which is $\dfrac{4}{32} = 0.125 \text{ mol}$324​=0.125 mol of $\text{O}_2$O2​ — exactly half the moles of Mg, as the $2:1$2:1 ratio of Mg to $\text{O}_2$O2​ demands. The numbers hang together.

Worked Example 5 — Decomposing a carbonate

What mass of calcium oxide is made when $50 \text{ g}$50 g of calcium carbonate is heated? Equation: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​. Using $M_r(\text{CaCO}_3) = 100$Mr​(CaCO3​)=100 and $M_r(\text{CaO}) = 40 + 16 = 56$Mr​(CaO)=40+16=56.

Step 1 — moles of $\text{CaCO}_3$CaCO3​: $\dfrac{50}{100} = 0.5 \text{ mol}$10050​=0.5 mol.

Step 2 — ratio $\text{CaCO}_3 : \text{CaO} = 1 : 1$CaCO3​:CaO=1:1.

Step 3 — moles of CaO $= 0.5 \text{ mol}$=0.5 mol.

Step 4 — mass of CaO $= 0.5 \times 56 = 28 \text{ g}$=0.5×56=28 g.

Answer: 28 g of calcium oxide.

Worked Example 6 — A 1 : 1 ratio with a small mass

What mass of hydrogen is produced when $4.8 \text{ g}$4.8 g of magnesium reacts with excess hydrochloric acid? Equation: $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$Mg+2HCl→MgCl2​+H2​. Using $A_r$Ar​: Mg = 24, and $M_r(\text{H}_2) = 2$Mr​(H2​)=2.

Step 1 — moles of Mg: $\dfrac{4.8}{24} = 0.2 \text{ mol}$244.8​=0.2 mol.

Step 2 — ratio $\text{Mg} : \text{H}_2 = 1 : 1$Mg:H2​=1:1.

Step 3 — moles of $\text{H}_2 = 0.2 \text{ mol}$H2​=0.2 mol.

Step 4 — mass of $\text{H}_2 = 0.2 \times 2 = 0.4 \text{ g}$H2​=0.2×2=0.4 g.

Answer: 0.4 g of hydrogen.

Worked Example 7 — Working out the mass of a reactant