Introducing the Mole

Higher tier only: This whole lesson is Higher tier. Foundation candidates do not need the mole, but should still be confident with relative formula mass from the previous lesson.

Atoms are unimaginably tiny — a single carbon atom has a mass of about $2 \times 10^{-23}$2×10−23 grams — so chemists never count them one by one. Instead they count them in enormous, fixed-size batches called moles, in the same way that a baker counts eggs in dozens. The mole lets us connect the masses we can actually weigh in the laboratory to the number of particles taking part in a reaction. This lesson, part of Topic C3 of OCR Gateway Science A, defines the mole and the Avogadro constant, shows how to convert between mass and moles using $M_r$Mr​, and uses balanced equations to calculate 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.

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What Is a Mole?

A mole is the unit chemists use for amount of substance. One mole of any substance 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 contains $6.02 \times 10^{23}$6.02×1023 carbon atoms; one mole of water molecules contains $6.02 \times 10^{23}$6.02×1023 water molecules. The mole is just a very large counting number, chosen so that the mass of one mole of a substance in grams is 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 $M_r$Mr​ (for a compound). So "the mass of one mole of $\text{CO}_2$CO2​" is simply its $M_r$Mr​, 44, expressed in grams.

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Relative Atomic and Formula Mass (recap)

The relative atomic mass ($A_r$Ar​) of an element is taken 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 are worth knowing because they appear in reacting-mass calculations again and again.

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Converting Between Mass and Moles

The central equation of the whole topic links the number of moles, the mass in grams and the relative formula mass:

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

This can be rearranged 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 simple triangle helps: cover the quantity you want, and the triangle shows the calculation.

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: Always check your $M_r$Mr​ first and write the three lines — equation, substitution, answer with unit. A wrong $M_r$Mr​ wrecks the whole calculation, so it pays to count the atoms carefully (and multiply everything inside any bracket).

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Calculating Reacting Masses

A balanced equation tells you the ratio of moles in which substances react. Combine that ratio with $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​ and you can work out the mass of any reactant or product. The method has four steps every time:

1. Work out the moles of the substance you are given.
2. Use the balanced equation to find the mole ratio to the substance you want.
3. Work out the moles of the substance you want.
4. Convert those moles to 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 is made when $6 \text{ g}$6 g of magnesium burns completely? Equation: $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO. Use $A_r$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 is $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​ requires. The numbers are consistent.

Worked Example 5 — Decomposing a carbonate

What mass of calcium oxide is produced by heating $50 \text{ g}$50 g of calcium carbonate? Equation: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​. Use $M_r(\text{CaCO}_3) = 100$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 2 : 1 ratio