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4.1 Relative Masses and Moles

GCSE Chemistry · AQA · Chapter 4: Chemical Calculations

A free GCSE Chemistry lesson for AQA (8462) from OBA School: relative atomic mass, relative formula mass, percentage by mass, the mole and the Avogadro constant. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Explain what relative atomic mass means and calculate it from the abundances of isotopes
  • Calculate the relative formula mass (Mr) of a substance from its formula
  • Calculate the percentage by mass of an element in a compound
  • (Higher tier) use moles = mass ÷ Mr to convert between mass and amount in moles
  • (Higher tier) use the Avogadro constant to work out numbers of atoms, molecules or ions

Key words

  • Relative atomic mass (Ar): The average mass of the atoms of an element compared with 1/12 of the mass of a carbon-12 atom. It takes account of the isotopes and how common each one is.
  • Relative formula mass (Mr): The sum of the relative atomic masses of all the atoms shown in the formula. It has no units.
  • Isotopes: Atoms of the same element with the same number of protons but different numbers of neutrons, so different mass numbers.
  • Mole (mol): The unit for amount of substance. One mole of a substance has a mass in grams equal to its Mr (or Ar). (Higher tier)
  • Avogadro constant: 6.02 × 10²³: the number of particles (atoms, molecules or ions) in one mole of any substance. (Higher tier)
  • Percentage by mass: The fraction of the mass of a compound that comes from one element, as a percentage.

Relative atomic mass and isotopes

A single atom is far too light to weigh, so chemists compare atoms with a standard: an atom of carbon-12, which is given a relative mass of exactly 12. The relative atomic mass, Ar, of an element tells you how heavy its atoms are on this scale. Magnesium atoms are twice as heavy as carbon-12 atoms, so Ar(Mg) = 24.

Most elements are a mixture of isotopes. The Ar is the mean mass of the atoms, taking into account how common (abundant) each isotope is. That is why some Ar values, such as Cl = 35.5 and Cu = 63.5, are not whole numbers – no atom has half a proton or neutron; it is an average.

Ar = (mass no. 1 × % 1) + (mass no. 2 × % 2), all ÷ 100

Worked example: relative atomic mass of boron

Imagine 100 boron atoms: 20 have a mass of 10 and 80 have a mass of 11.
total mass = (10 × 20) + (11 × 80) = 200 + 880 = 1080
mean mass = 1080 ÷ 100 = 10.8

Answer: Ar(B) = 10.8

Exam tip: The Ar is always closer to the mass number of the more abundant isotope. Use this to check your answer.

Relative formula mass, Mr

Atoms are far too light to weigh one at a time, so chemists compare masses. Each element has a relative atomic mass, Ar, which you read from the periodic table (for example H = 1, C = 12, O = 16, Cl = 35.5, Cu = 63.5). To find the relative formula mass, Mr, of a compound, add up the Ar values of every atom in its formula. This works for ionic compounds and for molecules (for a molecular substance Mr is also called the relative molecular mass).

Worked example: Mr of calcium hydroxide, Ca(OH)₂

The bracket is multiplied by 2, so there are 1 Ca, 2 O and 2 H.
Mr = 40 + 2 × (16 + 1) = 40 + 34 = 74

Answer: Mr = 74 (no units)

Worked example: Mr of ammonium sulfate, (NH₄)₂SO₄

Atoms: 2 N, 8 H, 1 S, 4 O.
Mr = (2 × 14) + (8 × 1) + 32 + (4 × 16) = 28 + 8 + 32 + 64 = 132

Answer: Mr = 132

Key fact: In a balanced equation the total Mr of the reactants equals the total Mr of the products (counting the big numbers in front). This is conservation of mass.

Percentage by mass of an element

To find what percentage of a compound’s mass is one element, use:

% by mass of element = (Ar × number of atoms of that element) ÷ Mr of compound × 100

Worked example: percentage of nitrogen in ammonium nitrate, NH₄NO₃

Mr = 14 + (4 × 1) + 14 + (3 × 16) = 80
Mass of nitrogen = 2 × 14 = 28
% N = 28 ÷ 80 × 100 = 35%

Answer: 35% nitrogen by mass

Farmers use calculations like this to compare fertilisers: the higher the percentage of nitrogen, the less fertiliser needs to be spread to supply the same mass of nitrogen.

The mole (Higher tier)

Chemists measure amounts of substance in moles (symbol mol). One mole of any substance has a mass in grams equal to its Ar or Mr. So 1 mol of carbon has a mass of 12 g, and 1 mol of water (Mr = 18) has a mass of 18 g.

Worked example: moles in 22 g of carbon dioxide, CO₂

Mr of CO₂ = 12 + (2 × 16) = 44
moles = mass ÷ Mr = 22 ÷ 44 = 0.5

Answer: 0.5 mol

Worked example: mass of 0.25 mol of magnesium oxide, MgO

Mr of MgO = 24 + 16 = 40
mass = moles × Mr = 0.25 × 40 = 10

Answer: 10 g

Exam tip: Make sure the mass is in grams. Convert kilograms first: 1 kg = 1000 g (and 1 tonne = 1 000 000 g).

Worked example: significant figures

How many moles are in 3.65 g of hydrogen chloride, HCl? (Mr = 36.5)
moles = 3.65 ÷ 36.5 = 0.1 on the calculator
The mass was given to 3 significant figures, so give the answer to 3 s.f.

Answer: 0.100 mol

The Avogadro constant (Higher tier)

One mole of any substance contains the same number of particles: 6.02 × 10²³. This is the Avogadro constant. So 1 mol of carbon contains 6.02 × 10²³ atoms, and 1 mol of water contains 6.02 × 10²³ molecules.

Worked example: number of molecules in 9.0 g of water

moles = 9.0 ÷ 18 = 0.50 mol
molecules = 0.50 × 6.02 × 10²³ = 3.01 × 10²³

Answer: 3.01 × 10²³ molecules

Worked example: number of oxygen atoms in 0.10 mol of CO₂

Each CO₂ molecule contains 2 O atoms, so moles of O atoms = 2 × 0.10 = 0.20 mol
atoms = 0.20 × 6.02 × 10²³ = 1.204 × 10²³

Answer: 1.20 × 10²³ oxygen atoms (3 s.f.)

Exam tip: Number of particles = moles × 6.02 × 10²³. Read the question carefully: does it want atoms, molecules or ions? One mole of NaCl contains 6.02 × 10²³ Na⁺ ions and 6.02 × 10²³ Cl− ions.

Common mistakes

  • Forgetting to multiply everything inside a bracket: Mg(NO₃)₂ has 2 N and 6 O, so Mr = 24 + 2 × 62 = 148.
  • Giving Mr units. Mr is a ratio and has no units; mass in grams does.
  • Using the atomic number instead of the relative atomic mass. Use the larger number on the AQA periodic table.
  • Dividing the wrong way: moles = mass ÷ Mr, not Mr ÷ mass.
  • Counting molecules when the question asks for atoms. One mole of CO₂ contains 3 moles of atoms.

Revision questions

Try these, then check your answers below each one.

Question 1. Use Ar: C = 12, O = 16, Ca = 40.<br/>Calculate the relative formula mass (Mr) of calcium carbonate, CaCO₃.

Answer: 100

Working: 40 + 12 + (3 × 16) = 100. Mr has no units.

Question 2. (Higher tier) What is the value of the Avogadro constant? (Choose from: 6.02 × 10²³, 6.02 × 10⁻²³, 2.40 × 10⁴, 1.00 × 10¹².)

Answer: 6.02 × 10²³

Working: The Avogadro constant is 6.02 × 10²³ per mole: one mole of any substance contains this many particles.

Question 3. Use Ar: O = 16, S = 32, Cu = 63.5.<br/>Calculate the percentage by mass of copper in copper sulfate, CuSO₄. Give your answer to 1 decimal place.

Answer: 39.8 %

Working: Mr = 63.5 + 32 + 64 = 159.5. % Cu = 63.5 ÷ 159.5 × 100 = 39.8%.

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