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5.3.1.2Relative formula mass

AQA GCSE Combined Science (8464), Higher tier · Chemistry › Quantitative chemistry › Chemical measurements, conservation of mass and the

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Relative formula mass (Mr) is the sum of the relative atomic masses of all the atoms in a formula. You need it for almost every calculation in quantitative chemistry. Questions also ask you to work out the percentage by mass of an element in a compound, or to show that the Mr values balance in an equation.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

  1. 3
    Read relative atomic masses from the periodic tableAr is the larger of the two numbers in each box, e.g. O = 16, Na = 23.
  2. 4
    Calculate Mr of a simple formulaAdd the Ar of every atom, e.g. CO2 = 12 + (2 × 16) = 44.
  3. 5
    Calculate Mr for formulae with bracketsMultiply everything inside the bracket, e.g. Ca(OH)2 = 40 + 2 × (16 + 1) = 74.
  4. 6
    Calculate percentage by mass of an elementDivide the total Ar of that element in the formula by the Mr, then multiply by 100.
  5. 7
    Show that Mr values balance in equationsThe total Mr of the reactants (using balancing numbers) equals the total Mr of the products.
  6. 8
    Use Mr to identify an unknown elementWork backwards, e.g. if XCl2 has Mr 111, X = 111 − 71 = 40, so X is calcium.

Notes

Ar and Mr

  • The relative atomic mass (Ar) of an element is an average mass of its atoms that takes account of the abundance of its isotopes. Read it from the periodic table.
  • The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of the atoms in the numbers shown in the formula.
  • Ar and Mr have no units, because they are relative values.
  • Elements made of molecules have an Mr too: O2 = 2 × 16 = 32.

Working out Mr

  • Count the atoms of each element, multiply by its Ar, then add everything up.
  • Example: H2SO4 = (2 × 1) + 32 + (4 × 16) = 98.
  • With brackets, work out the group first: Al2(SO4)3 = (2 × 27) + 3 × (32 + 4 × 16) = 54 + 288 = 342.

Percentage by mass of an element

  • % by mass of an element = (Ar × number of atoms of that element) ÷ Mr × 100
  • Example: nitrogen in ammonium nitrate, NH4NO3. Mr = 14 + (4 × 1) + 14 + (3 × 16) = 80. There are 2 nitrogen atoms, so % N = 28 ÷ 80 × 100 = 35%.
  • The percentages of all the elements in a compound add up to 100%.

Mr in balanced equations

  • In a balanced equation, the total Mr of the reactants (in the quantities shown) equals the total Mr of the products. This is conservation of mass again.
  • Example: 2Mg + O2 → 2MgO. Reactants: (2 × 24) + 32 = 80. Products: 2 × (24 + 16) = 80.
  • Work out the Mr of one formula first, then multiply by the balancing number.

Cheatsheet

  • Ar = relative atomic mass (from the periodic table)
  • Mr = sum of the Ar values of all the atoms in the formula
  • Ar and Mr have no units
  • % by mass of element = (Ar × number of atoms) ÷ Mr × 100
  • Balanced equation: total Mr of reactants = total Mr of products
  • Useful values: H2O = 18, O2 = 32, CO2 = 44, NaCl = 58.5, CaCO3 = 100

How to answer each type of question

Calculate the relative formula mass

1 to 2 marks4
  1. Count the atoms of each element, taking care with brackets.
  2. Multiply each count by the Ar given.
  3. Add them up and show the working. Do not give a unit.

Example. Calculate the relative formula mass (Mr) of magnesium nitrate, Mg(NO3)2.
Relative atomic masses (Ar): N = 14; O = 16; Mg = 24 [2 marks]

Show the model answer
24 + 2 × (14 + 3 × 16) (1)
= 24 + 124 = 148 (1)

Calculate the percentage by mass of an element

2 to 3 marks6
  1. Work out the Mr of the compound.
  2. Multiply the Ar of the element by the number of its atoms in the formula.
  3. Divide by the Mr, multiply by 100 and round as asked.

Example. Copper(II) sulfate has the formula CuSO4.
Calculate the percentage by mass of copper in copper(II) sulfate.
Give your answer to 3 significant figures.
Ar: O = 16; S = 32; Cu = 63.5 [3 marks]

Show the model answer
Mr = 63.5 + 32 + (4 × 16) = 159.5 (1)
63.5 ÷ 159.5 × 100 (1)
= 39.8% (1)

Show that the Mr values balance in an equation

2 marks7
  1. Work out the Mr of each substance.
  2. Multiply each by its balancing number and add up each side.
  3. Write both totals and state that they are equal.

Example. CH4 + 2O2 → CO2 + 2H2O
Show that the total Mr of the reactants is equal to the total Mr of the products.
Ar: H = 1; C = 12; O = 16 [2 marks]

Show the model answer
Reactants: 16 + (2 × 32) = 80 (1)
Products: 44 + (2 × 18) = 80, so the totals are equal (1)

Identify an element from an Mr

2 marks8
  1. Subtract the Ar values you know from the Mr.
  2. Divide by the number of atoms of the unknown element.
  3. Match the Ar to an element on the periodic table.

Example. Metal X forms an oxide with the formula X2O. The Mr of X2O is 62.
Identify metal X. Ar: O = 16 [2 marks]

Show the model answer
2 × Ar of X = 62 − 16 = 46, so Ar of X = 23 (1)
X is sodium (1)

Shortcuts and memory tricks

  • Learn the Mr values you use most (water 18, carbon dioxide 44, calcium carbonate 100) to save time and check answers.
  • Brackets first: work out the group (e.g. NO3 = 62), then multiply by the number outside.
  • Sense check for % by mass: it must be less than 100%, and all the elements together add up to 100%.
  • Type the whole calculation into your calculator in one go, using brackets, to avoid rounding errors.

Where marks are lost

  • Multiplying only the first atom after a bracket: in Ca(OH)2 both the O and the H are doubled.
  • Using the atomic number instead of the relative atomic mass from the periodic table.
  • Counting only one atom of the element for % by mass, e.g. using 14 instead of 28 for nitrogen in NH4NO3.
  • Including the balancing number when asked for the Mr of a substance: the Mr of water is 18, even if the equation shows 2H2O.
  • Rounding Ar values: use 35.5 for chlorine and 63.5 for copper, as printed.
  • Giving a unit such as g with an Mr.

Exam technique

  • Use the Ar values given in the question, or on the periodic table, exactly as printed.
  • Show each step: a mark is often given for the working even if the final answer is wrong.
  • Look for 'Give your answer to 3 significant figures' in percentage by mass questions.
  • In 'show that' questions, write out both totals in full and say that they are equal.

Quick recall

Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.

Calculate the relative formula mass (Mr) of calcium carbonate, CaCO3.
100
A metal chloride has the formula XCl3. Its relative formula mass is 133.5.
Calculate the relative atomic mass of X.
27
A metal, M, forms an oxide with the formula M2O3. The oxide contains 52.9% of M by mass.
Relative atomic masses (Ar): O = 16, Al = 27, Cl = 35.5, Cr = 52, Fe = 56
Identify metal M.
aluminium
Calculate the relative formula mass (Mr) of sodium chloride, NaCl.
58.5
Calculate the relative formula mass (Mr) of urea, CO(NH2)2.
60

Sample questions

Written for this site in the style of AQA exam questions. They are not taken from real past papers.

Question 1Easy5 marks
Relative atomic masses (Ar): H = 1, C = 12, O = 16, Mg = 24, Ca = 40
(a) Calculate the relative formula mass (Mr) of calcium carbonate, CaCO3.[2]
(b) Calculate the relative formula mass (Mr) of magnesium hydroxide, Mg(OH)2.[2]
(c) What is meant by the relative formula mass of a compound?[1]
Show the answer and mark scheme
(a) Answer: 100
  • 40 + 12 + (3 × 16)
  • 100
(b) Answer: 58
  • 24 + 2 × (16 + 1)
  • 58
(c)
  • the sum of the relative atomic masses of the atoms in the numbers shown in the formula
Question 2Medium7 marks
Ammonium sulfate, (NH4)2SO4, and ammonium nitrate, NH4NO3, are both used as fertilisers.
Relative atomic masses (Ar): H = 1, N = 14, O = 16, S = 32
(a) Calculate the relative formula mass (Mr) of ammonium sulfate.[2]
(b) Calculate the percentage by mass of nitrogen in ammonium sulfate.
Give your answer to 3 significant figures.[2]
(c) The percentage by mass of nitrogen in ammonium nitrate is 35.0%.
A farmer needs to add 70 kg of nitrogen to a field.
Calculate the mass of ammonium nitrate the farmer needs.[2]
(d) Suggest one reason why the farmer might choose ammonium nitrate rather than ammonium sulfate.[1]
Show the answer and mark scheme
(a) Answer: 132
  • (2 × 14) + (8 × 1) + 32 + (4 × 16)
  • 132
(b) Answer: 21.2%
  • (28 ÷ 132) × 100
  • 21.2 (%)
(c) Answer: 200 kg
  • 70 × (100 ÷ 35.0) / 70 ÷ 0.350
  • 200 (kg)
(d)
  • ammonium nitrate has a higher percentage of nitrogen (so a smaller mass is needed / less to transport and spread)
Question 3Hard7 marks
The relative formula mass of a compound can be used to work out the relative atomic mass of an unknown element in it.
Relative atomic masses (Ar): Li = 7, C = 12, O = 16, Na = 23, Cl = 35.5, K = 39, Rb = 85
(a) A Group 1 metal carbonate has the formula M2CO3. Its relative formula mass is 138.
Calculate the relative atomic mass of M and identify M.[3]
(b) A metal chloride has the formula XCl3. Its relative formula mass is 133.5.
Calculate the relative atomic mass of X.[2]
(c) Element Z forms an oxide with the formula Z2O5. The relative formula mass of the oxide is 142.
Calculate the relative atomic mass of Z.[2]
Show the answer and mark scheme
(a) Answer: Ar = 39, so M is potassium
  • mass of the CO3 part of the formula = 12 + (3 × 16) = 60
  • 2 × Ar = 138 − 60 = 78, so Ar = 39
  • potassium
(b) Answer: 27
  • 133.5 − (3 × 35.5)
  • 27
(c) Answer: 31
  • 142 − (5 × 16) = 62
  • 62 ÷ 2 = 31

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