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Relative Formula Masses (AQA GCSE Combined Science: Synergy: Physical Sciences): Revision Note

Exam code: 8465

Stewart Hird

Written by: Stewart Hird

Updated on

AQA Combined Science: Synergy: Physical SciencesAQA Combined Science: Trilogy: ChemistryAQA Chemistry

What is relative atomic mass?

  • The relative atomic mass (Ar) of an element compares the mass of atoms of that element with the mass of a carbon-12 atom

  • It is an average value that takes account of the different isotopes of the element and their abundances

  • Since it is a ratio, relative atomic mass has no units

  • The Ar of every element is given on the periodic table

    • It is always the larger of the two numbers shown for each element

Examiner Tips and Tricks

The definition of relative atomic mass compares atom masses to the 12C isotope

You are not required to calculate relative atomic mass from isotopic abundances for this course

Relative formula (molecular) mass

How to calculate relative formula mass

  • The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all the atoms in the formula

  • To calculate Mr, identify every atom in the formula and add up their Ar values

Examples

Substance

Atoms present

Calculation

Mr 

Hydrogen
H2 

2 x H

(2 x 1)

2

Water
H2O

(2 x H) + (1 x O)

(2 x 1) + (1 x 16)

18

Potassium carbonate
K2CO3 

(2 x K) + (1 x C) + (3 x O)

(2 x 39) + (1 x 12) + (3 x 16)

138

Calcium hydroxide
Ca(OH)2 

(1 x Ca) + (2 x O) + (2 x H)

(1 x 40) + (2 x 16) + (2 x 1)

74

Ammonium sulfate
(NH4)2SO4

(2 x N) + (8 x H) + (1 x S) + (4 x O)

(2 x 14) + (8 x 1) + (1 x 32) + (4 x 16)

132

Worked Example

Calculate the relative formula mass of:

  1. Sodium chloride, NaCl

  2. Copper oxide, CuO

  3. Magnesium nitrate, Mg(NO3)2

Answers:

  1. Sodium chloride

    • NaCl = 23 + 35.5 = 58.5

  2. Copper oxide

    • CuO = 63.5 + 16 = 79.5

  3. Magnesium nitrate

    • Mg(NO3)2 = 24 + (14 x 1 x 2) + (16 x 3 x 2) = 148

Examiner Tips and Tricks

The Ar of every element is on the periodic table, so you do not need to memorise them.

Remember that for formulae with brackets (e.g. Ca(NO3)2), the subscript outside the bracket multiplies everything inside it.

Percentage by mass

  • The percentage by mass of an element in a compound is calculated using:

% mass of an element = fraction numerator A subscript straight r cross times number space of space atoms space of space the space element over denominator M subscript straight r space of space the space compound end fraction x 100

  • For example, in water:

    • Water is a simple molecule with the chemical formula H2O

    • So, water is made of two hydrogen (H) atoms and one oxygen (O) atom

    • From the Periodic Table, the relative atomic mass of:

      • Hydrogen = 1

      • Oxygen = 16

    • Therefore, the total mass of water is:

      • (2 × 1) + 16 = 18

  • To find the percentage by mass of hydrogen:

    • Percentage of hydrogen = open parentheses fraction numerator 2 cross times 1 over denominator 18 end fraction close parentheses × 100 = 11.1%

  • Similarly, the percentage by mass of oxygen is:

    • Percentage of oxygen = open parentheses fraction numerator 1 cross times 16 over denominator 18 end fraction close parentheses × 100 = 88.9%

  • Note: The total percentage by mass of all the elements should add up to 100%, e.g. 11.1% + 88.9% = 100%

Worked Example

Calculate the percentage of carbon in the formula of glucose, C6H12O6.

Answer:

  • From the Periodic Table, the relative atomic masses are:

    • Carbon = 12

    • Hydrogen = 1

    • Oxygen = 16 

  • The total mass of glucose is:

    • (6 × 12) + (12 × 1) + (6 × 16) = 180

  • The equation for percentage composition is:

    • Percentage composition = stretchy left parenthesis fraction numerator total space mass space of space element over denominator total space space mass space of space compound end fraction stretchy right parenthesis × 100

  • So, the percentage composition of carbon in glucose is:

    • Percentage of carbon = open parentheses fraction numerator 6 cross times 12 over denominator 180 end fraction close parentheses × 100 = 40%

Worked Example

The chemical formula of the fertiliser ammonium nitrate is NH4NO3. Calculate the percentage by mass of nitrogen in ammonium nitrate.

Answer:

  • From the Periodic Table, the relative atomic masses are:

    • Nitrogen = 14

    • Hydrogen = 1

    • Oxygen = 16 

  • The total mass of ammonium nitrate is:

    • (1 x 14) + (4 × 1) + (1 x 14) + (3 × 16) = 80

  • The equation for percentage composition is:

    • Percentage Composition = stretchy left parenthesis fraction numerator total space mass space of space element over denominator total space space mass space of space compound end fraction stretchy right parenthesis × 100

  • So, the percentage composition of nitrogen in ammonium nitrate is:

    • Careful: There are two nitrogen atoms in ammonium nitrate

    • Percentage of nitrogen = open parentheses fraction numerator 2 cross times 14 over denominator 80 end fraction close parentheses × 100 = 35%

Examiner Tips and Tricks

The most common incorrect answer for percentage composition are values above 100%

This is not possible and usually happens when the numbers in the calculation are "upside down"

For example, the percentage composition of hydrogen in water:

  • CORRECT = open parentheses fraction numerator 2 cross times 1 over denominator 18 end fraction close parentheses x 100 = 11.1%

  • INCORRECT = stretchy left parenthesis fraction numerator 18 over denominator 2 cross times 1 end fraction stretchy right parenthesis x 100 = 900%

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Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.