National 5ChemistrySQARevision NotesChemical Changes & StructureFormulae & Reacting QuantitiesPercentage Composition

Percentage Composition (SQA National 5 Chemistry): Revision Note

Exam code: X813 75

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

Percentage Composition

  • The percentage composition of any compound expresses the mass of each element as a percentage of the total mass (the GFM)

How to calculate percentage composition

  1. Calculate the GFM of the entire compound using the RAMs from page 7 of the Data Booklet

  2. Find the total mass of the element you are interested in within the formula

    • Be careful to multiply the element's RAM by the number of times it appears

  3. Substitute these two numbers into the formula and calculate the percentage

  • The equation for percentage composition is:

Percentage composition = open parentheses fraction numerator total space mass space of space element over denominator GFM end fraction close parentheses × 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 page 7 of the Data Booklet:

      • Hydrogen = 1

      • Oxygen = 16

    • Therefore, the GFM of water is:

      • (2 × 1) + 16 = 18

  • To find the percentage composition of hydrogen:

    • Percentage of hydrogen = begin mathsize 14px style open parentheses fraction numerator 2 cross times 1 over denominator 18 end fraction close parentheses end style × 100 = 11.1%

  • Similarly, the percentage composition of oxygen is:

    • Percentage of oxygen = begin mathsize 14px style open parentheses fraction numerator 1 cross times 16 over denominator 18 end fraction close parentheses end style × 100 = 88.9%

  • Note: The total percentage by mass of all the elements should add up to 100%, e.g. 11.1% + 89.9% = 100%

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 = begin mathsize 14px style stretchy left parenthesis fraction numerator 18 over denominator 2 cross times 1 end fraction stretchy right parenthesis end style x 100 = 900%

Worked Example

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

[2]

Answer:

  • From page 7 of the Data Booklet, the RAMs are:

    • Carbon = 12

    • Hydrogen = 1

    • Oxygen = 16 

  • The GFM of glucose is:

    • (6 × 12) + (12 × 1) + (6 × 16) = 180 [1 mark]

  • The equation for percentage composition is:

    • Percentage composition = stretchy left parenthesis fraction numerator total space mass space of space element over denominator GFM 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% [1 mark]

Worked Example

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

[2]

Answer:

  • From page 7 of the Data Booklet, the RAMs are:

    • Nitrogen = 14

    • Hydrogen = 1

    • Oxygen = 16 

  • The GFM of ammonium nitrate is:

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

  • The equation for percentage composition is:

    • Percentage composition = stretchy left parenthesis fraction numerator total space mass space of space element over denominator GFM 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% [1 mark]

Examiner Tips and Tricks

  • Make sure that you account for all the atoms of the element in the question!

  • Examiners often pick chemicals with more than one atom of an element as both worked examples show

    • The NH4NO3 worked example is more challenging

    • The two nitrogen atoms are written in different parts of the compound

    • This is a common error, and many students forget to include both of them in the calculation

  • Another relatively common question was about SnF2, which contains 2 fluorine atoms, in toothpaste 

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Author
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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Richard Boole

Reviewer: Richard Boole

Expertise: Chemistry Content Creator

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.