IGCSEChemistryCIERevision Notes3. StoichiometryThe Mole & the Avogadro ConstantPercentage Yield & Purity

Percentage Yield & Purity (CIE IGCSE Chemistry)

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Calculating percentage yield, percentage by mass & percentage purity

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Percentage yield

  • Yield is the term used to describe the amount of product you get from a reaction
  • For economic reasons, the objective of every chemical producing company is to have as high a percentage yield as possible to increase profits and reduce costs and waste
  • In practice, you never get 100% yield in a chemical process for several reasons
  • These include:
    • Some reactants may be left behind in the equipment
    • The reaction may be reversible and in these reactions a high yield is never possible as the products are continually turning back into the reactants
    • Some products may also be lost during separation and purification stages such as filtration or distillation
    • There may be side reactions occurring where a substance reacts with a gas in the air or an impurity in one of the reactants
    • Products can also be lost during transfer from one container to another

How to calculate percentage yield

  • Percentage yield compares the actual yield to the theoretical yield
  • The equation for percentage yield is:

percentage yield = fraction numerator bold actual bold space bold yield over denominator bold theoretical bold space bold yield end fraction bold space bold cross times bold space bold 100

  • Actual yield is the recorded amount of product obtained
  • Theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions

Worked example

Copper(II) sulfate may be prepared by the reaction of dilute sulfuric acid with copper(II) oxide. A student prepared 1.6 g of dry copper(II) sulfate crystals.

Calculate the percentage yield if the theoretical yield is 2.0 g.

Answer

  • Actual yield of copper(II) sulfate = 1.6 g
  • Percentage yield of copper(II) sulfate = fraction numerator 1.6 over denominator 2.0 end fraction x 100 = 80%

Exam Tip

  • You are expected to remember the equation for percentage yield
  • If you remember it incorrectly and get a percentage yield greater than 100%, then you have made an error!
  • The most common error is to divide the theoretical yield by the actual yield
    • In this case, you just need to swap the numbers around in your calculation

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How to calculate percentage composition by mass 

  • The percentage composition of any compound is a way to express the mass of each element as a percentage of the total mass of the compound
  • The equation for percentage composition is:

percentage composition of an element = fraction numerator bold total bold space bold mass bold space bold of bold space bold the bold space bold element bold space bold in bold space bold the bold space bold compound over denominator bold relative bold space bold formula bold space bold mass bold space bold of bold space bold the bold space bold compound end fraction bold space bold cross times bold space bold 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 composition of hydrogen:
    • Percentage of hydrogen = fraction numerator 2 cross times 1 over denominator 18 end fraction × 100 = 11.1%
  • Similarly, the percentage composition of oxygen is:
    • Percentage of oxygen = fraction numerator 1 cross times 16 over denominator 18 end fraction × 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 by mass of iron in iron(III) oxide, Fe2O3.

Answer:

  • From the Periodic Table, the relative atomic masses are:
    • Fe = 56
    • Oxygen = 16 
  • The total mass of iron in iron(III) oxide is:
    • 2 x 56 = 112
  • The total mass of iron(III) oxide is:
    • (2 × 56) + (3 × 16) = 160
  • The equation for percentage composition is:
    • Percentage composition = fraction numerator total space mass space of space element over denominator total space mass space of space compound end fraction × 100
  • So, the percentage composition of iron in iron(III) oxide is:
    • Percentage of iron = 112 over 160 × 100 = 70%

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 
  • Careful: There are two nitrogen atoms in ammonium nitrate
  • The total mass of nitrogen in ammonium nitrate is:
    • 2 x 14 = 28
  • 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 = fraction numerator total space mass space of space element over denominator total space mass space of space compound end fraction × 100
  • So, the percentage composition of nitrogen in ammonium nitrate is:
  • Percentage of nitrogen = fraction numerator open parentheses 2 cross times 14 close parentheses over denominator 80 end fraction × 100 = 35%

Exam Tip

  • Make sure you calculate the percentage composition using the total mass of the element.
    • For example, a common mistake with ammonium nitrate is doing the calculation for only one atom of nitrogen.
    • This would lose a mark in an exam
  • Show ALL your working out. If you make a mistake in the calculation, you could still score a mark for your workings.

How to calculate percentage purity

  • A pure substance has nothing else mixed with it
  • Often, the product you are trying to obtain may become contaminated with unwanted substances such as unreacted reactants, catalysts and other impurities
  • To equation to calculate percentage purity is:

percentage purity = fraction numerator bold mass bold space bold of bold space bold pure bold space bold substance over denominator bold total bold space bold mass bold space bold of bold space bold substance end fraction bold space bold x bold space bold 100

Worked example

A sample of lead(II) bromide was made. It weighed 15 g.

The sample was found to be impure and only contained 13.5 g of lead(II) bromide. 

Calculate the percentage purity of the lead(II) bromide.

Answer:

  • The mass of the pure substance is 13.5 g
  • The total mass of the substance is 15 g
  • Percentage purity = fraction numerator mass space of space pure space substance over denominator total space mass space of space substance end fraction space straight x space 100
  • Percentage purity = fraction numerator 13.5 over denominator 15 end fraction cross times 100 = 90% 

Exam Tip

  • All of these calculations are to find a percentage so don't forget to multiply by 100 to convert your answer to a percentage.

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    Caroline

    Author: Caroline

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.