Percentage Yield & Purity (Cambridge International IGCSE Chemistry)

Calculating percentage yield, percentage by mass & percentage purity

Spec point — 3.3.8 · Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data

Extended tier only

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 =$\frac{\mathrm{actual} \mathrm{yield}}{\mathrm{theoretical} \mathrm{yield}}\times100$

• 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

  • Typically, this involves a reacting mass calculation based on the balanced symbol equation

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 = $\frac{1.6}{2.0}$ x 100 = 80%

Exam Hint

• 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

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 =$\frac{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{element} \mathrm{in} \mathrm{the} \mathrm{compound}}{\mathrm{relative} \mathrm{formula} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{compound}}\times100$

• 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 = $\frac{2\times1}{18}$ × 100 = 11.1%

• Similarly, the percentage composition of oxygen is:

  • Percentage of oxygen = $\frac{1\times16}{18}$ × 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 = $\frac{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{element}}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{compound}}$ × 100

• So, the percentage composition of iron in iron(III) oxide is:

  • Percentage of iron = $\frac{112}{160}$ × 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 = $\frac{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{element}}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{compound}}$ × 100

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

• Percentage of nitrogen = $\frac{( 2 \times 14 )}{80}$ × 100 = 35%

Exam Hint

• 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 =$\frac{\mathrm{mass} \mathrm{of} \mathrm{pure} \mathrm{substance}}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{substance}}x100$

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 = $\frac{\mathrm{mass} \mathrm{of} \mathrm{pure} \mathrm{substance}}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{substance}}x100$
• Percentage purity = $\frac{13.5}{15}\times100$ = 90%

Exam Hint

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

Percentage by Mass

Spec point — nan