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2.3.2Inorganic Chemistry

Practical- Determine the % of Oxygen in Air — Edexcel International GCSE Science Double Award 4SD0 - Chemistry Revision Notes & Topical Questions

5 min read 5 parts Lesson 8 of 52

📚 Edexcel International GCSE Science Double Award 4SD0 - Chemistry Hub / Topical Revision
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Part 1 of 5Method

2.3.2 Practical- Determine the % of Oxygen in Air
1
Measure the volume of air between the tap of a burette and the 50.0 cm³ mark by draining water from the burette into a measuring cylinder.
2
Moisten the inside of the burette with water, then close the tap.
3
Place iron filings or a plug of iron wool in the bottom of the burette, so that they sit in the trapped air.
4
Invert the burette into a trough of water, clamp it vertically, and record the starting water level on the scale.
5
Leave the apparatus for several days, then record the new water level and find the volume change.

Part 2 of 5Reasoning

Total air volume
  • The initial air volume is the 50.0 cm³ scale plus the space below the mark, minus the starting water reading on the scale.
Oxygen removed
  • Oxygen reacts with the iron, so the water level rises by the volume of oxygen that was used up.
Nitrogen remains
  • Nitrogen does not react with iron, so the gas left at the end is mostly nitrogen.
Accuracy
  • Reading the scale and the slow reaction introduce uncertainty, so repeats and an average improve reliability.
Expected value
  • Oxygen makes up about 21% of air, so a result close to 20 to 21% indicates a reliable experiment.

Part 3 of 5The Chemistry

The Chemistry
Rusting
  • Iron slowly reacts with oxygen and water over several days to form hydrated iron(III) oxide, which is rust.
Why iron
  • Iron reacts with oxygen but not with nitrogen, so the volume change matches the oxygen present in the air.
Phosphorus alternative
  • Phosphorus also removes oxygen, but it is toxic, so iron is used in school laboratories as the safer choice.

Part 4 of 5Worked Example

Question

  • At the start the burette reads 2.4 cm³ on the scale.
  • At the end it reads 13.6 cm³.
  • The air space between the tap and the 50.0 cm³ mark is 4.6 cm³.
  • Calculate the percentage of oxygen in the air.

  1. Initial air volume = 50.0 + 4.6 − 2.4 = 52.2 cm³.
  2. Final air volume = 50.0 + 4.6 − 13.6 = 41.0 cm³.
  3. Volume of oxygen used = 52.2 − 41.0 = 11.2 cm³.
  4. Percentage of oxygen = 11.2 ÷ 52.2 × 100 = 21.5%.

Answer 21.5% oxygen, close to the accepted value of about 21%.

Part 5 of 5Exam Checklist & Tips

2.3.2 – Key Points for the Exam

  • Burette A graduated glass tube with a tap, used to measure volumes of liquid or trapped gas.
  • Rusting The slow reaction of iron with oxygen and water to form hydrated iron(III) oxide.
  • Percentage oxygen The volume of oxygen removed divided by the initial volume of air, multiplied by 100.
  • Iron wool A fine mass of iron fibres used to react with the oxygen in trapped air.
  • Hydrated iron(III) oxide Rust, a flaky compound of iron, oxygen and water.

Common Mistakes

  • Using the final water level alone as the oxygen volume instead of subtracting the initial reading.
  • Stopping the experiment too early; the reaction must be left until the water level stops changing.
  • Stating that the result proves oxygen is exactly 21%; readings and an incomplete reaction introduce error.
  • Dividing by the scale volume instead of the total initial air volume when calculating the percentage.

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