Pearson Edexcel International GCSE in Chemistry · 4CH1 · Free · no account needed
The Gases in the Air
Four gases, one of which is doing almost all the chemistry — and how to measure it without ever touching it.
Topic 2 · Inorganic chemistry — one of 19 lessons in this topic, and one of 61 in Chemistry.
One of the 12 Chemistry required practicals · Its equation is on the Chemistry equations page with the other 16.
This lesson ends with Joseph Priestley. Learn the Gases in the Air, teach it, then defend it against Joseph Priestley.
What this lesson covers in the specification
Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.
- 2.9 — The approximate percentages by volume of the four most abundant gases in dry air
- 2.10 — Determining the percentage of oxygen in air using reactions of metals and non-metals
- 2.14 — Practical: determine the approximate percentage by volume of oxygen in air (required practical)
Specification section 2(d) Reactivity series — also taught in Combustion and Carbonates, The Reactivity Series, Rusting and its Prevention, Redox.
1 · Understand it
No exam language yet. The only question this section answers is: do I actually understand what is happening?
Dry air is a MIXTURE, not a compound, and four gases account for essentially all of it. NITROGEN is about 78%, OXYGEN about 21%, ARGON about 0.9%, and CARBON DIOXIDE about 0.04%. Those four come to over 99.9%, and everything else is a trace.
Two of those figures deserve a second look. Nitrogen dominates the air and does almost nothing chemically, because the triple bond in N₂ is exceptionally hard to break — which is fortunate, since a lower nitrogen and higher oxygen proportion would make fires very much harder to put out. And carbon dioxide is a strikingly small fraction, 0.04%, which is worth holding on to when its effects are discussed later, because a small percentage is not the same as a small effect.
Now the measurement, statement 2.10, and it rests on one idea. Oxygen cannot be picked out of a mixture and weighed. What you CAN do is give it something to react with, let it be removed entirely, and measure how much the volume of the air has fallen. The drop IS the oxygen.
Think of it like weighing a cat by weighing yourself twice
A cat will not sit on the scales. So you weigh yourself, then weigh yourself holding the cat, and subtract. You never measured the cat directly and you have its mass exactly. Every method on this page is that trick: you cannot measure the oxygen in a tube of air, so you arrange for something to take all of it away and measure the difference. The reagent is only there to remove the oxygen — it is the scales being read twice, not the thing being investigated.
The two routes, and why they are the same experiment
- USING A METAL — copper. Air is passed repeatedly over heated copper in a tube. The copper reacts with the oxygen to form black copper(II) oxide, removing it from the air.
- USING A NON-METAL — phosphorus. Phosphorus is burned in a sealed volume of air over water, and it reacts with the oxygen to form an oxide that dissolves, removing it.
- In both cases the reagent is chosen because it reacts with OXYGEN and not with nitrogen, so only the oxygen is taken.
- In both cases the total volume of gas falls, and the water level rises or the syringe reading drops to show it.
- The percentage of oxygen is the DECREASE in volume divided by the ORIGINAL volume, multiplied by 100.
- So the method is not two experiments but one, run with either a metal or a non-metal as the oxygen remover.
Calculating the percentage from two readings
A gas syringe contains 100 cm³ of air. The air is passed repeatedly over heated copper until there is no further change, and the final volume is 79 cm³. Calculate the percentage of oxygen in the air.
- Decrease in volume = 100 − 79 = 21 cm³.
- That decrease is the oxygen, because the copper removed it and nothing else.
- percentage of oxygen = (decrease ÷ original volume) × 100 = (21 ÷ 100) × 100.
- = 21%.
- The 79 cm³ remaining is mostly nitrogen, together with the argon and carbon dioxide, which the copper did not touch.
Answer: 21%, obtained without ever measuring the oxygen itself.
The phrase 'until there is no further change' is doing real work in that problem and is worth stating as a rule. The gas must be passed over the copper REPEATEDLY, backwards and forwards between two syringes, because a single pass leaves some oxygen unreacted. The experiment is finished only when the volume stops falling, which is the same 'to constant mass' logic met in the empirical formula practical.
One caution about the result. The figure obtained is always slightly LOW if the reaction is incomplete, and can be slightly high if the apparatus leaks air in during heating or if the gas is measured while still hot and expanded. The gas must be allowed to COOL before the final volume is read, because a hot gas occupies more room and would make the decrease look smaller than it is.
Finally, why 'DRY air' in statement 2.9. Ordinary air contains water vapour in an amount that varies enormously with the weather, from almost nothing to several per cent. Quoting fixed percentages therefore only makes sense once the water has been excluded, which is why the specification specifies dry air and why gas samples are passed through a drying agent before analysis.
2 · Grade 9 Notes
A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.
Statement 2.9 — the four gases of dry air
- Nitrogen
- About 78%. Very unreactive, because the triple bond in N₂ is hard to break.
- Oxygen
- About 21%. The reactive component — everything that burns, burns in this.
- Argon
- About 0.9%. A noble gas, so it takes part in no reactions at all.
- Carbon dioxide
- About 0.04%. A very small fraction, but the one whose amount is changing.
percentage of oxygen = (decrease in volume ÷ original volume) × 100
decrease in volume is the original volume minus the final volume; both volumes in the same units
Units: The units cancel, so cm³ throughout is fine. The decrease IS the oxygen, because only oxygen has been removed.
Statement 2.10 — the logic of both methods
- Oxygen cannot be measured directly in a mixture, so remove it entirely and measure the change.
- Choose a reagent that reacts with OXYGEN but not with nitrogen — heated copper, or burning phosphorus.
- Record the volume of air before.
- Let the reaction go to completion, so that all the oxygen is removed.
- Cool, then record the volume after; the decrease is the volume of oxygen.
Required practical 2.14 — the percentage of oxygen in air
- Connect two gas syringes to either end of a silica tube containing copper turnings.
- Set one syringe to a measured volume of air, such as 100 cm³, and the other to zero, and record the total volume.
- Heat the copper strongly with a Bunsen burner.
- Pass the air slowly backwards and forwards over the hot copper using the two syringes, so that all of it makes contact repeatedly.
- Observe the copper turning black as copper(II) oxide forms.
- Continue until the total volume stops decreasing, showing that all the oxygen has been used.
- Allow the apparatus to cool completely, then read the final total volume.
- Calculate the percentage as the decrease in volume divided by the original volume, multiplied by 100.
Variables
Independent (changed) — The volume of air passed over the copper
Dependent (measured) — The decrease in the total volume of gas
| Control variable | Why it must be held constant |
|---|---|
| Excess copper present | too little copper leaves oxygen unreacted, so the answer comes out too low |
| Apparatus cooled before the final reading | a hot gas is expanded, so the decrease would appear smaller than it is |
| Sealed apparatus with no leaks | air drawn in during heating adds oxygen that was never in the original sample |
Sources of error
| Type | What goes wrong | What to do |
|---|---|---|
| Systematic | The gas is passed over the copper too few times, so some oxygen never reacts. | Continue passing until the volume stops decreasing. |
| Systematic | The final volume is read while the apparatus is still hot, so the gas is expanded. | Cool the apparatus fully before the final reading. |
| Random | The syringe plunger sticks, so the volume readings vary between repeats. | Lubricate the plunger and repeat, taking a mean. |
Model answer [3 marks]
A 120 cm³ sample of air is passed repeatedly over heated copper until no further change occurs. The final volume is 95 cm³. Calculate the percentage of oxygen. [3]
The decrease in volume is 120 − 95 = 25 cm³, and this is the volume of oxygen removed by the copper. The percentage of oxygen is the decrease divided by the original volume, multiplied by 100, which gives (25 ÷ 120) × 100 = 20.8%, or about 21%.
Model answer [3 marks]
Explain why the air must be passed over the heated copper repeatedly rather than once. [3]
A single pass does not give the oxygen enough contact with the hot copper for all of it to react, so some oxygen would remain in the gas. Passing the air backwards and forwards ensures that all of the gas comes into contact with the hot copper many times, so that all the oxygen is removed. The process is continued until the volume stops decreasing, which shows that no oxygen is left and the reaction is complete.
Why an answer might come out wrong
- TOO LOW — the reaction was incomplete, so some oxygen remained unreacted
- TOO LOW — too little copper was present to take all the oxygen
- TOO SMALL A DECREASE — the final volume was read while the gas was still hot and expanded
- TOO HIGH — the apparatus leaked and drew in fresh air during the experiment
- The answer should come out near 21%; anything far from it points to one of these
Not this: The copper is what is being investigated in this experiment.
This: The copper is only a tool for REMOVING the oxygen. What is measured is the change in the volume of the AIR, and the copper is chosen simply because it reacts with oxygen and not with nitrogen.
Mark-losing trap. Nitrogen is about 78% and oxygen about 21% — not the other way round.
Mark-losing trap. The DECREASE in volume is the oxygen. Divide it by the ORIGINAL volume, not the final one.
Mark-losing trap. COOL the apparatus before the final reading, or the hot expanded gas hides part of the decrease.
3 · Prove it — the five questions
The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.
- Grade 6 · State [1 mark] — What is the approximate percentage of nitrogen in dry air?
- Grade 7 · Calculate [3 marks] — A 200 cm³ sample of air is passed over heated copper until no further change occurs. The final volume is 158 cm³. Calculate the percentage of oxygen in the air.
- Grade 8 · Explain [3 marks] — In the copper experiment, a student reads the final volume while the apparatus is still hot. Explain the effect on their answer.
- Grade 9 · Describe [6 marks] — Select every statement that belongs in a full-mark description of how to determine the percentage of oxygen in air using heated copper.
- 9+ · Analyse [6 marks] — A student uses the copper method and obtains 17% oxygen. A second student uses burning phosphorus over water and obtains 23%. Both conclude the other's method is unreliable. Select every statement that belongs in a full-mark analysis.
The people behind this science
Two ways into the same idea — the one who showed air is a mixture and found which part does the work, and the one who made the gas first and explained it wrongly to the end. Inside Incandio each of them answers knowing exactly which lesson you have just finished.
Antoine Lavoisier — the one who showed air is a mixture and found which part does the work
Air was regarded as one of the four elements until Lavoisier took it apart, and the experiment he used is essentially the one on this page. He heated mercury in a sealed volume of air for twelve days; a red powder formed on the surface and the volume of air fell by about a fifth. The remaining gas would not support a flame or a mouse. He then heated the red powder alone and recovered exactly the gas that had gone, which relit a glowing splint. Air was therefore a mixture of at least two things, and one of them was the part that made burning possible — which he named oxygen.
- “How did you prove that air is a mixture?”
- “Why did you heat the mercury for twelve days?”
- “What did the gas left behind fail to do?”
- “How did you recover the gas you had removed?”
- “Why did you name it oxygen?”
Joseph Priestley — the one who made the gas first and explained it wrongly to the end
Priestley isolated the gas in 1774, before Lavoisier, by focusing sunlight onto mercury oxide with a burning lens. A candle burned in it with a remarkably vigorous flame, and a mouse lived in a sealed portion far longer than in ordinary air. He tried breathing it himself and reported that his chest felt peculiarly light and easy. But he interpreted it through the phlogiston theory and called it dephlogisticated air — air with room to absorb more phlogiston — and he defended that account for the rest of his life, long after Lavoisier's explanation had prevailed.
- “How did you first make the new air?”
- “What did the candle do in it?”
- “What was it like to breathe?”
- “What did you think dephlogisticated air actually was?”
- “Why would you not accept Lavoisier's account?”
Think you understand it?
Reading is the start. On Incandio a topic counts as mastered only once you have taught it and defended it — the two things a revision guide cannot ask of you.
Incandio this topic → Prove it. Teach it. Defend it. What does “Incandio it” mean?
- Prove it — five questions on the Gases in the Air, marked on the spot, that tell you what was missing rather than just what was wrong.
- Teach it — Explain the Gases in the Air in your own words to Ember, an AI pupil who does not have it yet and asks the awkward question. Understanding is credited; wording is not required.
- Defend it — Argue the Gases in the Air against Joseph Priestley — three rounds, marked against descriptors you can read before you begin.
- Then the exam — read all 12 authored exam-style questions on this unit, each with its command word and its tariff, then have them marked on Incandio with the mistake you actually made named.
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Carry on through the course
- Previous lesson: Group 7 — the Halogens
- Next lesson: Combustion and Carbonates
- Group 1 — the Alkali Metals — Metals soft enough to cut with a knife, which set fire to water — and get worse as you go down.
- Group 7 — the Halogens — A green gas, a brown liquid and a grey solid — and a trend that runs the opposite way to Group 1.
- Combustion and Carbonates — Three things burning in oxygen, one carbonate breaking apart, and the 0.04% gas that changes the climate.
- The Reactivity Series — Putting eleven metals in order by watching what each one will and will not do.
- Rusting and its Prevention — Two things are needed to rust iron, and every method of stopping it removes one of them — or cheats.
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