Clean dry air: composition and approximate proportions
Use the required nitrogen and oxygen percentages while keeping the remaining gases and dry-air condition clear.
Exam scope and assessment notes
Core and Extended · Chemistry 0971, 2026–2028. The detailed greenhouse mechanism, catalytic-converter equation and photosynthesis symbol equation are taught in separate Extended lessons. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Distinguish a mixture of elements/compounds from a single substance.
- Calculate a percentage of a supplied gas volume under comparable conditions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does dry specify in the stated air composition?
What you will learn
- State the approximate composition of clean dry air with the required molecular formulas.
- Use approximate proportions without calling the remaining fraction entirely carbon dioxide.
Air is a mixture, not an oxygen molecule
Clean, dry air is approximately 78% nitrogen, N₂, and 21% oxygen, O₂. The remaining approximately 1% is a mixture including noble gases and carbon dioxide, CO₂. Do not label the whole remaining 1% carbon dioxide, or describe nitrogen and oxygen atoms as a single compound formula for air.
The values are approximate school proportions rather than an exact analysis of every air sample. Dry air excludes water vapour from this stated composition comparison. Humid air contains water vapour, whose amount varies; a polluted sample can contain other substances. Use the named clean/dry conditions rather than applying the proportions to every possible atmosphere.
Translate a proportion without changing the conditions
For a supplied volume of air in the same comparison conditions, an approximate constituent volume can be found by multiplying the total by its percentage fraction. In 1000 cm³ of the model clean dry air, about 780 cm³ is nitrogen and 210 cm³ oxygen, leaving about 10 cm³ of the other gases combined.
This is a composition calculation, not a description of three separately visible layers inside a container. Gases are mixed, and each constituent's volume fraction is considered at comparable temperature and pressure. If a later supplied experiment removes oxygen and asks for the fraction in the remaining gas, change the denominator rather than continuing to use the original total.
Pause and explain
Which description correctly identifies the remaining approximately 1%?
Worked example
Estimate nitrogen, oxygen and other gases in 500 cm³ of clean dry air using the school model.
Show the worked solution
- Nitrogen is 0.78×500=390 cm³.
- Oxygen is 0.21×500=105 cm³.
- The remaining gases together are 500−390−105=5 cm³, not all CO₂.
Answer About 390 cm³ N₂, 105 cm³ O₂ and 5 cm³ other gases combined.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe the required composition of clean dry air, naming formulas and the remainder.
Give me a hint
Use approximate proportions and distinguish the two main molecules.
Compare my reasoning
- About 78% is nitrogen, N₂.
- About 21% is oxygen, O₂.
- The remainder includes noble gases and CO₂, rather than all being carbon dioxide.
Approximately 78% N₂, 21% O₂ and a remainder including noble gases and CO₂.
Look for these in your work
- Both formulas and percentages match.
- The remainder is not assigned entirely to one gas.
Estimate the oxygen volume in 2000 cm³ of the model clean dry air and find the other-gas remainder after accounting for nitrogen.
Give me a hint
Use 21% oxygen, 78% nitrogen and the remaining approximate fraction.
Compare my reasoning
- Oxygen is 0.21×2000=420 cm³.
- Nitrogen is 0.78×2000=1560 cm³.
- Other gases together occupy the equivalent of 2000−420−1560=20 cm³ under comparable conditions.
About 420 cm³ O₂; about 20 cm³ other gases combined after 1560 cm³ N₂.
Look for these in your work
- All fractions use the original total.
- Approximate mixed-gas conditions remain explicit.
In a hypothetical same-condition model, all 21 cm³ oxygen is removed from 100 cm³ of clean dry air with no other change. Estimate nitrogen's percentage in the remaining gas.
Give me a hint
Keep 78 cm³ nitrogen, but use the reduced total rather than 100.
Compare my reasoning
- The remaining total is 100−21=79 cm³.
- The nitrogen amount remains equivalent to 78 cm³.
- Its new fraction is 78/79×100≈98.7%, an approximate model result.
Approximately 98.7% nitrogen in the remaining gas, using 79 cm³ as the denominator.
Look for these in your work
- The denominator changes after removal.
- The idealised no-other-change assumption is retained.
Common mistakes
- Calling the remaining 1% entirely carbon dioxide.
- Omitting the dry-air condition and ignoring water vapour.
- Writing N or O as the required molecular formulas.
- Treating approximate proportions as exact for every polluted or humid sample.
What can you explain now?
State the two main gas percentages and formulas in clean dry air.
Compare with the explanation
Approximately 78% N₂ and 21% O₂.
The remainder includes noble gases and CO₂, rather than CO₂ alone.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Recall the dry-air proportions and formulas, then estimate constituent volumes and recalculate a fraction after a stated removal.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.