Reacting masses by simple proportions
Use conserved mass ratios without needing the mole method for Core questions.
Exam scope and assessment notes
Core and Extended · Chemistry 0971, 2026–2028. 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
- Balance a symbol equation.
- Calculate a relative formula mass and scale a ratio.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If a ratio is 48:80, what is its value when the first part is 6?
What you will learn
- Calculate reacting masses using a stated or deduced simple proportion.
- Check whether a mass ratio includes reaction coefficients and conserved totals.
Build the mass ratio before scaling
A balanced equation gives whole-particle proportions. Combining its coefficients with relative masses gives a reacting mass ratio. For 2Mg+O₂→2MgO using Mg=24 and O=16, the relative contributions are 48:32:80. Therefore 48 g of magnesium reacts with 32 g of oxygen to produce 80 g of magnesium oxide, and smaller amounts scale in the same ratio if the reaction is complete with enough of the other reactant.
This Core route uses direct proportions, without calculating amounts in moles. Include coefficients before comparing masses: the equation's 2:1 magnesium-to-oxygen particle ratio is not a 2:1 mass ratio. Mass contributions on both sides must add to the same total.
State which reactant limits the prediction
If oxygen is supplied in excess, all the stated magnesium can react and its mass fixes the product prediction. If an amount of each reactant is supplied, compare their masses with the required ratio before assuming both disappear. A shortage of either reactant leaves some of the other unreacted.
For decomposition, the reactant mass can divide among products. CaCO₃→CaO+CO₂ has ratio 100:56:44 using Ca=40, C=12 and O=16. The sum of predicted product masses should match the reacted mass in a closed accounting of all products, including gases that might leave an experimental container.
Pause and explain
Why is the Mg:O₂ mass ratio not 2:1 in 2Mg+O₂→2MgO?
Worked example
6.0 g of magnesium burns completely in excess oxygen. Find the magnesium oxide mass using 2Mg+O₂→2MgO, Mg=24 and O=16.
Show the worked solution
- The mass ratio Mg:MgO is 48:80.
- Scale from 48 g Mg to 6.0 g using factor 6/48.
- Product mass is 80×6/48=10 g; the extra 4 g comes from oxygen.
Answer 10 g of magnesium oxide.
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Test this topic →From guided practice to a new situation
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.
12 g of magnesium reacts with excess oxygen. Use mass ratio 48:32:80 to find oxygen used and oxide produced.
Give me a hint
Scale every ratio part by 12/48.
Compare my reasoning
- The scale factor is one quarter.
- Oxygen used is 32/4=8 g; magnesium oxide is 80/4=20 g.
- Check conservation: 12+8=20 g.
8 g oxygen used and 20 g magnesium oxide produced.
Look for these in your work
- All masses use one scale factor.
- The reactant and product totals agree.
50 g CaCO₃ decomposes completely. Use mass ratio 100:56:44 to find both product masses.
Give me a hint
Halve all three ratio contributions.
Compare my reasoning
- The scale factor is 50/100=0.5.
- CaO mass is 56×0.5=28 g; CO₂ mass is 44×0.5=22 g.
- The product total 28+22 equals 50 g.
28 g CaO and 22 g CO₂.
Look for these in your work
- Both products are included.
- The mass check includes the gas.
A student heats 20 g CaCO₃ and measures 11.2 g solid CaO afterwards. Explain the apparent mass loss using complete reaction and the ratio 100:56:44.
Give me a hint
Account for a gaseous product leaving the container.
Compare my reasoning
- 20 g CaCO₃ corresponds to 20/100 of the stated reaction amounts.
- It produces 56×0.2=11.2 g CaO and 44×0.2=8.8 g CO₂.
- The CO₂ can leave the container, so the remaining solid is lighter while total product mass is conserved.
8.8 g CO₂ left; total product mass is still 20 g.
Look for these in your work
- The solid and gas masses fit the ratio.
- Apparent loss is distinguished from a failure of mass conservation.
Common mistakes
- Using equation coefficients as gram ratios without relative masses.
- Saying an oxide's mass must equal the starting metal's mass.
What can you explain now?
Using CaCO₃:CaO:CO₂ mass ratio 100:56:44, find the CO₂ mass from complete decomposition of 25 g CaCO₃.
Compare with the explanation
11 g
Scale by 25/100; the CO₂ contribution is 44×25/100=11 g.
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Make it stick
Write a complete mass ratio from one balanced equation, including coefficients. Scale it to a new amount and check the mass total before checking the solution.
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.