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DSE Chemistry | Molar Volume Calculations: A Tutor Guide | Moles First, Then 24
A student sees 10.0 g of calcium carbonate and writes 10 cubic decimetres, because 24 was remembered as a number to multiply by rather than a volume per mole. This lesson fixes the route into three steps: grams divided by molar mass gives moles, moles times 24 gives cubic decimetres, and the condition stated in the question decides whether 24 or 22.4 is used.
Objectives — Students can: (1) turn 10.0 g with a molar mass of 100 into 0.100 mol; (2) multiply by 24 for 2.40 cubic decimetres at room conditions and by 22.4 for 2.24 at standard conditions; (3) run backwards from 1.20 cubic decimetres to 0.050 mol; (4) identify the limiting reagent before applying a molar volume; (5) compute 3 minutes per question and 0.667 marks per minute for 9 questions worth 18 marks in 27 minutes.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Molar mass | What is the molar mass of calcium carbonate? | 100 grams per mole: 40 plus 12 plus 48 |
| Moles | How many moles are in 10.0 g of calcium carbonate? | 0.100 mol, since 10.0 divided by 100 is 0.1 |
| Molar volume | What volume does one mole of gas occupy at room conditions? | 24 cubic decimetres, against 22.4 at standard conditions |
| Limiting reagent | Two masses of reactants are given. What comes first? | Convert both to moles, divide by coefficients, and take the smaller |
02 Lesson flow (45 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Three boxes | Write grams, moles and cubic decimetres on the board with the number used at each step | Know that 24 is per mole, not per gram |
| 8-18 min | The carbonate example | 10.0 divided by 100 is 0.100, times 24 is 2.40 cubic decimetres | Both numbers carry units |
| 18-26 min | Change the condition | The same moles times 22.4 gives 2.24, and the two answers are compared | Explain the seven per cent gap |
| 26-34 min | Run backwards | 1.20 divided by 24 is 0.050 mol, which is then multiplied by molar mass | Direction is not reversed |
| 34-42 min | Limiting reagent | Convert to moles, divide by coefficients, keep the smaller | The larger one is never used |
| 42-45 min | Exit check | Four oral answers with notes closed | Three steps and two constants correct |
03 Teaching script (read it out)
Scene 1: Twenty four is per mole, not per gram
Nearly every error comes from this sentence. The unit of molar volume is cubic decimetres per mole, so a mole count must exist before multiplying; 10.0 g times 24 gives a number whose unit is meaningless. Make students write units inside the working, so a gram left unconverted stops the calculation by itself.
Ask the student to state that moles must come first, then show how the units cancel.
Scene 2: Room and standard conditions are two constants
The Hong Kong syllabus uses 24 at room temperature and pressure and 22.4 at standard temperature and pressure. The same 0.100 mol differs by 0.16 cubic decimetres, about seven per cent, and questions always state the condition. Before any gas volume, students should write 24 or 22.4 in the margin together with the condition quoted.
Ask for both constants and which condition selects each.
Scene 3: The reverse direction divides first
Going from volume back to mass means dividing by 24 to get moles, then multiplying by molar mass. The common error is 1.20 times 24, giving 28.8 mol. The self check is immediate: a volume below 24 cubic decimetres must mean fewer than one mole, so any answer above one is wrong.
Ask the student to apply the rule that below 24 means below one mole.
Scene 4: Mixtures need the limiting reagent first
When two reactant masses are given, convert both to moles, divide by the coefficients, and let the smaller decide the yield. 10.0 g of calcium carbonate is 0.100 mol and 4.0 g of sodium hydroxide is 0.100 mol, so a one to one equation uses both; other ratios need the division. Using the larger one overstates the yield and every later step inherits the error.
Ask why the coefficients matter when comparing moles.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Mass to gas volume | Mass and condition given | Divide by molar mass, multiply by 24 or 22.4, answer in cubic decimetres | Moles before volume |
| Volume to mass | Volume and condition given | Divide by 24 or 22.4, multiply by molar mass, answer in grams | Division, not multiplication |
| Limiting reagent | Two reactant masses given | Moles each, divide by coefficients, take the smaller | Coefficients cannot be skipped |
| Gas volume ratio | Two gases under one condition | Use the coefficient ratio directly, then check the total | Only valid at the same condition |
Sanity check (teach this): Order check: 10.0 divided by 100 is 0.100 mol, and times 24 that is 2.40 cubic decimetres, roughly a large water bottle and therefore plausible; 240 would mean the division by molar mass was skipped. Direction check: 1.20 is below 24, so the mole count must be below one, and 0.050 fits while 28.8 does not. Ratio check: 22.4 against 24 differs by 6.7 per cent, so the two answers cannot differ tenfold.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Volume of 10.0 g of calcium carbonate at room conditions | 2.40 cubic decimetres | Division by 100 first |
| Same sample at standard conditions | 2.24 cubic decimetres | Correct constant used |
| 1.20 cubic decimetres at room conditions is how many moles | 0.050 mol | Divided, not multiplied |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Multiplying grams by 24 | “10 grams times 24 is 240 cubic decimetres.” | “24 is per mole, not per gram. 10.0 divided by 100 is 0.100 mol, times 24 is 2.40 cubic decimetres.” |
| Mixing the two conditions | “The question did not say, so I used 22.4.” | “Room conditions means 24 and standard means 22.4. The gap is about seven per cent, so the wrong one costs marks.” |
| Multiplying in the reverse direction | “1.20 cubic decimetres times 24 is 28.8 mol.” | “Volume back to moles is a division: 1.20 over 24 is 0.050 mol. Below 24 means below one mole.” |
| Choosing the wrong limiting reagent | “Both are 0.100 mol, so either one will do.” | “Divide by the coefficients first. Only a one to one ratio uses both up, otherwise the smaller decides or the yield is overstated.” |
07 Homework (paper, section and question numbers)
- Chemistry chapter index — Revise moles and concentration, then do practice questions 1 and 2
- S4 Mole and stoichiometry — Question 3 applies the same coefficient comparison as the limiting reagent
- S4 Separation and purity — Questions 4 and 5 also start from a mole count before any percentage
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| How many moles in 10.0 g of calcium carbonate | 0.100 mol | Error: molar mass not used |
| Volume of 0.100 mol at room conditions | 2.40 cubic decimetres | Error: used 22.4 |
| How many moles in 1.20 cubic decimetres | 0.050 mol | Error: multiplied by 24 |
| How is the limiting reagent found | Divide by coefficients, take the smallest | Error: compared masses |
09 Tutor FAQ
Which constant should be memorised?
Both, because the question names the condition. A memory hook: room starts with r and pairs with 24, standard starts with s and pairs with 22.4.
Can coefficients be used directly on volumes?
Only when temperature and pressure match, since volume ratio then equals mole ratio. Otherwise convert each gas to moles first.
Why not use the ideal gas equation?
Molar volume covers the S4 syllabus with fewer errors. The ideal gas equation is reserved for questions where temperature or pressure changes during the process.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
