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DSE Chemistry | The Mole and Stoichiometry Teaching Guide (for tutors) | From grams to particles
Students memorise three formulae and fall apart the moment a limiting reagent appears. This lesson teaches one bridge only: grams to moles to particles or ratio, with units at every step and a mass check at the end.
Objectives — Students can: (1) use n = m / M to show that 22 g of carbon dioxide is 0.5 mol, or 3.01 x 10^23 molecules; (2) use c = n / V to get 1.00 mol/dm3 from 10 g of sodium hydroxide in 250 cm3; (3) use the coefficient ratio to identify oxygen as limiting when 3 mol of hydrogen reacts with 1 mol of oxygen, giving 36 g of water and 2 g of hydrogen left; (4) compute an 80 % yield from 43.2 g actual against 54 g theoretical.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Relative atomic masses | C = 12 and O = 16: what is the molar mass of CO2? | 44 g/mol |
| Coefficient ratio | In 2H2 + O2 -> 2H2O, what is the ratio of hydrogen to water? | 2 : 2, that is 1 : 1 |
| Unit conversion | How many dm3 is 250 cm3? | 0.250 dm3 |
| Percentages | What percentage is 43.2 out of 54? | 80 % |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | A dozen and a mole | Ask how many particles are in a mole if a dozen is twelve | Write an answer with units |
| 8-20 min | Molar mass | Add atomic masses from the periodic table for 22 g of CO2 | Compute once and write mol |
| 20-32 min | Concentration | Dissolve 10 g of NaOH in 250 cm3, find n then c | Write n and V before dividing |
| 32-44 min | Ratio and limiting reagent | 3 mol H2 with 1 mol O2: which runs out first? | Draw start / change / end rows |
| 44-52 min | Yield | Theoretical 54 g, actual 43.2 g: find the yield and two losses | Name two practical losses |
| 52-60 min | Exit check | Hand out four questions | Answer independently, redo errors at once |
03 Teaching script (read it out)
Scene 1: A mole counts particles, it does not weigh them
One mole is 6.02e+23 particles, the number in 12 g of carbon-12. The molar mass of CO2 is 12 + 2 x 16 = 44 g/mol, so n = 22 / 44 = 0.5 mol and the particle count is 3.01e+23. Make students say plainly that grams measure mass and moles count particles, with molar mass as the bridge.
Ask for a everyday analogy such as a dozen first, then let them invent one.
Scene 2: Concentration means moles first, then divide by volume
10 g of NaOH has molar mass 23 + 16 + 1 = 40 g/mol, so n = 0.25 mol. Write 250 cm3 as 0.25 dm3, then c = 0.25 / 0.25 = 1.00 mol/dm3. On dilution c1V1 = c2V2 gives 1.00 x 0.025 = 0.250 x c2, so 0.100 mol/dm3. Failing to convert cm3 is the usual factor-of-1000 error, so require the conversion line before any calculator work.
Make students read out “250 cm3 is 0.250 dm3” before touching the calculator.
Scene 3: The limiting reagent is decided by the coefficient ratio
3 mol of H2 needs only 1.5 mol of O2 by the 2 : 1 ratio, but only 1 mol of O2 is present, so oxygen limits. Water formed is 2 x 1 = 2 mol, or 36 g, and 1 mol of hydrogen, 2 g, is left. With excess hydrogen the theoretical yield is 54 g, so 43.2 g actual gives 80 % yield.
Insist on the three-row table rather than mental arithmetic.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Mass to moles | 22 g CO2, M = 44 | 1) n = m / M 2) write mol 3) multiply by 6.02 x 10^23 for particles | 0.5 mol |
| Concentration | 10 g NaOH in 250 cm3 | 1) find n 2) convert cm3 to dm3 3) c = n / V | 1.00 mol/dm3 |
| Limiting reagent | 3 mol H2 + 1 mol O2 | 1) find moles needed 2) identify the one that runs out 3) compute from it | 36 g water |
| Yield | Theoretical 54 g, actual 43.2 g | 1) actual / theoretical x 100 % 2) check it is at most 100 % | 80 % |
Sanity check (teach this): Mass check: 3 mol of H2 is 6 g and 1 mol of O2 is 32 g, 38 g in total; products are 36 g of water plus 2 g of hydrogen, also 38 g. Mass is conserved. A concentration of 1.00 mol/dm3 is typical for school labs and 80 % yield is below 100 %, as it must be.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| How many moles and molecules in 36 g of water? | 2 mol; 1.204 x 10^24 molecules | Dividing by 18, not 36 |
| 0.5 mol of NaCl in 500 cm3: concentration? | 1.00 mol/dm3 | Converting cm3 to dm3 |
| 2 mol H2 with 2 mol O2: which limits? | H2, since only 1 mol of O2 is needed | Using the ratio, not raw moles |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Treating grams as moles | “22 g of CO2 is 22 mol” | “Grams measure mass, moles count particles. Without dividing by 44 your number is forty-four times too big.” |
| Leaving cm3 unconverted | “250 cm3 means 250 dm3” | “1 dm3 is 1000 cm3, so writing 250 makes the concentration a thousand times too small.” |
| Comparing masses for the limiting reagent | “32 g of O2 beats 6 g of H2, so H2 runs out” | “Compare moles and coefficients, not grams. Convert everything to moles first.” |
| Submitting a yield above 100 % | “The yield is 120 %” | “Above 100 % means the theoretical yield is wrong or the product is wet. Go back and check the steps.” |
07 Homework (paper, section and question numbers)
- Chemistry chapter index — Review atomic structure, then questions 1 and 2
- Atomic structure and the periodic table — Question 3 needs molar mass
- Acids, bases and titration — Question 6 uses c1V1 = c2V2
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Moles in 22 g of CO2 | 0.5 mol | Wrong: forgot to divide by 44 |
| Concentration of 10 g NaOH in 250 cm3 | 1.00 mol/dm3 | Wrong: cm3 not converted |
| Water from 3 mol H2 and 1 mol O2 | 36 g | Wrong: computed from hydrogen |
| Yield from 54 g theoretical and 43.2 g actual | 80 % | Wrong: swapped actual and theoretical |
09 Tutor FAQ
Must students memorise Avogadro constant?
The paper normally supplies 6.02 x 10^23, so they only need to use it correctly and show the multiplication.
Do they memorise every molar mass?
No. They add relative atomic masses from the data given, but must show the addition so errors are traceable.
Yield versus atom economy?
Yield is what you actually collected; atom economy is how much of the reactant mass ends up in the target product. Do not mix the two.
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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.
