The shape of the lesson
By the end of the lesson, learners can
- A chemist's dozen: Pick a substance in the simulator and drag the mass slider.
- Mass, moles, particles: Work every conversion through the mole.
- Which particle?: The word mole means nothing until you say what it is a mole of.
- n = m ÷ M with m in grams and M in g mol−1.
- N = n × L with L = 6.02 × 1023 mol−1.
- Name the particle: atoms, molecules, ions or formula units.
- Multiply up by the number of atoms or ions in each formula.
One picture every idea returns to
One mole is the amount of substance that contains 6.02 × 1023 particles, the Avogadro constant L. That number is chosen so that one mole of carbon-12 atoms has a mass of exactly 12 g, which makes the molar mass M of any substance in g mol−1 numerically equal to its relative mass. So n = m ÷ M and N = n × L. Always say which particle you are counting: atoms, molecules, ions or formula units.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Counting by weighing.. Nobody can count 10 23 atoms, but anyone can weigh 12 grams of carbon. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: A chemist's dozen | Pick a substance in the simulator and drag the mass slider. Project tab 2 of the lab (Stoichiometry lab, the mole) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Mass, moles, particles | Work every conversion through the mole. | Pairs |
| 21 to 30 min | Teach and check: Which particle? | The word mole means nothing until you say what it is a mole of. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Which particle?, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| When a question asks for the number of atoms or ions, multiply the number of formula units by how many of that particle each one contains. 0.100 mol of MgCl 2 contains 0.200 mol of chloride ions, not 0.100 mol. | Return to Which particle?: ask a pair to state the correct version and show it on the lab before moving on. |
The mole and the Avogadro constant: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Define the mole.
For 4.40 g of carbon dioxide, CO2, calculate (a) the amount in moles, (b) the number of molecules and (c) the total number of atoms. (Ar: C 12.0, O 16.0)
Calculate the mass of 3.01 × 1022 atoms of magnesium. (Ar: Mg 24.3)
0.0200 mol of an oxide with the formula X2O3 has a mass of 3.20 g. Calculate the relative formula mass of the oxide and identify element X. (Ar: O 16.0)
Calculate the total number of ions in 2.00 g of calcium chloride, CaCl2. (Ar: Cl 35.5, Ca 40.1)
Calculate the mass, in grams, of a single molecule of water. (Ar: H 1.0, O 16.0)
Answer key and marking notes
The amount of substance that contains 6.02 × 1023 particles (the Avogadro constant), the same number as there are atoms in exactly 12 g of carbon-12.
- (a) Mr = 44.0, so n = 4.40 ÷ 44.0 = 0.100 mol.
- (b) 0.100 × 6.02 × 1023 = 6.02 × 1022 molecules.
- (c) Three atoms per molecule: 3 × 6.02 × 1022 = 1.81 × 1023 atoms.
- n = 3.01 × 1022 ÷ 6.02 × 1023 = 0.0500 mol.
- Mass = 0.0500 × 24.3 = 1.22 g.
- M = m ÷ n = 3.20 ÷ 0.0200 = 160 g mol−1, so the relative formula mass is 160.
- 2Ar(X) + 3 × 16.0 = 160, so Ar(X) = (160 − 48.0) ÷ 2 = 56.0.
- X is iron (Ar 55.8); the oxide is Fe2O3.
- Mr = 40.1 + 2 × 35.5 = 111.1, so n = 2.00 ÷ 111.1 = 0.0180 mol.
- Each formula unit gives one Ca2+ and two Cl−: 3 × 0.0180 = 0.0540 mol of ions.
- 0.0540 × 6.02 × 1023 = 3.25 × 1022 ions.
- One mole of water, 6.02 × 1023 molecules, has a mass of 18.0 g.
- Mass of one molecule = 18.0 ÷ 6.02 × 1023 = 2.99 × 10−23 g.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
How many molecules are there in 3.40 g of ammonia, NH3? (Ar: H 1.0, N 14.0; L = 6.02 × 1023 mol−1)
- A1.20 × 1023
- B4.82 × 1023
- C2.05 × 1024
- D3.01 × 1024
What is the mass of 0.0500 mol of calcium hydroxide, Ca(OH)2? (Ar: H 1.0, O 16.0, Ca 40.1)
- A3.71 g
- B2.86 g
- C1482 g
- D6.75 × 10−4 g
Which sample contains the same number of molecules as 16.0 g of oxygen, O2? (Ar: H 1.0, C 12.0, O 16.0)
- A22.0 g of carbon dioxide, CO2
- B16.0 g of methane, CH4
- C44.0 g of carbon dioxide, CO2
- D18.0 g of water, H2O
How many oxygen atoms are there in 0.0200 mol of aluminium sulfate, Al2(SO4)3? (L = 6.02 × 1023 mol−1)
- A1.44 × 1023
- B4.82 × 1022
- C1.20 × 1022
- D2.05 × 1023
Which statement about 1.00 mol of chlorine gas, Cl2, is correct? (Ar: Cl 35.5; L = 6.02 × 1023 mol−1)
- Ait has a mass of 71.0 g and contains 1.20 × 1024 chlorine atoms
- Bit has a mass of 35.5 g and contains 6.02 × 1023 chlorine atoms
- Cit has a mass of 71.0 g and contains 6.02 × 1023 chlorine atoms
- Dit has a mass of 35.5 g and contains 1.20 × 1024 chlorine atoms
1.51 × 1022 formula units of a metal oxide, MO, have a mass of 1.99 g. Which metal is M? (Ar: O 16.0, Mg 24.3, Ca 40.1, Fe 55.8, Cu 63.5; L = 6.02 × 1023 mol−1)
- Acopper
- Biron
- Ccalcium
- Dmagnesium
A drop of water has a volume of 0.050 cm3. Taking the density of water as 1.00 g cm−3, how many water molecules does the drop contain? (Ar: H 1.0, O 16.0; L = 6.02 × 1023 mol−1)
- A1.7 × 1021
- B5.0 × 1021
- C3.0 × 1022
- D1.7 × 1024
A mixture contains 5.30 g of sodium carbonate, Na2CO3 (Mr 106.0), and 5.85 g of sodium chloride, NaCl (Mr 58.5). How many sodium ions does it contain? (L = 6.02 × 1023 mol−1)
- A1.20 × 1023
- B9.03 × 1022
- C2.11 × 1023
- D1.51 × 1023
Quiz key
Mr of NH3 = 14.0 + 3 × 1.0 = 17.0, so n = 3.40 ÷ 17.0 = 0.200 mol. Molecules = 0.200 × 6.02 × 1023 = 1.20 × 1023. 4.82 × 1023 multiplies by 4 and so counts atoms, not molecules. 2.05 × 1024 multiplies the mass in grams by L without converting to moles. 3.01 × 1024 comes from dividing the wrong way, 17.0 ÷ 3.40 = 5.00 mol.
Mr = 40.1 + 2 × (16.0 + 1.0) = 74.1, so mass = 0.0500 × 74.1 = 3.71 g. 2.86 g uses CaOH (57.1), ignoring the 2 outside the bracket. 1482 g divides the molar mass by the amount instead of multiplying. 6.75 × 10−4 g divides the amount by the molar mass.
16.0 g of O2 (Mr 32.0) is 0.500 mol of molecules. 22.0 g of CO2 is 22.0 ÷ 44.0 = 0.500 mol, the same number of molecules. 16.0 g of CH4 has the same mass but is 1.00 mol, because methane molecules are lighter. 44.0 g of CO2 and 18.0 g of water are each 1.00 mol, which is the amount in 32.0 g of oxygen, not 16.0 g; they come from treating 16.0 g of O2 as one mole.
One formula unit contains 3 × 4 = 12 oxygen atoms, so n(O) = 12 × 0.0200 = 0.240 mol, and 0.240 × 6.02 × 1023 = 1.44 × 1023 atoms. 4.82 × 1022 counts only the 4 oxygen atoms written inside the bracket. 1.20 × 1022 is the number of formula units, not oxygen atoms. 2.05 × 1023 counts all 17 atoms in the formula unit (2 Al, 3 S and 12 O).
One mole of Cl2 is 6.02 × 1023 molecules with molar mass 2 × 35.5 = 71.0 g mol−1. Each molecule has two atoms, so there are 1.20 × 1024 atoms. A mass of 35.5 g is one mole of chlorine atoms, which is only half a mole of molecules. 6.02 × 1023 is the number of molecules, not atoms.
n = 1.51 × 1022 ÷ 6.02 × 1023 = 0.0251 mol. Molar mass of MO = 1.99 ÷ 0.0251 = 79.3 g mol−1, so Ar(M) = 79.3 − 16.0 = 63.3, which matches copper (CuO, 79.5). Check the others the same way: the same number of formula units of FeO (71.8) would weigh only 1.80 g, CaO (56.1) 1.41 g and MgO (40.3) 1.01 g. Choosing one of these usually means the moles were found incorrectly (for example multiplying by L instead of dividing) and the answer was then matched to the nearest familiar metal.
Mass = 0.050 cm3 × 1.00 g cm−3 = 0.050 g; n = 0.050 ÷ 18.0 = 2.78 × 10−3 mol; molecules = 2.78 × 10−3 × 6.02 × 1023 = 1.7 × 1021. 5.0 × 1021 multiplies by 3 and counts atoms. 3.0 × 1022 multiplies the mass in grams straight by L, skipping the molar mass. 1.7 × 1024 treats 0.050 cm3 as 0.050 dm3 (50 g), a factor of 1000 too large.
n(Na2CO3) = 5.30 ÷ 106.0 = 0.0500 mol, giving 2 × 0.0500 = 0.100 mol Na+. n(NaCl) = 5.85 ÷ 58.5 = 0.100 mol, giving 0.100 mol Na+. Total 0.200 mol, so 0.200 × 6.02 × 1023 = 1.20 × 1023 ions. 9.03 × 1022 (0.150 mol) forgets that each Na2CO3 has two sodium ions. 2.11 × 1023 (0.350 mol) counts every ion, carbonate and chloride included. 1.51 × 1023 (0.250 mol) doubles the sodium chloride instead of the sodium carbonate.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.