The shape of the lesson
By the end of the lesson, learners can
- One standard atom: Load an element into the simulator and look at its isotopes and their abundances.
- Adding up M r: Add the A r of every atom in the formula.
- Percentage by mass: The percentage of an element in a compound is the mass of that element in one formula, divided by M r, times 100.
- Standard: 1/12 of the mass of an atom of carbon-12.
- Mr: add the Ar of every atom; a number after a bracket multiplies everything inside.
- % by mass = (Ar × number of atoms) ÷ Mr × 100.
- No units: relative masses are ratios.
One picture every idea returns to
Relative atomic mass Ar is the average mass of the isotopes of an element compared with 1/12 of the mass of an atom of carbon-12. Relative molecular mass Mr is the sum of the Ar values of all the atoms in a formula; the same sum is used for ionic compounds (relative formula mass). Relative masses are ratios, so they have no 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, Weighed against carbon-12.. Atoms are far too light to weigh one at a time, so chemists compare them with one standard atom, carbon-12. 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: One standard atom | Load an element into the simulator and look at its isotopes and their abundances. Project tab 1 of the lab (Stoichiometry lab, mass spectrum) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Adding up M r | Add the A r of every atom in the formula. | Pairs |
| 21 to 30 min | Teach and check: Percentage by mass | The percentage of an element in a compound is the mass of that element in one formula, divided by M r, times 100. | Numbered Heads Together |
| 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 Percentage by mass, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Count every atom of the element. | Ammonium nitrate, NH 4 NO 3, has one N in the ammonium ion and one in the nitrate ion, so the nitrogen mass is 28, not 14. Using 14 gives 17.5%, exactly half the right answer. |
Relative masses: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Define relative atomic mass, Ar.
Calculate the relative formula masses of (a) calcium hydroxide, Ca(OH)2, and (b) magnesium nitrate, Mg(NO3)2. (Ar: H 1, N 14, O 16, Mg 24, Ca 40)
Calculate the relative formula mass of hydrated copper(II) sulfate, CuSO4·5H2O. (Ar: H 1, O 16, S 32, Cu 64)
Iron is extracted from iron(III) oxide, Fe2O3. Calculate the percentage by mass of iron in Fe2O3, and the maximum mass of iron that could be obtained from 400 tonnes of it. (Ar: O 16, Fe 56)
A farmer can buy urea, CO(NH2)2, or ammonium sulfate, (NH4)2SO4. Use percentages by mass of nitrogen to decide which supplies more nitrogen per kilogram. (Ar: H 1, C 12, N 14, O 16, S 32)
Extended. Chlorine consists of 75% chlorine-35 and 25% chlorine-37. Calculate the relative atomic mass of chlorine.
Answer key and marking notes
- The average mass of the isotopes of an element,
- compared with 1/12 of the mass of an atom of carbon-12.
- (a) 40 + 2 × (16 + 1) = 40 + 34 = 74.
- (b) 24 + 2 × (14 + 3 × 16) = 24 + 2 × 62 = 148.
- CuSO4: 64 + 32 + 4 × 16 = 160.
- Add 5 × 18 = 90 for the water of crystallisation: 160 + 90 = 250.
- Mr = 2 × 56 + 3 × 16 = 160.
- % Fe = 112 ÷ 160 × 100 = 70%.
- Mass of iron = 70% of 400 tonnes = 0.70 × 400 = 280 tonnes.
- Urea: Mr = 12 + 16 + 2 × 16 = 60; % N = 28 ÷ 60 × 100 = 46.7%.
- Ammonium sulfate: Mr = 132; % N = 28 ÷ 132 × 100 = 21.2%.
- Urea supplies more nitrogen per kilogram, more than twice as much.
- Total mass of 100 atoms = 35 × 75 + 37 × 25 = 2625 + 925 = 3550.
- Ar = 3550 ÷ 100 = 35.5.
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.
What is the relative formula mass of aluminium sulfate, Al2(SO4)3? (Ar: O 16, Al 27, S 32)
- A342
- B150
- C123
- D315
The relative atomic mass of magnesium is 24. Which statement is correct?
- AAn average magnesium atom is twice as heavy as an atom of carbon-12
- BOne magnesium atom has a mass of 24 g
- CA magnesium atom contains 24 protons
- DA magnesium atom is 24 times as heavy as an atom of carbon-12
What is the percentage by mass of oxygen in magnesium nitrate, Mg(NO3)2? (Ar: N 14, O 16, Mg 24)
- A64.9%
- B32.4%
- C10.8%
- D18.9%
Copper(I) sulfide, Cu2S, is a copper ore. What is the maximum mass of copper that can be obtained from 200 g of pure Cu2S? (Ar: S 32, Cu 64)
- A160 g
- B80 g
- C128 g
- D133 g
Which compound has the highest percentage by mass of oxygen? (Ar: H 1, C 12, O 16, Mg 24, S 32)
- Awater, H2O
- Bsulfur trioxide, SO3
- Ccarbon dioxide, CO2
- Dmagnesium oxide, MgO
What is the relative formula mass of hydrated sodium carbonate, Na2CO3·10H2O? (Ar: H 1, C 12, O 16, Na 23)
- A286
- B124
- C106
- D1240
The chloride of metal X has the formula XCl2 and a relative formula mass of 111. What is the relative atomic mass of X? (Ar: Cl 35.5)
- A40
- B75.5
- C55.5
- D182
The chart shows the isotopes of boron and their percentage abundances. What is the relative atomic mass of boron?
- A10.8
- B10.5
- C10.2
- D21
Quiz key
SO4 = 32 + 4 × 16 = 96. Mr = 2 × 27 + 3 × 96 = 54 + 288 = 342. 150 ignores the bracket (54 + 96); 123 uses one Al and one SO4 (27 + 96); 315 uses only one Al (27 + 288).
Ar compares the average mass of an atom with 1/12 of a carbon-12 atom. Carbon-12 has a relative mass of 12, so an atom of relative mass 24 is 24 ÷ 12 = 2 times as heavy. It is 24 times as heavy as 1/12 of a carbon-12 atom, not as a whole carbon-12 atom. A single atom weighs far less than 24 g, and magnesium has 12 protons.
Mr = 24 + 2 × (14 + 48) = 148. There are 6 O atoms: 6 × 16 = 96. 96 ÷ 148 × 100 = 64.9%. 32.4% uses only one nitrate (48 ÷ 148); 10.8% uses one O atom (16 ÷ 148); 18.9% is the nitrogen percentage (28 ÷ 148).
Mr = 2 × 64 + 32 = 160. Copper: 128 ÷ 160 × 100 = 80%. 80% of 200 g = 160 g. 80 g uses only one Cu atom (64 ÷ 160 = 40%). 128 g is just the mass of copper in one formula mass. 133 g treats the ore as CuS (64 ÷ 96 = 66.7%).
H2O: 16 ÷ 18 = 88.9%. CO2: 32 ÷ 44 = 72.7%. SO3: 48 ÷ 80 = 60%. MgO: 16 ÷ 40 = 40%. Water wins because the other element, hydrogen, is so light. SO3 has the most oxygen atoms but sulfur is heavy, so its percentage is lower.
Na2CO3 = 2 × 23 + 12 + 3 × 16 = 106. Water of crystallisation: 10 × 18 = 180. Total 106 + 180 = 286. 124 adds only one water (106 + 18); 106 ignores the water; 1240 multiplies everything by 10.
Two Cl atoms: 2 × 35.5 = 71. Ar(X) = 111 − 71 = 40, so X is calcium. 75.5 subtracts only one Cl atom; 55.5 halves 111; 182 adds the chlorine instead of subtracting it.
(10 × 20 + 11 × 80) ÷ 100 = (200 + 880) ÷ 100 = 10.8, close to 11 because boron-11 is much more common. 10.5 is a simple average that ignores abundance; 10.2 swaps the abundances; 21 just adds the two mass numbers.
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.