The shape of the lesson
By the end of the lesson, learners can
- Four steps to an empirical formula: Extended content.
- From percentages to molecular formulae: Percentages work exactly like masses.
- Ratios that are not whole: Sometimes step 3 gives a number like 1.5.
- Divide each mass or % by Ar, then by the smallest answer.
- Double if a value ends in .5; never round it away.
- Molecular formula: Mr ÷ empirical formula mass = the multiplier.
- Find a missing mass by subtraction: mass of oxide − mass of metal = mass of oxygen.
One picture every idea returns to
The empirical formula is the simplest whole-number ratio of the different atoms or ions in a compound. The molecular formula is the number and type of each atom in one molecule. To find the empirical formula, divide each mass (or percentage) by Ar, then divide by the smallest answer. To find the molecular formula, divide Mr by the empirical formula mass and multiply the empirical formula by the answer.
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, The simplest ratio.. Analysis tells you what a compound is made of, by mass. 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: Four steps to an empirical formula | Extended content. Project tab 3 of the lab (Stoichiometry lab, formulae) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: From percentages to molecular formulae | Percentages work exactly like masses. | Pairs |
| 21 to 30 min | Teach and check: Ratios that are not whole | Sometimes step 3 gives a number like 1.5. | 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, 15 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 Ratios that are not whole, 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 |
|---|---|
| Never round 1: 1.5 to 1: 2 (or to 1: 1). | A ratio ending in.5 means you must double every number. Rounding the iron oxide ratio gives FeO 2, which is wrong; doubling gives Fe 2 O 3. And never divide the masses themselves: 2.4 g Mg to 1.6 g O is a 1: 1 ratio of atoms, not 3: 2. |
Empirical and molecular formulae: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Define (a) empirical formula and (b) molecular formula.
1.4 g of nitrogen combines with 3.2 g of oxygen. Find the empirical formula of the oxide. (Ar: N 14, O 16)
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its Mr is 42. Find its empirical and molecular formulae. (Ar: H 1, C 12)
A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. Its Mr is 46. Find its empirical and molecular formulae. (Ar: H 1, C 12, O 16)
5.4 g of aluminium is heated in oxygen and forms 10.2 g of aluminium oxide. Find the empirical formula of the oxide. (Ar: O 16, Al 27)
A student finds the ratio of iron atoms to oxygen atoms in an oxide is 1 : 1.5 and writes the formula FeO2. Explain the mistake and give the correct formula.
Answer key and marking notes
- (a) The simplest whole-number ratio of the different atoms or ions in a compound.
- (b) The number and type of each atom in one molecule of the compound.
- N: 1.4 ÷ 14 = 0.1 mol; O: 3.2 ÷ 16 = 0.2 mol.
- Ratio 1 : 2, so NO2.
- C: 85.7 ÷ 12 = 7.14; H: 14.3 ÷ 1 = 14.3.
- Dividing by 7.14 gives 1 : 2, so the empirical formula is CH2 (mass 14).
- 42 ÷ 14 = 3, so the molecular formula is C3H6.
- C: 52.2 ÷ 12 = 4.35; H: 13.0 ÷ 1 = 13.0; O: 34.8 ÷ 16 = 2.175.
- Dividing by 2.175 gives 2 : 5.98 : 1, which is 2 : 6 : 1, so C2H6O (mass 24 + 6 + 16 = 46).
- 46 ÷ 46 = 1, so the molecular formula is also C2H6O.
- Mass of oxygen = 10.2 − 5.4 = 4.8 g.
- Al: 5.4 ÷ 27 = 0.2 mol; O: 4.8 ÷ 16 = 0.3 mol.
- Ratio 1 : 1.5, doubled to 2 : 3, so Al2O3.
- The student rounded 1.5 up to 2; a ratio ending in .5 is a real ratio and must not be rounded.
- Doubling both numbers gives 2 : 3, so the formula is Fe2O3.
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.
An oxide of sulfur contains 3.2 g of sulfur and 4.8 g of oxygen. What is its empirical formula? (Ar: O 16, S 32)
- ASO3
- BS2O3
- CSO2
- DS3O2
A compound contains 32.4% sodium, 22.5% sulfur and 45.1% oxygen by mass. What is its empirical formula? (Ar: O 16, Na 23, S 32)
- ANa2SO4
- BNa2SO3
- CNaSO4
- DNa2S2O3
A hydrocarbon has the empirical formula C2H5 and a relative molecular mass of 58. What is its molecular formula? (Ar: H 1, C 12)
- AC4H10
- BC2H5
- CC4H8
- DC5H12
Which pair of compounds has the same empirical formula?
- AC2H4 and C3H6
- BCH4 and C2H6
- CC2H2 and C2H4
- DH2O and H2O2
6.2 g of phosphorus burns to form an oxide, combining with 8.0 g of oxygen. What is the empirical formula of the oxide? (Ar: O 16, P 31)
- AP2O5
- BPO2
- CPO3
- DP4O10
1.12 g of iron is heated in chlorine and forms 3.25 g of an iron chloride. What is the empirical formula of the chloride? (Ar: Cl 35.5, Fe 56)
- AFeCl3
- BFe2Cl3
- CFeCl2
- DFeCl
2.46 g of hydrated magnesium sulfate, MgSO4·xH2O, is heated until all the water is lost. 1.20 g of anhydrous MgSO4 remains. What is the value of x? (Ar: H 1, O 16, Mg 24, S 32)
- A7
- B1
- C3
- D8
Which statement about the formula of the ionic compound calcium chloride, CaCl2, is correct?
- AIt is an empirical formula: it gives the ratio of Ca2+ to Cl− ions in the lattice
- BIt is a molecular formula: one molecule contains one Ca atom and two Cl atoms
- CIts empirical formula is CaCl
- DIts molecular formula is Ca2Cl4
Quiz key
S: 3.2 ÷ 32 = 0.1 mol. O: 4.8 ÷ 16 = 0.3 mol. Ratio 1 : 3, so SO3. S2O3 uses the mass ratio 3.2 : 4.8 = 2 : 3 without dividing by Ar; S3O2 turns that mass ratio upside down; SO2 is a different oxide of sulfur.
Na: 32.4 ÷ 23 = 1.41. S: 22.5 ÷ 32 = 0.703. O: 45.1 ÷ 16 = 2.82. Dividing by 0.703 gives 2.00 : 1 : 4.01, so Na2SO4. Na2SO3 would need only 38% oxygen; NaSO4 ignores the 2 : 1 ratio of Na to S; Na2S2O3 has twice as much sulfur as the data show.
C2H5 has mass 2 × 12 + 5 = 29. 58 ÷ 29 = 2, so the molecular formula is C4H10 (butane). C2H5 is only the empirical formula; C4H8 (Mr 56) does not simplify to C2H5; C5H12 has Mr 72.
C2H4 (2 : 4) and C3H6 (3 : 6) both simplify to CH2. CH4 stays CH4 but C2H6 simplifies to CH3. C2H2 gives CH and C2H4 gives CH2. H2O cannot be simplified, and H2O2 simplifies to HO.
P: 6.2 ÷ 31 = 0.2 mol. O: 8.0 ÷ 16 = 0.5 mol. Ratio 1 : 2.5, doubled to 2 : 5, so P2O5. PO2 and PO3 round 2.5 down or up. P4O10 is the molecular formula of this oxide, but it is not the simplest ratio, so it is not the empirical formula.
Mass of chlorine = 3.25 − 1.12 = 2.13 g. Fe: 1.12 ÷ 56 = 0.02 mol. Cl: 2.13 ÷ 35.5 = 0.06 mol. Ratio 1 : 3, so FeCl3. Fe2Cl3 divides by 71 (the Mr of Cl2) instead of 35.5, giving 0.03 mol. FeCl2 is iron(II) chloride, which does not fit the data.
Water lost = 2.46 − 1.20 = 1.26 g = 1.26 ÷ 18 = 0.07 mol. MgSO4 (Mr 120): 1.20 ÷ 120 = 0.01 mol. x = 0.07 ÷ 0.01 = 7. 1 compares the masses directly (1.26 : 1.20); 3 uses 2.46 g as the mass of MgSO4 (0.0205 mol); 8 divides the water mass by 16 instead of 18 (0.0788 mol).
Calcium chloride is a giant ionic lattice, not a collection of molecules, so it has only an empirical formula: the simplest ratio of ions, one Ca2+ to two Cl−. It has no molecules, so it has no molecular formula such as Ca2Cl4. CaCl does not balance the charges and is not the correct ratio.
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.