The shape of the lesson
By the end of the lesson, learners can
- Four steps to an empirical formula: Enter a composition by mass in the simulator and follow its working: divide by A r, divide by the smallest, then scale to whole numbers.
- From empirical to molecular: Several different compounds can share one empirical formula.
- Combustion data and water of crystallisation: Two classic variants: find carbon and hydrogen from the carbon dioxide and water formed on burning, and find x in a hydrated salt from the mass lost on heating.
- Divide by Ar, then by the smallest value.
- Scale ratios ending in .5, .33 or .67; never round them away.
- n = Mr ÷ empirical formula mass gives the molecular formula.
- Combustion: C from CO2 (12/44), H from H2O (2/18), O by difference.
One picture every idea returns to
The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound; the molecular formula is the actual number of atoms of each element in one molecule. To find the empirical formula, divide each mass or percentage by Ar and then divide by the smallest answer. The molecular formula is n × the empirical formula, where n = Mr ÷ empirical formula mass.
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 | Enter a composition by mass in the simulator and follow its working: divide by A r, divide by the smallest, then scale to whole numbers. 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 empirical to molecular | Several different compounds can share one empirical formula. | Pairs |
| 21 to 30 min | Teach and check: Combustion data and water of crystallisation | Two classic variants: find carbon and hydrogen from the carbon dioxide and water formed on burning, and find x in a hydrated salt from the mass lost on heating. | 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 Combustion data and water of crystallisation, 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 |
|---|---|
| Do not round a ratio like 1: 1.5 to 1: 2 or 1: 1.33 to 1: 1. | Those are real ratios of 2: 3 and 3: 4, and rounding them early gives the wrong formula (Fe 2 O 3 is 1: 1.5). In the hydrate experiment, heat to constant mass: if some water remains, the water lost is too small and x comes out too low. |
Empirical and molecular formulae: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
A compound contains 32.4% sodium, 22.6% sulfur and 45.0% oxygen by mass. Find its empirical formula. (Ar: O 16.0, Na 23.0, S 32.1)
A compound contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Its Mr is 88.0. Find its empirical and molecular formulae. (Ar: H 1.0, C 12.0, O 16.0)
An oxide of iron contains 70.0% iron by mass. Find its empirical formula. (Ar: O 16.0, Fe 55.8)
0.230 g of a compound containing carbon, hydrogen and oxygen burns completely to give 0.440 g of CO2 and 0.270 g of H2O. Find its empirical formula. (Ar: H 1.0, C 12.0, O 16.0)
2.86 g of hydrated sodium carbonate, Na2CO3·xH2O, is heated to constant mass. The anhydrous residue has a mass of 1.06 g. Find x. (Mr: Na2CO3 106.0, H2O 18.0)
In the experiment above, explain why the solid is heated to constant mass, and state the effect on the value of x if it is not.
Answer key and marking notes
- Na: 32.4 ÷ 23.0 = 1.41; S: 22.6 ÷ 32.1 = 0.704; O: 45.0 ÷ 16.0 = 2.81.
- Dividing by 0.704 gives 2.00 : 1.00 : 3.99, so Na2SO4.
- C: 54.5 ÷ 12.0 = 4.54; H: 9.1 ÷ 1.0 = 9.1; O: 36.4 ÷ 16.0 = 2.275.
- Dividing by 2.275 gives 2 : 4 : 1, so the empirical formula is C2H4O (mass 44.0).
- n = 88.0 ÷ 44.0 = 2, so the molecular formula is C4H8O2.
- Fe: 70.0 ÷ 55.8 = 1.25; O: 30.0 ÷ 16.0 = 1.875.
- Ratio 1 : 1.50; doubling gives 2 : 3, so Fe2O3.
- Mass of C = 0.440 × 12.0/44.0 = 0.120 g; mass of H = 0.270 × 2.0/18.0 = 0.030 g.
- Mass of O = 0.230 − 0.120 − 0.030 = 0.080 g.
- Moles: C 0.0100, H 0.030, O 0.0050, a ratio of 2 : 6 : 1, so C2H6O.
- Water lost = 2.86 − 1.06 = 1.80 g = 0.100 mol; Na2CO3 = 1.06 ÷ 106.0 = 0.0100 mol.
- x = 0.100 ÷ 0.0100 = 10.
- Heating to constant mass shows that all the water of crystallisation has been driven off.
- If some water remains, the mass lost is too small, so the calculated x is too low.
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 nitrogen contains 36.8% nitrogen and 63.2% oxygen by mass. What is its empirical formula? (Ar: N 14.0, O 16.0)
- AN2O3
- BNO
- CNO2
- DN3O2
A sample of a compound contains 2.4 g of carbon, 0.6 g of hydrogen and 1.6 g of oxygen. What is its empirical formula? (Ar: H 1.0, C 12.0, O 16.0)
- AC2H6O
- BCH3O
- CC4H12O
- DC2H3O
Which pair of compounds has the same empirical formula?
- AC2H4 and C3H6
- BC2H6 and C3H8
- CCH4 and C2H6
- DC2H2 and C2H4
A compound has the empirical formula CH2Cl and a relative molecular mass of 99.0. What is its molecular formula? (Ar: H 1.0, C 12.0, Cl 35.5)
- AC2H4Cl2
- BCH2Cl
- CC2H4Cl
- DC7H14Cl
What is the percentage by mass of nitrogen in ammonium nitrate, NH4NO3? (Ar: H 1.0, N 14.0, O 16.0)
- A35.0%
- B17.5%
- C60.0%
- D5.0%
0.450 g of a compound containing only carbon, hydrogen and oxygen is burned completely, giving 0.660 g of CO2 and 0.270 g of H2O. What is its empirical formula? (Ar: H 1.0, C 12.0, O 16.0)
- ACH2O
- BCHO
- CCH2
- DCH2O3
Gypsum is CaSO4·2H2O (Mr 172.2). A 3.44 g sample is heated gently and loses 0.54 g of water, forming CaSO4·xH2O. What is the value of x? (Mr of H2O = 18.0)
- A0.5
- B1.5
- C2
- D1
A hydrocarbon contains 90.0% carbon by mass and has a relative molecular mass of 40.0. What is its molecular formula? (Ar: H 1.0, C 12.0)
- AC3H4
- BC3H3
- CCH
- DC2H3
Quiz key
N: 36.8 ÷ 14.0 = 2.63; O: 63.2 ÷ 16.0 = 3.95. Dividing by 2.63 gives N : O = 1 : 1.50. A ratio of 1.5 must be doubled, giving 2 : 3, so the formula is N2O3. NO rounds 1.50 down and NO2 rounds it up; neither is allowed because 1.50 is not close to a whole number. N3O2 inverts the ratio.
C: 2.4 ÷ 12.0 = 0.20 mol; H: 0.6 ÷ 1.0 = 0.60 mol; O: 1.6 ÷ 16.0 = 0.10 mol. Dividing by 0.10 gives 2 : 6 : 1, so C2H6O. CH3O divides by the carbon value and then rounds the 0.5 for oxygen up to 1. C4H12O uses 32.0 for oxygen, the molar mass of O2 rather than Ar. C2H3O uses 2.0 for hydrogen, the molar mass of H2.
C2H4 and C3H6 both reduce to CH2 (1 : 2). C2H6 reduces to CH3 while C3H8 cannot be reduced, so they differ even though both are alkanes. CH4 and C2H6 (CH3) also differ. C2H2 (CH) and C2H4 (CH2) share the same number of carbon atoms, not the same ratio.
Empirical formula mass = 12.0 + 2 × 1.0 + 35.5 = 49.5. 99.0 ÷ 49.5 = 2, so every subscript doubles: C2H4Cl2. CH2Cl stops at the empirical formula. C2H4Cl doubles carbon and hydrogen but forgets chlorine. C7H14Cl comes from dividing 99.0 by the mass of CH2 alone (14.0) and ignoring chlorine.
Mr = 14.0 + 4.0 + 14.0 + 48.0 = 80.0. There are two N atoms (one in NH4+ and one in NO3−), so % N = 28.0 ÷ 80.0 × 100 = 35.0%. 17.5% counts only one nitrogen atom. 60.0% is the percentage of oxygen (48.0 ÷ 80.0). 5.0% is the percentage of hydrogen.
n(C) = n(CO2) = 0.660 ÷ 44.0 = 0.0150 mol (0.180 g). n(H) = 2 × 0.270 ÷ 18.0 = 0.0300 mol (0.030 g). Mass of O = 0.450 − 0.180 − 0.030 = 0.240 g, so n(O) = 0.0150 mol. Ratio 1 : 2 : 1 gives CH2O. CHO forgets that each water molecule contains two hydrogen atoms. CH2 assumes the compound has no oxygen and ignores the mass difference. CH2O3 counts all the oxygen in the products, much of which came from the oxygen used for burning.
n(gypsum) = 3.44 ÷ 172.2 = 0.0200 mol; n(water lost) = 0.54 ÷ 18.0 = 0.0300 mol, which is 1.5 H2O lost per formula unit. Water remaining = 2 − 1.5 = 0.5, so the product is CaSO4·0.5H2O (plaster of Paris). 1.5 is the water lost, not the water left. 2 is the starting hydrate, ignoring the mass loss. 1 comes from using 136.2 (anhydrous CaSO4) for the gypsum, which gives 0.0253 mol and about 1.2 water lost per unit, rounded to 1.
C: 90.0 ÷ 12.0 = 7.50; H: 10.0 ÷ 1.0 = 10.0. Ratio 1 : 1.33, which must be multiplied by 3 to give 3 : 4, so the empirical formula is C3H4 (mass 40.0). Since Mr = 40.0, the molecular formula is also C3H4. C3H3 rounds 1.33 down to CH and then multiplies by 40.0 ÷ 13.0 ≈ 3. CH stops after the rounding error. C2H3 multiplies by 2, which gives 2 : 2.67, and rounds that up; it does not match Mr 40.0 either.
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.