The shape of the lesson
By the end of the lesson, learners can
- Hydrated and anhydrous: The dot in a formula such as CuSO 4 ·5H 2 O shows water that is part of the crystal, in a fixed ratio.
- Finding the formula: Extended.
- Hydrated: contains water of crystallisation. Anhydrous: none.
- CuSO4: blue hydrated, white anhydrous.
- CoCl2: pink hydrated, blue anhydrous.
- x = mol of water ÷ mol of anhydrous salt.
One picture every idea returns to
A hydrated substance is chemically combined with water; an anhydrous substance contains no water. Water of crystallisation is the water molecules present in the crystals of a hydrated salt, as in CuSO4·5H2O. Heating a hydrated salt drives off this water: blue hydrated copper(II) sulfate becomes white anhydrous copper(II) sulfate. Extended: the number of water molecules, x, can be found from the masses before and after heating.
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, Water locked in a crystal.. Blue copper(II) sulfate crystals are not dry in the way they look: every formula unit holds five water molecules in its structure. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Hydrated and anhydrous | The dot in a formula such as CuSO 4 ·5H 2 O shows water that is part of the crystal, in a fixed ratio. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Finding the formula | Extended. | Jot Thoughts |
| 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 Finding the formula, 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.
Jot Thoughts
Teams write ideas on slips, one idea per slip, as many as they can in two minutes, saying each aloud as they place it in the middle. Then they sort the slips, which is where the structure of the topic appears.
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 |
|---|---|
| Hydrated crystals look and feel dry: the water of crystallisation is locked inside the crystal structure, not wetting the surface. | And removing it is a chemical change, shown by the colour change, not simply drying. |
Hydrated and anhydrous salts: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Define the terms hydrated, anhydrous and water of crystallisation.
Describe what is seen when blue hydrated copper(II) sulfate is heated strongly, and write the equation.
(Extended) 5.00 g of hydrated copper(II) sulfate, CuSO4·xH2O, is heated to constant mass. 3.20 g of anhydrous copper(II) sulfate (Mr 160) remains. Find x.
(Extended) 2.86 g of washing soda, Na2CO3·xH2O, is heated to constant mass, leaving 1.06 g of anhydrous sodium carbonate (Mr 106). Find x.
(Extended) A student heats a hydrated salt only once and not to constant mass. Explain the effect on the calculated value of x.
Describe a chemical test for water, and explain why it does not show that the water is pure.
Answer key and marking notes
- Hydrated: a substance that is chemically combined with water.
- Anhydrous: a substance that contains no water.
- Water of crystallisation: water molecules present in the crystals of a hydrated salt.
- The blue crystals turn to a white powder, and steam (water vapour) is given off.
- CuSO4·5H2O → CuSO4 + 5H2O
- Water lost = 5.00 − 3.20 = 1.80 g = 1.80 ÷ 18 = 0.100 mol.
- CuSO4 = 3.20 ÷ 160 = 0.0200 mol.
- x = 0.100 ÷ 0.0200 = 5.
- Water lost = 2.86 − 1.06 = 1.80 g = 0.100 mol.
- Na2CO3 = 1.06 ÷ 106 = 0.0100 mol.
- x = 0.100 ÷ 0.0100 = 10.
- Some water of crystallisation may remain in the solid, so the mass of water lost is too small.
- The calculated value of x is therefore too low.
- Add to anhydrous copper(II) sulfate: it turns from white to blue (or blue cobalt(II) chloride paper turns pink).
- Any liquid containing water gives this result; purity is shown by a boiling point of exactly 100 °C.
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.
Crystals of hydrated cobalt(II) chloride, CoCl2·6H2O, are heated gently until all the water of crystallisation is removed. What colour change is seen?
- Apink to blue
- Bblue to pink
- Cblue to white
- Dwhite to blue
Which formula represents a hydrated salt?
- AMgSO4·7H2O
- BMgSO4(aq)
- CMgSO4
- DMg(OH)2
12.5 g of blue hydrated copper(II) sulfate is heated in an evaporating basin until it has completely turned white. The white powder has a mass of 8.0 g. What mass of water was driven off?
- A4.5 g
- B8.0 g
- C12.5 g
- D20.5 g
When blue copper(II) sulfate crystals are heated in a test tube, drops of a colourless liquid collect near the top of the tube. How can you show that this liquid contains water?
- Aadd it to white anhydrous copper(II) sulfate, which turns blue
- Badd universal indicator, which turns green
- Cadd it to blue cobalt(II) chloride paper, which turns white
- Dbubble it through limewater, which turns milky
2.19 g of hydrated calcium chloride, CaCl2·xH2O, is heated to constant mass. 1.11 g of anhydrous calcium chloride remains. What is the value of x? (Ar: H 1, O 16, Cl 35.5, Ca 40)
- A6
- B2
- C1
- D4
What is the percentage by mass of water of crystallisation in hydrated magnesium sulfate, MgSO4·7H2O? (Ar: H 1, O 16, Mg 24, S 32)
- A51%
- B49%
- C7.3%
- D105%
While finding x in CuSO4·xH2O, a student heats the solid so strongly that some anhydrous copper(II) sulfate decomposes and gives off a gas. How does this affect the calculated value of x?
- Ax is too high, because the mass lost is larger than the mass of water alone
- Bx is too low, because the mass lost is larger than the mass of water alone
- Cx is unchanged, because all the water was still removed
- Dx is too low, because the mass of anhydrous salt recorded is too large
10.0 g of hydrated copper(II) sulfate, CuSO4·5H2O (Mr 250), is heated until all the water of crystallisation is removed. What mass of anhydrous copper(II) sulfate (Mr 160) remains?
- A6.40 g
- B3.60 g
- C9.28 g
- D10.0 g
Quiz key
Hydrated cobalt(II) chloride is pink and anhydrous cobalt(II) chloride is blue, so heating it turns it from pink to blue. Blue to pink is the reverse change, which happens when water is added (the basis of the cobalt(II) chloride test for water). Blue to white and white to blue are the colour changes of copper(II) sulfate, not cobalt(II) chloride.
The dot followed by 7H2O shows that seven water molecules are chemically combined with each MgSO4 in the crystal, so MgSO4·7H2O is a hydrated salt. MgSO4(aq) is the salt dissolved in water, a solution, not a hydrated solid. MgSO4 on its own is the anhydrous salt. Mg(OH)2 is a hydroxide, a base, not a salt containing water of crystallisation.
Heating removes only the water of crystallisation, so the loss in mass is the mass of water: 12.5 − 8.0 = 4.5 g. 8.0 g is the mass of anhydrous copper(II) sulfate that remains, and 12.5 g is the starting mass of the hydrated salt. 20.5 g comes from adding the two masses instead of subtracting.
Water turns white anhydrous copper(II) sulfate blue, so this is a chemical test for water. A green colour with universal indicator only shows that the liquid is neutral, and many neutral liquids are not water. Blue cobalt(II) chloride does respond to water, but it turns pink, not white. Limewater is the test for carbon dioxide.
Water lost = 2.19 − 1.11 = 1.08 g, and 1.08 ÷ 18 = 0.060 mol. Mr of CaCl2 = 40 + (2 × 35.5) = 111, so 1.11 ÷ 111 = 0.010 mol. The ratio is 0.060 ÷ 0.010 = 6, so the formula is CaCl2·6H2O. An answer of 1 comes from comparing the masses 1.08 and 1.11 directly instead of moles. 2 and 4 do not follow from the data.
Mr of MgSO4 = 24 + 32 + (4 × 16) = 120, and 7H2O = 7 × 18 = 126, so Mr of the hydrate = 246. Percentage of water = 126 ÷ 246 × 100 = 51%. 49% is the percentage of anhydrous MgSO4 (120 ÷ 246). 7.3% counts only one water molecule (18 ÷ 246). 105% comes from dividing by the anhydrous mass (126 ÷ 120), which cannot be right because a part cannot be more than the whole.
The calculation assumes all the lost mass is water. The extra gas makes the mass lost too large, so the moles of water are overestimated, while the remaining mass (and so the calculated moles of CuSO4) is too small. Both errors make the ratio, x, too high. A larger water value cannot make x too low. The result is not unchanged, because the lost mass includes more than water. The recorded anhydrous mass is too small, not too large.
Moles of hydrate = 10.0 ÷ 250 = 0.0400 mol, and each mole leaves one mole of CuSO4, so the mass is 0.0400 × 160 = 6.40 g. 3.60 g is the mass of water lost (0.0400 × 5 × 18), not the solid left. 9.28 g comes from removing only one water molecule per formula unit (10.0 − 0.0400 × 18). 10.0 g would mean no water was lost.
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.