Water of crystallisation and hydrate formulas
Read CuSO₄·5H₂O and CoCl₂·6H₂O as fixed water-to-salt ratios rather than extra salt molecules.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Distinguish combined water from surface moisture.
- Calculate formula masses and use mole ratios in the Extended stoichiometry sequence.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does 5H₂O after the dot in CuSO₄·5H₂O mean?
What you will learn
- Define water of crystallisation as water molecules present in hydrated crystals.
- Interpret both required hydrate formulas and apply their fixed ratios to supplied mass or mole data.
The dot introduces water molecules in the crystal
Water of crystallisation means water molecules present in hydrated crystals as part of the hydrate composition. CuSO₄·5H₂O has five water molecules for each copper(II) sulfate formula unit; CoCl₂·6H₂O has six for each cobalt(II) chloride formula unit. The number after the dot multiplies the entire H₂O group, not the salt formula or only oxygen.
For CuSO₄·5H₂O, the water part contributes ten hydrogen atoms and five oxygen atoms per salt unit in this formula accounting. The sulfate still contains its own four oxygen atoms, so total oxygen is nine. These ratios are not arbitrary amounts of surface water; they distinguish the named hydrate composition from an anhydrous salt or solution.
Use fixed ratios with supplied quantitative evidence
Using supplied relative atomic masses Cu=64, S=32, O=16 and H=1, anhydrous CuSO₄ has formula mass 160 and its five waters add 5×18=90. The hydrate total is 250, so water contributes 90/250=36% by mass in this stated approximation. Label the supplied atomic masses rather than treating the rounded value as every table's exact result.
When appropriate supervised dehydration data are supplied, water loss can be related to the hydrate's fixed ratio, provided the remaining salt identity is established and no other loss or decomposition occurred. A mass decrease alone is not proof that only water of crystallisation escaped. Formula interpretation is the syllabus objective; the linked quantitative practice also revisits already taught Extended mole and mass methods under explicit assumptions.
Sources for this lesson
Pause and explain
How many total oxygen atoms occur per CuSO₄·5H₂O formula unit?
Worked example
Using Cu=64, S=32, O=16 and H=1, calculate water's mass fraction in CuSO₄·5H₂O.
Show the worked solution
- CuSO₄ formula mass is 64+32+4×16=160.
- Five waters contribute 5×(2×1+16)=90, so hydrate total is 250.
- Water fraction is 90/250=0.36, or 36% by mass.
Answer 36% water by mass with the supplied rounded atomic masses.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Count hydrogen and total oxygen atoms per CuSO₄·5H₂O formula unit.
Give me a hint
Count salt and water parts separately, then combine.
Compare my reasoning
- Five waters contain 5×2=10 hydrogen atoms.
- The water part contains five oxygen atoms and sulfate four.
- Total oxygen is 5+4=9 per hydrate formula unit.
10 hydrogen atoms and 9 oxygen atoms.
Look for these in your work
- The whole water group is multiplied.
- Salt oxygen is included in the total.
Find the amount of water molecules associated with 0.20 mol CoCl₂·6H₂O hydrate formula units.
Give me a hint
Each hydrate unit has six waters.
Compare my reasoning
- The formula gives a hydrate-to-water ratio of 1:6.
- Water amount is 0.20×6=1.20 mol.
- This counts water molecules in the hydrate composition, not six moles of cobalt chloride per mole of water.
1.20 mol water molecules.
Look for these in your work
- The ratio is applied in the correct direction.
- The counted species is named.
A hypothetical 2.50 g CuSO₄·5H₂O sample is completely dehydrated without any salt loss or decomposition. Using Cu=64, S=32, O=16 and H=1, predict remaining CuSO₄ mass and water loss, then state why those assumptions matter.
Give me a hint
The hydrate formula mass is 250 and anhydrous salt 160 with these supplied values.
Compare my reasoning
- Hydrate amount is 2.50/250=0.0100 mol.
- CuSO₄ residue is 0.0100×160=1.60 g, leaving 0.90 g water loss.
- Salt loss or decomposition would make mass change include other material and invalidate pure-dehydration accounting.
1.60 g CuSO₄ remains; 0.90 g water is lost under the stated assumptions.
Look for these in your work
- Mass and ratio accounting agree.
- The interpretation is limited to the supplied pure-dehydration assumptions.
Common mistakes
- Multiplying CuSO₄ by five instead of H₂O.
- Counting only five hydrogen atoms in the five waters.
- Ignoring sulfate's oxygen when counting all oxygen atoms.
- Assuming any heating mass loss must be pure water loss.
What can you explain now?
How many water molecules correspond to two formula units of CoCl₂·6H₂O?
Compare with the explanation
12 water molecules.
Each formula unit has six water molecules; two units have 2×6=12.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Write both required hydrate formulas, explain the dot and count their water molecules. Use supplied masses to check a hydrate calculation and name its assumptions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.