Difficulty: Introductory | Prerequisites: Moles and Empirical Formulas study notes (Part A of this lab), periodic table familiarity
Many ionic salts crystallise with a fixed number of water molecules trapped in their crystal lattice. These are called hydrates, and the water they contain is called water of crystallisation. This section of LAB 11 teaches you how to heat a hydrate to drive off the water, measure the mass lost, and use that data to calculate the exact number of water molecules per formula unit of salt. The technique builds directly on the mole and empirical-formula skills from Part A.
Heat a hydrate to remove its water, weigh what is left, and use the mass difference to find moles of water and moles of anhydrous salt. Divide the moles of water by the moles of salt to get the coefficient in the hydrate formula. In this lab, the hydrate turns out to be MgSO₄ · 7H₂O (Epsom salt).
Hydrate
An ionic compound that has a definite number of water molecules chemically bound within its crystal structure. Think of it as a salt that comes pre-packaged with water molecules locked into its lattice.
Anhydrate (anhydrous salt)
The salt that remains after all water of crystallisation has been removed by heating. In simple terms, it is the "dry" form of the compound.
Water of crystallisation (water of hydration)
The water molecules incorporated into the crystal structure of a hydrate. These are not simply surface moisture; they occupy specific positions in the lattice and are present in a fixed whole-number ratio.
Percent water
The mass of water lost from a hydrate expressed as a percentage of the original hydrate mass. Calculated as (mass of water / mass of hydrate) x 100.
Dehydration
The process of removing water from a hydrate by heating, converting it to its anhydrous form. The reverse process, absorbing water from the environment, is called hydration.
Weigh the crucible empty, then with the hydrate sample.
Mass of hydrate = mass of (crucible + hydrate) minus mass of crucible.
Heat the crucible strongly to drive off all water. Allow it to cool, then weigh again.
Mass of anhydrate = mass of (crucible + anhydrate) minus mass of crucible.
A second heating confirms the mass is constant (all water removed).
Find the mass of water lost.
Mass of water = mass of hydrate minus mass of anhydrate.
Calculate percent water.
% water = (mass of water / mass of hydrate) x 100.
Convert to moles.
Moles of water = mass of water / 18.015 g/mol.
Moles of anhydrous salt = mass of anhydrate / molar mass of salt.
Find the ratio.
Divide moles of water by moles of salt. Round to the nearest whole number.
Write the hydrate formula.
Salt · nH₂O, where n is the whole-number ratio.
Mass of crucible = 32.16 g
Mass of crucible + hydrate = 40.94 g → mass of hydrate = 8.78 g
Mass of crucible + anhydrate = 36.48 g → mass of anhydrate = 4.32 g
Mass of water lost = 8.78 minus 4.32 = 4.46 g
Percent water = (4.46 / 8.78) x 100 = 50.80%
Moles of water = 4.46 g / 18.015 g/mol = 0.248 mol
Moles of MgSO₄ = 4.32 g / 120.366 g/mol = 0.036 mol
Ratio = 0.248 / 0.036 = 6.9 ≈ 7
Formula of hydrate: MgSO₄ · 7H₂O
This compound is commonly known as Epsom salt.
MgSO₄ · 7H₂O(s) → MgSO₄(s) + 7 H₂O(g)
The solid hydrate breaks down on heating into the solid anhydrous salt and water vapour. Note that the coefficient 7 in front of H₂O comes directly from the hydrate formula you determined.
\text{mass of water} = \text{mass of hydrate} - \text{mass of anhydrate}\%\;\text{water} = \frac{\text{mass of water}}{\text{mass of hydrate}} \times 100n = \frac{\text{moles of water}}{\text{moles of anhydrous salt}}where n is the coefficient in the hydrate formula: Salt · nH₂O.
Epsom salt (MgSO₄ · 7H₂O) is used in bath salts, agriculture as a magnesium supplement for plants, and in medicine as a laxative. Copper(II) sulphate pentahydrate (CuSO₄ · 5H₂O) is used as a fungicide and in electroplating. Understanding hydrate formulas matters in pharmacy and materials science, where the water content of a compound affects its mass, stability, and dosage calculations.
Students often think the water in a hydrate is simply absorbed surface moisture. It is not. The water occupies specific positions in the crystal lattice and is present in a fixed stoichiometric ratio.
Confusing "mass of hydrate" with "mass of anhydrate" when setting up the percent-water calculation. The denominator is always the original hydrate mass, not the anhydrate.
Assuming one heating is sufficient. If the second heating gives a different mass, the first heating did not remove all the water, and the lower mass should be used.
Writing the dehydration equation without matching the coefficient to the experimentally determined ratio. The 7 in front of H₂O is not a guess; it comes directly from the moles-of-water-to-moles-of-salt calculation.
⚠️ Calculating the formula of a hydrate from experimental data is a classic exam question. Practise the full sequence: masses → mass of water → moles of water and salt → ratio → formula.
⚠️ You may be given percent water and asked to work backwards to find the hydrate formula. Start with a 100 g sample, find grams of water and grams of salt, then convert to moles.
⚠️ Know the molar mass of water (18.015 g/mol) by heart. It comes up in nearly every hydrate problem.
True or false: A hydrate contains water molecules trapped in random positions throughout the crystal. (False, they occupy fixed, stoichiometric positions in the lattice.)
Fill in the blank: The mass of water lost equals the mass of ________ minus the mass of ________. (hydrate, anhydrate)
True or false: If the ratio of moles of water to moles of salt is 4.9, you should round to 5. (True, 4.9 is within rounding tolerance of 5.)
Fill in the blank: The molar mass of water is ________ g/mol. (18.015)
True or false: In the dehydration equation, water is produced as a liquid. (False, it is produced as a gas/vapour because the sample is heated.)
Q: A 5.00 g sample of a hydrate of CaCl₂ is heated until all water is removed. The anhydrous salt weighs 2.55 g. What is the formula of the hydrate?
A: Mass of water = 5.00 minus 2.55 = 2.45 g. Moles of water = 2.45 / 18.015 = 0.136 mol. Moles of CaCl₂ = 2.55 / 110.98 = 0.023 mol. Ratio = 0.136 / 0.023 = 5.9 ≈ 6. Formula: CaCl₂ · 6H₂O.
Q: A hydrate has a percent water of 36.0% and the anhydrous salt is Na₂CO₃ (molar mass 105.99 g/mol). How many water molecules are in the formula?
A: Assume 100 g sample. Water = 36.0 g, salt = 64.0 g. Moles of water = 36.0 / 18.015 = 2.00. Moles of Na₂CO₃ = 64.0 / 105.99 = 0.604. Ratio = 2.00 / 0.604 = 3.3 ≈ 10/3... recalculating: ratio ≈ 3.31. Multiply by 3 to check, but more precisely the known hydrate is Na₂CO₃ · 10H₂O with 62.9% water. At 36.0%, ratio ≈ 3. Formula: Na₂CO₃ · 3H₂O.
Q: Write the balanced equation for the dehydration of CuSO₄ · 5H₂O.
A: CuSO₄ · 5H₂O(s) → CuSO₄(s) + 5 H₂O(g).
Q: Why is a second heating performed in the hydrate experiment?
A: To confirm that all water has been removed. If the mass after the second heating matches the first, the sample has reached constant mass and the data are reliable.
Hydrate analysis connects to percent composition (you are effectively finding the percent water in a compound) and to empirical formulas (the ratio of water to salt is found the same way as element ratios). It also relates to thermochemistry: the energy required to break the bonds holding water in the lattice is an enthalpy of hydration, which you may encounter in later chapters.
Hydrate, anhydrate, anhydrous salt, water of crystallisation, water of hydration, percent water, dehydration, constant mass, Epsom salt, MgSO₄ · 7H₂O, magnesium sulphate heptahydrate, CuSO₄ · 5H₂O, copper sulphate pentahydrate, CaCl₂ · 6H₂O, moles of water, hydrate formula, LAB 11, general chemistry, MSU chemistry