How to Find Moles of Water in a Hydrated Salt

A blue hydrated-salt sample is heated, and its mass decreases by 0.90 g. How many moles of water were originally in the crystals?

If water vapour is the only substance that escaped, the sample contained 0.90 g of water. Since the molar mass of water is \(18.02\ \text{g mol}^{-1}\):

\[ n(\ce{H2O})=\frac{0.90}{18.02} =0.0499\ \text{mol} \]

That’s the essential method: find the mass of water lost during heating, then divide by the molar mass of water.

The more interesting part is working out why this mass loss represents water, how to measure it reliably, and how to use the result to determine the number of water molecules in the hydrate’s formula.

01Why heating reveals the amount of water

Some ionic compounds form crystals that contain water molecules as part of their crystal structure. These compounds are called hydrated salts.

The water incorporated into the crystal is called water of crystallisation, or water of hydration.

For example, hydrated copper(II) sulfate has the formula:

\[ \ce{CuSO4.5H2O} \]

The number 5 tells you that there are five water molecules for every one formula unit of copper(II) sulfate.

Notice that this formula contains two different types of chemical entity:

• \(\ce{CuSO4}\) represents an ionic compound, counted in formula units.
• \(\ce{H2O}\) represents water molecules, counted in molecules.

The water is part of the crystal structure. It isn’t simply liquid water sitting on the surface.

Predict what happens when the salt is heated

Suppose you place blue hydrated copper(II) sulfate crystals in a crucible and heat them.

What should happen to their mass?

The expected observation is a decrease in mass. Heating drives water molecules out of the crystal as water vapour, leaving anhydrous copper(II) sulfate behind.

Anhydrous means that the compound no longer contains water of crystallisation.

The dehydration reaction is:

\[ \ce{CuSO4.5H2O(s) ->[\Delta] CuSO4(s) + 5H2O(g)} \]

The blue crystals become white as they lose their water.

The colour change is useful evidence that dehydration has occurred, but it doesn’t tell you how many moles of water were removed.

For that, you need mass measurements.

02How the heating experiment works

The experiment uses a balance to measure the mass of the hydrated salt before and after heating.

The apparatus consists of:

• A clean, dry porcelain crucible containing the hydrated salt.
• A crucible lid, positioned slightly ajar.
• A pipe-clay triangle supporting the crucible.
• A tripod and Bunsen burner providing heat.
• A balance for measuring mass.
• A desiccator for cooling the sample without exposing it to moist air.

The experimental procedure

Step 1: Measure the empty crucible.

Weigh the clean, dry crucible and its lid.

Record this mass so you can subtract it from later measurements.

Step 2: Add the hydrated salt.

Place a known quantity of hydrated salt into the crucible and weigh the crucible, lid, and sample together.

Step 3: Heat the sample.

Heat the crucible gently at first, then increase the heating as appropriate.

Keep the lid slightly ajar. This allows water vapour to escape while reducing the chance of solid particles being lost through spitting.

Step 4: Cool and weigh.

Allow the crucible to cool in a desiccator before weighing it again.

A desiccator provides a dry environment, helping prevent the anhydrous salt from absorbing moisture from the air.

Do not place a hot crucible directly on the balance. Heating creates air currents that can affect the reading, and hot equipment can damage the balance.

Step 5: Repeat until constant mass is reached.

Heat, cool, and weigh the sample again.

Continue until consecutive mass readings agree within the balance’s measurement resolution.

This is called heating to constant mass.

Why isn’t one heating enough?

An initial experiment involving just one heating and one weighing might show that the sample loses mass.

However, it cannot establish whether all the water has been removed.

A second heating might cause another decrease in mass, revealing that dehydration was incomplete.

Repeated heating and weighing improves the experiment by checking whether further mass loss occurs.

There is one important limitation: constant mass does not automatically prove that the residue is anhydrous. Some hydrated salts can form stable, partially dehydrated compounds under particular heating conditions.

You must use suitable heating conditions to remove all the water without decomposing the salt.

The experiment therefore relies on three things:

• Water must be completely removed.
• The salt itself must remain in the crucible.
• The remaining solid must be the anhydrous salt.

These assumptions matter when interpreting the final calculation.

03Step 1: Calculate the mass of water lost

Consider the following illustrative experimental data for hydrated copper(II) sulfate.

First, calculate the initial mass of hydrated salt.

\[ m_{\text{hydrate}}=27.10-24.60 \]

\[ m_{\text{hydrate}}=2.50\ \text{g} \]

Next, calculate the mass of the anhydrous copper(II) sulfate remaining.

\[ m_{\text{salt}}=26.20-24.60 \]

\[ m_{\text{salt}}=1.60\ \text{g} \]

The difference between these two masses is the mass of water released.

\[ m_{\ce{H2O}}=2.50-1.60 \]

\[ \boxed{m_{\ce{H2O}}=0.90\ \text{g}} \]

You can also obtain this directly by subtracting the two combined crucible readings:

\[ 27.10-26.20=0.90\ \text{g} \]

The crucible mass cancels because it remains unchanged.

Our sign convention is that mass lost is positive:

\[ m_{\text{lost}}=m_{\text{before}}-m_{\text{after}} \]

We use the positive mass difference when calculating the amount of water released.

04Step 2: Convert the mass of water into moles

Now use the relationship between mass, molar mass, and chemical amount:

\[ n=\frac{m}{M} \]

Where:

• \(n\) is the chemical amount, measured in mol.
• \(m\) is the mass, measured in g.
• \(M\) is the molar mass, measured in \(\text{g mol}^{-1}\).

For water:

\[ M(\ce{H2O})=2(1.008)+16.00 \]

\[ M(\ce{H2O})=18.02\ \text{g mol}^{-1} \]

Substitute the mass of water lost:

\[ n(\ce{H2O})= \frac{0.90\ \text{g}} {18.02\ \text{g mol}^{-1}} \]

\[ \boxed{n(\ce{H2O})=0.0499\ \text{mol}} \]

Notice what happens to the units:

\[ \frac{\text{g}}{\text{g mol}^{-1}} =\text{mol} \]

The grams cancel, leaving moles.

This dimensional check is useful in exams. If your calculation for chemical amount produces grams instead of moles, something has gone wrong.

For more practice with this conversion, see Moles from Mass and Molar Mass: HSC Chemistry Guide.

05Step 3: Find the number of water molecules per formula unit

Finding the moles of water answers the title question.

However, HSC Chemistry questions often go one step further and ask you to determine the value of \(x\) in a hydrate formula:

\[ \ce{CuSO4.xH2O} \]

For this, you need to compare the amount of water with the amount of anhydrous salt.

The molar mass of anhydrous copper(II) sulfate is:

\[ M(\ce{CuSO4})= 63.55+32.06+4(16.00) \]

\[ M(\ce{CuSO4})=159.61\ \text{g mol}^{-1} \]

The remaining salt has a mass of 1.60 g, so:

\[ n(\ce{CuSO4})= \frac{1.60}{159.61} \]

\[ n(\ce{CuSO4})=0.0100\ \text{mol} \]

We now have both amounts:

The formula contains one \(\ce{CuSO4}\) formula unit for every \(x\) water molecules.

Therefore:

\[ x=\frac{n(\ce{H2O})}{n(\ce{CuSO4})} \]

Substituting:

\[ x=\frac{0.0499}{0.0100} \]

\[ x=4.99\approx5 \]

The experimental result is sufficiently close to the whole number 5.

Therefore, the hydrate formula is:

\[ \boxed{\ce{CuSO4.5H2O}} \]

The result makes physical sense: each formula unit of copper(II) sulfate is associated with five water molecules.

Why the ratio has no units

Both amounts are measured in moles:

\[ x=\frac{\text{mol of water}} {\text{mol of salt}} \]

The mole units cancel:

\[ \frac{\text{mol}}{\text{mol}}=1 \]

Therefore, \(x\) is a dimensionless ratio.

It tells you how many water molecules are associated with each salt formula unit, not how many grams of water are present.

What exactly are we counting?

This distinction is easy to overlook.

When you calculate \(n(\ce{H2O})\), you are finding the amount of water molecules.

When you calculate \(n(\ce{CuSO4})\), you are finding the amount of copper(II) sulfate formula units.

You are not counting individual copper ions or sulfate ions.

If you wanted the actual number of water molecules, you could use Avogadro’s constant:

\[ N=nN_A \]

Where:

• \(N\) is the number of specified entities.
• \(n\) is their amount in moles.
• \(N_A=6.022\times10^{23}\ \text{mol}^{-1}\) is Avogadro’s constant.

For our sample:

\[ N(\ce{H2O})= 0.0499\times6.022\times10^{23} \]

\[ N(\ce{H2O})=3.00\times10^{22} \]

So the sample contained approximately \(3.00\times10^{22}\) water molecules.

For a refresher on the relationship between moles and particles, see The Mole and Avogadro Constant: HSC Chemistry Guide.

06The common mistake: using the starting mass as the mass of water

Imagine a student sees that the original hydrated salt had a mass of 2.50 g.

They calculate:

\[ n(\ce{H2O})=\frac{2.50}{18.02} \]

\[ n(\ce{H2O})=0.139\ \text{mol} \]

This is incorrect.

The 2.50 g represents the entire hydrated sample. It includes both the salt and the water.

Only the 0.90 g that disappeared during heating represents water, assuming the experimental conditions are valid.

There is another tempting mistake:

\[ x=\frac{0.90}{1.60} \]

This is also incorrect.

The calculation compares the masses of water and salt, not their chemical amounts. Water and copper(II) sulfate have different molar masses, so equal masses do not represent equal numbers of particles.

Always convert both substances into moles before calculating their ratio.

08Experimental errors and how they affect your result

The calculation is straightforward, but the experiment has limitations.

Understanding how different errors affect the result helps you evaluate experimental data rather than blindly trusting a neat-looking answer.

Consider why these errors matter.

Incomplete dehydration: Some water remains in the residue. You underestimate the mass of water lost and overestimate the mass of anhydrous salt. Both effects push the calculated ratio down.

Loss of solid through spitting: Some salt escapes from the crucible. You mistakenly count this lost salt as water, while the remaining salt mass is too small. Both effects push the calculated ratio up.

Moisture absorption: The anhydrous salt absorbs water from the atmosphere while cooling. The final mass becomes too high, making the calculated mass of water lost too small.

Non-volatile impurities: An impurity remains in the residue and is incorrectly counted as anhydrous salt. This overestimates the amount of salt and makes \(x\) too small.

Decomposition: If the salt itself decomposes, gases other than water may escape. The mass difference can no longer be treated as the mass of water alone.

A useful habit is to check whether your calculated ratio is reasonably close to a small whole number.

However, don’t automatically round an experimental result to the nearest integer. A result far from a whole number may indicate incomplete dehydration, contamination, or another experimental problem.

09Questions and solutions

The following questions progress from straightforward mass-loss calculations to more demanding questions about experimental reliability and what the measurements can actually establish.

Question 1

A 0.720 g sample of hydrated sodium carbonate is heated until all its water of crystallisation has been removed. The remaining anhydrous sodium carbonate has a mass of 0.540 g.

Assuming no solid is lost and no decomposition occurs, calculate the number of moles of water released.

Solution 1

The sample releases approximately \(9.99\times10^{-3}\) mol of water.

The first step is to calculate the mass lost during heating:

\[ m_{\ce{H2O}}=0.720-0.540 \]

\[ m_{\ce{H2O}}=0.180\ \text{g} \]

The molar mass of water is:

\[ M(\ce{H2O})=18.02\ \text{g mol}^{-1} \]

Use the mass-to-moles relationship:

\[ n=\frac{m}{M} \]

Substitute:

\[ n(\ce{H2O})=\frac{0.180}{18.02} \]

\[ \boxed{n(\ce{H2O})=0.00999\ \text{mol}} \]

The important point is that the calculation uses the mass lost, not the original mass of hydrated sodium carbonate.

Question 2

A 3.05 g sample of hydrated barium chloride has the general formula \(\ce{BaCl2.xH2O}\).

The sample is heated until constant mass is reached. The remaining anhydrous barium chloride has a mass of 2.60 g.

Calculate the value of \(x\) and determine the formula of the hydrated salt.

Use:

• \(M(\ce{Ba})=137.33\ \text{g mol}^{-1}\)
• \(M(\ce{Cl})=35.45\ \text{g mol}^{-1}\)
• \(M(\ce{H2O})=18.02\ \text{g mol}^{-1}\)

Assume all the mass lost is water and the residue is pure anhydrous barium chloride.

Solution 2

The hydrated salt has the formula \(\ce{BaCl2.2H2O}\), so \(x=2\).

First, calculate the mass of water lost:

\[ m_{\ce{H2O}}=3.05-2.60 \]

\[ m_{\ce{H2O}}=0.45\ \text{g} \]

Calculate the molar mass of water:

\[ M(\ce{H2O})=18.02\ \text{g mol}^{-1} \]

Therefore:

\[ n(\ce{H2O})=\frac{0.45}{18.02} \]

\[ n(\ce{H2O})=0.02497\ \text{mol} \]

Next, calculate the molar mass of anhydrous barium chloride:

\[ M(\ce{BaCl2})=137.33+2(35.45) \]

\[ M(\ce{BaCl2})=208.23\ \text{g mol}^{-1} \]

Calculate the amount of barium chloride:

\[ n(\ce{BaCl2})=\frac{2.60}{208.23} \]

\[ n(\ce{BaCl2})=0.01249\ \text{mol} \]

Now calculate the mole ratio:

\[ x=\frac{n(\ce{H2O})}{n(\ce{BaCl2})} \]

\[ x=\frac{0.02497}{0.01249} \]

\[ x=2.00 \]

Therefore:

\[ \boxed{\ce{BaCl2.2H2O}} \]

The result means that every formula unit of barium chloride in the hydrate is associated with two water molecules.

Question 3

A 1.23 g sample of magnesium sulfate heptahydrate, \(\ce{MgSO4.7H2O}\), is heated until all water of crystallisation has been removed.

Calculate the total number of water molecules released.

Use:

• \(M(\ce{MgSO4})=120.37\ \text{g mol}^{-1}\)
• \(M(\ce{H2O})=18.02\ \text{g mol}^{-1}\)
• \(N_A=6.022\times10^{23}\ \text{mol}^{-1}\)

Solution 3

The sample releases approximately \(2.10\times10^{22}\) water molecules.

The important detail is that the sample contains a known hydrate, so we can use its formula directly.

First, calculate the molar mass of magnesium sulfate heptahydrate:

\[ M(\ce{MgSO4.7H2O}) =120.37+7(18.02) \]

\[ M(\ce{MgSO4.7H2O}) =246.51\ \text{g mol}^{-1} \]

Calculate the amount of hydrated salt:

\[ n(\ce{MgSO4.7H2O}) =\frac{1.23}{246.51} \]

\[ n(\ce{MgSO4.7H2O}) =0.00499\ \text{mol} \]

Each formula unit contains seven water molecules.

Therefore, the amount of water is:

\[ n(\ce{H2O})=7(0.00499) \]

\[ n(\ce{H2O})=0.0349\ \text{mol} \]

Now convert the amount of water into the number of molecules:

\[ N=nN_A \]

\[ N=0.0349(6.022\times10^{23}) \]

\[ \boxed{N=2.10\times10^{22}\text{ molecules}} \]

Notice the distinction between the two entities:

• \(0.00499\) mol represents magnesium sulfate formula units.
• \(0.0349\) mol represents water molecules.

The seven-to-one ratio comes directly from the chemical formula.

Question 4

A student heats a sample containing hydrated magnesium sulfate and 0.200 g of inert silica impurity.

After heating to constant mass, the following measurements are obtained:

An independent test confirms that all water has been removed, the silica remains unchanged, and the magnesium sulfate has not decomposed.

The student assumes that the entire 1.404 g residue is anhydrous magnesium sulfate and calculates a hydrate formula.

Explain why this assumption is incorrect, then calculate:

(a) The actual value of \(x\) in \(\ce{MgSO4.xH2O}\).

(b) The incorrect value of \(x\) obtained if the silica impurity is ignored.

Solution 4

The actual hydrate has the formula \(\ce{MgSO4.7H2O}\), while ignoring the silica impurity produces an incorrect result of approximately \(x=6\).

The key issue is that the final residue contains two substances:

• Anhydrous magnesium sulfate.
• The original silica impurity.

The silica does not evaporate during heating, so it must be subtracted before calculating the amount of magnesium sulfate.

Part (a): Calculate the actual value of \(x\).

First, calculate the amount of water released:

\[ n(\ce{H2O})=\frac{1.261}{18.02} \]

\[ n(\ce{H2O})=0.0700\ \text{mol} \]

Next, calculate the actual mass of magnesium sulfate remaining.

\[ m(\ce{MgSO4})=1.404-0.200 \]

\[ m(\ce{MgSO4})=1.204\ \text{g} \]

The molar mass of magnesium sulfate is:

\[ M(\ce{MgSO4})=120.37\ \text{g mol}^{-1} \]

Therefore:

\[ n(\ce{MgSO4})=\frac{1.204}{120.37} \]

\[ n(\ce{MgSO4})=0.0100\ \text{mol} \]

Calculate the ratio:

\[ x=\frac{0.0700}{0.0100} \]

\[ \boxed{x=7.00} \]

The actual formula is:

\[ \boxed{\ce{MgSO4.7H2O}} \]

Part (b): Calculate the incorrect value.

If the student treats the entire residue as magnesium sulfate, they calculate:

\[ n_{\text{apparent}}(\ce{MgSO4}) =\frac{1.404}{120.37} \]

\[ n_{\text{apparent}}(\ce{MgSO4}) =0.0117\ \text{mol} \]

The apparent mole ratio becomes:

\[ x_{\text{apparent}}= \frac{0.0700}{0.0117} \]

\[ \boxed{x_{\text{apparent}}\approx6} \]

The student obtains a neat whole number, but the answer is still wrong.

The silica has increased the apparent mass of anhydrous magnesium sulfate. This makes the calculated amount of salt too large and the water-to-salt ratio too small.

This question highlights an important experimental principle: a result close to a whole number does not automatically mean the method was correct.

Question 5

An unknown magnesium sulfate hydrate is heated until a constant mass is reached.

The following illustrative measurements are recorded:

• Initial mass of hydrated sample: 1.925 g.
• Final mass of anhydrous magnesium sulfate: 1.204 g.

Assume that:

• The original sample contains no impurities.
• No magnesium sulfate is lost during heating.
• The final residue is completely anhydrous.
• The only substance released during heating is water.

A student calculates \(x=4.00\) and concludes that every formula unit in the original sample must have contained exactly four water molecules.

(a) Verify the student’s calculated bulk mole ratio.

(b) Is the student’s conclusion about every formula unit justified by the measurements?

(c) Construct an alternative composition containing a mixture of magnesium sulfate monohydrate, \(\ce{MgSO4.H2O}\), and magnesium sulfate heptahydrate, \(\ce{MgSO4.7H2O}\), that produces exactly the same initial and final masses.

(d) Explain what the heating experiment can establish, and what it cannot establish about the original solid.

Use \(M(\ce{MgSO4})=120.37\ \text{g mol}^{-1}\) and \(M(\ce{H2O})=18.02\ \text{g mol}^{-1}\).

Solution 5

The bulk mole ratio is approximately four water molecules per magnesium sulfate formula unit, but this does not prove that every formula unit originally contained exactly four water molecules.

A mixture of monohydrate and heptahydrate can produce precisely the same mass measurements.

Part (a): Verify the bulk mole ratio.

First, calculate the mass of water released:

\[ m_{\ce{H2O}}=1.925-1.204 \]

\[ m_{\ce{H2O}}=0.721\ \text{g} \]

Calculate the amount of water:

\[ n(\ce{H2O})=\frac{0.721}{18.02} \]

\[ n(\ce{H2O})=0.0400\ \text{mol} \]

Now calculate the amount of anhydrous magnesium sulfate:

\[ n(\ce{MgSO4})=\frac{1.204}{120.37} \]

\[ n(\ce{MgSO4})=0.0100\ \text{mol} \]

Therefore:

\[ x_{\text{average}}= \frac{0.0400}{0.0100} \]

\[ \boxed{x_{\text{average}}=4.00} \]

The student’s calculation is correct.

However, the interpretation requires more thought.

Part (b): Determine whether the conclusion is justified.

No. The measurements establish the average ratio of water molecules to magnesium sulfate formula units across the entire sample.

They do not establish that every crystal has the same hydration state.

This is because a bulk mass measurement combines the contributions of all the particles in the sample. Different combinations of hydrates can produce the same overall ratio.

Part (c): Construct an alternative composition.

Consider a sample containing:

• \(0.00500\) mol of \(\ce{MgSO4.H2O}\).
• \(0.00500\) mol of \(\ce{MgSO4.7H2O}\).

The total amount of magnesium sulfate formula units is:

\[ n_{\text{total}}(\ce{MgSO4}) =0.00500+0.00500 \]

\[ n_{\text{total}}(\ce{MgSO4}) =0.0100\ \text{mol} \]

Now calculate the total amount of water.

The monohydrate contributes one mole of water per mole of magnesium sulfate:

\[ n_1(\ce{H2O})=0.00500\ \text{mol} \]

The heptahydrate contributes seven moles of water per mole of magnesium sulfate:

\[ n_2(\ce{H2O})=7(0.00500) \]

\[ n_2(\ce{H2O})=0.0350\ \text{mol} \]

Therefore, the total amount of water is:

\[ n_{\text{total}}(\ce{H2O}) =0.00500+0.0350 \]

\[ n_{\text{total}}(\ce{H2O}) =0.0400\ \text{mol} \]

The overall ratio is:

\[ x_{\text{average}}= \frac{0.0400}{0.0100}=4.00 \]

This mixture has exactly the same bulk mole ratio as the student’s proposed tetrahydrate.

We can also check the masses.

The total anhydrous magnesium sulfate mass is:

\[ m(\ce{MgSO4})=0.0100(120.37) \]

\[ m(\ce{MgSO4})=1.204\ \text{g} \]

The total water mass is:

\[ m(\ce{H2O})=0.0400(18.02) \]

\[ m(\ce{H2O})=0.721\ \text{g} \]

The initial mass is therefore:

\[ m_{\text{initial}}=1.204+0.721 \]

\[ m_{\text{initial}}=1.925\ \text{g} \]

These are exactly the same measurements as the original experiment, to the stated precision.

Part (d): Explain what the experiment establishes.

The experiment establishes that the sample contains approximately four moles of water for every mole of magnesium sulfate formula units.

It does not establish whether the sample is:

• A uniform hydrate with four water molecules per formula unit.
• A mixture of different hydrated forms with the same average water-to-salt ratio.

The mass-loss method measures the overall composition of the sample. It does not directly identify the individual crystal structures or prove that all crystals have an identical hydration state.

Additional structural or phase-identification evidence would be needed to distinguish the two possibilities.

This is an important distinction in experimental chemistry: a correct bulk mole ratio does not always prove a unique microscopic structure.

10What to learn next

Once you can determine the moles of water in a hydrated salt, the next step is using the complete hydrate formula in quantitative calculations.

For example, knowing that a salt is \(\ce{MgSO4.7H2O}\) allows you to calculate its molar mass, determine the mass percentage of water, and use its formula in chemical reaction calculations.

The same mass-to-moles relationship applies throughout these problems. The extra challenge is identifying exactly which chemical entities you are counting and making sure the experimental measurements support the formula you calculate.