How to Find a Hydrated Salt Formula from Heating Data

A crucible containing blue copper(II) sulfate crystals is weighed, heated, and weighed again. The balance reading falls by 0.90 g. Has some copper(II) sulfate disappeared, or has something else escaped from the crystals?

If heating removes only water and leaves the copper(II) sulfate intact, that 0.90 g is the mass of water originally held within the crystals. We can use it to work out exactly how much water was present for every formula unit of the salt.

To find the formula of a hydrated salt from heating data, subtract the final mass from the initial mass to find the water lost. Convert that water mass and the remaining anhydrous salt mass into moles, then divide the moles of water by the moles of salt. The resulting ratio gives the hydration number, \(x\), in a formula such as \(\ce{CuSO4}\cdot x\ce{H2O}\).

There is one important condition: the mass lost must really be water. A good experimental method is designed to test that assumption, not just produce a convenient number.

01What does heating a hydrated salt actually remove?

Some ionic compounds form crystals that incorporate water molecules into their crystal structures. This is called water of crystallisation, or water of hydration.

For example, hydrated copper(II) sulfate can have the formula

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

This tells us that the crystal contains five moles of water for every mole of copper(II) sulfate formula units.

It does not mean that the crystal consists of separate molecules of \(\ce{CuSO4}\) with five water molecules loosely attached. Copper(II) sulfate is an ionic compound, and the water is incorporated into its extended crystal structure.

You can picture the basic mass change using a wet towel. When the towel dries, its mass decreases because water leaves, while the fabric remains.

The comparison is useful for understanding the subtraction. But it breaks down chemically: water of crystallisation is part of a solid’s crystal structure, not simply liquid water trapped between fibres.

When a suitable hydrated salt is heated sufficiently, water escapes as vapour and an anhydrous salt remains. Anhydrous means without water.

For copper(II) sulfate pentahydrate, complete dehydration can be represented by

\[
\ce{CuSO4}\cdot 5\ce{H2O}
\rightarrow
\ce{CuSO4}+5\ce{H2O}
\]

The hydrated solid is blue, while anhydrous copper(II) sulfate is pale, approximately white. This colour change is useful supporting evidence, although colour alone cannot prove that dehydration is complete.

02How the heating experiment produces useful data

The experimental question is simple: if water escapes but the salt remains, can we measure the masses accurately enough to determine the original ratio?

The apparatus and the path of the water

A typical school laboratory arrangement uses a porcelain crucible, a lid, a clay triangle, a tripod, a Bunsen burner, and a balance.

The sequence is:

1. Weigh the clean, dry crucible and its lid together.
2. Add the hydrated salt and weigh the crucible, lid, and sample together.
3. Place the crucible on a clay triangle supported above the Bunsen burner.
4. Heat with the lid slightly ajar, allowing water vapour to escape while reducing the chance of solid particles being thrown out.
5. Allow the crucible to cool in dry conditions, preferably in a desiccator, before weighing.
6. Repeat the heating, cooling, and weighing process until successive readings agree within the balance’s measurement resolution.

The lid must not seal the crucible during heating. Water vapour needs a way out.

Why the experiment needs repeated heating

Suppose the first heating leaves some water in the solid.

The balance reading decreases, but the residue is not fully anhydrous. If we calculate the formula immediately, we underestimate the amount of water originally present.

A better design repeats the heating and weighing.

If successive cooled readings become constant, that suggests the process has reached a stable endpoint under those heating conditions.

But constant mass does not prove that the residue is the intended anhydrous salt. The solid might contain a stable partially dehydrated form, or excessive heating might cause decomposition.

The experimental conclusion therefore depends on two things:

• The mass has stopped changing under suitable drying conditions.
• The chemical identity of the remaining solid is justified.

We will see why that second condition becomes important.

03Turn balance readings into a hydration number

The balance measures the combined mass of the container and its contents. It does not directly display the mass of the salt.

That means we must subtract the mass of the crucible and lid before calculating moles.

Let:

• \(m_c\) = mass of the empty crucible and lid, in grams (g)
• \(m_i\) = mass of the crucible, lid, and hydrated salt before heating, in grams (g)
• \(m_f\) = mass of the crucible, lid, and anhydrous salt after heating, in grams (g)

Assuming the crucible and lid do not change mass, and only water is lost:

\[
\begin{aligned}
m_{\mathrm{hydrate}}&=m_i-m_c\\
m_{\mathrm{salt}}&=m_f-m_c\\
m_{\mathrm{water}}&=m_i-m_f
\end{aligned}
\]

Notice that the container’s mass cancels when we calculate the water lost.

If we define mass change as \(\Delta m=m_f-m_i\), dehydration produces a negative value. The positive mass of water lost is therefore \(m_{\mathrm{water}}=-\Delta m\).

A quick check using percentage water

There is another way to test whether the answer makes sense.

The experimental percentage by mass of water is

\[
\begin{aligned}
\%\ce{H2O}
&=\frac{0.90}{2.50}\times100\\
&=36.0\%
\end{aligned}
\]

For \(\ce{CuSO4}\cdot 5\ce{H2O}\), the predicted percentage is

\[
\begin{aligned}
\%\ce{H2O}
&=\frac{5(18.02)}{159.61+5(18.02)}
\times100\\
&=36.1\%
\end{aligned}
\]

The experimental and predicted percentages agree closely.

This is a useful consistency check. It does not independently prove that the lost material was water, because both calculations depend on that assumption.

If you need more practice converting a formula into mass percentages, see How to Calculate Element Percentages from Chemical Formulas.

04What if the sample has not finished losing water?

Imagine heating a hydrated salt once, recording its mass, and deciding that the experiment is finished.

What if the residue still contains water?

The calculated anhydrous salt mass would be too large, while the calculated water loss would be too small. Both errors push the hydration number downwards.

This is why the pattern across successive weighings matters.

05How close to a whole number is close enough?

A common mistake is rounding any calculated hydration number to the nearest integer.

That can turn poor experimental data into a neat-looking but unjustified formula.

Consider three hypothetical outcomes:

Why might a fractional ratio be chemically meaningful?

Some crystalline compounds contain fewer than one water molecule per salt formula unit on average. Calcium sulfate hemihydrate, for example, is written as

\[
\ce{CaSO4}\cdot\tfrac12\ce{H2O}
\]

The half does not mean that an individual water molecule has been cut in two. It describes the overall stoichiometric ratio in the solid.

In many school problems, \(x\) is expected to be a small whole number. But the experimental evidence must justify that interpretation.

A sensible rounding decision asks:

• Is the ratio very close to a plausible simple value?
• Are the balance readings sufficiently precise?
• Has the residue reached a reliable endpoint?
• Is the starting material a single pure hydrate?
• Is the mass loss known to be water alone?

Keep extra digits during the calculation, and round only after considering these questions.

Three ways a convincing result can still be wrong

These are not merely errors in calculation. They challenge the chemical model used to interpret the balance readings.

For instance, repeating heating until a mass becomes constant will not repair the results if some solid has already escaped from the crucible.

Likewise, if the original salt decomposes into a different solid and a gas, the final residue must not be treated as the original anhydrous salt.

This is a crucial distinction between a correct calculation and a valid experimental conclusion.

A hydration formula is not a molecular identity

Heating data can establish an overall mole ratio, assuming the composition and mass-loss model are valid.

It cannot, on its own, prove that the original sample contained only one crystalline phase.

A mixture of two different hydrates of the same salt can sometimes produce the same average hydration ratio as a single pure hydrate.

That is closely related to a limitation of empirical formula calculations: composition ratios do not necessarily identify a unique substance. The same principle appears in How to Find an Empirical Formula from Percentage Composition.

06Questions and solutions

The questions below move from a standard hydration calculation to problems where interpreting the experiment becomes harder than calculating the mole ratio.

Unless otherwise stated, masses are sample masses excluding the container, measurements are illustrative, and molar masses are expressed in \(\mathrm{g\,mol^{-1}}\).

Question 1

A sample of hydrated sodium carbonate, \(\ce{Na2CO3}\cdot x\ce{H2O}\), is heated until its mass becomes constant.

Assuming the salt remains chemically unchanged apart from losing its water of crystallisation, calculate \(x\) and write the formula of the hydrate.

Use \(M(\ce{Na2CO3})=105.99\) and \(M(\ce{H2O})=18.02\).

Solution 1

The hydrated salt is sodium carbonate decahydrate, \(\ce{Na2CO3}\cdot 10\ce{H2O}\).

First calculate the anhydrous salt mass:

\[
\begin{aligned}
m_{\mathrm{salt}}
&=30.86-29.80\\
&=1.06\ \mathrm{g}
\end{aligned}
\]

The water lost is

\[
\begin{aligned}
m_{\mathrm{water}}
&=32.66-30.86\\
&=1.80\ \mathrm{g}
\end{aligned}
\]

Convert both masses into moles:

\[
\begin{aligned}
n(\ce{Na2CO3})
&=\frac{1.06}{105.99}\\
&=0.0100\ \mathrm{mol}\\[4pt]
n(\ce{H2O})
&=\frac{1.80}{18.02}\\
&=0.0999\ \mathrm{mol}
\end{aligned}
\]

The mole ratio is

\[
\begin{aligned}
x&=\frac{0.0999}{0.0100}\\
&\approx9.99
\end{aligned}
\]

Therefore, \(x=10\), giving

\[
\ce{Na2CO3}\cdot 10\ce{H2O}
\]

The result means there are approximately ten water molecules per sodium carbonate formula unit in the original pure hydrate.

The common error is using 30.86 g as the mass of sodium carbonate. That reading includes the crucible and lid. Subtracting the container mass is essential before converting to moles.

Question 2

A student heats 2.50 g of \(\ce{CuSO4}\cdot x\ce{H2O}\), producing 1.78 g of solid.

A separate chemical analysis establishes that this residue is pure copper(II) sulfate monohydrate, \(\ce{CuSO4}\cdot\ce{H2O}\), rather than anhydrous copper(II) sulfate.

Assume that no solid escapes and that the only substance lost is water.

(a) Determine the hydration number \(x\) of the original crystals.

(b) Calculate the additional mass that should be lost if the monohydrate is completely dehydrated.

(c) Explain why treating the 1.78 g residue as anhydrous \(\ce{CuSO4}\) would give the wrong formula.

Use \(M(\ce{CuSO4})=159.61\) and \(M(\ce{H2O})=18.02\).

Solution 2

(a) The original hydrate has \(x=5\). (b) Another approximately 0.18 g of water must be removed. (c) The 1.78 g residue still contains water, so it cannot be treated as anhydrous salt.

Start with the composition of the residue.

One mole of copper(II) sulfate monohydrate has a molar mass of

\[
159.61+18.02
=177.63\ \mathrm{g\,mol^{-1}}
\]

Therefore, the amount of monohydrate remaining is

\[
\begin{aligned}
n(\mathrm{monohydrate})
&=\frac{1.78}{177.63}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

There is one copper(II) sulfate formula unit per monohydrate formula unit, so the original sample contained approximately 0.0100 mol of copper(II) sulfate units.

The water lost during the first heating has mass

\[
2.50-1.78=0.72\ \mathrm{g}
\]

Its amount is

\[
\begin{aligned}
n(\ce{H2O})_{\mathrm{lost}}
&=\frac{0.72}{18.02}\\
&=0.0400\ \mathrm{mol}
\end{aligned}
\]

Thus, approximately four waters were lost per salt formula unit.

But the residue is a monohydrate. One water per salt formula unit remains.

Therefore,

\[
x=4+1=5
\]

The original formula is

\[
\ce{CuSO4}\cdot5\ce{H2O}
\]

For part (b), the remaining monohydrate contains approximately 0.0100 mol of water.

The additional water mass is

\[
\begin{aligned}
m_{\mathrm{extra}}
&=0.0100\times18.02\\
&\approx0.18\ \mathrm{g}
\end{aligned}
\]

The predicted final anhydrous mass is consequently

\[
1.78-0.18\approx1.60\ \mathrm{g}
\]

For part (c), the tempting method is to divide 1.78 g by the molar mass of anhydrous \(\ce{CuSO4}\).

That is invalid because 1.78 g includes both salt and crystal water.

It would overestimate the amount of anhydrous salt while underestimating how much water was originally present.

This question exposes an assumption that is often left unstated: the final solid must actually be anhydrous before the usual hydrate calculation works.

Question 3

Two equal 2.87 g portions of the same pure hydrated zinc sulfate, \(\ce{ZnSO4}\cdot x\ce{H2O}\), are heated separately.

Both are heated, cooled, and weighed repeatedly until their readings become constant.

Assume Student A’s sample was completely dehydrated without decomposition or loss of solid. Student B’s retained residue is also completely dehydrated, but some solid escaped at an unknown stage of heating.

(a) Calculate the apparent hydration number from each experiment.

(b) Decide which value represents the original hydrate, explaining why constant mass does not settle the disagreement.

(c) The two final residue masses differ by 0.09 g. Must Student B have ejected exactly 0.09 g of material? Calculate the mass ejected in the two limiting cases where it escaped entirely before dehydration or entirely after dehydration.

Use \(M(\ce{ZnSO4})=161.44\) and \(M(\ce{H2O})=18.02\).

Solution 3

(a) Student A obtains \(x\approx7\), while Student B obtains an apparent \(x\approx8\). (b) The supported formula is \(\ce{ZnSO4}\cdot7\ce{H2O}\). (c) The escaped material need not have weighed 0.09 g because it may still have contained water.

For Student A, the water mass is

\[
2.87-1.61=1.26\ \mathrm{g}
\]

Convert to moles:

\[
\begin{aligned}
n(\ce{H2O})
&=\frac{1.26}{18.02}\\
&=0.0699\ \mathrm{mol}\\[4pt]
n(\ce{ZnSO4})
&=\frac{1.61}{161.44}\\
&=0.00997\ \mathrm{mol}
\end{aligned}
\]

Therefore,

\[
x=\frac{0.0699}{0.00997}
\approx7.01
\]

Student A’s result supports

\[
\ce{ZnSO4}\cdot7\ce{H2O}
\]

For Student B, the apparent water loss is

\[
2.87-1.52=1.35\ \mathrm{g}
\]

Using the ordinary calculation:

\[
\begin{aligned}
x
&=\frac{1.35/18.02}{1.52/161.44}\\
&\approx7.96
\end{aligned}
\]

This gives an apparent hydration number of eight.

It is tempting to prefer Student B’s result because the sample lost more mass, and repeated heating eventually produced a stable reading.

But some solid escaped. Student B’s 1.35 g mass decrease includes material other than evaporated water.

The smaller final residue also makes the calculated number of moles of zinc sulfate too small.

Both effects increase the apparent water-to-salt ratio.

The supported value is therefore Student A’s \(x=7\).

For part (c), notice what the 0.09 g difference actually tells us.

The retained anhydrous residues differ by

\[
1.61-1.52=0.090\ \mathrm{g}
\]

So Student B has 0.090 g less anhydrous zinc sulfate than Student A.

If the material escaped after complete dehydration, the ejected mass would indeed be 0.090 g.

But if hydrated crystals escaped before any dehydration, the ejected material would have contained both zinc sulfate and water.

The molar mass of zinc sulfate heptahydrate is

\[
\begin{aligned}
M_{\mathrm{hydrate}}
&=161.44+7(18.02)\\
&=287.58\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The mass of hydrate containing 0.090 g of anhydrous zinc sulfate is

\[
\begin{aligned}
m_{\mathrm{ejected}}
&=0.090\times\frac{287.58}{161.44}\\
&=0.160\ \mathrm{g}
\end{aligned}
\]

Thus:

• Ejection after complete dehydration: approximately 0.090 g of material escaped.
• Ejection before dehydration: approximately 0.160 g of hydrated material escaped.

If ejection occurred during partial dehydration, the actual mass could fall between these values.

The final residue masses alone do not reveal when the solid escaped.

This is why experimental reasoning matters: a constant final mass can coexist with a systematic error, and even a visible error does not always tell us its exact size.

Question 4

A student investigates a hydrated magnesium carbonate, \(\ce{MgCO3}\cdot x\ce{H2O}\).

For this idealised experiment, independent chemical tests establish that gentle heating removes only water, leaving pure \(\ce{MgCO3}\). Stronger heating then decomposes the magnesium carbonate completely into \(\ce{MgO}\) and carbon dioxide.

No solid is lost mechanically, and no other reactions occur.

The recorded sample masses are:

(a) Determine the formula of the original hydrated magnesium carbonate.

(b) Verify quantitatively that the second mass decrease is consistent with the formation of carbon dioxide.

(c) Explain why using the strongly heated residue as the ‘anhydrous salt’ would produce an invalid hydration calculation.

Use:

• \(M(\ce{MgCO3})=84.31\)
• \(M(\ce{MgO})=40.31\)
• \(M(\ce{CO2})=44.01\)
• \(M(\ce{H2O})=18.02\)

Solution 4

(a) The original hydrate is \(\ce{MgCO3}\cdot3\ce{H2O}\). (b) The second loss is consistent with one mole of carbon dioxide per mole of magnesium carbonate. (c) The final residue is magnesium oxide, not anhydrous magnesium carbonate.

The crucial decision is to use the intermediate mass of 0.843 g to determine the amount of magnesium carbonate.

That is the stage where the chemical analysis establishes that dehydration is complete and decomposition has not yet occurred.

For part (a), the mass of water lost during gentle heating is

\[
\begin{aligned}
m_{\mathrm{water}}
&=1.384-0.843\\
&=0.541\ \mathrm{g}
\end{aligned}
\]

The amount of water is

\[
\begin{aligned}
n(\ce{H2O})
&=\frac{0.541}{18.02}\\
&=0.0300\ \mathrm{mol}
\end{aligned}
\]

The amount of magnesium carbonate is

\[
\begin{aligned}
n(\ce{MgCO3})
&=\frac{0.843}{84.31}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

Therefore,

\[
\begin{aligned}
x&=\frac{0.0300}{0.0100}\\
&=3.00
\end{aligned}
\]

The formula is

\[
\ce{MgCO3}\cdot3\ce{H2O}
\]

For part (b), the assumed second reaction is

\[
\ce{MgCO3(s) -> MgO(s) + CO2(g)}
\]

The second mass loss is

\[
\begin{aligned}
m(\ce{CO2})
&=0.843-0.403\\
&=0.440\ \mathrm{g}
\end{aligned}
\]

The amount of carbon dioxide released is

\[
\begin{aligned}
n(\ce{CO2})
&=\frac{0.440}{44.01}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

This agrees with the original amount of magnesium carbonate.

We can also check the final residue:

\[
\begin{aligned}
n(\ce{MgO})
&=\frac{0.403}{40.31}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

The three amounts are consistent with the 1:1:1 mole relationship between \(\ce{MgCO3}\), \(\ce{MgO}\), and \(\ce{CO2}\).

For part (c), suppose someone used the total mass loss:

\[
1.384-0.403=0.981\ \mathrm{g}
\]

and assumed it was entirely water.

They would then divide 0.981 g by the molar mass of water and 0.403 g by the molar mass of magnesium carbonate.

That would give

\[
x=\frac{0.981/18.02}{0.403/84.31}
\approx11.4
\]

This answer is invalid, not merely badly rounded.

The mass lost includes both water and carbon dioxide, while the final residue is \(\ce{MgO}\), not \(\ce{MgCO3}\).

The correct calculation uses the gentle-heating stage, where the assumed reaction and the identity of the residue match the hydration model.

A further limitation is worth recognising: real thermal decomposition can involve intermediate compounds and overlapping reactions. The clean two-stage pathway was explicitly stipulated for this problem. It must not be assumed for every hydrated carbonate.

Question 5

Two samples, A and B, both contain copper(II) sulfate and water of crystallisation, with no impurities or free liquid water.

Sample A is independently known to be a single, pure crystalline hydrate, \(\ce{CuSO4}\cdot x\ce{H2O}\).

Sample B is independently known to contain a mixture of exactly two crystalline phases:

• Copper(II) sulfate monohydrate, \(\ce{CuSO4}\cdot\ce{H2O}\)
• Copper(II) sulfate pentahydrate, \(\ce{CuSO4}\cdot5\ce{H2O}\)

The proportions of these two phases in B are unknown.

The two samples are subjected to the same carefully controlled heating sequence. In both samples, the first treatment converts every crystalline phase completely into pure copper(II) sulfate monohydrate. The second treatment removes the remaining water, giving pure anhydrous copper(II) sulfate.

No copper(II) sulfate is lost or decomposed.

Remarkably, the samples give identical masses at every stage:

Use \(M(\ce{CuSO4})=159.61\) and \(M(\ce{H2O})=18.02\).

(a) Determine the hydration number of pure sample A.

(b) Determine the number of moles of monohydrate and pentahydrate originally present in sample B. Are their masses equal?

(c) Suppose the independent information about the original crystal phases had not been supplied. Would the three mass measurements prove that A and B originally contained the same hydrate? Explain why or why not.

(d) Would repeating the intermediate heating until the mass was perfectly constant resolve that uncertainty? Identify what kind of additional evidence would be needed.

Solution 5

(a) Pure sample A is \(\ce{CuSO4}\cdot3\ce{H2O}\). (b) Sample B initially contained equal numbers of moles of monohydrate and pentahydrate, but unequal masses. (c) The heating data alone cannot establish that the original samples had the same crystalline composition. (d) Repeated constant-mass measurements cannot remove that ambiguity.

This problem turns on a distinction that is easy to overlook: a measured average hydration ratio is not always the formula of a single pure hydrate.

Part (a): Find the hydration number for A.

The second treatment leaves 1.596 g of anhydrous copper(II) sulfate.

Therefore,

\[
\begin{aligned}
n(\ce{CuSO4})
&=\frac{1.596}{159.61}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

The total mass of water originally present was

\[
2.137-1.596=0.541\ \mathrm{g}
\]

This corresponds to

\[
\begin{aligned}
n(\ce{H2O})
&=\frac{0.541}{18.02}\\
&=0.0300\ \mathrm{mol}
\end{aligned}
\]

The hydration ratio is

\[
\begin{aligned}
x&=\frac{0.0300}{0.0100}\\
&=3.00
\end{aligned}
\]

Because sample A is independently known to be one pure crystalline hydrate, its formula is

\[
\ce{CuSO4}\cdot3\ce{H2O}
\]

Part (b): Find the composition of B.

Sample B also contains a total of 0.0100 mol of copper(II) sulfate formula units and 0.0300 mol of water.

But B is not a single hydrate.

Let:

• \(p\) = moles of copper(II) sulfate pentahydrate initially present
• \(q\) = moles of copper(II) sulfate monohydrate initially present

Each hydrate formula unit contributes one copper(II) sulfate formula unit.

Therefore,

\[
p+q=0.0100
\]

The pentahydrate contributes five moles of water per mole of hydrate, while the monohydrate contributes one.

So the water balance is

\[
5p+q=0.0300
\]

Subtract the first equation from the second:

\[
\begin{aligned}
4p&=0.0200\\
p&=0.00500\ \mathrm{mol}
\end{aligned}
\]

Substituting back:

\[
\begin{aligned}
q&=0.0100-0.00500\\
&=0.00500\ \mathrm{mol}
\end{aligned}
\]

Thus, the original mixture contained:

• 0.00500 mol of pentahydrate
• 0.00500 mol of monohydrate

They contribute equal numbers of copper(II) sulfate formula units, but their masses are not equal.

The molar mass of pentahydrate is

\[
159.61+5(18.02)
=249.71\ \mathrm{g\,mol^{-1}}
\]

Its mass in B is

\[
\begin{aligned}
m_5&=0.00500(249.71)\\
&=1.249\ \mathrm{g}
\end{aligned}
\]

The molar mass of monohydrate is

\[
159.61+18.02
=177.63\ \mathrm{g\,mol^{-1}}
\]

Its mass in B is

\[
\begin{aligned}
m_1&=0.00500(177.63)\\
&=0.888\ \mathrm{g}
\end{aligned}
\]

Their combined mass is approximately

\[
1.249+0.888=2.137\ \mathrm{g}
\]

which agrees with the initial measurement.

The two hydrates are present in a 1:1 mole ratio, not a 1:1 mass ratio.

Part (c): Why the equal masses do not prove identical original compositions.

The original mass measurements tell us that each sample contains an overall ratio of three water molecules per copper(II) sulfate formula unit.

For A, this represents one pure trihydrate.

For B, it is an average of two different hydrates.

In B, half of the copper(II) sulfate formula units originally belonged to monohydrate and half to pentahydrate.

Their average hydration number is

\[
\frac{1+5}{2}=3
\]

Therefore, the mixture has exactly the same overall water-to-salt ratio as pure copper(II) sulfate trihydrate.

This is why the initial and final masses agree.

But what about the intermediate mass? Surely that gives us extra information?

Not enough, because the question states that both samples are completely converted into the same monohydrate.

In either sample, the intermediate mass must be

\[
\begin{aligned}
m_{\mathrm{intermediate}}
&=0.0100(177.63)\\
&\approx1.776\ \mathrm{g}
\end{aligned}
\]

Both samples therefore have the same initial mass, the same intermediate mass, and the same final mass, despite starting with different crystalline compositions.

The extra weighing does not distinguish the original phases.

Part (d): Why constant mass cannot resolve the ambiguity.

Repeating the intermediate treatment can strengthen the evidence that both samples have reached a stable monohydrate endpoint.

It cannot recover information about the original crystal structures once both have been converted into the same product.

The mass data are compatible with multiple original compositions.

To distinguish the samples without being told their identities, we would need independent evidence about their initial crystalline phases. A technique that distinguishes crystal structures, such as X-ray diffraction, could provide that information.

The reasoning is the difficult part of this question.

A student who calculates \(x=3\) for both samples has performed the arithmetic correctly. The error would be claiming that both samples must therefore have been pure copper(II) sulfate trihydrate.

Heating data determine an overall composition ratio. They do not automatically establish that the starting material was a single pure compound.

06What to check before reporting a hydrate formula

For a normal HSC-style calculation, the working can be condensed into a short sequence.

First, use the balance measurements to find the masses of water and anhydrous salt:

\[
\begin{aligned}
m_{\mathrm{water}}&=m_i-m_f\\
m_{\mathrm{salt}}&=m_f-m_c
\end{aligned}
\]

Next, convert each mass into moles, using the appropriate molar mass.

Then calculate

\[
x=\frac{n_{\mathrm{water}}}{n_{\mathrm{salt}}}
\]

Finally, write the formula with a justified hydration number.

But before reporting the result, return to the physical experiment. Ask whether the water really escaped completely, whether the salt remained intact, and whether the final material was genuinely anhydrous.

Those checks determine whether the mole ratio represents the original hydrate or merely describes what happened to the mass.

This also points towards the next useful concept in experimental chemistry: a measurement can be precise without being chemically valid. A perfectly repeatable balance reading is valuable evidence, but only when the reaction model connecting that reading to the formula is correct.