How to Tell Chemical Changes from Physical Changes Using Evidence

You heat a blue solution, and it turns pale green. Has a chemical reaction occurred, or has the solution simply changed physically? The colour change is evidence that something may have happened, but it doesn't prove that new substances have formed. To distinguish a chemical change from a physical change, you need to work out whether the chemical species themselves have changed. The key is to connect what you can observe with what must be happening at the particle level.

## The key distinction: did the chemical identity change?

A physical change alters a substance's state, shape, size, or arrangement without changing its chemical identity. Melting ice, crushing a crystal, and dissolving sugar in water are examples.

A chemical change occurs when atoms are rearranged to form different chemical species. The atoms are conserved, but the substances they make up are different.

[IMAGE: Two contrasting particle-level scenes. On the left, ice melting into liquid water, with identical H2O molecules before and after, but different spacing and arrangement. On the right, hydrogen and oxygen molecules rearranging to form water molecules, with clear before-and-after particle representations. Include a simple macroscopic view above each particle model.]

Consider melting ice. Before melting, the particles are water molecules, \(\ce{H2O}\). Afterwards, they are still water molecules. Their arrangement and movement have changed, but their chemical identity has not.

Now consider hydrogen reacting with oxygen to form water:

\[
\ce{2H2 + O2 -> 2H2O}
\]

The hydrogen and oxygen molecules have been rearranged into water molecules. A new chemical species has formed, so this is a chemical change.

The distinction is not whether something looks different. It is whether the chemical species present before and after the process are the same.

## Separate the observation from the inference

This is one of the most important skills in experimental chemistry: knowing the difference between what you actually observed and what you think happened at the particle level.

An observation is something you can see, measure, or detect. An inference is an explanation of that observation using a scientific model.

For example:

- Observation: Bubbles form at an electrode.
- Inference: A gas is being produced.
- Further evidence: A gas test identifies the gas as hydrogen.
- Conclusion: Hydrogen molecules, \(\ce{H2}\), have formed.

Notice the difference. You cannot identify hydrogen just by looking at bubbles.

Use this table to avoid jumping to conclusions.

| Observation | What it tells you | What it does not prove |
|---|---|---|
| Bubbles appear | A gas phase is forming or escaping | That a new gas has been produced |
| A solid disappears | The solid is no longer visible | That it has reacted rather than dissolved |
| A solution changes colour | Its appearance has changed | That a new chemical species exists |
| Temperature increases | Energy has been transferred | That a chemical reaction occurred |
| A solid forms | Solid particles are present | That a new compound has formed |
| A gas passes an identification test | The test supports the identity of the gas | The test alone does not establish how it formed |

The strongest explanations connect an observation to a specific chemical species.

For example, saying "a gas formed because bubbles appeared" is incomplete. Saying "a gas was collected and identified as carbon dioxide, supporting the conclusion that a chemical reaction produced \(\ce{CO2}\)" is much stronger.

[IMAGE: An annotated comparison between observable evidence and inferred particles. Show a boiling beaker with bubbles and water molecules in the vapour, alongside an electrolysis vessel with gases forming at separate electrodes. Clearly distinguish what an observer sees from the particle-level explanation.]

## Follow an experiment: boiling water versus electrolysis

Here's a particularly useful comparison because both processes produce bubbles.

Imagine two experiments involving water.

In the first, water is heated until it boils. In the second, an electric current is passed through water and gases are collected at two electrodes.

Both produce gas. Are both chemical changes?

Before reading on, predict whether the particles in each experiment have the same chemical identity before and after.

### Experiment A: boiling water

Heat water in a beaker. As its temperature reaches its boiling point under the experimental pressure, bubbles form throughout the liquid.

The water changes from liquid to gas:

\[
\ce{H2O(l) -> H2O(g)}
\]

The particles are still water molecules. The molecules move further apart and move more freely, but their chemical identity has not changed.

If the vapour is cooled, it condenses back into liquid water.

This is a physical change.

### Experiment B: electrolysis of water

Now consider an electrolysis experiment.

The apparatus contains water with a small amount of dilute sulfuric acid added to improve electrical conductivity. Two inert electrodes are placed in the solution and connected to a direct-current power supply.

Above each electrode sits a separate graduated gas-collection tube. Each tube is initially filled with liquid, so gas entering the tube displaces the liquid and can be measured.

[IMAGE: A scientifically accurate electrolysis apparatus showing a central water reservoir, dilute sulfuric acid electrolyte, two inert electrodes connected to a DC power supply, and separate inverted graduated gas-collection tubes above the electrodes. Label the negative cathode and positive anode. Show bubbles rising into each separate tube, with approximately twice the gas volume at the cathode. Include the connecting wires and physical layout.]

The first observation is that bubbles form at both electrodes.

But that observation alone is not enough. The bubbles could theoretically be caused by something other than the formation of new chemical species.

A better experimental design collects the gases separately.

The gases can then be tested using standard identification tests:

- Hydrogen produces a squeaky pop when a small sample is tested with a burning splint.
- Oxygen relights a glowing splint.

These tests support the identification of hydrogen at the cathode and oxygen at the anode. Such tests should be performed on small samples under appropriate laboratory supervision.

There is another useful observation. The volume of hydrogen collected is approximately twice the volume of oxygen.

This is exactly what the chemical equation predicts:

\[
\ce{2H2O(l) -> 2H2(g) + O2(g)}
\]

Two water molecules decompose to produce two hydrogen molecules and one oxygen molecule.

The original water molecules have been converted into different chemical species.

This is a chemical change.

### Why was the improved apparatus necessary?

Imagine the original experiment only involved placing two electrodes into an open beaker and watching the bubbles.

You would know that bubbles were appearing, but you couldn't reliably identify the gases or compare their volumes.

The revised apparatus addresses both problems:

1. Separate collection prevents the gases from being mixed together.
2. Graduated tubes allow their volumes to be measured.
3. Gas identification tests provide evidence about their chemical identities.
4. The measured volume ratio can be compared with the balanced equation.

The experiment therefore provides much stronger evidence than simply observing bubbles.

### Compare the two processes

| Feature | Boiling water | Electrolysis of water |
|---|---|---|
| Starting species | \(\ce{H2O}\) | \(\ce{H2O}\) |
| Final species | \(\ce{H2O}\) | \(\ce{H2}\) and \(\ce{O2}\) |
| What changes? | Physical state | Chemical identity |
| New chemical species? | No | Yes |
| Classification | Physical change | Chemical change |

The important difference is not that one process produces bubbles and the other doesn't. Both do.

The difference is that boiling changes the state of water, while electrolysis changes its chemical composition.

## Translate the names into formulas

You cannot reliably compare chemical species if you haven't written their formulas correctly.

Start by identifying the type of substance.

For molecular substances, use the correct molecular formula. For ionic compounds, identify the ions and balance their charges.

Some useful examples from the electrolysis experiment are:

| Chemical name | Formula | Important detail |
|---|---|---|
| Water | \(\ce{H2O}\) | Two hydrogen atoms per oxygen atom |
| Hydrogen gas | \(\ce{H2}\) | Elemental hydrogen is diatomic |
| Oxygen gas | \(\ce{O2}\) | Elemental oxygen is diatomic |
| Sulfuric acid | \(\ce{H2SO4}\) | Used as the electrolyte |
| Sulfate ion | \(\ce{SO4^{2-}}\) | Remains in solution |

A common mistake is writing hydrogen as \(\ce{H}\) or oxygen as \(\ce{O}\) when describing the elemental gases. Under ordinary conditions, these elements exist as diatomic molecules, \(\ce{H2}\) and \(\ce{O2}\).

For ionic compounds, use the ion charges.

For example, magnesium forms \(\mathrm{Mg}^{2+}\), while oxide forms \(\mathrm{O}^{2-}\).

Their charges balance in a 1:1 ratio, giving:

\[
\ce{MgO}
\]

You don't write \(\ce{Mg2O2}\) because the simplest whole-number ratio is 1:1.

### State symbols are not new chemical identities

You will often see these symbols in chemical equations:

| Symbol | Meaning |
|---|---|
| (s) | Solid |
| (l) | Liquid |
| (g) | Gas |
| (aq) | Dissolved in water |

Compare:

\[
\ce{H2O(s) -> H2O(l) -> H2O(g)}
\]

The state changes, but the formula stays the same.

Now compare:

\[
\ce{2H2O(l) -> 2H2(g) + O2(g)}
\]

The formulas change because new chemical species have formed.

This gives you a reliable first decision rule:

- Same chemical species, different state or arrangement: physical change.
- Different chemical species formed through rearrangement of atoms: chemical change.

Remember that atoms must still be conserved. A chemical reaction rearranges atoms; it does not create or destroy them.

Also, never change a chemical formula's subscripts just to balance an equation. Subscripts define the substance itself. Coefficients are used to balance the number of particles.

## Worked example: Do the gas volumes support electrolysis?

During the electrolysis of water, a student collects \(18.0\ \mathrm{mL}\) of gas at the cathode and \(9.0\ \mathrm{mL}\) at the anode. Both gas samples pass their respective identification tests for hydrogen and oxygen.

Explain whether the observations support a chemical change. Check whether the measured gas volumes agree with the balanced equation.

**Step 1**: Write the balanced chemical equation.

\[
\ce{2H2O(l) -> 2H2(g) + O2(g)}
\]

The equation predicts two moles of hydrogen for every one mole of oxygen.

**Step 2**: Connect the mole ratio to the gas volumes.

At the same temperature and pressure, equal volumes of gases contain equal numbers of moles.

Therefore, the expected volume ratio is:

\[
V(\ce{H2}):V(\ce{O2})=2:1
\]

**Step 3**: Calculate the experimental ratio.

\[
\begin{aligned}
\text{Volume ratio}
&=\frac{18.0}{9.0}\\
&=2.00:1
\end{aligned}
\]

The measured ratio agrees with the predicted ratio.

**Step 4**: Interpret the evidence.

The gas identification tests support the conclusion that hydrogen and oxygen have formed.

The volume ratio is also consistent with the decomposition equation.

Water molecules have been converted into different chemical species, so the process is a chemical change.

The important point is that the volume ratio supports the particle-level explanation. The bubbles alone would not establish it.

## Worked example: How much carbon dioxide forms when baking soda decomposes?

A student heats \(0.840\ \mathrm{g}\) of sodium hydrogen carbonate until it completely decomposes.

The gases are cooled so that water condenses, and the dry carbon dioxide is collected at \(25^\circ\mathrm{C}\) and \(100\ \mathrm{kPa}\).

Calculate the expected volume of carbon dioxide.

Use:

- \(M(\ce{NaHCO3})=84.0\ \mathrm{g\,mol^{-1}}\)
- Molar gas volume at the stated conditions: \(24.8\ \mathrm{L\,mol^{-1}}\)

[IMAGE: Experimental arrangement for thermal decomposition of sodium hydrogen carbonate. Show a heated hard-glass test tube containing a white solid, connected by a delivery tube to a cooled condensation trap and then a gas-collection syringe. Show the direction of gas flow and label the apparatus components without displaying the calculated result.]

**Step 1**: Translate the chemical names into formulas and write the reaction.

Sodium hydrogen carbonate is \(\ce{NaHCO3}\). The products are sodium carbonate, carbon dioxide, and water.

The balanced equation is:

\[
\ce{2NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(g)}
\]

This is a decomposition reaction.

**Step 2**: Calculate the initial amount of sodium hydrogen carbonate.

Use:

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

Where:

- \(n\) is the amount in moles.
- \(m\) is the mass in grams.
- \(M\) is the molar mass in grams per mole.

Substitute:

\[
\begin{aligned}
n(\ce{NaHCO3})
&=\frac{0.840}{84.0}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Use the mole ratio to calculate the amount of carbon dioxide.

The balanced equation gives a 2:1 ratio between sodium hydrogen carbonate and carbon dioxide.

Therefore:

\[
\begin{aligned}
n(\ce{CO2})
&=\frac{0.0100}{2}\\
&=0.00500\ \mathrm{mol}
\end{aligned}
\]

**Step 4**: Calculate the gas volume.

Use:

\[
V=nV_m
\]

Where \(V_m\) is the molar volume of the gas.

\[
\begin{aligned}
V(\ce{CO2})
&=0.00500\times24.8\\
&=0.124\ \mathrm{L}\\
&=124\ \mathrm{mL}
\end{aligned}
\]

The expected carbon dioxide volume is \(124\ \mathrm{mL}\).

**Step 5**: Connect the calculation to the classification.

The original sodium hydrogen carbonate has decomposed into sodium carbonate, carbon dioxide, and water.

These are different chemical species, so this is a chemical change.

Notice that the water vapour is condensed before measuring the dry carbon dioxide. Without that assumption, the collected gas volume would not represent carbon dioxide alone.

## What other evidence can mislead you?

The water experiments establish the basic distinction, but some observations are less straightforward.

Three common traps are worth remembering.

### 1. A solid appearing does not always mean a reaction

Imagine a hot, saturated solution of potassium nitrate cooling down.

White crystals begin to appear at the bottom of the beaker.

It is tempting to say that a chemical reaction has occurred because a new solid is visible.

But potassium nitrate was already present in the solution. As the solution cools, less potassium nitrate can remain dissolved, so some of it crystallises.

The particles are still potassium ions and nitrate ions. They have returned to an ordered solid lattice.

This is crystallisation, a physical change.

[IMAGE: Two beakers side by side. In the first, a clear hot saturated potassium nitrate solution cools and white crystals form at the bottom. In the second, two separate clear ionic solutions are mixed and a cloudy white precipitate forms. Show the different starting arrangements and resulting solids without using generic process boxes.]

Compare this with mixing barium chloride and sodium sulfate solutions:

\[
\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}
\]

A new solid, barium sulfate, forms.

The net ionic equation is:

\[
\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}
\]

Here, dissolved barium and sulfate ions combine to form a new insoluble compound.

This is a chemical change.

Both experiments produce white solids. Their classification depends on the chemical species involved, not the appearance of the final mixture.

### 2. A mass increase does not mean atoms have appeared from nowhere

When magnesium burns in air, it reacts with oxygen to form magnesium oxide.

\[
\ce{2Mg(s) + O2(g) -> 2MgO(s)}
\]

The white product can have a greater mass than the original magnesium strip.

That makes sense because oxygen atoms from the surrounding air have become part of the product.

The system boundary matters. If you only weigh the magnesium and its product, oxygen has entered from outside that system. If you include the surrounding air in a closed system, total mass is conserved.

### 3. Reversibility does not automatically mean a physical change

Some chemical reactions can be reversed.

For example, anhydrous copper(II) sulfate can react with water to form hydrated copper(II) sulfate. Heating the hydrated compound can remove the water again.

The forward and reverse processes involve changes in chemical identity.

So don't use reversibility as your deciding test. Compare the chemical species before and after.

## Questions and solutions

### Question 1

A student removes an ice cube from a freezer and allows it to melt completely. The resulting liquid is then placed in a freezer until it becomes solid again.

Classify both processes as physical or chemical changes. Explain your answer using chemical formulas.

### Solution 1

Both processes are physical changes because the chemical species remains water throughout.

During melting:

\[
\ce{H2O(s) -> H2O(l)}
\]

During freezing:

\[
\ce{H2O(l) -> H2O(s)}
\]

The water molecules remain \(\ce{H2O}\). Only their arrangement, movement, and physical state change.

A common mistake is to assume that breaking apart the solid structure means chemical bonds have broken. In ice, the water molecules are held in a solid arrangement by intermolecular attractions. Melting changes this arrangement without converting water into a different chemical substance.

### Question 2

A student performs two experiments.

**Experiment A:** A hot, saturated potassium nitrate solution is cooled. A mass of \(4.0\ \mathrm{g}\) of white potassium nitrate crystals is recovered. No water evaporates during the process.

**Experiment B:** Equal volumes of clear aqueous barium chloride and sodium sulfate solutions are mixed. A white solid forms immediately.

(a) Classify each process as a physical or chemical change.

(b) Write a balanced chemical equation for Experiment B.

(c) Explain why the appearance of a white solid is insufficient to classify both processes in the same way.

### Solution 2

Experiment A is a physical change, while Experiment B is a chemical change.

**Experiment A:**

Potassium nitrate has the formula \(\ce{KNO3}\).

In the original solution, its ions are dispersed throughout the water:

\[
\ce{KNO3(s) -> K+(aq) + NO3-(aq)}
\]

When the solution cools, the ions return to an ordered solid lattice.

\[
\ce{K+(aq) + NO3-(aq) -> KNO3(s)}
\]

The potassium and nitrate ions remain the same chemical species. Only their arrangement and physical state change.

Therefore, crystallisation is a physical change.

**Experiment B:**

Barium chloride is \(\ce{BaCl2}\), and sodium sulfate is \(\ce{Na2SO4}\).

Barium ions and sulfate ions form insoluble barium sulfate:

\[
\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}
\]

The balanced molecular equation is:

\[
\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}
\]

A new compound, barium sulfate, has formed. This is a chemical change.

The tempting conclusion is that both experiments must be chemical because both produce white solids.

However, Experiment A simply recovers the original dissolved compound. Experiment B combines ions from different dissolved compounds to produce a new substance.

The starting chemical species determine the classification, not the colour or appearance of the final solid.

### Question 3

Molten sodium chloride is used in two separate experiments.

In Experiment A, the molten sodium chloride is allowed to cool until it becomes solid again.

In Experiment B, an electric current is passed through molten sodium chloride using inert electrodes. Sodium metal is identified at the cathode, and chlorine gas is identified at the anode.

(a) Classify each process.

(b) Write a balanced equation for Experiment B.

(c) Explain the particle-level difference between the two experiments.

[IMAGE: Two experimental arrangements using molten sodium chloride. On the left, a container of molten sodium chloride cooling into a solid. On the right, an electrolytic cell containing molten sodium chloride with two inert electrodes connected to a DC power supply, showing the electrode positions and separate product collection locations. Do not include the reaction equation or completed particle explanation.]

### Solution 3

Experiment A is a physical change, while Experiment B is a chemical change.

**Experiment A:**

Cooling changes the state of sodium chloride:

\[
\ce{NaCl(l) -> NaCl(s)}
\]

Sodium chloride remains sodium chloride. Its ions become arranged in a solid ionic lattice rather than moving freely through the liquid.

No new chemical species are formed.

**Experiment B:**

The balanced equation is:

\[
\ce{2NaCl(l) -> 2Na(l) + Cl2(g)}
\]

Molten sodium chloride contains sodium ions, \(\mathrm{Na}^{+}\), and chloride ions, \(\mathrm{Cl}^{-}\).

During electrolysis, the ions undergo electron transfer:

At the cathode:

\[
\ce{Na+ + e- -> Na}
\]

At the anode:

\[
\ce{2Cl- -> Cl2 + 2e-}
\]

Sodium ions become neutral sodium atoms, while chloride ions form chlorine molecules.

The chemical identities of the particles have changed, so electrolysis is a chemical change.

The key difference is that cooling changes the arrangement of existing ions, whereas electrolysis converts those ions into different chemical species.

### Question 4

A student burns a clean magnesium ribbon in air.

The initial magnesium ribbon has a mass of \(0.120\ \mathrm{g}\). After burning is complete, the white magnesium oxide product has a mass of \(0.199\ \mathrm{g}\).

Assume all the magnesium reacts, no solid product is lost, and the product is pure magnesium oxide.

Given:

- \(M(\ce{Mg})=24.3\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{MgO})=40.3\ \mathrm{g\,mol^{-1}}\)

(a) Calculate the mass of oxygen incorporated into the product.

(b) Calculate the amount of magnesium and magnesium oxide in moles.

(c) Explain why the increase in mass is consistent with conservation of mass.

[IMAGE: A laboratory crucible containing a magnesium ribbon being heated in air, with its lid slightly ajar to allow oxygen access while reducing product loss. Include a separate balance showing the initial and final mass measurements. Do not show the calculated oxygen mass.]

### Solution 4

The magnesium undergoes a chemical change because it reacts with oxygen to form magnesium oxide. The increase in mass is caused by oxygen entering the solid product from the air.

**(a) Calculate the mass of oxygen.**

Subtract the initial magnesium mass from the final product mass:

\[
\begin{aligned}
m(\mathrm{O})
&=0.199-0.120\\
&=0.079\ \mathrm{g}
\end{aligned}
\]

Therefore, \(0.079\ \mathrm{g}\) of oxygen has been incorporated.

**(b) Calculate the amounts in moles.**

For magnesium:

\[
\begin{aligned}
n(\ce{Mg})
&=\frac{0.120}{24.3}\\
&=0.00494\ \mathrm{mol}
\end{aligned}
\]

For magnesium oxide:

\[
\begin{aligned}
n(\ce{MgO})
&=\frac{0.199}{40.3}\\
&=0.00494\ \mathrm{mol}
\end{aligned}
\]

The amounts are equal to three significant figures.

This agrees with the balanced equation:

\[
\ce{2Mg(s) + O2(g) -> 2MgO(s)}
\]

The equation gives a 1:1 mole ratio between magnesium and magnesium oxide.

**(c) Explain conservation of mass.**

The magnesium has combined with oxygen from the surrounding air.

The solid product therefore has a greater mass than the original magnesium ribbon.

There is no violation of conservation of mass. The magnesium ribbon and its product are an open system because oxygen enters from the surroundings.

If the system boundary included all the oxygen in the surrounding air, the total mass would remain conserved.

The mass increase is consistent with the formation of a new chemical substance, rather than evidence that matter has been created.

### Question 5

An experiment is conducted using anhydrous copper(II) sulfate, a white solid.

A student begins with \(1.60\ \mathrm{g}\) of pure anhydrous copper(II) sulfate. The sample is exposed to water vapour until its mass becomes constant. It turns blue and has a final mass of \(2.50\ \mathrm{g}\).

The blue solid is then heated in a sealed apparatus. It turns white again, while \(0.90\ \mathrm{g}\) of water condenses in a cooler section of the apparatus. The total mass of the sealed apparatus remains unchanged.

Assume:

- All the mass gained during hydration is water incorporated into one pure hydrated compound.
- No unbound surface water remains.
- All the water released during heating is collected.
- The anhydrous copper(II) sulfate is completely converted into the hydrated compound initially.

Given:

- \(M(\ce{CuSO4})=159.6\ \mathrm{g\,mol^{-1}}\)
- \(M(\ce{H2O})=18.0\ \mathrm{g\,mol^{-1}}\)

(a) Calculate the amount of anhydrous copper(II) sulfate initially present.

(b) Calculate the amount of water incorporated into the blue solid.

(c) Determine the formula of the hydrated compound.

(d) Classify the hydration and dehydration processes. Explain why the reversibility of the process and conservation of mass do not make it a physical change.

[IMAGE: A sealed horizontal glass apparatus containing a white anhydrous copper sulfate sample at one end and a cooler collection section at the other. Show the initial white solid, the blue hydrated solid after exposure to water vapour, and the water condensate collecting in the cooler section during subsequent heating. Include the heating position and sealed apparatus boundaries. Do not show the calculated hydrate formula.]

### Solution 5

Both hydration and dehydration are chemical changes because the chemical composition of the solid changes. The blue compound contains a definite number of water molecules per formula unit, rather than simply being wet.

**(a) Calculate the initial amount of anhydrous copper(II) sulfate.**

Use:

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

Substitute:

\[
\begin{aligned}
n(\ce{CuSO4})
&=\frac{1.60}{159.6}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

The initial amount of copper(II) sulfate is \(0.0100\ \mathrm{mol}\).

**(b) Calculate the amount of incorporated water.**

The mass gained during hydration is:

\[
\begin{aligned}
m(\ce{H2O})
&=2.50-1.60\\
&=0.90\ \mathrm{g}
\end{aligned}
\]

Convert this mass into moles:

\[
\begin{aligned}
n(\ce{H2O})
&=\frac{0.90}{18.0}\\
&=0.0500\ \mathrm{mol}
\end{aligned}
\]

**(c) Determine the hydrate formula.**

Divide the amount of water by the amount of anhydrous copper(II) sulfate:

\[
\begin{aligned}
\text{Mole ratio}
&=\frac{0.0500}{0.0100}\\
&=5.00:1
\end{aligned}
\]

There are five water molecules incorporated per formula unit of copper(II) sulfate.

The hydrated compound is:

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

This is copper(II) sulfate pentahydrate.

**(d) Explain the classification.**

The hydration process is:

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

The original anhydrous copper(II) sulfate has been converted into a hydrated compound with a different chemical composition.

This is a chemical change.

When the blue compound is heated, the reverse process occurs:

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

The hydrated compound loses its water and returns to anhydrous copper(II) sulfate.

This is also a chemical change because the chemical composition of the solid changes again.

There are two tempting arguments for calling the process physical, but neither works.

First, the process is reversible. However, some chemical reactions are reversible. Being able to reverse a change does not mean that the chemical species remained unchanged.

Second, the total mass of the sealed apparatus remains constant. But mass conservation applies to both physical and chemical changes. In this experiment, water moves between the solid and the cooler section of the apparatus without leaving the system.

The decisive evidence is the formation of a compound with a different composition. The measured 5:1 mole ratio establishes the number of water molecules incorporated per formula unit.

If the sample had simply been wet, the mass increase alone would not establish the formation of a definite hydrate. That is why the assumptions about complete conversion and the absence of unbound water matter.

## What to remember when answering HSC questions

When you are asked whether an observed process is a physical or chemical change, don't start by listing indicators.

Start with the chemical species.

1. Identify the starting substances and write their formulas.
2. Identify the final substances using the available experimental evidence.
3. Compare the chemical species before and after.
4. Decide whether only the physical state or arrangement changed, or whether new chemical species formed.
5. Explain how the observations support your conclusion.
6. Check that your chemical equation conserves atoms.

A strong HSC response connects the macroscopic evidence to the particle-level explanation. It doesn't simply say that a colour change, gas, or precipitate proves a reaction.

For more detail on interpreting individual observations, read [Indicators of Chemical Change: Evidence for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-indicators-of-chemical-change/).

If you're unsure whether a disappearing solid has dissolved or reacted, the next useful distinction is covered in [Dissolving vs Reacting: How to Tell the Difference in HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-explain-the-difference-between-dissolving-and-reacting/).

Once you can identify which chemical species have changed, writing and balancing the corresponding chemical equation becomes much more straightforward.