Chemistry 9701/35 — October/November 2025

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis

Q1Medium-EasyManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Hydrated ethanedioic acid is a diprotic acid with the formula $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ where $x$ is an integer.

Ethanedioic acid reacts with manganate(VII) ions when heated.

5(\text{COOH})_2(\text{aq}) + 2\text{MnO}_4^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

You will determine the value of $x$ in $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ by titrating a solution containing ethanedioic acid with manganate(VII) ions.

FA 1 is $6.20 \text{ g dm}^{-3}$ aqueous ethanedioic acid, $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ .
FA 2 is $0.0200 \text{ mol dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .
FA 3 is $1.0 \text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .

(a)

Method

Fill the burette with FA 2.
Pipette $25.0 \text{ cm}^3$ of FA 1 into a conical flask.
Use the measuring cylinder to add approximately $20 \text{ cm}^3$ of FA 3 to the conical flask.
Place the conical flask on a tripod and gauze and heat carefully until the temperature of the solution is approximately $70 \text{ }^\circ\text{C}$ .
Remove the flame.
Carefully lift the hot conical flask and place it on the white tile under the burette.
Add FA 2 drop-wise for the first $2\text{--}3 \text{ cm}^3$ . Any initial pink colouring may take several seconds to disappear.
If the reaction mixture turns brown, reheat it to about $70 \text{ }^\circ\text{C}$ . If the brown colour disappears, continue the titration. If the brown colour remains, discard the contents of the flask and begin a new titration.
The end-point is reached when a permanent pale pink colour is formed.
Perform a rough titration with FA 2. Record your burette readings in the space below.

The rough titre is .............................. $\text{cm}^3$ .

Carry out as many titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record all your burette readings and the volume of FA 2 added in each accurate titration.

Results

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example results:

Rough titre: 24.70 $\text{cm}^3$ .

Accurate titrations:

	1	2
initial burette reading / $\text{cm}^3$	0.00	24.60
final burette reading / $\text{cm}^3$	24.60	49.20
titre / $\text{cm}^3$	24.60	24.60

All burette readings recorded to 0.05 $\text{cm}^3$ . The two accurate titres are concordant (within 0.10 $\text{cm}^3$ of each other).

Final answer

Example results: rough 24.70 cm3; accurate titres 24.60 and 24.60 cm3 (candidate-dependent)

Detailed explanation

Background Concept

In a titration the burette is read before and after each addition. The titre is the volume delivered, found by subtracting the initial reading from the final reading. Burettes are graduated so that readings can be estimated to the nearest 0.05 $\text{cm}^3$ . A rough titration gives an approximate end-point; accurate titrations are then done carefully, often drop-wise near the end-point, and are repeated until two or more agree closely.

Understanding the Question

This part asks you to carry out the redox titration and record your results. The marks are awarded for the quality of the recorded data: two burette readings and a titre for the rough run, initial and final readings for at least two accurate runs, correct headings with units, readings to 0.05 $\text{cm}^3$ , and accurate titres that agree within 0.10 $\text{cm}^3$ .

Approach

Record the rough titre first. Then perform accurate titrations, adding FA 2 drop-wise near the end-point. For each accurate run write down the initial and final burette readings and subtract to find the titre. Present the results in a clear table with headings and units. Check that the accurate titres are concordant.

Step-by-Step Reasoning

The rough titre gives an approximate idea of the volume needed, so subsequent titrations can be done more quickly until near the end-point.
For each accurate titration, read the burette before and after delivery. Example: initial 0.00 $\text{cm}^3$ , final 24.60 $\text{cm}^3$ , titre 24.60 $\text{cm}^3$ .
Record every reading to 0.05 $\text{cm}^3$ . For example, 24.60 is acceptable, but 24.6 is not precise enough.
Repeat until two titres agree within 0.10 $\text{cm}^3$ . In the example, 24.60 and 24.60 are concordant.
The table must include headings such as 'initial burette reading / $\text{cm}^3$ ', 'final burette reading / $\text{cm}^3$ ' and 'titre / $\text{cm}^3$ '.

Key Takeaways

Good titration records show precision, correct units, and concordant results. The examiner is looking for evidence that you can use a burette properly and record data honestly.

Common Mistakes

Recording readings to only 1 decimal place instead of 0.05 $\text{cm}^3$ .
Forgetting to include units in table headings.
Not showing a rough titre.
Using accurate titres that differ by more than 0.10 $\text{cm}^3$ .
Recording the final reading as the titre without subtracting the initial reading.

Things to Be Careful About

Always read the burette at eye level to avoid parallax error.
The end-point is a permanent pale pink colour, not a transient pink that disappears.
If the mixture turns brown, reheat; if brown remains, discard and start again.
Make sure the accurate titres you select for the mean are within 0.20 $\text{cm}^3$ of each other overall.

Techniques used

record burette readings to 0.05 cm3calculate titre by subtractioncheck concordance of accurate titres

(b)

From your accurate titration results, calculate a suitable mean value to be used in your calculations.

Show clearly how you obtained this value.

$25.0 \text{ cm}^3$ of FA 1 required .............................. $\text{cm}^3$ of FA 2.

1M Similar

(c)

(i)

Give your answers to (c)(ii), (c)(iii) and (c)(iv) to the appropriate number of significant figures.

1M Similar

(ii)

Calculate the amount, in mol, of manganate(VII) ions, $\text{MnO}_4^-$ , in the volume of FA 2 calculated in (b).

amount of $\text{MnO}_4^-$ = .............................. mol

1M Similar

(iii)

Calculate the amount, in mol, of ethanedioic acid that reacts with the manganate(VII) ions in (c)(ii).

amount of $(\text{COOH})_2$ = .............................. mol

Hence calculate the concentration, in $\text{mol dm}^{-3}$ , of ethanedioic acid in FA 1.

concentration of $(\text{COOH})_2$ = .............................. $\text{mol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

amount of $(\text{COOH})_2 = 4.92 \times 10^{-4} \times \frac{5}{2} = 1.23 \times 10^{-3}$ mol

concentration of $(\text{COOH})_2 = \frac{1.23 \times 10^{-3}}{25.0/1000} = 0.0492$ mol dm $^{-3}$

Answer

amount = $1.23 \times 10^{-3}$ mol; concentration = $0.0492$ mol dm $^{-3}$

Final answer

amount = 1.23 × 10^-3 mol; concentration = 0.0492 mol dm^-3

Detailed explanation

Background Concept

The balanced equation shows that 5 mol of ethanedioic acid react with 2 mol of manganate(VII) ions:

5(\text{COOH})_2(\text{aq}) + 2\text{MnO}_4^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

So the mole ratio of acid to manganate(VII) is 5:2. Once the amount of acid is known, its concentration in FA 1 is found by dividing by the volume of FA 1 used, 25.0 cm $^3$ , converted to dm $^3$ .

Understanding the Question

This part has two calculations: first the amount of ethanedioic acid that reacted, then the concentration of ethanedioic acid in FA 1. The amount of manganate(VII) comes from part (c)(ii).

Approach

Use the 5:2 ratio to convert moles of MnO $_4^-$ to moles of acid. Then divide by the volume of FA 1 in dm $^3$ to get concentration.

Step-by-Step Reasoning

From (c)(ii), amount of MnO $_4^-$ = 4.92 × 10 $^{-4}$ mol.
Amount of acid = 4.92 × 10 $^{-4}$ × 5/2 = 1.23 × 10 $^{-3}$ mol.
Volume of FA 1 = 25.0 cm $^3$ = 0.0250 dm $^3$ .
Concentration = 1.23 × 10 $^{-3}$ / 0.0250 = 0.0492 mol dm $^{-3}$ .

Key Takeaways

The stoichiometric ratio from the balanced equation is essential. Always convert volumes to dm $^3$ before calculating concentration.

Common Mistakes

Using the ratio 2:5 instead of 5:2.
Forgetting to convert 25.0 cm $^3$ to dm $^3$ .
Quoting the amount as the concentration.

Things to Be Careful About

The amount is in mol; the concentration is in mol dm $^{-3}$ .
Use the exact volume of FA 1 pipetted, 25.0 cm $^3$ , not the titre.

Techniques used

apply 5:2 stoichiometric ratiocalculate concentration from moles and volume

(iv)

Calculate the relative molecular mass, $M_r$ , of the ethanedioic acid in FA 1.

$M_r$ = ..............................

1M Similar

(v)

Calculate the value of $x$ in $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ .

Show your working.

$x$ = ..............................

1M Similar

(d)

Explain why it is necessary to add FA 3 in each titration.

1M Similar

Q2Medium-EasyManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Hydrated zinc sulfate has the formula $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ where $y$ is an integer.

Hydrated zinc sulfate decomposes when heated, losing only its water of crystallisation and becoming anhydrous.

You will determine the value of $y$ in $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ by heating the hydrated salt until it becomes anhydrous.

FA 4 is hydrated zinc sulfate, $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ .

(a)

Method

Weigh the crucible with its lid. Record the mass in the space for Results.
Add between $3.20 \text{ g}$ and $3.40 \text{ g}$ of FA 4 to the crucible.
Weigh the crucible with its lid and FA 4. Record the mass.
Place the crucible on the pipeclay triangle. Gently heat the crucible and contents for approximately 2 minutes with the lid on.
Remove the lid. Heat the crucible and contents strongly for approximately 4 minutes.
Replace the lid and leave the crucible and residue to cool for at least 5 minutes.

You may wish to begin work on Question 3 while the crucible is cooling.

Weigh the crucible with its lid and its contents. Record the mass.
Remove the lid. Heat the crucible strongly for approximately 3 minutes.
Replace the lid and leave the crucible and residue to cool for at least 5 minutes.
Weigh the crucible with its lid and its contents. Record the mass.
Calculate the mass of FA 4 used and the mass of residue obtained. Record the masses.

Results

5M Similar

DifficultyMedium-Easy

Worked solution

Answer

A correct results table should contain the following headings and readings. Use the readings you actually obtain in the experiment; the numbers below are a complete example.

Description	Mass / g
Crucible + lid	15.00
Crucible + lid + FA 4	18.20
Crucible + lid + residue after first heating	16.82
Crucible + lid + residue after second heating	16.80
Mass of FA 4 used	3.20
Mass of residue	1.80

All four balance readings are recorded to the same number of decimal places (2 decimal places here).

Mass of FA 4 used $=18.20-15.00=3.20\ \text{g}$

Mass of residue $=16.80-15.00=1.80\ \text{g}$

The residue mass after the second heating, 16.80 g, differs from the residue mass after the first heating by only -0.02 g (allow +0.02 to -0.05), so the residue has reached constant mass. This mass ratio also lies within the accuracy range: $3.20/1.80=1.78$ .

Final answer

A results table with six headings and units, all four weighings to the same decimal place, constant mass reached on second heating; for the example values, masses of FA4 and residue are 3.20 g and 1.80 g (candidate-dependent).

Detailed explanation

Background Concept

When a hydrated salt is heated, only the water of crystallisation is usually removed if the temperature is controlled carefully:

$\text{ZnSO}_4 \cdot y\text{H}_2\text{O(s)} \rightarrow \text{ZnSO}_4\text{(s)} + y\text{H}_2\text{O(g)}$

The mass before heating includes both the anhydrous salt and its water of crystallisation; after heating, only anhydrous $\text{ZnSO}_4$ remains. The mass difference is the mass of water lost. To be sure that all water has been lost, the salt is heated, cooled, and reweighed until the mass is constant. This is analogous to the idea behind concordant titres: one result alone is unreliable.

Understanding the Question

You are carrying out the classic gravimetric determination of $y$ in $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ . The required outcomes in part (a) are practical: recording the detail of the weighing and heating process properly, using consistent headings and units, calculating the masses, and obtaining an accurate mass ratio between the hydrated salt and the residue.

The mark scheme awards:

B1 for six correct table headings and units,
B1 for recording all four readings to the same number of decimal places and showing that the second-heating mass is close enough to the first-heating mass,
B1 for correct subtractions and for using 2.90-3.40 g of FA4,
two accuracy marks based on the ratio (mass hydrated salt : mass residue).

Approach

Record every balance reading as it is read, not after doing arithmetic. Use a table with a heading and unit for each measurement. Calculate the hydrated mass and anhydrous residue mass by subtraction. Then perform a second heating and reweighing; the fact that the second reading agrees within +0.02 to -0.05 g with the first shows that the sample has not changed further, so it gives you confidence that actual constant mass has been reached.

Step-by-Step Reasoning

Set up six table rows: crucible + lid; crucible + lid + FA4; crucible + lid + residue after first heating; crucible + lid + residue after second heating; mass of FA4; mass of residue. Use units / g or (g).
Example: if the crucible + lid is 15.00 g, then crucible + lid + FA4 is 18.20 g, so FA4 = 3.20 g. This is in the allowed range 2.90-3.40 g.
After the first heating, suppose the cooled mass is 16.82 g. After the change from 16.82 g to 16.80 g on the second heating is only -0.02 g; this is within the mark-scheme tolerance and supports the constant-mass claim.
Residue = 16.80 - 15.00 = 1.80 g.
The accuracy check in mark scheme IV and V uses the ratio: mass hydrated / mass residue = 3.20/1.80 = 1.78. This is within the top accuracy range 1.60-1.96, so it would score both accuracy marks.

The sample values are educational; in the real practical, use and record your own readings, to the same number of decimals throughout.

Key Takeaways

Heat the hydrated salt to constant mass to make sure all water is removed.
Record all weighings to the same number of decimal places.
The mass of water lost is not itself a direct measurement; it is the initial hydrated mass minus the residue mass.
A good mass ratio must be based on correct subtraction and accurate heating.

Common Mistakes

Using inconsistent decimal places among the four readings.
Writing the units only outside some values and not all.
Forgetting that the residue after heating is anhydrous zinc sulfate, not the hydrated salt.
Heating for too short a time or not checking that the second heating gives effectively the same mass.
Using a third/sample mass outside the allowed range of FA4.

Things to Be Careful About

The mark scheme states all four weighings must be recorded to the same decimal places; if you mix 15.0 g and 18.24 g you lose accuracy. It also insists the reading after the second heating be within +0.02 and -0.05 g of the first reading; use the same balance and handling procedure each time. Do not write "not accurate" as a conclusion. The units can be written as / g, (g) or in grams, but they must be present.

Techniques used

weigh a crucible and lid before and after adding FA 4heat the hydrated salt to constant mass using repeated heating cyclesrecord all four weighings in a table with headings and unitscalculate the masses of hydrated salt and anhydrous residue by subtractionrepeat heating and reweighing until the mass change is within the allowed range

(b)

(i)

Calculate the amount, in mol, of anhydrous zinc sulfate residue formed in the decomposition of FA 4.

amount of $\text{ZnSO}_4$ = .............................. mol

Calculate the amount, in mol, of water of crystallisation lost.

amount of $\text{H}_2\text{O}$ = .............................. mol

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using the sample results from part (a):

Mass of residue, which is anhydrous $\text{ZnSO}_4$ : $1.80\ \text{g}$

Mass of water lost: $=3.20-1.80=1.40\ \text{g}$

$n(\text{ZnSO}_4)=\frac{1.80}{161.5}=0.0111\ \text{mol}$

$n(\text{H}_2\text{O})=\frac{1.40}{18.0}=0.0778\ \text{mol}$

Answer

$n(\text{ZnSO}_4)=0.0111\ \text{mol}$ and $n(\text{H}_2\text{O})=0.0778\ \text{mol}$ for the sample masses. Replace these with the values from your own experiment.

Final answer

For the worked example: n(ZnSO4) = 0.0111 mol and n(H2O) = 0.0778 mol (candidate-dependent).

Detailed explanation

Background Concept

The amount (number of moles) of a pure substance is connected to its measured mass by

$n = \frac{m}{M_r}$

where the molar mass has units of g mol^-1. For $\text{ZnSO}_4$ , using the periodic table: $M_r$ (ZnSO4) = 65.4 + 32.1 + 4(16.0) = 161.5 g mol^-1. For water: 18 g mol^-1.

The product is heated until it loses only its water of crystallisation. The residue in the crucible is anhydrous $\text{ZnSO}_4$ only. The difference between the original FA4 mass and the residue mass is the mass of water that was lost.

Getting Started

This question starts with your own measured masses. In part (b)(i), you are asked to convert those masses into the amount of anhydrous residue and the amount of water lost. The question therefore assumes you have correctly done part (a), so it is important to use the residue mass from your own result.

The command word "Calculate" means show the substitution and final numerical answer with units. It is not enough to write the formula.

Approach

First get the mass of anhydrous $\text{ZnSO}_4$ : that is the residue mass after heating. Then subtract that from the initial FA4 mass to get the mass of water lost. Use the equation $n = m/M_r$ separately for zinc sulfate and water.

Good questions to keep the work in 2-4 significant figures. The mark scheme rewards answers to 2-4 s.f.: 0.0111 is fine; 0.01 is too rough.

Step-by-Step Reasoning

Identify the residue mass as anhydrous $\text{ZnSO}_4$ . For the sample: $m(\text{ZnSO}_4) = 1.80\ \text{g}$ .
Identify the water mass: the mass loss on heating, $3.20 - 1.80 = 1.40\ \text{g}$ .
Convert to moles of $\text{ZnSO}_4$ :

$n = \frac{1.80}{161.5} = 0.01115..., \quad \boxed{0.0111\ \text{mol}}$

Convert to moles of $\text{H}_2\text{O}$ :

$n = \frac{1.40}{18.0} = 0.0778..., \quad \boxed{0.0778\ \text{mol}}$

Both answers are given to 3 significant figures, which is fine.

If your own masses differ, simply substitute your own values into the same divisions.

Key Takeaways

Always convert mass to moles with the correct molar mass for the substance asked for.
The mass of water lost is a difference, so it depends on both the initial hydrated mass and the residue mass.
Give results with units.

Common Mistakes

Mistaking the residue mass, or using the initial hydrated mass as the $\text{ZnSO}_4$ mass.
Calculating the water mass incorrectly, e.g. residue - initial, giving a negative amount.
Using 24 for water due to confusion with gas volume, instead of 18.0.
Quoting too few significant figures, such as 0.01 mol, which costs an accuracy mark.

Things to Careful About

Use 161.5 g mol^-1, not the hydrated salt molar mass. Keep the two decimals in mass to two but the moles to an appropriate number of sig figs. Do not round intermediate amounts too soon; keeping those to at least 3 sig. figs before dividing is good. Provide both the working and the final boxed amount in Your writing.

Techniques used

calculate the mass of water lost by subtracting the residue mass from the hydrated salt massconvert mass to moles using the molar mass of anhydrous zinc sulfateconvert mass of water to moles using the molar mass of waterkeep the answers to an appropriate number of significant figures

(ii)

Calculate the value of $y$ in the formula $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ .

Show your working.

$y$ = ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Ratio = moles water / moles anhydrous salt

$y = \frac{n(\text{H}_2\text{O})}{n(\text{ZnSO}_4)}= \frac{0.0778}{0.0111}=7.0$

Since $y$ is an integer, $y=7$ .

Answer

$y = 7$

For the sample data, $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ .

Final answer

y = 7

Detailed explanation

Background Concept

In the formula $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ , the integer $y$ tells you how many moles of water are attached to each mole of anhydrous salt. Once you know the amounts of anhydrous residue and water in moles, the ratio water-to-salt directly gives $y$ .

Understanding

The question asks you to find the value of $y$ using the amounts you calculated in part (b)(i). Since those are used, this is a data-analysis step: take the ratio of $n(\text{H}_2\text{O})$ to $n(\text{ZnSO}_4)$ .

Approach

Use the number of moles you already worked out. The formula ratio is the mole ratio, so:

$y = \frac{n(\text{H}_2\text{O})}{n(\text{ZnSO}_4)}$

You should not recalculate the masses unless you want to use the exact mole values with more decimal places. In practice, rounding the moles value and then dividing is usually accurate enough; using exact fractions gives the cleanest result.

Step-by-Step Reasoning

From part (b)(i), $n(\text{H}_2\text{O}) = 0.0778\ \text{mol}$ and $n(\text{ZnSO}_4)=0.0111\ \text{mol}$ .
Divide:

$y = \frac{0.0778}{0.0111} = 7.0$

Since the empirical formula must involve whole numbers of water molecules, $y$ is the nearest integer, $7$ .
Hence the hydrated salt is $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ (zinc(II) sulfate heptahydrate).

The sample result in two decimal fractions gave 7.0 exactly. If your mole ratio comes out around 6.9 or 7.2, the experiment error must be a random) rounding the nearest integer is the expected final step.

Key Takeaways

The formula unit requires integer moles of water in the crystal hydrate.
Use the mole ratio directly, not the mass ratio.
Give $y$ as an integer, after rounding - do not leave it as a decimal wherever it must be whole.

Common Mistakes

Leaving $y$ as 6.98 or 7.0 rather than writing 7.
Dividing $n(\text{ZnSO}_4)$ by $n(\text{H}_2\text{O})$ instead of $n(\text{H}_2\text{O})$ by $n(\text{ZnSO}_4)$ .
Using masses directly instead of moles.
Rounding too early and getting e.g. 8 instead of 7.

Things to Be Careful About

No need to provide an entire formula reaction; just show the ratio calculation. Keep the final numerator to 3 decimal figures when writing the division to make the result clearer. The value of $y$ has no unit. If your empirical ratio is not a clean integer, this is expected; round to the nearest integer.

Techniques used

divide the amount of water by the amount of zinc sulfate to obtain water-to-salt mole ratioround the ratio to the nearest integer

(c)

A student suggests using this thermal decomposition method to investigate the number of moles of water of crystallisation in hydrated ethanedioic acid. The teacher says that this method is unsuitable.

Suggest why this method is unsuitable.

1M Similar

DifficultyEasy

Worked solution

Answer

The method is unsuitable because ethanedioic (oxalic) acid would decompose, or catch fire, when heated. The mass loss on heating would therefore not only be due to loss of water of crystallisation, so the result would be unreliable among the heating decomposition.

Final answer

The acid itself decomposes / is flammable when heated, so the loss in mass is not solely due to water of crystallisation.

Detailed explanation

Background Concept

For a thermally stable salt like $\text{ZnSO}_4$ , heating removes only water of water crystallisation. The salt itself molten zinc sulfate, so mass change is a faithful measure of water lost. For organic acids such as ethanedioic acid, heating is different. Ethanodioic acid (oxalic acid) decomposes at relatively low temperatures, and it can also be flammable of its decomposition to $\text{CO}$ , $\text{CO}_2$ and water. Such a decomposition changes the mass of the remaining solid, not merely through loss of water.

Understanding the Question

This part asks you to explain why the same thermal method proposed for zinc sulfate can't be used for hydrated ethanedethaned. The question is not about calculation; it wants a single chemical reason that the mass of the acid itself would change on heating. The phrase "method is unsuitable" is a red flag to find a limitation.

ApproachData

The key is: what happens to ethanedioic acid before or during the removal of water? It decomposes. That means the solid cannot be assumed to be just the anhydrous acid when the mass stabilises, so the loss in mass is a mixture of water loss and decomposition. Consequently, you cannot use the mass-loss value as the water of crystallisation.

Step-by-Step Reasoning

The proposed thermal method assumes only water leaves the sample.
For ethanedioic acid, heating causes decomposition/oxidation (with risk of flaming) of the acid itself.
Therefore the residue is not simply the anhydrous ethanedioic acid, and the observed loss in mass is not entirely due to water.
Because you are measuring y from water mass, the result would be invalid.

The mark scheme accepts either "the acid would decompose" or "the acid is flammable"." You only need one clear idea. It is not a heating temperature issue alone.

KeyInsights

Thermal analysis of hydration water is only valid when the anhydrous form does not itself decompose at the heating temperature.
A mass-loss method works only if the only change is water removal.
Flammability of organic compounds is a real limitation for methods that use strong heating.

Common Mistakes

Writing "ethanedioic acid is acidic" is irrelevant.
Saying that it is soluble or needs a higher temperature is not the core reason.
Saying that it will contain water in its crystallization but not chemically combine is insufficient.

Things to Be Careful About

Do not propose a different method unless asked. Do not say "it has water in its hydrous form" because that misses the point. The mark is for the decomposition (or flammability) of ethanedioic acid when heated.

Techniques used

identify an organic acid that decomposes before it can safely lose only waterexplain why additional mass loss makes the method invalid

Q3Medium-HardQualitative AnalysisManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

Qualitative analysis

For each test you should record all your observations in the spaces provided.

Examples of observations include:

colour changes seen
the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added
the formation of any gas and its identification (where appropriate) by a suitable test.

You should record clearly at what stage in a test an observation is made.

Where no change is observed, you should write 'no change'.

Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given.

If any solution is warmed, a boiling tube must be used. If a solid is heated, a hard-glass test-tube must be used.

Rinse and reuse test-tubes and boiling tubes where possible.

No additional tests should be attempted.

(a)

A bottle labelled FA 5 is thought to contain hydrated zinc sulfate. It would therefore contain zinc ions and sulfate ions as well as water of crystallisation.

(i)

Devise and carry out tests to investigate whether zinc ions, sulfate ions and water of crystallisation are present.

Record the tests you carry out and the observations you see in the space provided.

5M Similar

DifficultyMedium

Worked solution

Answer

Test for water of crystallisation:

Heat a sample of solid FA 5 in a hard-glass test-tube.
Observation: condensation / droplets / steam forms on the cooler parts of the tube.

Test for cation and anion:

Make an aqueous solution of FA 5 by dissolving some solid in distilled water.
Test for cation: To a portion of the solution, add aqueous ammonia dropwise, then add excess.
- Observation: a white precipitate forms, which is insoluble in excess aqueous ammonia.
- (This indicates Mg²⁺, not Zn²⁺, as zinc hydroxide is soluble in excess ammonia.)
Test for sulfate: To a second portion of the solution, add aqueous barium chloride (or barium nitrate).
- Observation: a white precipitate forms.
Confirm sulfate (distinguish from sulfite): Add dilute hydrochloric acid (or nitric acid) to the white precipitate.
- Observation: the precipitate is insoluble.
- (Alternatively, add aqueous potassium manganate(VII) to the solution: the purple colour remains, confirming no sulfite is present.)

Final answer

See working for test procedures and observations.

Detailed explanation

Background Concept

Qualitative analysis in Paper 3 requires the candidate to design and execute tests based on the suspected identity of the sample, but remain open to the possibility that the sample is not what is labelled. Water of crystallisation is detected by heating the solid and observing condensation. Cations are identified using aqueous ammonia or sodium hydroxide; the solubility of the precipitate in excess reagent is a key distinguishing feature (e.g., Zn²⁺ gives a white ppt soluble in excess NH₃, whereas Mg²⁺ gives a white ppt insoluble in excess NH₃). Sulfate ions are identified by adding barium chloride (or nitrate) in acidic conditions to give a white precipitate of barium sulfate; this distinguishes sulfate from sulfite, which would dissolve in acid and decolourise potassium manganate(VII).

Understanding the Question

The candidate is given a bottle labelled FA 5, suspected to be hydrated zinc sulfate (ZnSO₄·7H₂O). They must devise and carry out tests to check for Zn²⁺, SO₄²⁻, and H₂O. The mark scheme reveals FA 5 is actually MgSO₄·7H₂O. The candidate must record accurate observations and use them to conclude that Zn²⁺ is absent.

Approach

Water of crystallisation: Heat the solid directly and look for condensation.
Prepare test solutions: Dissolve the solid in water to test for ions.
Cation test: Use aqueous ammonia. A white precipitate insoluble in excess rules out Zn²⁺ and points to Mg²⁺.
Anion test: Use barium chloride followed by dilute acid to confirm SO₄²⁻ and rule out SO₃²⁻.

Step-by-Step Reasoning

M1: Heating the solid FA 5 will drive off water of crystallisation. The observation is condensation, droplets, or steam on the cooler parts of the test-tube.
M2: To test for ions, the solid must be dissolved in distilled water to form an aqueous solution.
M3: Adding aqueous ammonia to a portion of the solution produces a white precipitate. Because it is insoluble in excess ammonia, the cation is Mg²⁺, not Zn²⁺ (which would form a soluble complex [Zn(NH₃)₄]²⁺).
M4: Adding aqueous barium chloride (BaCl₂) or barium nitrate (Ba(NO₃)₂) to another portion produces a white precipitate of BaSO₄, indicating sulfate ions.
M5: To ensure the white precipitate is sulfate and not sulfite (which would also form a white ppt with Ba²⁺ but dissolve in acid), add dilute HCl or HNO₃. The precipitate remains insoluble. Alternatively, adding KMnO₄(aq) to the original solution should leave the purple colour unchanged, confirming no sulfite is present.

Key Takeaways

Always test for water of crystallisation by heating the solid first.
The solubility of precipitates in excess reagent (especially NH₃) is critical for distinguishing between similar cations like Mg²⁺ and Zn²⁺.
Sulfate confirmation requires acidification to rule out sulfite and carbonate interference.

Common Mistakes

Writing 'zinc ions present' without justification, or missing the fact that the white ppt is insoluble in excess ammonia.
Forgetting to acidify the barium chloride test, which could lead to a false positive from carbonate or sulfite.
Not recording the observation of condensation/steam when heating the solid.

Things to Be Careful About

The question says 'thought to contain', so the candidate must let the observations dictate the conclusion, not the label.
State symbols are not strictly required for qualitative observations, but the reagents must be named correctly (e.g., 'aqueous ammonia', not just 'ammonia').
'No change' must be written where appropriate; vague answers like 'nothing happens' may not score.

Techniques used

heat solid to detect water of crystallisationdissolve sample to prepare test solutionsuse ammonia to identify cation and distinguish from zincuse barium chloride to test for sulfateuse acid and potassium manganate(VII) to distinguish sulfate from sulfite

(ii)

Use your observations in (a)(i) to complete Table 3.1 to show whether each species is present in FA 5.

Use a tick (✓) if the species is present.
Use a cross (✗) if the species is not present.

Table 3.1

Species	Presence
$\text{Zn}^{2+}$
$\text{SO}_4^{2-}$
$\text{H}_2\text{O}$

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Species	Presence
Zn²⁺	✗
SO₄²⁻	✓
H₂O	✓

Final answer

Zn²⁺: ✗, SO₄²⁻: ✓, H₂O: ✓

Detailed explanation

Background Concept

In Paper 3, candidates must use their recorded observations to draw valid conclusions. If a test expected for a specific ion is negative (or indicates a different ion), that ion is absent. The presence of water of crystallisation is confirmed by condensation on heating. Sulfate is confirmed by the acid-insoluble barium sulfate precipitate.

Understanding the Question

Using the observations from (a)(i), the candidate must complete a table indicating whether Zn²⁺, SO₄²⁻, and H₂O are present in FA 5.

Approach

Zn²⁺: The ammonia test gave a white ppt insoluble in excess. Zn(OH)₂ is soluble in excess NH₃. Therefore, Zn²⁺ is absent (✗).
SO₄²⁻: The barium chloride test gave a white ppt insoluble in acid. Therefore, SO₄²⁻ is present (✓).
H₂O: Condensation was observed on heating. Therefore, H₂O is present (✓).

Step-by-Step Reasoning

Zn²⁺: The observation of a white precipitate insoluble in excess aqueous ammonia is characteristic of Mg²⁺, not Zn²⁺. Thus, Zn²⁺ is not present.
SO₄²⁻: The formation of a white precipitate with barium chloride that is insoluble in dilute acid confirms the presence of sulfate ions.
H₂O: The condensation of water droplets on the cooler parts of the test-tube during heating confirms the presence of water of crystallisation.

Key Takeaways

Conclusions must directly follow from the recorded observations.
A negative result for the suspected ion (with a positive result for an alternative) means the suspected ion is absent.

Common Mistakes

Assuming the label is correct and writing '✓' for Zn²⁺ despite the ammonia test indicating Mg²⁺.
Forgetting to mark H₂O as present because it is 'obvious' from the heating test.

Things to Be Careful About

Use the exact symbols requested: ✓ for present, ✗ for not present.

Techniques used

interpret qualitative test resultsdraw logical conclusions about sample composition

(b)

You are provided with solid FA 6.

(i)

Heat a few crystals of FA 6 in a hard-glass test-tube until no further gas is evolved. Record all your observations.

Leave the test-tube until it is cool.

Keep the cooled residue for use in (b)(ii).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Observations on heating FA 6:

The solid is purple / dark purple.
The solid jumps / moves around / fizzes in the test-tube.
A black residue is left after heating.
A gas is evolved that relights a glowing splint (or makes it glow more brightly).
(The gas is oxygen.)

Final answer

Purple solid jumps, leaves black residue, gas relights glowing splint.

Detailed explanation

Background Concept

Potassium permanganate(VII), KMnO₄, is a purple solid that decomposes on heating to give potassium manganate(VI) (K₂MnO₄, green), manganese(IV) oxide (MnO₂, black), and oxygen gas. The oxygen is identified by relighting a glowing splint. The 'jumping' or 'fizzing' is due to the rapid evolution of oxygen gas from the solid.

Understanding the Question

The candidate heats solid FA 6 (which is KMnO₄) and must record all observations, including the initial colour, physical changes during heating, the colour of the residue, and a test for the evolved gas.

Approach

Note the initial colour of the solid (purple).
Describe the physical behaviour during heating (jumping/moving).
Note the colour of the final residue (black MnO₂).
Test the gas with a glowing splint (it relights, confirming O₂).

Step-by-Step Reasoning

Initial colour: KMnO₄ is a deep purple solid.
During heating: The rapid release of oxygen gas causes the solid to jump or move around in the tube.
Residue: MnO₂ is a black solid, so a black residue is left (along with green K₂MnO₄, but the black is most visually prominent and specifically credited).
Gas test: Oxygen supports combustion and relights a glowing splint.

Key Takeaways

Thermal decomposition of KMnO₄ produces O₂, a black solid (MnO₂), and a green soluble salt (K₂MnO₄).
Always test evolved gases with appropriate tests (glowing splint for O₂).

Common Mistakes

Describing the residue as 'colourless' or 'white'.
Forgetting to test the gas or misidentifying it (e.g., saying it extinguishes a flame).
Not mentioning the initial purple colour.

Things to Be Careful About

The question asks to 'record all your observations'. Include colour, physical changes, and gas tests.
Use a hard-glass test-tube for heating solids, as specified in the question instructions.

Techniques used

heat solid in hard-glass test-tubetest evolved gas with glowing splint

(ii)

To the cooled residue from (b)(i), add approximately $3\text{ cm}$ depth of distilled water and stir. Filter the solution formed into a test-tube.

The colour of the solution is .............................. .

1M Similar

(c)

You are provided with aqueous solutions FA 7 and FA 8 and with solid FA 9.

FA 7 is an aqueous solution of FA 6.
FA 7, FA 8 and FA 9 contain compounds which all have one metal that is the same but which may be in different oxidation states.

(i)

Carry out the following tests on FA 7, FA 8 and FA 9 and record your observations in Table 3.2. For each test use a $1\text{ cm}$ depth of a solution or a spatula measure of solid.

Table 3.2

4M Similar

DifficultyMedium-Hard

Worked solution

Answer

Table 3.2 Observations:

Test	FA 7 (acidified KMnO₄)	FA 8 (MnSO₄)	FA 9 (MnO₂)
Test 1: Add H₂O₂	Bubbles / effervescence; purple solution turns colourless (or yellow); gas relights glowing splint	No change	Bubbles / effervescence; gas relights glowing splint
Test 2: Add NaOH, leave to stand	(Not required / crossed out)	Off-white precipitate forms; turns brown on standing	No change
Test 3: Add aqueous FeSO₄	Solution turns colourless / yellow	No change	(Not required / crossed out)

Final answer

See table above for observations.

Detailed explanation

Background Concept

This part tests the chemistry of manganese in different oxidation states:

FA 7: Acidified KMnO₄ contains Mn(VII) (MnO₄⁻), a strong oxidising agent (purple).
FA 8: MnSO₄ contains Mn(II) (Mn²⁺), which forms a pale pink/off-white precipitate with NaOH that is rapidly oxidised by air to brown MnO(OH) or MnO₂.
FA 9: MnO₂ contains Mn(IV), which is insoluble and acts as a catalyst for the decomposition of H₂O₂.

Reactions:

H₂O₂ with MnO₄⁻: MnO₄⁻ is reduced to Mn²⁺ (colourless), and H₂O₂ is oxidised to O₂ (bubbles).
H₂O₂ with MnO₂: MnO₂ catalyses the decomposition of H₂O₂ to O₂ and H₂O.
NaOH with Mn²⁺: Mn²⁺ + 2OH⁻ → Mn(OH)₂ (off-white ppt). On standing, 2Mn(OH)₂ + O₂ → 2MnO(OH)₂ (brown).
FeSO₄ with MnO₄⁻: MnO₄⁻ oxidises Fe²⁺ to Fe³⁺ (yellow) and is reduced to Mn²⁺ (colourless), so the purple colour disappears and the solution may turn yellow due to Fe³⁺.

Understanding the Question

The candidate must perform three tests on three different manganese compounds and record the observations in a table. Some cells are crossed out in the question paper (Test 2 on FA 7, Test 3 on FA 9), meaning those tests are not required.

Approach

Test 1 (H₂O₂): FA 7 (Mn VII) oxidises H₂O₂ → O₂ bubbles, purple to colourless. FA 8 (Mn II) no reaction. FA 9 (Mn IV) catalyses H₂O₂ decomposition → O₂ bubbles.
Test 2 (NaOH): FA 8 (Mn II) gives off-white ppt turning brown. FA 9 (Mn IV) no change.
Test 3 (FeSO₄): FA 7 (Mn VII) oxidises Fe²⁺ → purple fades, solution may turn yellow (Fe³⁺). FA 8 (Mn II) no reaction.

Step-by-Step Reasoning

Test 1, FA 7: MnO₄⁻(aq) is a strong oxidising agent. It oxidises H₂O₂ to O₂ gas (bubbles/effervescence, relights glowing splint) and is itself reduced to Mn²⁺(aq), which is colourless. The purple colour of the solution disappears. (It may turn yellow briefly due to Fe³⁺ if impurities exist, but 'colourless' is the primary observation for MnO₄⁻ reduction in acid).
Test 1, FA 8: Mn²⁺ cannot oxidise H₂O₂, and H₂O₂ does not react with Mn²⁺ under these conditions. Observation: no change.
Test 1, FA 9: MnO₂ is a well-known catalyst for the decomposition of H₂O₂ into H₂O and O₂. Observation: bubbles/effervescence, gas relights glowing splint.
Test 2, FA 8: Adding NaOH to Mn²⁺ gives Mn(OH)₂, an off-white (or pale pink) precipitate. On standing in air, it is oxidised to brown manganese(III) oxide-hydroxide (MnO(OH)) or MnO₂.
Test 2, FA 9: MnO₂ is insoluble and does not react with NaOH. Observation: no change.
Test 3, FA 7: MnO₄⁻ oxidises Fe²⁺ to Fe³⁺. The purple MnO₄⁻ is reduced to colourless Mn²⁺. The solution turns from purple to colourless, and may appear yellow due to the formation of Fe³⁺(aq).
Test 3, FA 8: Mn²⁺ cannot oxidise Fe²⁺. Observation: no change.

Key Takeaways

Mn(VII) is a strong oxidising agent (purple, reduced to colourless Mn²⁺).
Mn(II) forms a characteristic off-white precipitate with NaOH that turns brown on standing due to oxidation by air.
Mn(IV) (MnO₂) is a catalyst for H₂O₂ decomposition.
Transition metals in high oxidation states can oxidise Fe²⁺ to Fe³⁺.

Common Mistakes

Saying FA 7 turns 'brown' with H₂O₂ (it turns colourless; brown is for MnO₂ formation in neutral conditions, but here it is acidified).
Forgetting to mention the gas test result (relights glowing splint) for Test 1.
Describing the Mn(OH)₂ precipitate as 'white' without noting it turns brown on standing.
Writing observations for crossed-out cells.

Things to Be Careful About

'No change' must be written explicitly where there is no reaction.
For Test 3, FA 7, the solution may turn yellow due to Fe³⁺; both 'colourless' and 'yellow' are acceptable as the purple disappears.
Ensure observations are specific: 'bubbles' is better than 'gas evolved'; 'relights glowing splint' is required for O₂.

Techniques used

use hydrogen peroxide to test for oxidising agents and catalystsuse sodium hydroxide to identify manganese(II) ionsuse iron(II) sulfate to test for oxidising agents

(ii)

Suggest the identity of the metal in FA 6/FA 7, FA 8 and FA 9.

The metal is .............................. .

1M Similar

(iii)

Complete Table 3.3 to suggest the oxidation state of the metal in FA 6/FA 7 and FA 8.

Table 3.3

	FA 6 / FA 7	FA 8
oxidation state

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

	FA 6 / FA 7	FA 8
oxidation state	+7	+2

Final answer

FA 6/FA 7: +7, FA 8: +2

Detailed explanation

Background Concept

FA 6 is KMnO₄ and FA 7 is acidified KMnO₄(aq). In MnO₄⁻, oxygen is -2, so Mn + 4(-2) = -1 → Mn = +7.
FA 8 is MnSO₄ (or MnCl₂). In Mn²⁺, the oxidation state is +2.
FA 9 is MnO₂. Mn + 2(-2) = 0 → Mn = +4.

Understanding the Question

The candidate must state the oxidation state of the metal (manganese) in FA 6/FA 7 and FA 8.

Approach

FA 6/FA 7 contains MnO₄⁻, where Mn is in the +7 oxidation state.
FA 8 contains Mn²⁺ (from MnSO₄), where Mn is in the +2 oxidation state.

Step-by-Step Reasoning

In KMnO₄, the manganate(VII) ion is MnO₄⁻. With O at -2, Mn must be +7 to give a net charge of -1.
In MnSO₄, the manganese ion is Mn²⁺, so the oxidation state is +2.

Key Takeaways

Manganate(VII) is MnO₄⁻ (Mn is +7).
Manganese(II) salts contain Mn²⁺ (Mn is +2).
Oxidation states can be deduced from the formula and known ion charges.

Common Mistakes

Writing '7' instead of '+7' (oxidation states must include the sign).
Confusing the oxidation state with the charge of the ion (though they are the same for monatomic ions, it is good practice to write +7).
Assigning +4 to FA 6/FA 7 (that is FA 9, MnO₂).

Things to Be Careful About

Always include the '+' or '-' sign for oxidation states.
The question asks for FA 6/FA 7 and FA 8 only; do not include FA 9 unless asked.

Techniques used

determine oxidation states from chemical behaviour