Chemistry 9701/52 — February/March 2025

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q1Medium-EasyPlanningAnalysis, Conclusions and Evaluation

Brass is an alloy of copper and zinc.
An experiment is completed to find the percentage by mass of copper in a sample of brass.

In the experiment, the sample of brass is reacted with an excess of concentrated nitric acid, $\text{HNO}_3$ . This forms a solution containing $\text{Cu}^{2+}\text{(aq)}$ ions. The amount of $\text{Cu}^{2+}\text{(aq)}$ ions formed is determined by titration.

A student uses the following method.

step 1 Weigh a glass beaker and record the mass.
step 2 Add approximately $1.00\text{ g}$ of powdered brass to the beaker and record the mass of the beaker and the brass.
step 3 Transfer the brass into conical flask A which contains excess concentrated $\text{HNO}_3$ .
step 4 Reweigh the glass beaker and record the mass.
step 5 Add aqueous sodium carbonate, $\text{Na}_2\text{CO}_3\text{(aq)}$ , dropwise to flask A until a precipitate of copper(II) carbonate, $\text{CuCO}_3\text{(s)}$ , appears. Then add dilute ethanoic acid dropwise until the precipitate is fully dissolved.
step 6 Transfer all the contents of flask A to a $100.0\text{ cm}^3$ volumetric flask and make up to the mark with distilled water. This is solution B.
step 7 Transfer $10.0\text{ cm}^3$ of solution B into conical flask C.
step 8 Add $10\text{ cm}^3$ , an excess, of aqueous potassium iodide, $\text{KI(aq)}$ , to flask C.
step 9 Titrate the contents of flask C against $0.0600\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ , using starch solution as an indicator.
step 10 Repeat steps 7 to 9 until concordant results are obtained.

(a)

Both copper and zinc in brass react with concentrated $\text{HNO}_3$ in step 3.

\begin{aligned} \text{reaction 1} &\quad \text{Zn(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow \text{Zn(NO}_3)_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \\ \text{reaction 2} &\quad \text{Cu(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \end{aligned}

Suggest why step 3 in the experiment should be completed in a fume cupboard.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{NO}_2$ is a toxic gas.

Final answer

NO2 is a toxic gas

Detailed explanation

Background Concept

Concentrated nitric acid is a powerful oxidising agent. When it reacts with metals such as copper and zinc, it is reduced to nitrogen dioxide, $\text{NO}_2$ , rather than producing hydrogen gas. $\text{NO}_2$ is a brown, toxic, pungent gas that is harmful if inhaled. Laboratory safety requires that any reaction producing a toxic gas be carried out in a fume cupboard, which extracts the gas away from the user and prevents it from entering the laboratory atmosphere.

Understanding the Question

Step 3 involves adding the brass (a mixture of copper and zinc) to excess concentrated $\text{HNO}_3$ . The two equations provided (reaction 1 for zinc and reaction 2 for copper) both show $\text{NO}_2\text{(g)}$ as a product. The question asks why this step should be done in a fume cupboard. The answer is that the reaction produces a toxic gas.

Approach

Look at the products of the reactions in step 3. Identify the gaseous product and state its hazard. The equations are given, so the gaseous product is clearly $\text{NO}_2\text{(g)}$ . The fume cupboard is needed because this gas is toxic.

Step-by-Step Reasoning

From reaction 1: $\text{Zn(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow \text{Zn(NO}_3)_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$

From reaction 2: $\text{Cu(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$

Both reactions produce $\text{NO}_2\text{(g)}$ . Nitrogen dioxide is a toxic gas — it is harmful if inhaled and can cause serious respiratory damage. A fume cupboard removes the gas from the laboratory atmosphere, protecting the person carrying out the experiment. This is the single marking point: $\text{NO}_2$ is a toxic gas.

Key Takeaways

When a reaction produces a toxic, harmful, or irritating gas, it must be carried out in a fume cupboard. Recognising hazardous products from a balanced equation is a core safety skill in chemistry.

Common Mistakes

Saying "the acid is dangerous" — while true, the specific point is the gas produced by the reaction.
Saying "it produces a gas" without specifying that the gas is toxic.

Things to Be Careful About

The mark requires the word "toxic" (or an equivalent such as "poisonous" or "harmful"). Simply saying "it produces $\text{NO}_2$ " without stating the hazard may not score the mark.

Techniques used

identify a toxic gas hazardrelate a reaction product to a safety requirement

(b)

Explain why the glass beaker is reweighed in step 4.

1M Similar

DifficultyEasy

Worked solution

Answer

To determine the accurate mass of brass transferred to the conical flask by difference: mass of brass = (mass of beaker + brass) − (mass of beaker after transfer).

Final answer

To determine the mass of brass transferred by difference

Detailed explanation

Background Concept

Weighing by difference is a standard analytical technique. Instead of trying to weigh a powder directly (which is difficult because powder can be lost or stick to surfaces), you weigh the container before and after transferring the powder, and subtract. This gives the exact mass of material actually transferred.

Understanding the Question

In steps 1–4, the student weighs the beaker, then the beaker with brass, then the beaker after transferring the brass to the conical flask. Step 4 asks why the beaker is reweighed. The answer is that the difference between the second and third weighings gives the exact mass of brass that was actually transferred.

Approach

Recognise that the mass of brass added is not simply "about 1.00 g" — it must be determined precisely. The reweighing in step 4 allows the mass of brass transferred to be calculated by difference.

Step-by-Step Reasoning

Mass of beaker + brass = 26.65 g (from step 2).
Mass of beaker after transfer = 25.65 g (from step 4).
Mass of brass transferred = 26.65 − 25.65 = 1.00 g.

The reweighing in step 4 is essential because not all the brass may be transferred — some powder may remain stuck to the beaker. Weighing by difference accounts for this and gives the accurate mass of brass that actually reacted.

Key Takeaways

Weighing by difference is a fundamental technique in quantitative chemistry. It eliminates errors from powder loss during transfer and gives the exact mass of material used.

Common Mistakes

Saying "to check the mass" without explaining that it is to find the mass of brass by difference.
Confusing the purpose: it is not to check the beaker's mass but to determine how much brass was transferred.

Things to Be Careful About

The mark scheme requires the idea of "by difference" — the mass of brass transferred is the difference between the two weighings. Make this explicit in your answer.

Techniques used

use weighing by differenceexplain the purpose of a procedural step

(c)

Name a suitable piece of apparatus to transfer the $10.0\text{ cm}^3$ of solution B in step 7.

1M Similar

DifficultyEasy

Worked solution

Answer

A 10.0 $\text{cm}^3$ volumetric pipette.

Final answer

A 10.0 cm3 volumetric pipette

Detailed explanation

Background Concept

A volumetric pipette is designed to deliver a single, precise, fixed volume of liquid (e.g. 10.0 $\text{cm}^3$ , 25.0 $\text{cm}^3$ ). It is the correct apparatus for transferring an exact aliquot of a solution. A measuring cylinder is less precise, and a burette is used for variable volumes in titrations.

Understanding the Question

Step 7 requires transferring exactly 10.0 $\text{cm}^3$ of solution B into conical flask C. The question asks for a suitable piece of apparatus. The answer is a volumetric pipette calibrated to 10.0 $\text{cm}^3$ .

Approach

Recall the apparatus used for transferring a precise fixed volume: a volumetric pipette. It is the standard choice for taking an aliquot in a titration procedure.

Step-by-Step Reasoning

The experiment requires an exact 10.0 $\text{cm}^3$ aliquot of solution B. A volumetric pipette is manufactured to deliver a precise volume (here 10.0 $\text{cm}^3$ ) with high accuracy. A measuring cylinder would only give an approximate volume, and a burette is not designed for transferring a single fixed volume quickly. Therefore, a 10.0 $\text{cm}^3$ volumetric pipette is the correct choice.

Key Takeaways

Volumetric pipettes are used for transferring exact fixed volumes (aliquots). They are essential in titration procedures for ensuring accurate and reproducible results.

Common Mistakes

Saying "measuring cylinder" — this is not precise enough for a titration aliquot.
Saying "burette" — a burette is for delivering variable volumes during a titration, not for transferring a fixed aliquot.
Forgetting the capacity: "a volumetric pipette" alone may not score; the size (10.0 $\text{cm}^3$ ) should be stated.

Things to Be Careful About

The mark scheme specifies "(10 $\text{cm}^3$ ) volumetric pipette". Both the type of apparatus and its capacity are required.

Techniques used

select volumetric apparatus for a fixed volume

(d)

Identify the substance used to rinse the burette before step 9 is done for the first time.

1M Similar

DifficultyEasy

Worked solution

Answer

$0.0600\text{ mol dm}^{-3}$ aqueous sodium thiosulfate (the solution to be used in the titration).

Final answer

0.0600 mol dm-3 aqueous sodium thiosulfate

Detailed explanation

Background Concept

A burette must be rinsed with the solution that will be placed in it before filling. This prevents the solution from being diluted by residual water (or contaminated by a different solution) left from a previous rinse or wash. Rinsing with the titrant itself ensures the concentration of the solution in the burette is exactly as intended.

Understanding the Question

Step 9 involves titrating against $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ . The question asks what substance should be used to rinse the burette before the first titration. The answer is the same solution that will be used in the burette.

Approach

Recall the rule: rinse the burette with the solution it will contain. Here, that is $0.0600\text{ mol dm}^{-3}$ aqueous sodium thiosulfate.

Step-by-Step Reasoning

If the burette is rinsed with distilled water, a thin film of water would remain on the walls and dilute the thiosulfate solution when it is added, changing its effective concentration. Rinsing with the $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ solution ensures that the concentration of the solution in the burette is exactly $0.0600\text{ mol dm}^{-3}$ .

Key Takeaways

Always rinse a burette with the solution it will contain. This is a standard technique in volumetric analysis.

Common Mistakes

Saying "distilled water" — this would dilute the titrant and give incorrect results.
Saying "deionised water" — same problem.
Saying "the solution being titrated" (i.e. the contents of the conical flask) — the burette must be rinsed with what goes IN the burette, not what is in the flask.

Things to Be Careful About

The answer must specify the concentration ( $0.0600\text{ mol dm}^{-3}$ ) and the substance (aqueous sodium thiosulfate). The mark scheme accepts " $0.0600\text{ mol dm}^{-3}$ aqueous sodium thiosulfate".

Techniques used

identify the rinsing solution for a burette

(e)

In step 9, $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ is used.
This solution is prepared from $0.200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ before the experiment.

Describe how to make a $100.0\text{ cm}^3$ standard solution of $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ from the $0.200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ solution.

Give the name and capacity of any apparatus you would use.

Write your answer in a series of numbered steps.

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Volume of $0.200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ needed:
$c_1V_1 = c_2V_2 \Rightarrow V_1 = \frac{0.0600 \times 100.0}{0.200} = 30.00\text{ cm}^3$

Answer

Using a burette, transfer $30.00\text{ cm}^3$ of $0.200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ into a $100.0\text{ cm}^3$ volumetric flask.
Add distilled water up to the $100.0\text{ cm}^3$ graduation mark on the volumetric flask.
Stopper the flask and invert it several times to mix the solution thoroughly.

Final answer

Dilute 30.00 cm3 of 0.200 mol dm-3 Na2S2O3 to 100.0 cm3 with distilled water in a volumetric flask

Detailed explanation

Background Concept

A standard solution is one of accurately known concentration. When a more concentrated solution is available, a dilution is performed: a measured volume of the concentrated solution is transferred to a volumetric flask and made up to the mark with solvent (here distilled water). The dilution equation $c_1V_1 = c_2V_2$ relates the initial and final concentrations and volumes.

Understanding the Question

The student needs $100.0\text{ cm}^3$ of $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ from a stock solution of $0.200\text{ mol dm}^{-3}$ . The question asks for a series of numbered steps describing how to do this, naming the apparatus and its capacity.

Approach

Use the dilution equation to find the volume of stock solution needed.
Describe the practical steps: measure the stock solution with a burette, transfer to a volumetric flask, make up to the mark with distilled water, and mix.

Step-by-Step Reasoning

Step 1 — Calculate the volume needed.
Using $c_1V_1 = c_2V_2$ :
$0.200 \times V_1 = 0.0600 \times 100.0$
$V_1 = \frac{0.0600 \times 100.0}{0.200} = 30.00\text{ cm}^3$

Step 2 — Transfer the stock solution.
A burette is used because it can deliver $30.00\text{ cm}^3$ accurately. The $30.00\text{ cm}^3$ of $0.200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ is run into a $100.0\text{ cm}^3$ volumetric flask.

Step 3 — Make up to the mark.
Distilled water is added until the bottom of the meniscus sits exactly on the $100.0\text{ cm}^3$ graduation mark. The flask is stoppered and inverted several times to ensure the solution is homogeneous.

The three marks correspond to: (M1) the volume calculation = $30.00\text{ cm}^3$ ; (M2) using a burette to transfer it to a volumetric flask; (M3) making up to the mark with distilled water.

Key Takeaways

The dilution equation $c_1V_1 = c_2V_2$ is used to find the volume of stock solution required.
A burette delivers accurate volumes; a volumetric flask makes up a precise final volume.
Thorough mixing (inverting the flask) is essential for a homogeneous solution.

Common Mistakes

Using a measuring cylinder instead of a burette for the $30.00\text{ cm}^3$ — not accurate enough.
Forgetting to name the volumetric flask as $100.0\text{ cm}^3$ .
Saying "add water to make up to 100 cm³" without specifying "up to the mark" or using distilled water.
Not showing the calculation for the volume of stock solution.

Things to Be Careful About

The mark scheme gives three distinct marks: the calculated volume ( $30.00\text{ cm}^3$ ), the use of a burette to transfer it to a volumetric flask, and making up to the mark with distilled water. All three must appear in the answer. The question asks for numbered steps, so present them as such.

Techniques used

calculate dilution volumedescribe preparation of a standard solution by dilution

(f)

The measurements collected during steps 1 to 4 are shown in Table 1.1.

Table 1.1


mass of glass beaker/g	25.55
mass of glass beaker containing powdered brass/g	26.65
mass of glass beaker after transferring brass to conical flask A/g	25.65

Determine the mass of powdered brass added to conical flask A.

1M Similar

DifficultyEasy

Worked solution

Working

Mass of brass transferred = mass of (beaker + brass) − mass of beaker after transfer
$= 26.65 - 25.65 = 1.00\text{ g}$

Answer

$1.00\text{ g}$

Final answer

1.00 g

Detailed explanation

Background Concept

Weighing by difference: the mass of material transferred is obtained by subtracting the mass of the container after transfer from the mass of the container plus material before transfer. This accounts for any powder that may have remained in the beaker.

Understanding the Question

Table 1.1 gives three masses: the empty beaker (25.55 g), the beaker plus brass (26.65 g), and the beaker after transferring the brass (25.65 g). The question asks for the mass of brass added to conical flask A. This is the difference between the second and third masses.

Approach

Subtract the mass of the beaker after transfer from the mass of the beaker containing the brass. The empty beaker mass (25.55 g) is not needed for this calculation — it is a check that the beaker's mass is constant.

Step-by-Step Reasoning

Mass of beaker + brass = 26.65 g
Mass of beaker after transfer = 25.65 g
Mass of brass transferred = 26.65 − 25.65 = 1.00 g

Note: the mass of the empty beaker (25.55 g) is not used in this subtraction. It was recorded in step 1 as a check, but the difference between steps 2 and 4 directly gives the mass of brass transferred.

Key Takeaways

Weighing by difference is a reliable way to determine the mass of a transferred solid. The empty container mass is often recorded as a control but may not be needed in the final calculation.

Common Mistakes

Subtracting the wrong values, e.g. 26.65 − 25.55 = 1.10 g (this includes the empty beaker, which is wrong).
Forgetting units or giving the answer to too many decimal places.

Things to Be Careful About

The correct subtraction is 26.65 − 25.65 = 1.00 g. The answer should be given to two decimal places consistent with the data.

Techniques used

calculate mass by difference

(g)

The volumes measured in each of the titrations are shown in Table 1.2.

Table 1.2

	rough titration	titration 1	titration 2	titration 3
final burette reading/ $\text{cm}^3$	24.05	24.80	45.35	22.50
initial burette reading/ $\text{cm}^3$	3.25	4.50	24.80	2.10
titre/ $\text{cm}^3$

(i)

Complete Table 1.2.

1M Similar

DifficultyEasy

Worked solution

Answer

	rough	1	2	3
titre/ $\text{cm}^3$	20.80	20.30	20.55	20.40

Working: titre = final reading − initial reading.

rough: $24.05 - 3.25 = 20.80$
1: $24.80 - 4.50 = 20.30$
2: $45.35 - 24.80 = 20.55$
3: $22.50 - 2.10 = 20.40$

Final answer

Titres: rough 20.80, 1: 20.30, 2: 20.55, 3: 20.40 cm3

Detailed explanation

Background Concept

In a titration, the titre is the volume of titrant delivered from the burette. It is calculated as the difference between the final and initial burette readings: titre = final reading − initial reading.

Understanding the Question

Table 1.2 gives final and initial burette readings for four titrations (rough, 1, 2, 3). The task is to complete the table by calculating each titre.

Approach

For each column, subtract the initial reading from the final reading.

Step-by-Step Reasoning

Rough: $24.05 - 3.25 = 20.80\text{ cm}^3$
Titration 1: $24.80 - 4.50 = 20.30\text{ cm}^3$
Titration 2: $45.35 - 24.80 = 20.55\text{ cm}^3$
Titration 3: $22.50 - 2.10 = 20.40\text{ cm}^3$

Each titre is recorded to two decimal places, consistent with the burette readings.

Key Takeaways

The titre is always final reading minus initial reading. Burette readings are recorded to two decimal places (typically to the nearest $0.05\text{ cm}^3$ ).

Common Mistakes

Subtracting in the wrong order (initial − final), which gives a negative value.
Forgetting to include units.
Recording the titre to the wrong number of decimal places.

Things to Be Careful About

All titres should be positive values. Check that each subtraction is final − initial. The rough titration is included in the table and must also be completed.

Techniques used

calculate titre from burette readings

(ii)

Calculate a suitable mean titre to use in the calculations.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Concordant titres: 20.30 and 20.40 (within $0.10\text{ cm}^3$ of each other).

Mean titre $= \frac{20.30 + 20.40}{2} = 20.35\text{ cm}^3$

Answer

$20.35\text{ cm}^3$

Final answer

20.35 cm3

Detailed explanation

Background Concept

Concordant titres are those that agree within $0.10\text{ cm}^3$ of each other. When calculating a mean titre, only concordant results are averaged — the rough titration and any outliers are excluded. This improves the reliability of the result.

Understanding the Question

From part (g)(i), the titres are: rough 20.80, 1: 20.30, 2: 20.55, 3: 20.40. The question asks for a suitable mean titre to use in the calculations.

Approach

Identify which titres are concordant (agree within $0.10\text{ cm}^3$ ), then average only those values.

Step-by-Step Reasoning

Looking at the titres:

20.80 (rough) — this is a rough estimate, never used in the mean.
20.30 and 20.40 — these agree within $0.10\text{ cm}^3$ , so they are concordant.
20.55 — this differs from 20.30 by $0.25\text{ cm}^3$ and from 20.40 by $0.15\text{ cm}^3$ , so it is not concordant with the others.

Mean titre $= \frac{20.30 + 20.40}{2} = 20.35\text{ cm}^3$

Key Takeaways

Only concordant titres are averaged. The rough titration is never included in the mean. Outliers (titres that differ by more than $0.10\text{ cm}^3$ from the others) are also excluded.

Common Mistakes

Averaging all four titres including the rough: $\frac{20.80 + 20.30 + 20.55 + 20.40}{4} = 20.51$ — this is wrong.
Including the non-concordant 20.55 in the mean.
Averaging 20.30, 20.55 and 20.40 (three values) instead of just the two concordant ones.

Things to Be Careful About

The mark scheme specifically uses $(20.30 + 20.40) \div 2 = 20.35$ . Only these two values are concordant. The rough titration is never used in the mean.

Techniques used

select concordant titrescalculate a mean

(iii)

Calculate the percentage error in the titre in titration 3.

Show your working.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Each burette reading has an uncertainty of $\pm 0.05\text{ cm}^3$ . A titre involves two readings (final and initial), so the total uncertainty is $2 \times 0.05 = 0.10\text{ cm}^3$ .

Percentage error $= \frac{0.10}{20.40} \times 100 = 0.490\%$

Answer

$0.490\%$

Final answer

0.490%

Detailed explanation

Background Concept

Percentage error measures the uncertainty of a measurement relative to its magnitude. For a burette, each reading is accurate to $\pm 0.05\text{ cm}^3$ . Since a titre is the difference between two readings (final and initial), the maximum uncertainty in the titre is the sum of the uncertainties of the two readings: $2 \times 0.05 = 0.10\text{ cm}^3$ .

Understanding the Question

The question asks for the percentage error in the titre of titration 3, which is 20.40 $\text{cm}^3$ . Working must be shown.

Approach

Determine the absolute uncertainty in the titre: $2 \times 0.05 = 0.10\text{ cm}^3$ .
Divide by the titre value and multiply by 100 to get the percentage.

Step-by-Step Reasoning

Absolute uncertainty in the titre = $2 \times 0.05\text{ cm}^3 = 0.10\text{ cm}^3$ .

Titre in titration 3 = 20.40 $\text{cm}^3$ .

Percentage error:
$\frac{0.10}{20.40} \times 100 = 0.490\%$

The answer is given to three significant figures, consistent with the data.

Key Takeaways

The uncertainty in a titre is double the uncertainty of a single burette reading, because two readings (initial and final) are involved. Percentage error = (absolute uncertainty / measured value) × 100.

Common Mistakes

Using only one burette reading uncertainty ( $0.05$ instead of $0.10$ ).
Using the wrong titre value (e.g. using the mean 20.35 instead of 20.40).
Forgetting to multiply by 100 to get a percentage.
Not showing working — the mark scheme requires it.

Things to Be Careful About

The mark scheme formula is $\frac{2 \times 0.05}{20.40} \times 100 = 0.490$ . The "2 × 0.05" accounts for the two burette readings. The titre used is that of titration 3 (20.40), not the mean.

Techniques used

calculate percentage erroraccount for two burette readings

(h)

In step 8, $\text{Cu}^{2+}\text{(aq)}$ ions react with $\text{I}^-\text{(aq)}$ ions. The ionic equation for the reaction is shown.

\text{reaction 3} \quad 2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}

In step 9, $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ reacts with $\text{I}_2\text{(aq)}$ formed in reaction 3. The equation for the reaction is shown.

\text{reaction 4} \quad \text{I}_2\text{(aq)} + 2\text{Na}_2\text{S}_2\text{O}_3\text{(aq)} \rightarrow \text{Na}_2\text{S}_4\text{O}_6\text{(aq)} + 2\text{NaI(aq)}

(i)

Using a second sample of brass, another student determined the mean titre to be $17.35\text{ cm}^3$ of $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ .

Calculate the amount, in mol, of $\text{Na}_2\text{S}_2\text{O}_3$ in this student's mean titre.

1M Similar

DifficultyEasy

Worked solution

Working

Amount of $\text{Na}_2\text{S}_2\text{O}_3 = c \times V = 0.0600 \times \frac{17.35}{1000}$
$= 1.041 \times 10^{-3}\text{ mol}$

Answer

$1.04 \times 10^{-3}\text{ mol}$ (or $0.00104\text{ mol}$ )

Final answer

1.04 × 10^-3 mol

Detailed explanation

Background Concept

The amount of substance (in moles) in a solution is given by $n = cV$ , where $c$ is the concentration in $\text{mol dm}^{-3}$ and $V$ is the volume in $\text{dm}^3$ . Since volumes are often given in $\text{cm}^3$ , they must be converted by dividing by 1000.

Understanding the Question

The mean titre is $17.35\text{ cm}^3$ of $0.0600\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$ . The question asks for the amount, in mol, of $\text{Na}_2\text{S}_2\text{O}_3$ in this titre.

Approach

Use $n = cV$ . Convert the volume from $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000, then multiply by the concentration.

Step-by-Step Reasoning

$n = 0.0600\text{ mol dm}^{-3} \times \frac{17.35}{1000}\text{ dm}^3$
$n = 0.0600 \times 0.01735 = 1.041 \times 10^{-3}\text{ mol}$

This is the amount of sodium thiosulfate that reacted with the iodine in the titration.

Key Takeaways

The formula $n = cV$ is fundamental in volumetric analysis. Always convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000.

Common Mistakes

Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ (dividing by 1000).
Using the wrong concentration or volume.
Giving the answer to too many or too few significant figures.

Things to Be Careful About

The mark scheme gives $1.04 \times 10^{-3}$ / $0.00104$ . The answer should be given to three significant figures (consistent with the data).

Techniques used

calculate moles from concentration and volume

(ii)

Use the equation for reaction 4 and your answer to (h)(i) to determine the amount, in mol, of $\text{I}_2$ that reacted with the $\text{Na}_2\text{S}_2\text{O}_3$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From reaction 4: $\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}$

Mole ratio $\text{I}_2 : \text{Na}_2\text{S}_2\text{O}_3 = 1 : 2$

Amount of $\text{I}_2 = \frac{1.041 \times 10^{-3}}{2} = 5.21 \times 10^{-4}\text{ mol}$

Answer

$5.21 \times 10^{-4}\text{ mol}$

Final answer

5.21 × 10^-4 mol

Detailed explanation

Background Concept

Stoichiometry uses the coefficients in a balanced equation to relate the amounts of reactants and products. In reaction 4, one mole of $\text{I}_2$ reacts with two moles of $\text{Na}_2\text{S}_2\text{O}_3$ . Therefore, the amount of $\text{I}_2$ is half the amount of $\text{Na}_2\text{S}_2\text{O}_3$ .

Understanding the Question

From part (h)(i), the amount of $\text{Na}_2\text{S}_2\text{O}_3$ is $1.041 \times 10^{-3}$ mol. The question asks for the amount of $\text{I}_2$ that reacted, using the stoichiometry of reaction 4.

Approach

Identify the mole ratio between $\text{I}_2$ and $\text{Na}_2\text{S}_2\text{O}_3$ from the balanced equation, then divide the amount of $\text{Na}_2\text{S}_2\text{O}_3$ by 2.

Step-by-Step Reasoning

Reaction 4: $\text{I}_2\text{(aq)} + 2\text{Na}_2\text{S}_2\text{O}_3\text{(aq)} \rightarrow \text{Na}_2\text{S}_4\text{O}_6\text{(aq)} + 2\text{NaI(aq)}$

The coefficient of $\text{I}_2$ is 1 and the coefficient of $\text{Na}_2\text{S}_2\text{O}_3$ is 2. So:
$n(\text{I}_2) = \frac{n(\text{Na}_2\text{S}_2\text{O}_3)}{2} = \frac{1.041 \times 10^{-3}}{2} = 5.205 \times 10^{-4} \approx 5.21 \times 10^{-4}\text{ mol}$

Key Takeaways

Balanced equations give the mole ratios needed to convert between amounts of different species. Always check the coefficients carefully.

Common Mistakes

Using the wrong ratio (e.g. multiplying by 2 instead of dividing by 2).
Not carrying forward the value from part (h)(i) correctly.
Rounding too early in the calculation.

Things to Be Careful About

The mark scheme answer is $5.21 \times 10^{-4}$ / $0.000521$ . Error carried forward (ecf) applies: if part (h)(i) was wrong but the ratio is applied correctly, this mark can still be earned.

Techniques used

use stoichiometric ratio from a balanced equation

(iii)

Use the equation for reaction 3 and your answer to (h)(ii) to determine the amount, in mol, of $\text{Cu}^{2+}$ in $10.0\text{ cm}^3$ of their solution B.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From reaction 3: $2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} + \text{I}_2$

Mole ratio $\text{Cu}^{2+} : \text{I}_2 = 2 : 1$

Amount of $\text{Cu}^{2+} = 2 \times 5.21 \times 10^{-4} = 1.04 \times 10^{-3}\text{ mol}$ (in $10.0\text{ cm}^3$ of solution B)

Answer

$1.04 \times 10^{-3}\text{ mol}$

Final answer

1.04 × 10^-3 mol

Detailed explanation

Background Concept

Reaction 3 shows that two moles of $\text{Cu}^{2+}$ ions produce one mole of $\text{I}_2$ : $2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}$ . Therefore, the amount of $\text{Cu}^{2+}$ is twice the amount of $\text{I}_2$ .

Understanding the Question

From part (h)(ii), the amount of $\text{I}_2$ is $5.21 \times 10^{-4}$ mol. The question asks for the amount of $\text{Cu}^{2+}$ in the $10.0\text{ cm}^3$ aliquot of solution B, using the stoichiometry of reaction 3.

Approach

Identify the mole ratio between $\text{Cu}^{2+}$ and $\text{I}_2$ from reaction 3, then multiply the amount of $\text{I}_2$ by 2.

Step-by-Step Reasoning

Reaction 3: $2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}$

The coefficient of $\text{Cu}^{2+}$ is 2 and the coefficient of $\text{I}_2$ is 1. So:
$n(\text{Cu}^{2+}) = 2 \times n(\text{I}_2) = 2 \times 5.205 \times 10^{-4} = 1.041 \times 10^{-3}\text{ mol}$

This is the amount of $\text{Cu}^{2+}$ in the $10.0\text{ cm}^3$ aliquot of solution B.

Key Takeaways

The stoichiometric chain is: $\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{I}_2 \rightarrow \text{Cu}^{2+}$ . Each step uses the mole ratio from the relevant balanced equation.

Common Mistakes

Using the wrong ratio (e.g. dividing by 2 instead of multiplying by 2).
Confusing the direction of the conversion.
Forgetting that this is the amount in the $10.0\text{ cm}^3$ aliquot, not the whole $100.0\text{ cm}^3$ solution.

Things to Be Careful About

Error carried forward applies: if part (h)(ii) was wrong but the ratio is applied correctly, this mark can still be earned. The answer is $1.04 \times 10^{-3}$ mol.

Techniques used

use stoichiometric ratio from a balanced equation

(iv)

Calculate the mass of copper present in the second sample of powdered brass.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Amount of $\text{Cu}^{2+}$ in $10.0\text{ cm}^3$ of solution B $= 1.041 \times 10^{-3}\text{ mol}$

Solution B is $100.0\text{ cm}^3$ , so total $\text{Cu}^{2+}$ in solution B:
$= 1.041 \times 10^{-3} \times \frac{100.0}{10.0} = 1.041 \times 10^{-2}\text{ mol}$

Mass of copper $= n \times M_r = 1.041 \times 10^{-2} \times 63.5$
$= 0.661\text{ g}$

Answer

$0.661\text{ g}$

Final answer

0.661 g

Detailed explanation

Background Concept

The $10.0\text{ cm}^3$ aliquot taken in step 7 is only one-tenth of the $100.0\text{ cm}^3$ solution B. To find the total amount of $\text{Cu}^{2+}$ in the whole solution (and hence in the brass sample), the amount in the aliquot must be multiplied by the dilution factor $\frac{100.0}{10.0} = 10$ .

The mass of a substance is calculated from $m = n \times M_r$ , where $M_r$ of copper is 63.5.

Understanding the Question

From part (h)(iii), the amount of $\text{Cu}^{2+}$ in $10.0\text{ cm}^3$ of solution B is $1.041 \times 10^{-3}$ mol. The question asks for the mass of copper in the second sample of brass.

Approach

Scale up the amount from the $10.0\text{ cm}^3$ aliquot to the full $100.0\text{ cm}^3$ solution B (multiply by 10).
Convert moles of copper to mass using $m = n \times M_r$ with $M_r(\text{Cu}) = 63.5$ .

Step-by-Step Reasoning

Step 1 — Scale up to the full solution.
The $10.0\text{ cm}^3$ aliquot is $\frac{10.0}{100.0} = \frac{1}{10}$ of solution B. So the total amount of $\text{Cu}^{2+}$ in solution B is:
$1.041 \times 10^{-3} \times 10 = 1.041 \times 10^{-2}\text{ mol}$

Step 2 — Convert to mass.
Mass of copper $= n \times M_r = 1.041 \times 10^{-2} \times 63.5 = 0.661\text{ g}$

This is the mass of copper in the second sample of brass.

Key Takeaways

When an aliquot is taken from a larger solution, the amount in the aliquot must be multiplied by the dilution factor to find the total amount. Mass is calculated from $m = n \times M_r$ .

Common Mistakes

Forgetting to scale up from $10.0\text{ cm}^3$ to $100.0\text{ cm}^3$ .
Using the wrong molar mass for copper.
Confusing the dilution factor (multiplying by 10 vs dividing by 10).

Things to Be Careful About

The mark scheme shows: $1.041 \times 10^{-3} \times 10 \times 63.5 = 0.661$ . The scaling factor is 10 because $100.0/10.0 = 10$ . The molar mass of copper is 63.5.

Techniques used

scale up from aliquot to original solutionconvert moles to mass

(v)

The mass of the second sample of powdered brass was $1.05\text{ g}$ .

Calculate the percentage by mass of copper in the second sample of powdered brass.

Give your answer to three significant figures.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Percentage by mass of copper $= \frac{\text{mass of copper}}{\text{mass of brass}} \times 100$
$= \frac{0.661}{1.05} \times 100 = 62.95\%$

To three significant figures: $63.0\%$

Answer

$63.0\%$

Final answer

63.0%

Detailed explanation

Background Concept

Percentage by mass is calculated as: (mass of component / total mass of sample) × 100. In this case, the component is copper and the sample is the brass.

Understanding the Question

From part (h)(iv), the mass of copper is 0.661 g. The mass of the brass sample is 1.05 g. The question asks for the percentage by mass of copper, to three significant figures.

Approach

Divide the mass of copper by the mass of brass and multiply by 100. Then round to three significant figures.

Step-by-Step Reasoning

$\text{Percentage by mass of Cu} = \frac{0.661}{1.05} \times 100 = 62.95\%$

To three significant figures: 63.0%.

Note: 62.95 rounds to 63.0 (three significant figures). The zero after the decimal point is significant because it is a placeholder that indicates the precision of the measurement.

Key Takeaways

Percentage composition = (mass of component / total mass) × 100. Always check the required number of significant figures in the final answer.

Common Mistakes

Giving the unrounded answer (62.95%) instead of 63.0%.
Rounding to the wrong number of significant figures (e.g. 63% has only two significant figures).
Using the wrong mass of brass (e.g. 1.00 g from part (f) instead of 1.05 g).

Things to Be Careful About

The mark scheme gives 63.0. Three significant figures means the answer must be written as 63.0, not 63 or 62.95. The mass of brass here is 1.05 g (given in the question), not the 1.00 g from part (f) — this is a different sample.

Techniques used

calculate percentage compositionexpress answer to the required significant figures

Q2Medium-HardPlanningAnalysis, Conclusions and Evaluation

Ester X has the formula $\text{CH}_3\text{COOR}$ .
$\text{R}$ is an alkyl group with the general formula $\text{C}_n\text{H}_{2n+1}$ .
Ester X undergoes alkaline hydrolysis with aqueous potassium hydroxide, $\text{KOH(aq)}$ .
The resulting mixture is acidified with dilute hydrochloric acid, $\text{HCl(aq)}$ .
The organic products of the hydrolysis after acidification are ethanoic acid, $\text{CH}_3\text{COOH}$ , and an alcohol, $\text{ROH}$ . Once the identity of $\text{ROH}$ is found, the structure of ester X can then be determined.

A student uses the following steps.

step 1 Equal molar quantities of ester X and $\text{KOH(aq)}$ are placed in a round-bottomed flask.
step 2 A few drops of a suitable indicator are added to show whether a reaction has occurred.
step 3 A substance is added to promote smooth boiling.
step 4 The reaction mixture is set up for reflux and heated for 30 minutes.
step 5 After 30 minutes, the reaction mixture in the round-bottomed flask is acidified by adding $\text{HCl(aq)}$ dropwise.
step 6 Thin-layer chromatography is carried out on the reaction mixture.

(a)

(i)

Complete the diagram in Fig. 2.1 to show the apparatus used for reflux in step 4. Label the diagram.

Fig. 2.1

2M Similar

DifficultyEasy

Worked solution

Answer

A Liebig condenser is fitted vertically into the neck of the round-bottomed flask. The condenser is labelled. The lower side-arm is labelled "water in" with an arrow pointing inwards, and the upper side-arm is labelled "water out" with an arrow pointing outwards.

Final answer

Liebig condenser fitted vertically with water in at bottom and water out at top

Detailed explanation

Background Concept

Reflux is a technique used to heat a reaction mixture for an extended period without losing volatile components. The mixture is boiled in a flask, and the vapours are condensed back into the flask using a condenser.

Understanding the Question

The question asks to complete a diagram showing a round-bottomed flask being heated, to illustrate the setup for reflux. You need to add the condenser and label the water flow correctly.

Approach

Recall the standard reflux apparatus: a round-bottomed flask with a vertical condenser (usually a Liebig condenser) attached to the top. Cold water must enter at the bottom and exit at the top to ensure the condenser jacket is always full of water, providing efficient cooling.

Step-by-Step Reasoning

Draw a Liebig condenser vertically on top of the round-bottomed flask neck.
Label the condenser as "(Liebig) condenser".
Draw an arrow pointing into the lower side-arm and label it "water in".
Draw an arrow pointing out of the upper side-arm and label it "water out".

Key Takeaways

Reflux setups always have the condenser vertical and water flowing counter-current (in at bottom, out at top) for maximum efficiency.

Common Mistakes

Forgetting to label the water in/out arrows or getting the direction wrong (water in at top would leave the jacket unfilled).
Drawing a different type of condenser (e.g., vertical but not labelled as Liebig, or a condenser at an angle).

Things to Be Careful About

Ensure all labels are clear and the water flow direction is correct (bottom to top).

Techniques used

draw reflux apparatuslabel condenser and water flow

(ii)

Suggest the type of substance added to promote smooth boiling in step 3.

1M Similar

(b)

As the reaction proceeds in step 4, the indicator changes colour.

Table 2.1 shows the colours of three different indicators at pH 1.0 and at pH 14.0 and the pH range over which the indicators change colour.

Table 2.1

indicator	colour at pH 1.0	pH range over which it changes colour	colour at pH 14.0
thymolphthalein	colourless	9.5–10.5	blue
methyl orange	red	3.0–4.5	yellow
bromocresol green	yellow	4.0–5.5	blue

Use the table to identify a suitable indicator.
Explain your choice.

2M Similar

DifficultyMedium

Worked solution

Answer

thymolphthalein
As the KOH is used up during the reaction, the pH of the mixture decreases from 14 to around 9 (the pH of the potassium ethanoate solution), which falls within the colour change range of thymolphthalein (9.5–10.5).

Final answer

thymolphthalein; pH drops from 14 to ~9, within range 9.5-10.5

Detailed explanation

Background Concept

Alkaline hydrolysis of an ester (saponification) uses a strong base like KOH. The reaction is: $\text{CH}_3\text{COOR} + \text{KOH} \rightarrow \text{CH}_3\text{COO}^-\text{K}^+ + \text{ROH}$ . Initially, the solution is strongly alkaline due to excess or stoichiometric KOH (pH ~14). As the reaction proceeds, KOH is consumed, and the product is a salt of a weak acid (potassium ethanoate), which hydrolyses to give a slightly alkaline solution (pH ~8-9). When the reaction is complete, the pH will be around 9.

Understanding the Question

An indicator is added to show whether the reaction has occurred. We need to choose an indicator that changes colour as the pH changes from ~14 to ~9.

Approach

Look at the pH ranges of the given indicators. The pH change is from 14 down to ~9. The indicator must change colour within this range.

Step-by-Step Reasoning

Initial pH is ~14 (KOH solution). Thymolphthalein is blue, methyl orange is yellow, bromocresol green is blue.
Final pH (when KOH is used up) is ~9 (potassium ethanoate solution). Thymolphthalein is colourless, methyl orange is red, bromocresol green is blue.
Thymolphthalein changes from blue to colourless in the range 9.5–10.5. This range is within the pH change of the mixture (14 to 9).
Methyl orange (3.0–4.5) and bromocresol green (4.0–5.5) change colour at much lower pH values, which the mixture will not reach during the hydrolysis (it only goes down to ~9).
Therefore, thymolphthalein is the suitable indicator.

Key Takeaways

When selecting an indicator for a reaction that changes pH, ensure the indicator's colour change range overlaps with the pH change of the reaction mixture.

Common Mistakes

Choosing an indicator that changes colour in acidic range (methyl orange, bromocresol green) without realising the mixture remains alkaline.
Not explaining that the pH range of the indicator must match the pH change of the mixture.

Things to Be Careful About

The product is a salt of a weak acid and strong base, so the final pH is alkaline (~9), not neutral (7) or acidic. The indicator must change colour between 9 and 14.

Techniques used

select indicator based on pH rangecalculate or estimate pH changes during reaction

(c)

In step 6, a small sample of the reaction mixture is analysed along with samples of ester X and ethanoic acid.

Fig. 2.2 shows the chromatogram produced.

Fig. 2.2

State what feature of the chromatogram shows that the hydrolysis is incomplete.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

There is a spot in the reaction mixture chromatogram with the same $R_f$ value (or same vertical position / height from the baseline) as the spot on the ester X chromatogram.

Final answer

spot in reaction mixture has same Rf as ester X

Detailed explanation

Background Concept

Thin-layer chromatography (TLC) separates components of a mixture based on their affinity for the stationary phase (silica) and the mobile phase (solvent). Each compound has a characteristic $R_f$ value (ratio of distance moved by spot to distance moved by solvent front). If a reaction is incomplete, the starting material will still be present in the product mixture.

Understanding the Question

The chromatogram shows lanes for 'ester X' (starting material), 'reaction mixture' (products + unreacted material), and 'ethanoic acid' (product). We need to identify the feature showing hydrolysis is incomplete.

Approach

Incomplete hydrolysis means some ester X has not reacted. Therefore, the reaction mixture lane should contain a spot corresponding to ester X.

Step-by-Step Reasoning

Look at the 'ester X' lane: there is one spot near the baseline.
Look at the 'reaction mixture' lane: there are three spots. One is near the baseline, aligned with the ester X spot. One is in the middle. One is near the top, aligned with the ethanoic acid spot.
The spot in the reaction mixture lane that is at the same height (same $R_f$ value) as the spot in the ester X lane indicates the presence of unreacted ester X.
This proves the hydrolysis is incomplete.

Key Takeaways

In TLC analysis of a reaction, a spot in the product mixture lane aligning with the starting material lane indicates incomplete reaction.

Common Mistakes

Saying 'there are more spots' without specifying which one indicates incomplete reaction.
Not mentioning $R_f$ value or same vertical position/height.

Things to Be Careful About

Be precise: 'same $R_f$ value' or 'same distance from baseline'.

Techniques used

interpret TLC chromatogramcompare Rf values

(d)

Suggest an experimental process that could be used to extract the alcohol, $\text{ROH}$ , from the reaction mixture.

1M Similar

(e)

Fig. 2.3 shows an infrared spectrum of the $\text{ROH}$ extracted in (d).

Fig. 2.3

Table 2.2

bond	functional groups containing the bond	characteristic infrared absorption range (in wavenumbers)/ $\text{cm}^{-1}$
C–O	hydroxy, ester	1040–1300
C=C	aromatic compound, alkene	1500–1680
C=O	amide carbonyl, carboxyl ester	1640–1690 1670–1740 1710–1750
C $\equiv$ N	nitrile	2200–2250
C–H	alkane	2850–2950
N–H	amine, amide	3300–3500
O–H	carboxyl hydroxy	2500–3000 3200–3650

Use Table 2.2 to explain how the infrared spectrum in Fig. 2.3 shows that the $\text{ROH}$ extracted does not contain any ester X.

1M Similar

(f)

Fig. 2.4 shows the proton ( $^1\text{H}$ ) NMR spectrum of compound $\text{ROH}$ .

Fig. 2.4

Table 2.3 shows some relevant ( $^1\text{H}$ ) NMR information.

Use Table 2.3 to complete Table 2.4, and state the name of $\text{ROH}$ .

Table 2.3

environment of proton	example	chemical shift range $\delta$ /ppm
alkane	$-\text{CH}_3$ , $-\text{CH}_2-$ , $>\text{CH}-$	0.9–1.7
alkyl next to C=O	$\text{CH}_3-\text{C=O}$ , $-\text{CH}_2-\text{C=O}$ , $>\text{CH}-\text{C=O}$	2.2–3.0
alkyl next to aromatic ring	$\text{CH}_3-\text{Ar}$ , $-\text{CH}_2-\text{Ar}$ , $>\text{CH}-\text{Ar}$	2.3–3.0
alkyl next to electronegative atom	$\text{CH}_3-\text{O}$ , $-\text{CH}_2-\text{O}$ , $-\text{CH}_2-\text{Cl}$	3.2–4.0
attached to alkene	$=\text{CHR}$	4.5–6.0
attached to aromatic ring	$\text{H}-\text{Ar}$	6.0–9.0
aldehyde	$\text{HCOR}$	9.3–10.5
alcohol	$\text{ROH}$	0.5–6.0
phenol	$\text{Ar}-\text{OH}$	4.5–7.0
carboxylic acid	$\text{RCOOH}$	9.0–13.0

Table 2.4

chemical shift $\delta$ /ppm	splitting pattern	relative peak area	structure responsible for the peak
1.2	doublet	6	$-\text{CH}_3$
		1
	multiplet	1

Name of $\text{ROH}$

3M Similar

DifficultyMedium-Hard

Worked solution

Answer

chemical shift $\delta$ /ppm	splitting pattern	relative peak area	structure responsible for the peak
1.2	doublet	6	$-\text{CH}_3$
2.2	singlet	1	$-\text{OH}$
4.0	multiplet	1	$>\text{CH}-\text{O}$

Name of ROH: propan-2-ol

Final answer

Row 2: 2.2, singlet, 1, -OH; Row 3: 4.0, multiplet, 1, >CH-O; propan-2-ol

Detailed explanation

Background Concept

Proton ( $^1$ H) NMR spectroscopy provides information about the environment of hydrogen atoms in a molecule:

Chemical shift ( $\delta$ ): indicates the electronic environment (e.g., next to electronegative atoms, aromatic rings, etc.).
Splitting pattern: indicates the number of neighbouring protons ( $n+1$ rule).
Relative peak area: indicates the number of protons in that environment.

Understanding the Question

We are given an NMR spectrum of ROH with three signals: a doublet at 1.2 ppm (6H), a singlet at 2.2 ppm (1H), and a multiplet at 4.0 ppm (1H). We need to complete the table and identify ROH.

Approach

Use the chemical shift ranges in Table 2.3 to assign each signal to a proton environment. Use splitting patterns to confirm neighbouring protons. Combine the fragments to deduce the structure.

Step-by-Step Reasoning

Signal at 1.2 ppm (doublet, 6H):
- Shift 1.2 ppm is in the alkane range (0.9–1.7 ppm).
- Area 6H suggests two equivalent $-\text{CH}_3$ groups.
- Splitting is a doublet, so each $-\text{CH}_3$ is next to 1 proton ( $n+1 = 2 \rightarrow n=1$ ). This fits $-\text{CH}(-\text{CH}_3)_2$ .
- Table entry: 1.2, doublet, 6, $-\text{CH}_3$ .
Signal at 2.2 ppm (singlet, 1H):
- Shift 2.2 ppm could be alkyl next to C=O or alcohol. Since we have an alcohol (ROH), the -OH proton often appears as a singlet (no splitting due to exchange) and can vary widely (0.5–6.0 ppm). 2.2 ppm is reasonable for an alcohol -OH.
- Area 1H, singlet (no neighbouring protons to split it, or exchange broadens it).
- Table entry: 2.2, singlet, 1, $-\text{OH}$ .
Signal at 4.0 ppm (multiplet, 1H):
- Shift 4.0 ppm is in the range for alkyl next to electronegative atom (3.2–4.0 ppm), specifically $-\text{CH}_2-\text{O}$ or $>\text{CH}-\text{O}$ .
- Area 1H, multiplet (complex splitting due to neighbouring protons: the 6H from two methyls and 1H from OH, though OH often doesn't split, the 6H from methyls split the CH into a septet, which is a multiplet).
- This fits a $>\text{CH}-\text{O}$ group (methine group attached to oxygen).
- Table entry: 4.0, multiplet, 1, $>\text{CH}-\text{O}$ .
Deduce structure:
- Fragments: two $-\text{CH}_3$ groups, one $-\text{OH}$ , one $>\text{CH}-$ .
- Combine: $(\text{CH}_3)_2\text{CH}-\text{OH}$ .
- This is propan-2-ol (isopropyl alcohol).
- Formula: $\text{C}_3\text{H}_8\text{O}$ , which fits $\text{C}_n\text{H}_{2n+1}\text{OH}$ with $n=3$ ( $\text{C}_3\text{H}_7\text{OH}$ ).

Key Takeaways

NMR data (shift, splitting, integration) must be combined to deduce molecular fragments and then the full structure. Remember that -OH protons often appear as singlets and can have variable chemical shifts.

Common Mistakes

Assigning the 2.2 ppm peak to a $-\text{CH}_2-$ next to C=O (but there is no C=O in an alcohol, and area is 1H).
Forgetting that the -OH proton gives a singlet.
Misinterpreting the multiplet at 4.0 ppm.

Things to Be Careful About

The chemical shift for -OH can vary widely (0.5–6.0 ppm), so don't rule it out just because 2.2 ppm is also in the 'alkyl next to C=O' range; the absence of C=O in the IR spectrum (part e) confirms it's an alcohol.
The multiplet at 4.0 ppm is a septet (split by 6 equivalent methyl protons), but 'multiplet' is acceptable as per the mark scheme.

Techniques used

interpret 1H NMR spectrumdetermine splitting patterns and chemical shiftsdeduce molecular structure from NMR data

(g)

Draw the displayed formula for ester X.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Displayed formula of isopropyl ethanoate ( $\text{CH}_3\text{COOCH}(\text{CH}_3)_2$ ):

Left: $\text{CH}_3-$ group (C bonded to 3 H's).
Middle: Carbonyl group $\text{C=O}$ (C double bonded to O, single bonded to left C and right O).
Right: Ester oxygen $-\text{O}-$ bonded to a central CH group.
Central CH: bonded to 1 H (up), and two $\text{CH}_3$ groups (right and down, each C bonded to 3 H's).

Final answer

Displayed formula of isopropyl ethanoate

Detailed explanation

Background Concept

Ester X has the formula $\text{CH}_3\text{COOR}$ . From part (f), we identified ROH as propan-2-ol, so R is the propan-2-yl group (isopropyl group), $-\text{CH}(\text{CH}_3)_2$ .

Understanding the Question

Draw the displayed formula for ester X. A displayed formula must show all atoms and all bonds.

Approach

Combine the ethanoate part ( $\text{CH}_3\text{COO}-$ ) with the isopropyl group ( $-\text{CH}(\text{CH}_3)_2$ ). Draw all C-H, C-C, C-O, and C=O bonds.

Step-by-Step Reasoning

Ethanoate part: $\text{CH}_3-\text{C}(=\text{O})-\text{O}-$ . Draw C bonded to 3 H's, double bonded to O, single bonded to another O.
Isopropyl part: $-\text{CH}(\text{CH}_3)_2$ . Draw the central C bonded to 1 H, and two $\text{CH}_3$ groups (each C bonded to 3 H's).
Connect the ester oxygen to the central C of the isopropyl group.
Ensure all bonds are shown as lines.

Key Takeaways

Displayed formulas show every atom and every bond. For esters, ensure the ester linkage $-\text{COO}-$ is correctly drawn with all hydrogens on the alkyl groups.

Common Mistakes

Drawing a structural formula instead of a displayed formula (e.g., $\text{CH}_3\text{COOCH}(\text{CH}_3)_2$ without showing all bonds).
Forgetting to show all C-H bonds.
Drawing the wrong isomer (e.g., propyl ethanoate instead of isopropyl ethanoate).

Things to Be Careful About

The question asks for a displayed formula, so every single bond must be drawn as a line. Do not use condensed formulae.

Techniques used

draw displayed formulaconstruct ester structure from components

(h)

Ester X will undergo hydrolysis with water in the presence of $\text{H}_2\text{SO}_4\text{(aq)}$ under reflux, using a similar procedure.

Suggest why none of the indicators in Table 2.1 would change colour in this experiment.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The pH remains below 3.0 (or the solution remains acidic) throughout the reaction because dilute $\text{H}_2\text{SO}_4$ is present as a catalyst, so none of the indicators change colour.

Final answer

pH remains below 3.0

Detailed explanation

Background Concept

Esters can be hydrolysed under acidic conditions (acid hydrolysis) or alkaline conditions (alkaline hydrolysis/saponification). Acid hydrolysis is reversible and uses a dilute strong acid (like $\text{H}_2\text{SO}_4$ ) as a catalyst. The reaction is: $\text{CH}_3\text{COOR} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{ROH}$ .

Understanding the Question

In acid-catalysed hydrolysis, the student uses dilute $\text{H}_2\text{SO}_4$ and reflux. The question asks why none of the indicators in Table 2.1 would change colour.

Approach

Look at the pH ranges of the indicators. They all change colour at pH > 3.0 (methyl orange starts at 3.0). In acid hydrolysis, the solution is acidic due to the catalyst and the product (ethanoic acid). The pH will not rise above 3.0.

Step-by-Step Reasoning

The reaction uses dilute $\text{H}_2\text{SO}_4$ as a catalyst. This means the solution is strongly acidic initially (pH < 1).
As the reaction proceeds, ethanoic acid is produced, but the strong acid catalyst remains, so the pH stays low (below 3.0).
The indicators in Table 2.1 change colour at pH 3.0 or above (methyl orange: 3.0-4.5, bromocresol green: 4.0-5.5, thymolphthalein: 9.5-10.5).
Since the pH never rises above 3.0, none of the indicators will change colour.

Key Takeaways

In acid-catalysed reactions, the pH remains acidic. Indicators that change colour in alkaline or neutral ranges will not show a colour change.

Common Mistakes

Saying 'the pH doesn't change' (it does change slightly as acid is produced, but it stays low).
Not referencing the specific pH range of the indicators (below 3.0).

Things to Be Careful About

The key point is that the pH remains below the lowest colour change range of the given indicators (3.0 for methyl orange).

Techniques used

explain pH behaviour in acid-catalysed hydrolysis