Chemistry 9701/52 — October/November 2024

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

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q1MediumAnalysis, Conclusions and EvaluationPlanning

Solution X is dilute hydrochloric acid of unknown concentration. A student uses titration with aqueous sodium carbonate, $\text{Na}_2\text{CO}_3(\text{aq})$ , to determine the concentration of solution X.

The student plans to prepare $250.0\text{ cm}^3$ of $0.500\text{ mol dm}^{-3}$ aqueous sodium carbonate, $\text{Na}_2\text{CO}_3(\text{aq})$ .

(a)

(i)

Calculate the mass of $\text{Na}_2\text{CO}_3(\text{s})$ needed to make $250.0\text{ cm}^3$ of $0.500\text{ mol dm}^{-3}$ $\text{Na}_2\text{CO}_3(\text{aq})$ , using a two decimal place balance.

1M Similar

DifficultyEasy

Worked solution

Working

n = c \times V = 0.500 \times 0.250 = 0.125 \text{ mol}

M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0

\text{mass} = 0.125 \times 106.0 = 13.25 \text{ g}

Answer

13.25 g

Final answer

13.25 g

Detailed explanation

Background Concept

To prepare a solution of known concentration, the required mass of solute is determined from the relationship $n = cV$ (where $n$ is amount in mol, $c$ is concentration in mol dm⁻³, and $V$ is volume in dm³), followed by converting moles to mass using $m = n \times M_r$ . The relative formula mass is calculated by summing the atomic masses of all atoms in the formula.

Understanding the Question

The question asks for the mass of anhydrous sodium carbonate ( $\text{Na}_2\text{CO}_3$ ) needed to prepare exactly 250.0 cm³ of a 0.500 mol dm⁻³ solution. The answer must be given to two decimal places (as would be read from a two-decimal-place balance).

Approach

Convert volume to dm³, calculate moles using $n = cV$ , then multiply by $M_r$ to get mass.

Step-by-Step Reasoning

Volume conversion: $250.0 \text{ cm}^3 = 0.2500 \text{ dm}^3$
Moles needed: $n = 0.500 \times 0.2500 = 0.125 \text{ mol}$
$M_r(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 46.0 + 12.0 + 48.0 = 106.0$
Mass = $0.125 \times 106.0 = 13.25 \text{ g}$

Key Takeaways

Always convert cm³ to dm³ before using $n = cV$ .
The relative formula mass of sodium carbonate is 106.0 (anhydrous form).
A two-decimal-place balance reads to 0.01 g, so 13.25 g is appropriate.

Common Mistakes

Using 250.0 instead of 0.2500 in the $n = cV$ calculation (forgetting to convert cm³ to dm³).
Calculating $M_r$ incorrectly, e.g. forgetting there are two sodium atoms or three oxygen atoms.
Giving the answer as 13.3 g (only one decimal place) instead of 13.25 g.

Things to Be Careful About

The question specifies a two-decimal-place balance, so the answer must have exactly two decimal places.
Ensure you use the anhydrous formula mass (106.0), not the decahydrate ( $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ , $M_r = 286$ ).

Techniques used

calculate moles from concentration and volumecalculate mass from moles and relative formula mass

(ii)

Describe how the student should make $250.0\text{ cm}^3$ of $0.500\text{ mol dm}^{-3}$ $\text{Na}_2\text{CO}_3(\text{aq})$ starting from the mass of $\text{Na}_2\text{CO}_3(\text{s})$ calculated in (i) supplied in a $50\text{ cm}^3$ beaker.

Give the name and size of any key apparatus to be used.

Write your answer using a series of numbered steps.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Add a small volume of distilled water to the beaker and stir to dissolve the $\text{Na}_2\text{CO}_3$ completely.
Transfer the solution into a $250\text{ cm}^3$ volumetric flask, rinsing the beaker with distilled water and adding the rinsings to the flask.
Add distilled water to the volumetric flask until the bottom of the meniscus is level with the calibration mark.

Final answer

Dissolve in beaker with small volume of distilled water; transfer to 250 cm³ volumetric flask rinsing the beaker; top up to the mark with distilled water.

Detailed explanation

Background Concept

Preparing a standard solution (one of accurately known concentration) requires quantitative transfer of the solute into a volumetric flask and dilution to a precise volume. The key principles are: complete dissolution before transfer, rinsing to ensure no solute is left behind, and topping up to the calibration mark so the final volume is exactly correct.

Understanding the Question

The question asks for a numbered-step description of how to prepare 250.0 cm³ of 0.500 mol dm⁻³ Na₂CO₃(aq) from the solid already weighed into a 50 cm³ beaker. Three marks are available, each corresponding to a key procedural step.

Approach

Follow the standard method: dissolve → transfer (with rinsing) → make up to mark. Name the volumetric flask and its size.

Step-by-Step Reasoning

M1 — Dissolution: The solid must be dissolved in a small volume of distilled water in the beaker. Using only a small volume ensures the final volume can be brought exactly to the mark without overflowing. Distilled water is specified to avoid impurities.

M2 — Transfer and rinse: The solution is poured into a 250 cm³ volumetric flask. The beaker must then be rinsed with distilled water and the rinsings added to the flask. This ensures quantitative transfer — all the solute ends up in the flask, not left adhering to the beaker walls.

M3 — Make up to mark: Distilled water is added until the bottom of the meniscus sits on the calibration mark. This gives exactly 250.0 cm³ of solution, ensuring the concentration is precisely 0.500 mol dm⁻³.

Key Takeaways

Always rinse the dissolving vessel and add rinsings to the volumetric flask for quantitative transfer.
The volumetric flask must be named with its size (250 cm³).
The final adjustment must reference the meniscus and the calibration mark.

Common Mistakes

Omitting the rinsing step (loses M2 mark).
Not specifying the size of the volumetric flask.
Saying 'add water to 250 cm³' instead of 'add water to the mark' — the mark scheme requires reference to the calibration mark.
Using 'deionised water' instead of 'distilled water' (usually accepted but distilled is the mark scheme term).

Things to Be Careful About

The answer must be in numbered steps as specified.
Each mark requires a distinct action; combining steps into one sentence may lose a mark.
The word 'rinsing' must appear in connection with the transfer step.

Techniques used

describe a standard solution preparation procedureidentify key apparatus and rinsing technique

(b)

The student incorrectly makes up the $250.0\text{ cm}^3$ of $\text{Na}_2\text{CO}_3(\text{aq})$ and the concentration is not $0.500\text{ mol dm}^{-3}$ . The student calls this solution Y.

The student uses the following method to determine both the concentration of solution X and the concentration of solution Y.

step 1 Transfer $25.0\text{ cm}^3$ of solution Y into a conical flask using a volumetric pipette. Add a few drops of methyl orange.

step 2 Titrate the sample in the conical flask with solution X.

step 3 Transfer a fresh $25.0\text{ cm}^3$ portion of solution Y into a second conical flask, but do not add methyl orange.

step 4 Use the burette to add the volume of solution X used in step 2 to the second conical flask.

step 5 Measure and record the mass of a dry evaporating basin.

step 6 Transfer the contents of the second conical flask into the evaporating basin.

step 7 Use a water bath to heat the solution in the evaporating basin until all the water is evaporated and only solid remains.

step 8 Measure and record the mass of the evaporating basin and solid residue.

(i)

In step 1, $25.0\text{ cm}^3$ of solution Y is transferred using a volumetric pipette.

State what the volumetric pipette should be rinsed with before carrying out step 1.

1M Similar

DifficultyEasy

Worked solution

Answer

Solution Y

Final answer

solution Y

Detailed explanation

Background Concept

Before using a volumetric pipette, it must be rinsed with the solution it is about to contain. This is because any residual water (from washing) would dilute the solution drawn into the pipette, causing the actual concentration delivered to be lower than intended. Rinsing with the solution itself removes this water and ensures the pipette delivers the correct concentration.

Understanding the Question

The pipette is being used to transfer 25.0 cm³ of solution Y (the sodium carbonate solution) into the conical flask. The question asks what the pipette should be rinsed with beforehand.

Approach

Apply the standard rule: rinse the pipette with the solution it will contain.

Step-by-Step Reasoning

The pipette will contain solution Y. Therefore, it must be rinsed with solution Y to prevent dilution from residual water. Rinsing with distilled water would be incorrect because it would leave water in the pipette that dilutes the solution Y, giving a lower concentration of Na₂CO₃ in the flask and an inaccurate titre.

Key Takeaways

Pipettes are always rinsed with the solution they will contain.
Burettes are rinsed with the solution they will contain (solution X in this case).
Conical flasks are NOT rinsed with the solution — they should be clean and dry (or rinsed with distilled water only), because any extra solution in the flask would change the amount of solute being titrated.

Common Mistakes

Saying 'distilled water' — this would dilute the solution and is wrong.
Saying 'solution X' — the pipette contains solution Y, not X.
Saying 'the acid' — the pipette is used for the alkali (solution Y).

Things to Be Careful About

The mark scheme specifically requires 'solution Y' as the answer. Simply saying 'the sodium carbonate solution' may also be accepted, but 'solution Y' is the precise term used in the question.

Techniques used

identify correct rinsing solution for volumetric pipette

(ii)

Describe how the student detects the end-point of the titration in step 2.

1M Similar

(iii)

Explain why the evaporating basin is not heated directly with a Bunsen burner in step 7.

1M Similar

(iv)

Suggest what the student should do to ensure all of the water has evaporated from the residue before completing step 8.

1M Similar

(c)

The student’s burette readings taken in step 2 are shown in Fig. 1.1.

Fig. 1.1

(i)

Use Fig. 1.1 to complete Table 1.1.

Table 1.1


burette reading (final) / $\text{cm}^3$
burette reading (initial) / $\text{cm}^3$
volume of solution X added / $\text{cm}^3$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer


burette reading (final) / cm³	14.85
burette reading (initial) / cm³	1.35
volume of solution X added / cm³	13.50

The volume is calculated as $14.85 - 1.35 = 13.50 \text{ cm}^3$ .

Final answer

Final: 14.85 cm³, Initial: 1.35 cm³, Volume added: 13.50 cm³

Detailed explanation

Background Concept

A burette is graduated from top to bottom (0 cm³ at the top, 50 cm³ at the bottom). Readings are taken from the bottom of the meniscus at eye level. Burette readings must be recorded to two decimal places (nearest 0.05 cm³), as the smallest graduation is typically 0.1 cm³ and the uncertainty is ±0.05 cm³. The titre volume is the difference between the final and initial readings.

Understanding the Question

The student has taken two burette readings (shown in Fig. 1.1) and these must be recorded in a table along with the calculated volume of solution X added.

Approach

Read each burette to the nearest 0.05 cm³, record both readings, then subtract to find the titre.

Step-by-Step Reasoning

Initial reading: The meniscus is between the 1 and 2 cm³ graduations. The bottom of the meniscus sits at 1.35 cm³ (three small divisions below the 1 mark, each division being 0.1 cm³, with the meniscus at the midpoint between 1.3 and 1.4, giving 1.35).

Final reading: The meniscus is between the 14 and 15 cm³ graduations. The bottom of the meniscus sits at 14.85 cm³ (eight and a half small divisions below the 14 mark).

Volume added: $14.85 - 1.35 = 13.50 \text{ cm}^3$

Both readings and the titre must be recorded to two decimal places.

Key Takeaways

Always read from the bottom of the meniscus.
Record to two decimal places (e.g. 1.35, not 1.4 or 1.3).
The titre = final reading − initial reading (not the other way around).

Common Mistakes

Reading from the top of the meniscus instead of the bottom.
Recording to only one decimal place (e.g. 1.4 instead of 1.35).
Reversing the subtraction (getting a negative number).
Misreading the scale by one division.

Things to Be Careful About

The burette scale increases downward, so a reading near the top is a small number and near the bottom is a large number.
All values must have exactly two decimal places, including the titre (13.50, not 13.5).
M1 requires both readings correct; M2 requires the subtraction correct.

Techniques used

read a burette to two decimal placescalculate titre volume from initial and final readings

(ii)

Calculate the percentage error in the volume of solution X calculated in (i).

Show your working.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Each burette reading has an uncertainty of $\pm 0.05 \text{ cm}^3$ . Two readings are taken, so the total uncertainty in the titre is:

2 \times 0.05 = 0.10 \text{ cm}^3

\text{percentage error} = \frac{0.10}{13.50} \times 100 = 0.741\%

Answer

0.741%

Final answer

0.741%

Detailed explanation

Background Concept

When a titre volume is calculated as the difference between two burette readings, the uncertainties in both readings combine. Each individual reading has an uncertainty of ±0.05 cm³ (half the smallest graduation of 0.1 cm³). Since the titre involves two readings (initial and final), the total uncertainty is $2 \times 0.05 = 0.10 \text{ cm}^3$ . The percentage error is then the absolute uncertainty divided by the measured value, multiplied by 100.

Understanding the Question

The question asks for the percentage error in the titre volume of 13.50 cm³ calculated in part (c)(i). Working must be shown.

Approach

Determine the total uncertainty in the titre (from two readings), then divide by the titre volume and multiply by 100.

Step-by-Step Reasoning

Uncertainty per reading = ±0.05 cm³
Number of readings used = 2 (initial and final)
Total uncertainty in titre = $2 \times 0.05 = 0.10 \text{ cm}^3$
Percentage error = $\frac{0.10}{13.50} \times 100 = 0.741\%$

Note: The mark scheme writes this as $\frac{2 \times 0.05}{13.50} \times 100 = 0.741$ , combining steps 3 and 4.

Key Takeaways

The percentage error in a titre uses $2 \times 0.05$ in the numerator (not $1 \times 0.05$ ), because two readings contribute uncertainty.
The denominator is the titre volume itself (13.50 cm³).
Working must be shown — the answer alone does not score.

Common Mistakes

Using only one uncertainty (0.05 instead of 0.10) — this gives 0.370%, which is wrong.
Dividing by 25.0 (the pipette volume) instead of 13.50 (the titre).
Not showing working (the mark scheme explicitly requires it).
Using ±0.1 cm³ per reading instead of ±0.05 cm³.

Things to Be Careful About

The factor of 2 must appear in the working to show understanding of why two readings contribute.
The answer should be given to 3 significant figures (0.741%) as shown in the mark scheme.

Techniques used

calculate percentage uncertainty from burette reading precision

(d)

A second student repeats the experiment in (b) using solution X and $25.0\text{ cm}^3$ of solution Y.

The results are shown in Table 1.2.

Table 1.2


volume of solution X added in titration / $\text{cm}^3$	13.35
mass of dry evaporating basin / g	44.52
mass of evaporating basin and solid residue / g	45.69

The reaction that takes place in the titration is:

2\text{HCl}(\text{aq}) + \text{Na}_2\text{CO}_3(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

Using the results in Table 1.2, calculate the following:

the amount, in mol, of sodium chloride
the concentration of solution X and the concentration of solution Y.

3M Similar

DifficultyMedium

Worked solution

Working

Amount of NaCl:

\text{mass of NaCl} = 45.69 - 44.52 = 1.17 \text{ g}

M_r(\text{NaCl}) = 58.5

n(\text{NaCl}) = \frac{1.17}{58.5} = 0.0200 \text{ mol}

Concentration of solution X (HCl):
From the equation, $n(\text{HCl}) = n(\text{NaCl}) = 0.0200 \text{ mol}$ (ratio 2:2)

c(\text{HCl}) = \frac{0.0200}{0.01335} = 1.50 \text{ mol dm}^{-3}

Concentration of solution Y (Na₂CO₃):
From the equation, $n(\text{Na}_2\text{CO}_3) = \frac{n(\text{NaCl})}{2} = \frac{0.0200}{2} = 0.0100 \text{ mol}$

c(\text{Na}_2\text{CO}_3) = \frac{0.0100}{0.0250} = 0.400 \text{ mol dm}^{-3}

Answer

Amount of NaCl = 0.0200 mol
Concentration of solution X = 1.50 mol dm⁻³
Concentration of solution Y = 0.400 mol dm⁻³

Final answer

n(NaCl) = 0.0200 mol; c(X) = 1.50 mol dm⁻³; c(Y) = 0.400 mol dm⁻³

Detailed explanation

Background Concept

This question combines gravimetric analysis (finding moles from mass of a product) with titration stoichiometry. The balanced equation $2\text{HCl} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ gives the mole ratios between all species. The NaCl produced comes from both reactants, and its amount allows back-calculation of the amounts of HCl and Na₂CO₃ that reacted. Concentration is then found from $c = n/V$ .

Understanding the Question

The second student's data gives: titre volume of X = 13.35 cm³, mass of empty basin = 44.52 g, mass of basin + residue = 45.69 g. From these, calculate n(NaCl), c(X), and c(Y). The pipette volume of Y is 25.0 cm³ (from the method in part b).

Approach

Find mass of NaCl by difference, then moles using Mr.
Use the 2:2 ratio (HCl:NaCl) to find moles of HCl, then divide by titre volume for c(X).
Use the 1:2 ratio (Na₂CO₃:NaCl) to find moles of Na₂CO₃, then divide by 25.0 cm³ for c(Y).

Step-by-Step Reasoning

M1 — Moles of NaCl:

Mass of NaCl = 45.69 − 44.52 = 1.17 g
Mr(NaCl) = 23.0 + 35.5 = 58.5
n(NaCl) = 1.17 / 58.5 = 0.0200 mol

M2 — Concentration of X (HCl):

From the equation: 2HCl → 2NaCl, so the ratio is 1:1
n(HCl) = n(NaCl) = 0.0200 mol
Volume of HCl = 13.35 cm³ = 0.01335 dm³
c(HCl) = 0.0200 / 0.01335 = 1.498... ≈ 1.50 mol dm⁻³

M3 — Concentration of Y (Na₂CO₃):

From the equation: 1Na₂CO₃ → 2NaCl, so the ratio is 1:2
n(Na₂CO₃) = n(NaCl) / 2 = 0.0200 / 2 = 0.0100 mol
Volume of Na₂CO₃ = 25.0 cm³ = 0.0250 dm³
c(Na₂CO₃) = 0.0100 / 0.0250 = 0.400 mol dm⁻³

Key Takeaways

The mole ratio from the balanced equation is essential for converting between species.
The 2:2 ratio between HCl and NaCl means they have equal moles.
The 1:2 ratio between Na₂CO₃ and NaCl means Na₂CO₃ has half the moles of NaCl.
Always convert cm³ to dm³ before calculating concentration.

Common Mistakes

Using the wrong mole ratio (e.g. dividing n(NaCl) by 2 to get n(HCl) instead of n(Na₂CO₃)).
Forgetting to convert cm³ to dm³ (dividing by 13.35 instead of 0.01335).
Using the wrong volume for c(Y) — it's 25.0 cm³ (the pipette volume), not 13.35 cm³.
Not recognising that the NaCl comes from BOTH reactants and using it to find both concentrations.

Things to Be Careful About

The volume of solution Y used is 25.0 cm³ (from the pipette), not the titre volume.
The titre volume (13.35 cm³) is the volume of solution X (HCl) used.
Answers should be to 3 significant figures to match the data precision.

Techniques used

calculate moles from mass and Mruse stoichiometric ratios to find moles of reactantscalculate concentration from moles and volume

(e)

State what happens to the value obtained for the concentration of solution Y if not all the water is evaporated in step 7.

Explain your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The calculated concentration of solution Y would be higher than the true value.

If not all the water is evaporated, the measured mass of the residue is too high. This gives a calculated $n(\text{NaCl})$ that is too high, which in turn gives a calculated $n(\text{Na}_2\text{CO}_3)$ that is too high, so the calculated concentration of Y is too high.

Final answer

The calculated concentration of Y will be higher than the true value because the mass of residue (and hence n(NaCl) and n(Na₂CO₃)) will be too high.

Detailed explanation

Background Concept

In gravimetric analysis, the mass of a dried residue is used to determine the amount of a product, and from that, the amounts of reactants via stoichiometry. If the residue is not fully dry, the measured mass includes water, inflating the apparent mass of the product. This error propagates through all subsequent calculations.

Understanding the Question

The student uses the mass of NaCl residue to work back to the concentration of solution Y (Na₂CO₃). The question asks what happens to the calculated c(Y) if some water remains in the residue.

Approach

Trace the error: incomplete evaporation → mass too high → moles of NaCl too high → moles of Na₂CO₃ too high → concentration of Y too high.

Step-by-Step Reasoning

M1 — Direction of error: The calculated concentration of Y will be higher than the true value.

M2 — Explanation: If water remains in the residue, the measured mass of 'NaCl' is actually NaCl + water. This means:

Mass of residue is too high
$n(\text{NaCl}) = \text{mass}/58.5$ is too high
$n(\text{Na}_2\text{CO}_3) = n(\text{NaCl})/2$ is too high
$c(\text{Y}) = n(\text{Na}_2\text{CO}_3)/0.025$ is too high

Therefore the calculated concentration of Y is greater than the true concentration.

Key Takeaways

Systematic errors in gravimetric analysis propagate through stoichiometric calculations.
An overestimated mass leads to an overestimated concentration.
Always state both the direction of the error (higher/lower) AND the reason.

Common Mistakes

Saying the concentration would be lower — the error direction is wrong.
Only stating the direction without explaining why (loses M2).
Only explaining the mechanism without stating the direction (loses M1).
Confusing the effect on c(X) vs c(Y) — the question specifically asks about c(Y).

Things to Be Careful About

Both marks are needed: M1 for the direction (higher) and M2 for the explanation (mass/amount of NaCl too high).
The explanation must connect the water remaining to the mass being too high, not just say 'the result is wrong'.

Techniques used

analyse the effect of incomplete evaporation on gravimetric resulttrace error propagation through stoichiometric calculation

Q2Medium-EasyPlanningAnalysis, Conclusions and Evaluation

A group of students uses the following method to investigate the change in mass that takes place when metal M combines with sulfur to form a metal sulfide.

step 1 Weigh a clean crucible and lid using a balance. Record the mass.

step 2 Place a coiled length of metal wire into the crucible and weigh the crucible, lid and wire.

step 3 Cover the metal wire in the crucible with a large quantity of powdered sulfur and replace the lid.

step 4 Heat the crucible in a fume hood until the crucible glows red. Continue heating strongly until no more sulfur can be seen.

step 5 Allow the apparatus to cool and weigh the crucible, lid and residue.

Fig. 2.1

(a)

(i)

Fig. 2.1 shows how the apparatus is set up when heating the crucible in step 4.

Identify the missing piece of apparatus used to support the crucible during heating.

1M Similar

DifficultyEasy

Worked solution

Answer

pipeclay triangle

Final answer

pipeclay triangle

Detailed explanation

Background Concept

When heating a crucible strongly, it cannot be placed directly on a tripod because the tripod bars are too thick and would not support the crucible evenly, potentially causing it to crack or tip over. A pipeclay triangle is a standard piece of apparatus consisting of a metal frame covered with a heat-resistant ceramic (pipeclay) coating. It sits on the tripod and provides a stable, heat-resistant platform for the crucible.

Understanding the Question

The question asks to identify the missing piece of apparatus in Fig. 2.1 that supports the crucible during heating. The diagram shows a tripod and a crucible, but the crucible is floating above the tripod bars. We need to name the component that bridges the two.

Approach

Recall standard laboratory procedures for heating crucibles. The missing link between the tripod and the crucible is the pipeclay triangle.

Step-by-Step Reasoning

Observe the apparatus in Fig. 2.1: a tripod is present, and a crucible is being heated above it.
Recall that a crucible requires a stable, heat-resistant support that allows heat from the Bunsen burner to reach the bottom of the crucible.
The correct apparatus for this is a pipeclay triangle, which rests on the tripod ring and holds the crucible securely.

Key Takeaways

Always ensure crucibles are supported by a pipeclay triangle on a tripod when strong heating is required. Direct contact with metal tripod bars is incorrect and dangerous.

Common Mistakes

Writing 'wire gauze': wire gauze is used for flasks and beakers to distribute heat evenly, not for supporting crucibles under strong direct heat.
Writing 'ceramic mat': this is used to protect the bench, not to support the crucible during heating.

Things to Be Careful About

Use the exact terminology 'pipeclay triangle'. 'Triangle' alone is not sufficient; 'wire gauze' is incorrect.

Techniques used

identify laboratory apparatus

(ii)

Suggest why the students heat their crucibles in a fume hood in step 4.

1M Similar

(iii)

Suggest why a large quantity of powdered sulfur is used in step 3.

1M Similar

DifficultyEasy

Worked solution

Answer

A large excess of sulfur is used to ensure that all of the metal M reacts completely, making M the limiting reactant.

Final answer

to ensure all of the metal M reacts

Detailed explanation

Background Concept

In a reaction between two reactants, the limiting reactant is the one that is completely consumed first and determines the amount of product formed. To ensure a quantitative reaction (where all of a specific reactant is converted to product), the other reactant is typically added in excess.

Understanding the Question

The students want to find out how much sulfur reacts with a known mass of metal M. To do this accurately, they need to be sure that every atom of M has reacted. Using a large quantity of sulfur guarantees this.

Approach

Explain the purpose of using an excess of one reactant in a synthesis experiment.

Step-by-Step Reasoning

The goal is to combine metal M with sulfur to form a metal sulfide.
If sulfur were the limiting reactant, some metal M would be left unreacted, and the final mass would not accurately reflect the stoichiometry of the sulfide.
By using a large quantity (excess) of powdered sulfur, sulfur is in excess and metal M is the limiting reactant.
This ensures all of the metal M reacts, allowing accurate calculation of the mass of sulfur that combined with it.

Key Takeaways

In quantitative synthesis experiments, always use an excess of the reagent that is not being measured or is easier to remove later (like volatile sulfur, which can be driven off or is in excess but doesn't affect the final mass of the metal sulfide if excess is removed, though here excess sulfur is likely just not fully reacting or is driven off as gas).

Common Mistakes

Saying 'to make the reaction go faster': while powdered sulfur increases surface area and speed, the 'large quantity' specifically ensures completion.
Saying 'to produce more product': the amount of product is limited by the metal M anyway.

Things to Be Careful About

The key phrase is 'ensure all of the metal M reacts'. Do not just say 'excess sulfur' without explaining why.

Techniques used

justify use of excess reagent

(b)

Five students each use a different mass of M.

Their results are shown in Table 2.1.

Complete Table 2.1 by inserting values for the mass of M and the mass of sulfur which reacts for each student.

Table 2.1

	student 1	student 2	student 3	student 4	student 5
mass of crucible and lid / g	34.15	38.28	35.68	33.70	36.84
mass of crucible, lid and M / g	35.58	39.42	36.54	34.27	37.13
mass of crucible, lid and residue after heating / g	36.04	39.70	36.74	34.42	37.19
mass of M / g
mass of sulfur which reacts / g

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Mass of M = (mass of crucible, lid and M) − (mass of crucible and lid)
Mass of S reacting = (mass of crucible, lid and residue) − (mass of crucible, lid and M)

	student 1	student 2	student 3	student 4	student 5
mass of M / g	1.43	1.14	0.86	0.57	0.29
mass of sulfur which reacts / g	0.46	0.28	0.20	0.15	0.06

Example calculation (student 1):
Mass of M = $35.58 - 34.15 = 1.43 \text{ g}$
Mass of S = $36.04 - 35.58 = 0.46 \text{ g}$

Answer

See completed table above.

Final answer

mass of M: 1.43, 1.14, 0.86, 0.57, 0.29; mass of S: 0.46, 0.28, 0.20, 0.15, 0.06

Detailed explanation

Background Concept

In a mass-balance experiment like this, the mass of a specific component is found by subtracting the mass of the container (crucible + lid) from the total mass. The mass of sulfur that actually reacted is found by comparing the final mass of the solid residue (metal sulfide + unreacted sulfur if any, but ideally just metal sulfide) with the initial mass of the metal.

Actually, a simpler way:
Mass of residue (metal sulfide) = (mass of crucible, lid and residue) - (mass of crucible and lid).
Mass of sulfur reacted = (Mass of residue) - (Mass of metal M).
Alternatively: Mass of sulfur reacted = (mass of crucible, lid and residue) - (mass of crucible, lid and M). This works because the only thing added to the 'crucible + lid + M' system that remains is the sulfur that reacted (assuming excess sulfur is removed or doesn't add to the final weighed mass, though the method says 'until no more sulfur can be seen', implying excess is removed or volatilized).

Understanding the Question

We are given three mass measurements for each student and need to calculate the mass of metal M used and the mass of sulfur that reacted with it.

Approach

Use simple subtraction to isolate the mass of M and the mass of S from the total masses.

Step-by-Step Reasoning

For each student:

Mass of M: Subtract the empty crucible mass from the crucible + M mass.
- Student 1: $35.58 - 34.15 = 1.43 \text{ g}$
- Student 2: $39.42 - 38.28 = 1.14 \text{ g}$
- Student 3: $36.54 - 35.68 = 0.86 \text{ g}$
- Student 4: $34.27 - 33.70 = 0.57 \text{ g}$
- Student 5: $37.13 - 36.84 = 0.29 \text{ g}$
Mass of S reacting: Subtract the 'crucible + lid + M' mass from the 'crucible + lid + residue' mass. This gives the mass of sulfur that is now part of the residue (i.e., reacted).
- Student 1: $36.04 - 35.58 = 0.46 \text{ g}$
- Student 2: $39.70 - 39.42 = 0.28 \text{ g}$
- Student 3: $36.74 - 36.54 = 0.20 \text{ g}$
- Student 4: $34.42 - 34.27 = 0.15 \text{ g}$
- Student 5: $37.19 - 37.13 = 0.06 \text{ g}$

Key Takeaways

Always carefully track which masses correspond to which components. The difference between 'crucible + M' and 'crucible + residue' directly gives the mass of sulfur incorporated into the product.

Common Mistakes

Subtracting the wrong values (e.g., residue - empty crucible gives total mass of product, not just sulfur).
Forgetting to align the columns correctly for each student.

Things to Be Careful About

Ensure calculations are done to the correct number of decimal places (2 d.p. as given in the data). The final answers should match the precision of the input data.

Techniques used

calculate mass of reactants from mass differences

(c)

Identify the independent variable in this experiment.

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(d)

(i)

Plot a graph on the grid in Fig. 2.2 to show the relationship between mass of sulfur which reacts and mass of M.

Use a cross ( $\times$ ) to plot each data point. Draw a straight line of best fit which includes the origin.

Fig. 2.2

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Data points to plot (x, y):

Student 1: $(1.43, 0.46)$
Student 2: $(1.14, 0.28)$
Student 3: $(0.86, 0.20)$
Student 4: $(0.57, 0.15)$
Student 5: $(0.29, 0.06)$

Answer

A graph with the above 5 points plotted as crosses ( $\times$ ) and a straight line of best fit drawn passing through the origin $(0,0)$ and through the cluster of points.

Final answer

See diagram for plotted points and line of best fit through origin

Detailed explanation

Background Concept

A graph of mass of sulfur reacting against mass of metal M should ideally be a straight line through the origin, because the mass of sulfur reacting is directly proportional to the mass of metal M (Law of Definite Proportions). The gradient of this line represents the mass ratio of sulfur to metal in the compound.

Understanding the Question

We need to plot the data from part (b) on the provided grid (Fig. 2.2) and draw a line of best fit that includes the origin $(0,0)$ .

Approach

Plot each $(x, y)$ pair where $x$ = mass of M and $y$ = mass of S.
Draw a straight line that best represents the trend, ensuring it passes through $(0,0)$ because if $0\text{ g}$ of metal is used, $0\text{ g}$ of sulfur will react.

Step-by-Step Reasoning

Plotting: Mark each point accurately on the grid.
- $(1.43, 0.46)$ : x=1.43, y=0.46
- $(1.14, 0.28)$ : x=1.14, y=0.28
- $(0.86, 0.20)$ : x=0.86, y=0.20
- $(0.57, 0.15)$ : x=0.57, y=0.15
- $(0.29, 0.06)$ : x=0.29, y=0.06
Line of best fit: Draw a straight line. It must go through $(0,0)$ . It should have roughly equal numbers of points on either side, minimizing the distance to all points. Note that $(1.43, 0.46)$ is somewhat high compared to the others, so the line might pass slightly below it.

Key Takeaways

Lines of best fit through the origin are common in stoichiometry experiments. The gradient gives the constant mass ratio.

Common Mistakes

Forgetting to include the origin in the line of best fit.
Drawing a curve instead of a straight line.
Plotting points inaccurately (e.g., misreading the axis scales).

Things to Be Careful About

The x-axis goes from 0.0 to 1.6 in steps of 0.2. The y-axis goes from 0.00 to 0.50 in steps of 0.05. Ensure points are plotted within these ranges accurately.

Techniques used

plot data points on a graphdraw line of best fit through origin

(ii)

Circle the point on the graph you consider to be most anomalous.

Suggest one reason why this anomaly may have occurred during this experimental procedure.

Assume no error was made in the measurement of any mass.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Anomalous point: Student 1 (1.43, 0.46) or the point furthest from the line of best fit.
Reason: Not all of the excess sulfur had been removed/volatilised in step 4, so the final mass includes unreacted sulfur, making the mass of sulfur appearing to react too high.

Final answer

Anomalous point: Student 1; Reason: excess sulfur not fully removed

Detailed explanation

Background Concept

An anomalous result (or outlier) is a data point that does not fit the general trend of the data. In a proportional relationship, this would be a point that is far from the line of best fit. We must also consider experimental errors that could cause such a deviation, keeping in mind the specific steps of the method.

Understanding the Question

We need to identify which point on the graph is most anomalous and suggest a reason for it, assuming no measurement errors (i.e., the balance readings are correct).

Approach

Look at the ratios (y/x) or visually inspect the graph to find the point furthest from the line through the origin.
Relate the deviation to the experimental steps. If the y-value (mass of S) is too high, it means the final residue was heavier than it should be. This could happen if unreacted sulfur was not removed.

Step-by-Step Reasoning

Calculate ratios (mass S / mass M) for each student:
- S1: $0.46 / 1.43 \approx 0.321$
- S2: $0.28 / 1.14 \approx 0.246$
- S3: $0.20 / 0.86 \approx 0.233$
- S4: $0.15 / 0.57 \approx 0.263$
- S5: $0.06 / 0.29 \approx 0.207$
  Student 1 has a significantly higher ratio (0.321) than the others (which cluster around 0.23-0.26). Thus, Student 1's point is anomalous.
Reason: The mass of sulfur 'reacting' is calculated as (final mass) - (initial mass of M). If the final mass is too high, it means there is extra mass in the residue. Since we assume no measurement error, the extra mass must be unreacted sulfur that was not removed during heating. Step 4 says 'Continue heating strongly until no more sulfur can be seen', but visually it's hard to tell if all sulfur gas has escaped, especially if the crucible is covered or if sulfur sublimed and recondensed.

Key Takeaways

Anomalies often arise from incomplete reactions or failure to remove excess reagents. Always link the error to a specific step in the method.

Common Mistakes

Suggesting 'balance was not zeroed': the question says 'Assume no error was made in the measurement of any mass'.
Suggesting 'metal was impure': possible, but 'excess sulfur not removed' is the more direct link to the method described.

Things to Be Careful About

The reason must specifically address why the mass of sulfur appears to be too high (y-value too high). If you say 'metal was wet', that would make the initial mass of M too high, affecting the x-value, which is not the primary issue here (though it could be). Focus on the final residue mass being too high.

Techniques used

identify anomalous result in datapropose experimental error explanation

(e)

Determine the gradient of your line of best fit.

State the coordinates of both points you used in your calculation.

These must be selected from your line of best fit.

Give your gradient to three significant figures.

\text{coordinates 1} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\quad \text{coordinates 2} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Select two points on the line of best fit (not necessarily data points).
Example points: $(0.00, 0.00)$ and $(1.00, 0.24)$

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{0.24 - 0.00}{1.00 - 0.00} = 0.240

(Note: Actual coordinates will vary based on the drawn line. Typical acceptable gradient is between 0.230 and 0.250.)

\text{coordinates 1} = (0.00, 0.00) \quad \text{coordinates 2} = (1.00, 0.240)

Answer

Gradient $\approx 0.240$ (to 3 s.f.)

Final answer

gradient = 0.240 (example values)

Detailed explanation

Background Concept

The gradient of a line on a graph is calculated as $\frac{\text{change in y}}{\text{change in x}}$ . For a line through the origin, it is simply $\frac{y}{x}$ for any point on the line. The gradient represents the constant ratio between the two variables.

Understanding the Question

We need to calculate the gradient of the line of best fit drawn in part (d)(i). We must state the coordinates used and give the answer to 3 significant figures.

Approach

Choose two points on the line of best fit that are far apart to minimize reading errors.
Read their $(x, y)$ coordinates.
Calculate $\frac{y_2 - y_1}{x_2 - x_1}$ .

Step-by-Step Reasoning

Use the origin $(0,0)$ as one point if the line passes through it (which it should).
Choose another point on the line, e.g., where $x = 1.00$ . Read the corresponding $y$ value from the line (not the data point). Let's say $y = 0.240$ .
Gradient $= \frac{0.240 - 0}{1.00 - 0} = 0.240$ .
Ensure the answer is to 3 significant figures: $0.240$ .

Key Takeaways

Always use points on the line of best fit, not the actual data points, for gradient calculations. Use a large triangle (far apart points) to reduce percentage error in reading coordinates.

Common Mistakes

Using data points instead of points on the line (especially the anomalous point).
Not giving the answer to 3 significant figures.
Reading the wrong axis or misinterpreting the scale.

Things to Be Careful About

The coordinates must be 'selected from your line of best fit'. Do not just copy a data point unless it lies exactly on your line. The mark scheme accepts a range of gradients (typically 0.230 - 0.250) depending on how the line was drawn.

Techniques used

calculate gradient from line of best fit

(f)

Use your line of best fit in Fig. 2.2 to determine if the results obtained by the students are reliable.

Explain your answer.

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(g)

Another student suggests that the metal M used in the experiment is strontium, Sr, which forms strontium sulfide, SrS, when heated with sulfur.

(i)

Deduce the gradient of the line of best fit for the graph of mass of sulfur which reacts against mass of strontium for the compound SrS.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For strontium sulfide, $\text{SrS}$ :
Molar mass of $\text{S} = 32.1 \text{ g mol}^{-1}$
Molar mass of $\text{Sr} = 87.6 \text{ g mol}^{-1}$

The mass ratio of S to Sr in the compound is:

\text{gradient} = \frac{\text{mass of S}}{\text{mass of Sr}} = \frac{32.1}{87.6} = 0.366

Answer

Gradient $= 0.366$

Final answer

0.366

Detailed explanation

Background Concept

The gradient of the graph (mass of S reacting / mass of M) represents the mass ratio of sulfur to metal in the formed compound. For a known compound like $\text{SrS}$ , we can calculate this theoretical mass ratio using the relative atomic masses ( $A_r$ ) of the elements.

Understanding the Question

We are given that the metal might be strontium (Sr), forming SrS. We need to calculate what the gradient of the graph would be if this were true.

Approach

Find the molar mass of S and Sr.
Calculate the ratio $\frac{A_r(\text{S})}{A_r(\text{Sr})}$ .

Step-by-Step Reasoning

From the data sheet (or standard values): $A_r(\text{S}) = 32.1$ , $A_r(\text{Sr}) = 87.6$ .
In SrS, 1 mole of Sr reacts with 1 mole of S.
Mass of 1 mole of S = $32.1 \text{ g}$ .
Mass of 1 mole of Sr = $87.6 \text{ g}$ .
Gradient $= \frac{\text{mass of S}}{\text{mass of Sr}} = \frac{32.1}{87.6} \approx 0.3664$ .
To 3 significant figures, gradient $= 0.366$ .

Key Takeaways

The gradient of a mass-mass graph for a binary compound is equal to the ratio of the molar masses of the elements (adjusted for stoichiometry).

Common Mistakes

Using the wrong atomic masses.
Calculating the ratio the wrong way round (mass of M / mass of S instead of mass of S / mass of M). Remember the y-axis is mass of S and x-axis is mass of M.

Things to Be Careful About

Use the correct $A_r$ values. Ensure the ratio is $\frac{y}{x}$ , i.e., $\frac{\text{mass of S}}{\text{mass of Sr}}$ .

Techniques used

calculate theoretical mass ratio from formula

(ii)

Use your answer to (i) to explain if the results of the experiment support the student’s suggestion.

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