Chemistry 9701/33 — February/March 2017

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

questions

marks

120

minutes

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

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

The concentration of hydrogen peroxide may be given in $\text{mol dm}^{-3}$ or as 'volume strength'. You will determine the concentration of hydrogen peroxide in $\text{mol dm}^{-3}$ and in 'volume strength' by a gas collection method.

Hydrogen peroxide decomposes to form water and oxygen. The reaction is much faster in the presence of a catalyst such as manganese(IV) oxide.

2\mathrm{H}_2\mathrm{O}_2\mathrm{(aq)} \rightarrow 2\mathrm{H}_2\mathrm{O}\mathrm{(l)} + \mathrm{O}_2\mathrm{(g)}

'Volume strength' is defined as the volume of oxygen in $\text{cm}^3$ produced from the decomposition of $1.0\text{ cm}^3$ of hydrogen peroxide at room temperature and pressure. For example, $1.0\text{ cm}^3$ of '100 volume' hydrogen peroxide will produce $100\text{ cm}^3$ of oxygen.

FA 1 is a solution of hydrogen peroxide, $\mathrm{H}_2\mathrm{O}_2$ .
FA 2 is manganese(IV) oxide, $\mathrm{MnO}_2$ .

(a)

Method

Read the whole method before starting any practical work.

The diagram below may help you in setting up your apparatus.

Fig. 1.1

Fill the tub with water to a depth of about $5\text{ cm}$ .
Fill the $250\text{ cm}^3$ measuring cylinder completely with water. Hold a piece of paper towel firmly over the top, invert the measuring cylinder and place it in the water in the tub.
Remove the paper towel and clamp the inverted measuring cylinder so that the open end is in the water just above the base of the tub.
Rinse the $50\text{ cm}^3$ measuring cylinder with a little FA 1 then use it to transfer $150\text{ cm}^3$ of FA 1 into the reaction flask labelled X.
Check that the bung fits tightly in the neck of flask X, clamp flask X and place the end of the delivery tube into the inverted $250\text{ cm}^3$ measuring cylinder.
Remove the bung from the neck of the flask. Tip FA 2 into the hydrogen peroxide and replace the bung immediately. Remove the flask from the clamp and swirl it to mix the contents. Swirl the flask occasionally until no more gas is given off. Replace the flask in the clamp.
Measure and record the final volume of gas in the measuring cylinder in the space below.

Keep FA 1 for use in Question 2.

Result

DifficultyEasy

Worked solution

Answer

Record the final volume of oxygen gas collected in the measuring cylinder (e.g., V cm³). The value must be within 10% of the supervisor's provided value.

Final answer

See working / candidate-dependent (must be within 10% of supervisor's value)

Detailed explanation

Background Concept

In gas collection experiments using water displacement, the volume of gas evolved is measured by the volume of water displaced from an inverted measuring cylinder. The apparatus must be airtight so that all generated gas is directed into the measuring cylinder. Accuracy depends on ensuring the system is sealed before the reaction begins and reading the meniscus at eye level.

Understanding the Question

Part (a) asks you to perform the practical method described and record the final volume of oxygen gas collected in the measuring cylinder. You are not given a specific numerical answer; instead, you must record your experimental result and ensure it is reasonable (within 10% of the supervisor's expected value).

Approach

Simply follow the method, ensure the bung is replaced immediately after adding the catalyst, and read the final volume of gas in the measuring cylinder. Record this value with the correct unit (cm³ or dm³).

Step-by-Step Reasoning

Perform the experiment as described: add 150 cm³ of FA 1 to flask X, add FA 2, seal immediately, and swirl until gas evolution stops.
Read the final volume of gas in the inverted 250 cm³ measuring cylinder. This is the volume of oxygen collected.
Record the value with units (e.g., 145 cm³). Mark schemes for Paper 3 practicals award marks for unambiguous recording (M1) and for the value being within a reasonable range (typically ±10%) of the supervisor's value (M2).

Key Takeaways

Always record practical observations with clear units. In gas collection, ensure the system is sealed before the reaction starts to prevent gas loss, which would lead to an artificially low volume reading.

Common Mistakes

Forgetting to include units (cm³ or dm³) with the recorded volume.
Recording a value that is obviously outside the 10% tolerance of the supervisor's value (e.g., due to a leak or premature gas escape).

Things to Be Careful About

Ensure the measuring cylinder is fully filled with water and inverted without trapping air bubbles. Read the meniscus at eye level. The volume recorded is the final reading; since the cylinder was initially full (0 cm³ gas), the final reading is the total volume of oxygen collected.

Techniques used

record gas volume from water displacementassess concordancy with supervisor's value

(b)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Use the information on page 2 to calculate the 'volume strength' of FA 1.

'volume strength' of FA 1 = ............................

DifficultyMedium-Easy

Worked solution

Working

Volume strength = $\frac{\text{Volume of } \text{O}_2 \text{ collected (cm}^3\text{)}}{150}$

Answer

Volume strength = $\frac{\text{V(a)}}{150}$ (to 2–4 s.f.)

Final answer

V(O₂) / 150 (to 2-4 s.f.)

Detailed explanation

Background Concept

'Volume strength' is a commercial measure of hydrogen peroxide concentration. It is defined as the volume of oxygen gas (in cm³) produced at room temperature and pressure from the decomposition of 1.0 cm³ of the hydrogen peroxide solution. For example, '100 volume' H₂O₂ produces 100 cm³ of O₂ per 1 cm³ of solution.

Understanding the Question

Part (b)(i) asks you to calculate the volume strength of FA 1 using the volume of oxygen gas you collected in part (a). You added 150 cm³ of FA 1 to the flask.

Approach

Use the definition of volume strength: divide the total volume of oxygen gas collected (in cm³) by the volume of hydrogen peroxide solution decomposed (150 cm³). Ensure the answer is given to 2–4 significant figures.

Step-by-Step Reasoning

Let the volume of oxygen collected in part (a) be $V$ cm³.
Volume of FA 1 used = 150 cm³.
Volume strength = $\frac{V}{150}$ .
Calculate the value and round to 2–4 significant figures. For example, if $V = 150$ cm³, volume strength = 1.0.

Key Takeaways

Volume strength is a direct ratio of gas volume to solution volume. Always ensure the units for both volumes are the same (cm³) before dividing.

Common Mistakes

Forgetting to divide by 150 and just writing the gas volume.
Using incorrect significant figures (the mark scheme requires 2–4 s.f.).

Things to Be Careful About

Ensure the volume of oxygen is in cm³. If you recorded it in dm³, convert it to cm³ by multiplying by 1000 before dividing by 150.

Techniques used

calculate volume strength from gas volume and sample volume

(ii)

Calculate the number of moles of oxygen collected in the measuring cylinder.
[Assume 1 mole of gas occupies $24.0\text{ dm}^3$ under these conditions.]

moles of $\mathrm{O}_2$ = ............................ mol

DifficultyMedium-Easy

Worked solution

Working

Moles of $\text{O}_2$ = $\frac{\text{Volume of } \text{O}_2 \text{ (cm}^3\text{)}}{24.0 \times 1000}$

Answer

Moles of $\text{O}_2$ = $\frac{\text{V(a)}}{24000}$ mol (to 2–4 s.f.)

Final answer

V(O₂) / 24000 mol

Detailed explanation

Background Concept

At room temperature and pressure (RTP), 1 mole of any gas occupies 24.0 dm³ (or 24,000 cm³). This allows conversion between the volume of a gas and the number of moles using the equation: $n = \frac{V}{24000}$ (when $V$ is in cm³) or $n = \frac{V}{24.0}$ (when $V$ is in dm³).

Understanding the Question

Part (b)(ii) asks you to calculate the number of moles of oxygen gas collected in the measuring cylinder. You are given that 1 mole of gas occupies 24.0 dm³ under these conditions.

Approach

Convert the volume of oxygen collected (in cm³) to dm³ by dividing by 1000, then divide by the molar volume (24.0 dm³ mol⁻¹). Alternatively, divide the volume in cm³ directly by 24,000 cm³ mol⁻¹.

Step-by-Step Reasoning

Let the volume of oxygen collected be $V$ cm³.
Volume in dm³ = $\frac{V}{1000}$ dm³.
Moles of $\text{O}_2$ = $\frac{V / 1000}{24.0} = \frac{V}{24000}$ mol.
Calculate the value and round to 2–4 significant figures.

Key Takeaways

Always check the units of volume against the units of molar volume. 24.0 dm³ = 24,000 cm³. Mixing these up is a common source of error.

Common Mistakes

Forgetting to convert cm³ to dm³, resulting in an answer that is 1000 times too large.
Using 22.4 dm³ (which is for STP, not RTP).

Things to Be Careful About

The mark scheme specifically allows 2–4 significant figures. Do not over-round intermediate values; keep full precision in your calculator and round only at the end.

Techniques used

calculate moles of gas from volume using molar volume

(iii)

Using your answer to (ii) calculate the number of moles of hydrogen peroxide in the volume of FA 1 added to flask X.

moles of $\mathrm{H}_2\mathrm{O}_2$ = ............................ mol

DifficultyMedium-Easy

Worked solution

Working

From the equation: $2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$
Moles of $\text{H}_2\text{O}_2$ = $2 \times$ moles of $\text{O}_2$ (from part b(ii))

Answer

Moles of $\text{H}_2\text{O}_2$ = $2 \times \text{answer to (ii)}$ (to 2–4 s.f.)

Final answer

2 × moles of O₂

Detailed explanation

Background Concept

Stoichiometry allows us to relate the amounts of reactants and products in a chemical reaction using the balanced chemical equation. The coefficients in the balanced equation represent the molar ratio.

Understanding the Question

Part (b)(iii) asks for the moles of hydrogen peroxide that decomposed. You have already calculated the moles of oxygen produced in part (b)(ii).

Approach

Use the balanced equation to find the molar ratio between H₂O₂ and O₂, then multiply the moles of O₂ by this ratio.

Step-by-Step Reasoning

Balanced equation: $2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$ .
Molar ratio of H₂O₂ to O₂ is 2 : 1.
Moles of H₂O₂ = $2 \times$ moles of O₂.
Substitute the answer from part (b)(ii) and calculate to 2–4 significant figures.

Key Takeaways

Always write the balanced equation first to determine the correct molar ratio. Do not assume a 1:1 ratio unless the equation supports it.

Common Mistakes

Using a 1:1 ratio instead of 2:1.
Carrying forward an unrounded value from part (b)(ii) without noting that error carried forward (ecf) is allowed, but final answers must be to 2–4 s.f.

Things to Be Careful About

The mark scheme requires you to use your answer to (ii). If your answer to (ii) was wrong, you can still get this mark if you used your wrong value correctly (error carried forward). However, the final answer here must be to 2–4 significant figures.

Techniques used

use stoichiometry to find moles of reactant

(iv)

Calculate the concentration of hydrogen peroxide, FA 1, in $\text{mol dm}^{-3}$ .

concentration of $\mathrm{H}_2\mathrm{O}_2$ , FA 1 = ............................ $\text{mol dm}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

Concentration = $\frac{\text{Moles of } \text{H}_2\text{O}_2}{\text{Volume of FA 1 (dm}^3\text{)}}$
Volume of FA 1 = $\frac{150}{1000}$ dm³ = 0.150 dm³
Concentration = $\frac{\text{answer to (iii)} \times 1000}{150}$

Answer

Concentration of $\text{H}_2\text{O}_2$ , FA 1 = $\frac{\text{answer to (iii)} \times 1000}{150}$ mol dm⁻³ (to 2–4 s.f.)

Final answer

(moles of H₂O₂ × 1000) / 150 mol dm⁻³

Detailed explanation

Background Concept

Concentration in mol dm⁻³ (molarity) is defined as the number of moles of solute per cubic decimetre (litre) of solution. The formula is $c = \frac{n}{V}$ , where $V$ must be in dm³. If $V$ is given in cm³, you must divide by 1000 or multiply $n$ by 1000.

Understanding the Question

Part (b)(iv) asks for the concentration of hydrogen peroxide in FA 1 in mol dm⁻³. You have the moles of H₂O₂ from part (b)(iii) and the volume of FA 1 added was 150 cm³.

Approach

Divide the moles of H₂O₂ by the volume of FA 1 in dm³. Alternatively, use the formula: Concentration = $\frac{n \times 1000}{V(\text{cm}^3)}$ .

Step-by-Step Reasoning

Moles of H₂O₂ = answer to part (b)(iii).
Volume of FA 1 = 150 cm³ = 0.150 dm³.
Concentration = $\frac{\text{moles}}{0.150} = \frac{\text{moles} \times 1000}{150}$ mol dm⁻³.
Calculate the value and round to 2–4 significant figures.

Key Takeaways

Always ensure volume is in dm³ when calculating concentration in mol dm⁻³. Multiplying moles by 1000 and dividing by volume in cm³ is a quick and reliable method.

Common Mistakes

Forgetting to convert 150 cm³ to dm³, resulting in an answer 1000 times too small.
Using the volume of oxygen gas instead of the volume of hydrogen peroxide solution.

Things to Be Careful About

The mark scheme explicitly requires showing the working $\frac{\text{(iii)} \times 1000}{150}$ . Do not just write the final number; show the substitution to earn the method mark.

Techniques used

calculate concentration from moles and volume

(c)

(i)

A source of error in this experiment is that some oxygen escapes before the bung can be inserted.

Suggest a change to the practical procedure given in (a) to reduce this source of error. You may draw a diagram as part of your answer.

DifficultyMedium-Easy

Worked solution

Answer

Use a dropping funnel or syringe to add FA 1 (hydrogen peroxide) to the flask, and subtract the volume of liquid added from the total gas volume collected. OR add FA 2 (MnO₂) using an ignition tube or weighing boat without removing the bung from the flask.

Working

No change in volume of gas collected; the volume of liquid added displaces an equal volume of gas, which must be accounted for if FA 1 is added via a funnel.

Final answer

Use a dropping funnel/syringe for FA 1 (and subtract its volume) OR add FA 2 via an ignition tube/weighing boat without removing the bung.

Detailed explanation

Background Concept

In gas collection experiments, any delay between starting the reaction and sealing the system results in gas escaping into the atmosphere, leading to an artificially low volume reading. To eliminate this error, the reactants must be brought together without opening the system.

Understanding the Question

Part (c)(i) identifies a source of error: oxygen escapes before the bung can be inserted. You are asked to suggest a change to the procedure to reduce this error.

Approach

Propose a method to add the catalyst (or the reactant) without removing the bung. Alternatively, add the reactant via a dropping funnel so the system remains sealed, and correct for the volume of liquid added.

Step-by-Step Reasoning

The error occurs because the bung is removed to add MnO₂, allowing O₂ to escape.
Improvement 1: Use a dropping funnel or syringe to add FA 1 (H₂O₂) to the flask containing MnO₂. The system remains sealed. Since adding liquid displaces gas, subtract the volume of FA 1 added from the total gas volume collected.
Improvement 2: Add FA 2 (MnO₂) without removing the bung. This can be done by placing MnO₂ in a small ignition tube or a floating weighing boat, lowering it into the flask through the bung, and then sealing it immediately.
Both methods prevent gas loss before the system is sealed.

Key Takeaways

When evaluating gas collection methods, consider how reactants are introduced. If the system must be opened, gas will escape. Use sealed addition methods (dropping funnels, ignition tubes) to eliminate this error.

Common Mistakes

Suggesting 'do it faster' without providing a practical apparatus change.
Forgetting to mention subtracting the volume of liquid added if a dropping funnel is used for the liquid reactant.

Things to Be Careful About

The mark scheme accepts either adding the solid catalyst without removing the bung OR adding the liquid via a dropping funnel (with the volume correction). Ensure your suggestion is practically feasible.

Techniques used

propose improvement to reduce gas loss before sealing

(ii)

The error in reading a $50\text{ cm}^3$ measuring cylinder is $\pm 0.5\text{ cm}^3$ .

Calculate the maximum percentage error in the volume of hydrogen peroxide added to flask X in (a).

maximum percentage error in volume of $\mathrm{H}_2\mathrm{O}_2$ = ............................ %

DifficultyMedium-Easy

Worked solution

Working

Absolute error in 50 cm³ measuring cylinder = $\pm 0.5$ cm³
Percentage error for 50 cm³ = $\frac{0.5}{50} \times 100 = 1.0\%$
Volume of H₂O₂ added = 150 cm³ (measured three times using 50 cm³ cylinder)
Total absolute error = $0.5 \times 3 = 1.5$ cm³
Maximum percentage error = $\frac{1.5}{150} \times 100 = 3.0\%$

Answer

Maximum percentage error in volume of H₂O₂ = 3.0%

Final answer

3.0%

Detailed explanation

Background Concept

Percentage error is calculated as $\frac{\text{absolute error}}{\text{measured value}} \times 100$ . When a volume is measured in multiple steps using the same apparatus, the absolute errors add up. For example, measuring 150 cm³ using a 50 cm³ cylinder three times means the absolute error is $3 \times$ the error of one measurement.

Understanding the Question

Part (c)(ii) asks for the maximum percentage error in the volume of hydrogen peroxide added to flask X. The volume added is 150 cm³, measured using a 50 cm³ measuring cylinder with an error of $\pm 0.5$ cm³.

Approach

Calculate the percentage error for a single 50 cm³ measurement, then multiply by 3 (since 150 cm³ requires three 50 cm³ measurements) to get the total percentage error for 150 cm³.

Step-by-Step Reasoning

Absolute error per measurement = 0.5 cm³.
Number of measurements = $\frac{150}{50} = 3$ .
Total absolute error = $0.5 \times 3 = 1.5$ cm³.
Percentage error = $\frac{1.5}{150} \times 100 = 1.0\% \times 3 = 3.0\%$ .
Alternatively: Percentage error for 50 cm³ = $\frac{0.5}{50} \times 100 = 1.0\%$ . For 150 cm³, it is $1.0\% \times 3 = 3.0\%$ .

Key Takeaways

When measuring a large volume using a smaller apparatus multiple times, the absolute errors accumulate. Always multiply the single measurement error by the number of measurements.

Common Mistakes

Calculating the percentage error for 50 cm³ (1.0%) and forgetting to multiply by 3 for the total volume of 150 cm³.
Using 150 cm³ as the denominator with 0.5 cm³ as the numerator, giving 0.33%.

Things to Be Careful About

The mark scheme awards 2 marks: one for the 1.0% calculation (or equivalent working) and one for the final 3.0% answer. If you show 3.0% with no working, you still get both marks, but showing working is safer.

Techniques used

calculate percentage error from absolute error

(iii)

Explain why the presence of $20\text{ cm}^3$ of air in the $250\text{ cm}^3$ measuring cylinder before the start of the experiment would decrease the accuracy of the results obtained in (a).

DifficultyMedium

Worked solution

Answer

Two readings are needed to find the volume of gas evolved (final reading minus initial reading). The presence of air means the initial reading is not zero, introducing an additional reading error. Thus, there is twice the percentage error in the gas volume reading.

Working

Volume of gas = Final reading - Initial reading
Error = Error in final reading + Error in initial reading

Final answer

Two readings are needed to find the volume of gas evolved, so there is twice the percentage error in the gas volume reading.

Detailed explanation

Background Concept

In water displacement gas collection, the volume of gas produced is calculated as the difference between the final reading and the initial reading on the measuring cylinder. If the cylinder is initially full of water, the initial reading is 0 cm³, and only one reading (the final one) is needed. If there is air in the cylinder, both an initial and a final reading must be taken.

Understanding the Question

Part (c)(iii) asks why the presence of 20 cm³ of air in the measuring cylinder before the experiment would decrease the accuracy of the results.

Approach

Explain that gas volume is found by subtracting the initial reading from the final reading. Two readings mean two potential errors, doubling the percentage error compared to starting from zero.

Step-by-Step Reasoning

Normally, the measuring cylinder is filled completely with water, so the initial gas volume reading is 0.0 cm³.
If 20 cm³ of air is present, the initial reading is 20.0 cm³ (or some non-zero value).
The volume of oxygen evolved = Final reading - Initial reading (e.g., 145.0 - 20.0 = 125.0 cm³).
Both the final and initial readings have an associated reading error (e.g., $\pm 0.5$ cm³ or $\pm 1$ division).
When subtracting two readings, the absolute errors add: Total error = Error in final + Error in initial.
This means there are two readings to make instead of one, so the percentage error in the calculated gas volume is approximately twice as large as if the initial reading was zero.

Key Takeaways

When a measurement requires a difference between two values, the absolute errors from both readings combine. Always try to start from zero if possible to minimize the number of readings and reduce cumulative error.

Common Mistakes

Saying 'the air takes up space' without explaining the effect on readings.
Failing to mention that two readings are needed to find the volume of gas evolved.

Things to Be Careful About

The mark scheme specifically looks for the point that 'two readings are needed' leading to 'twice the percentage error'. Ensure your explanation links the presence of air to the need for an initial reading.

Techniques used

evaluate effect of initial air volume on gas volume measurement

(d)

If you repeated the method described using half the mass of FA 2, what volume of gas would you expect to collect? Explain your answer.

DifficultyEasy

Worked solution

Answer

No change in the volume of gas collected.
MnO₂ (FA 2) is a catalyst; it increases the rate of reaction but does not affect the amount (yield) of oxygen produced.

Working

Catalysts lower the activation energy, providing an alternative pathway, but do not change the stoichiometry or the limiting reactant.

Final answer

No change; MnO₂ is a catalyst and does not affect the yield/amount of product.

Detailed explanation

Background Concept

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the overall reaction. It works by providing an alternative reaction pathway with a lower activation energy. Importantly, a catalyst does not change the position of equilibrium or the total amount of product formed from a given amount of reactant.

Understanding the Question

Part (d) asks what would happen to the volume of gas collected if half the mass of FA 2 (MnO₂) was used instead of the full mass.

Approach

Recognize that MnO₂ is a catalyst. Reducing the amount of catalyst will decrease the rate of reaction (it will take longer to finish), but it will not change the total volume of oxygen produced, as the amount of limiting reactant (H₂O₂) is unchanged.

Step-by-Step Reasoning

Identify the role of FA 2 (MnO₂): it is a catalyst for the decomposition of H₂O₂.
Effect of catalyst on rate: less catalyst means a slower rate of reaction (longer time to collect all gas).
Effect of catalyst on yield: catalysts do not affect the stoichiometry or the theoretical yield. The volume of O₂ depends only on the moles of H₂O₂ decomposed.
Since the volume of H₂O₂ (150 cm³) is unchanged, the total moles of O₂ produced, and thus the total volume of gas collected, will be the same.
Conclusion: No change in the final volume of gas collected.

Key Takeaways

Catalysts affect the rate of reaction, not the yield. Students often confuse catalysts with reactants and incorrectly assume that less catalyst means less product.

Common Mistakes

Stating that less catalyst means less gas is produced.
Saying 'the reaction will be slower' but forgetting to explicitly state that the final volume of gas is unchanged.

Things to Be Careful About

The question asks for the volume of gas expected and an explanation. Both parts are needed for the mark. Ensure you clearly state 'no change' and identify MnO₂ as a catalyst.

Techniques used

understand catalyst function and effect on yield

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations12M
Q3Qualitative Analysis17M

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