Chemistry 9701/35 — October/November 2019

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

questions

marks

120

minutes

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

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

In this experiment you will determine the concentration of a sample of hydrochloric acid. You will do this by measuring the volume of hydrogen produced when an excess of magnesium reacts with the acid.

\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}

FA 1 is magnesium powder, Mg.
FA 2 is hydrochloric acid, HCl.

(a)

Method

Weigh the container with FA 1. Record the mass.
Fill the tub with water to a depth of approximately $5\text{ cm}$ .
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 just above the base of the tub.
Use the $25\text{ cm}^3$ measuring cylinder to place $25.0\text{ cm}^3$ of FA 2 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 flask X. Tip all of FA 1 into flask X and replace the bung immediately. Remove the flask from the clamp and swirl to mix the contents.
Swirl the flask occasionally until no more gas is evolved. Replace the flask in the clamp.
Measure and record the final volume of gas in the measuring cylinder.
Weigh and record the mass of the container with any residual solid.
Calculate and record the mass of FA 1 used.

Keep FA 2 for use in Question 2.

DifficultyEasy

Worked solution

Answer

Record the initial mass of the container + FA 1 in a results table, with a clear heading and unit (g).
After the reaction, record the final mass of the container + any residual solid, with heading and unit (g).
Calculate and record the mass of FA 1 used:

\text{mass used} = \text{initial mass} - \text{final mass}

Record the final volume of gas collected in the inverted measuring cylinder, with unit (cm³).

For example, if initial mass = 25.40 g and final mass = 25.00 g, mass used = 0.40 g; if the final gas volume = 60.0 cm³, record 60.0 cm³.

Final answer

Record initial and final masses with units; calculate mass used by difference; record gas volume with unit (candidate-dependent).

Detailed explanation

Background Concept

This experiment measures the volume of hydrogen gas produced when an excess of magnesium reacts with hydrochloric acid. The equation is

\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}

The gas is collected over water in an inverted measuring cylinder. Because 1 mol of gas occupies 24.0 dm³ at the temperature of the experiment, the measured volume can be converted into moles of H₂, and the stoichiometry of the equation then gives the amount of HCl that reacted. The accuracy of the final concentration depends entirely on how carefully the masses and the gas volume are recorded.

Understanding the Question

Part (a) is a practical recording task, not a calculation. You are asked to carry out the method and record the data needed for the later calculations: the mass of FA 1 before reaction, the mass of the container with any residual solid after reaction, the mass of FA 1 used, and the final volume of gas collected. The marks are for correct recording with headings and units, and for calculating the mass used correctly.

Approach

Set up a clear results table before starting. Record every measurement with its unit. Calculate the mass of FA 1 used by subtracting the final mass from the initial mass. Record the final gas volume exactly as read from the measuring cylinder.

Step-by-Step Reasoning

Weigh the container with FA 1 and record the mass, e.g. 25.40 g.
After the reaction, weigh the container with any residual solid and record the mass, e.g. 25.00 g.
Calculate the mass used: 25.40 − 25.00 = 0.40 g. This is the mass of magnesium that actually reacted.
Record the final volume of gas in the 250 cm³ measuring cylinder, e.g. 60.0 cm³.
The mark scheme requires the volume to be within ±10% of the supervisor's value and all masses to have unambiguous headings and units.

Key Takeaways

Accurate recording with units is essential in practical work. Mass used is found by difference. The gas volume is a candidate-dependent reading that must be recorded with its unit.

Common Mistakes

Leaving out units or headings in the results table.
Recording the total mass instead of the mass used.
Not waiting until no more gas is evolved before reading the volume.
Reading the measuring cylinder without checking that it is vertical and at eye level.

Things to Be Careful About

Use the same balance for both mass readings. Make sure the bung is replaced immediately after adding FA 1 so that little gas escapes. Record the volume to the precision of the measuring cylinder (usually 1 cm³ or 0.5 cm³).

Techniques used

record masses with units in a results tablecalculate mass used by differencerecord gas volume with correct units

(b)

Calculations

(i)

Calculate the number of moles of hydrogen gas produced.
(Assume $1\text{ mol}$ of gas occupies $24.0\text{ dm}^3$ at this temperature.)

DifficultyEasy

Worked solution

Working

Using the example gas volume $V = 60.0\text{ cm}^3 = 0.0600\text{ dm}^3$ :

n(\text{H}_2) = \frac{V}{24.0} = \frac{0.0600}{24.0} = 0.00250\text{ mol}

In general, $n(\text{H}_2) = \dfrac{V(\text{cm}^3)}{24\,000}$ .

Answer

$0.00250$ mol (using $V = 60.0\text{ cm}^3$ ).

Final answer

0.00250 mol (for V = 60.0 cm3)

Detailed explanation

Background Concept

At the temperature and pressure of the experiment, one mole of any gas occupies 24.0 dm³. Therefore the number of moles of a gas can be found from its volume using

n = \frac{V}{24.0}

where V is in dm³. Since the gas volume is measured in cm³, it must first be divided by 1000 to convert to dm³, or the volume in cm³ can be divided directly by 24 000.

Understanding the Question

Using the final gas volume recorded in part (a), calculate the number of moles of hydrogen gas produced. The volume is candidate-dependent; in this worked solution a representative volume of 60.0 cm³ is used.

Approach

Convert the recorded volume from cm³ to dm³, then divide by the molar gas volume 24.0 dm³ mol⁻¹.

Step-by-Step Reasoning

Using V = 60.0 cm³:

V = 60.0\text{ cm}^3 = 0.0600\text{ dm}^3

n(\text{H}_2) = \frac{0.0600}{24.0} = 0.00250\text{ mol}

In general, n(H₂) = V(cm³)/24 000. The answer should be given to 2–4 significant figures.

Key Takeaways

The molar gas volume converts gas volume to moles. Always check units: volume must be in dm³ when using 24.0.

Common Mistakes

Forgetting to convert cm³ to dm³ and dividing 60.0 by 24.0 to get 2.5 mol.
Using 24 000 incorrectly.
Giving too many significant figures, e.g. 0.002500000.

Things to Be Careful About

Use the actual volume you recorded, not the example. Give the final answer to 2–4 significant figures.

Techniques used

convert gas volume from cm3 to dm3divide volume by molar gas volume to find moles

(ii)

Calculate the concentration of hydrochloric acid in FA 2.

DifficultyMedium-Easy

Worked solution

Working

From the equation, $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$ , $n(\text{HCl}) = 2n(\text{H}_2)$ .

n(\text{HCl}) = 2 \times 0.00250 = 0.00500\text{ mol}

Volume of FA 2 = $25.0\text{ cm}^3 = 0.0250\text{ dm}^3$ .

c(\text{HCl}) = \frac{0.00500}{0.0250} = 0.200\text{ mol dm}^{-3}

Answer

$0.200\text{ mol dm}^{-3}$ (using $V = 60.0\text{ cm}^3$ ).

Final answer

0.200 mol dm^-3 (for V = 60.0 cm3)

Detailed explanation

Background Concept

The balanced equation shows that 1 mol Mg produces 1 mol H₂ and reacts with 2 mol HCl. Therefore the amount of HCl that reacted is twice the amount of H₂ produced:

n(\text{HCl}) = 2n(\text{H}_2)

Concentration is amount of solute divided by volume of solution in dm³:

c = \frac{n}{V}

Understanding the Question

Using the moles of H₂ from part (b)(i), calculate the concentration of HCl in FA 2. The volume of FA 2 used is 25.0 cm³.

Approach

First find moles of HCl using the stoichiometric ratio 2:1. Then divide by the volume of FA 2 in dm³.

Step-by-Step Reasoning

From part (b)(i), n(H₂) = 0.00250 mol.

n(\text{HCl}) = 2 \times 0.00250 = 0.00500\text{ mol}

Volume of FA 2 = 25.0 cm³ = 0.0250 dm³.

c(\text{HCl}) = \frac{0.00500}{0.0250} = 0.200\text{ mol dm}^{-3}

The answer should be given to 2–4 significant figures.

Key Takeaways

Stoichiometry links the amount of H₂ to the amount of HCl. Concentration requires volume in dm³.

Common Mistakes

Forgetting the factor of 2.
Using 25.0 cm³ instead of 0.0250 dm³ in the concentration formula.
Writing the unit as mol/dm³ without the dm⁻³ or using wrong units.

Things to Be Careful About

Use the actual n(H₂) from your own reading. Keep the answer to 2–4 significant figures.

Techniques used

use stoichiometric ratio to find moles of HCldivide moles by volume to find concentration

(iii)

In this experiment the magnesium powder was in excess.

Calculate the mass of magnesium powder needed for complete reaction with all the hydrochloric acid in $25.0\text{ cm}^3$ of FA 2.

DifficultyMedium-Easy

Worked solution

Working

Moles of Mg needed = moles of H₂ = $0.00250\text{ mol}$ (since 1 mol Mg gives 1 mol H₂).

m(\text{Mg}) = 0.00250 \times 24.3 = 0.0608\text{ g}

Answer

$0.0608$ g (using $V = 60.0\text{ cm}^3$ ).

Final answer

0.0608 g (for V = 60.0 cm3)

Detailed explanation

Background Concept

From the equation, 1 mol Mg produces 1 mol H₂. Hence the moles of Mg needed for complete reaction with the HCl in 25.0 cm³ of FA 2 equals the moles of H₂ produced when all that HCl reacts, i.e. n(Mg) = n(H₂). The mass is found from

m = n \times M_r

where M_r(Mg) = 24.3.

Understanding the Question

Calculate the mass of magnesium powder needed so that all the HCl in 25.0 cm³ of FA 2 reacts. This is the minimum mass, not an excess.

Approach

Use n(H₂) from part (b)(i) as the moles of Mg, then multiply by the molar mass of Mg.

Step-by-Step Reasoning

Using n(H₂) = 0.00250 mol:

n(\text{Mg}) = 0.00250\text{ mol}

m(\text{Mg}) = 0.00250 \times 24.3 = 0.0608\text{ g}

The answer should be given to 2–4 significant figures.

Key Takeaways

Stoichiometry can be used to find the exact amount of a reactant needed. Mass = moles × molar mass.

Common Mistakes

Multiplying by 2 instead of using the 1:1 Mg:H₂ ratio.
Using the molar mass of Mg as 24.0 instead of 24.3.
Using the mass of HCl rather than Mg.

Things to Be Careful About

Use the actual n(H₂) from your reading. Give the answer in grams to 2–4 significant figures.

Techniques used

use stoichiometric ratio to find moles of Mgconvert moles to mass using molar mass

(c)

A student suggested two modifications to the method in (a) to give a more accurate value for the concentration.

For each suggestion, state whether you agree with the student and explain your answer.

Suggestion 1: Use magnesium ribbon rather than powdered magnesium; keep the rest of the experiment the same.

Suggestion 2: Use twice the mass of magnesium powder; keep the rest of the experiment the same.

DifficultyMedium

Worked solution

Answer

Suggestion 1: Agree. Magnesium ribbon has a smaller surface area than powder, so the reaction is slower. Less hydrogen is lost while the bung is being replaced, so the measured gas volume is more accurate.

Suggestion 2: Disagree. The magnesium is already in excess, so doubling its mass does not change the amount of HCl that reacts or the volume of hydrogen produced. (It would also make the reaction faster, so more gas could be lost before the bung is fitted.)

Final answer

Suggestion 1: agree; Suggestion 2: disagree.

Detailed explanation

Background Concept

The rate of a reaction between a solid and a solution depends on the surface area of the solid. Powdered magnesium has a much larger surface area than ribbon, so it reacts faster. In this gas-collection method, a little gas can escape while the bung is being replaced after adding the magnesium. A slower reaction gives more time to replace the bung before much gas is lost, so the measured volume is closer to the true volume.

The amount of hydrogen produced is determined by the limiting reagent. Magnesium is added in excess, so all the HCl reacts and the amount of H₂ depends only on the amount of HCl. Adding more magnesium than is needed does not increase the amount of H₂.

Understanding the Question

Two modifications are suggested to improve accuracy. For each, decide whether you agree and explain using chemistry. Suggestion 1 changes the form of magnesium (ribbon instead of powder). Suggestion 2 doubles the mass of magnesium powder.

Approach

For each suggestion, think about its effect on the rate of reaction and on the amount of gas collected. Consider whether the measured gas volume would be closer to or further from the true value.

Step-by-Step Reasoning

Suggestion 1: Magnesium ribbon has a smaller surface area than powder, so the reaction is slower. Less hydrogen escapes while the bung is being fitted, so the measured volume is more accurate. Agree.

Suggestion 2: The magnesium is already in excess, so doubling its mass does not change the amount of HCl that reacts or the volume of H₂ produced. In fact, a larger mass of powder could make the reaction even faster, causing more gas to be lost before the bung is replaced. Disagree.

Key Takeaways

Surface area affects rate, not the equilibrium amount of product. The limiting reagent controls the amount of product. Gas loss during the method is a source of error that can be reduced by slowing the reaction.

Common Mistakes

Thinking that more magnesium always produces more hydrogen.
Saying the ribbon is better because it is 'purer' or 'more accurate' without linking to rate and gas loss.
Forgetting that the reaction with powder is fast enough to lose gas while fitting the bung.

Things to Be Careful About

The mark scheme accepts either reason for Suggestion 2: magnesium is in excess, or the reaction is faster so more gas is lost. State at least one clearly.

Techniques used

compare surface area and rate of reactionevaluate effect of excess reagent on gas volume

(d)

Another student carried out the experiment in (a) but used less magnesium than that calculated in (b)(iii).

State and explain the effect this would have on the calculated concentration of hydrochloric acid in FA 2.

DifficultyMedium-Easy

Worked solution

Answer

Using less magnesium than the calculated amount means the magnesium is no longer in excess, so not all the HCl reacts. Less hydrogen is produced, so the calculated moles of H₂ (and hence the calculated concentration of HCl) is lower than the true value.

Final answer

Calculated concentration is lower.

Detailed explanation

Background Concept

In a reaction, the limiting reagent determines how much product is formed. If less magnesium is used than is needed to react with all the HCl, then magnesium becomes the limiting reagent and some HCl remains unreacted. The volume of H₂ produced is therefore smaller than it would be if all the HCl reacted.

The calculation in part (b) assumes that all the HCl in FA 2 reacted, because magnesium was in excess. If that assumption is false, the calculated amount of HCl is too low.

Understanding the Question

A student uses less magnesium than the mass calculated in part (b)(iii). State and explain the effect on the calculated concentration of HCl.

Approach

Trace the effect through the calculation: less Mg → less H₂ produced → smaller n(H₂) → smaller calculated n(HCl) → smaller calculated concentration.

Step-by-Step Reasoning

If the magnesium is insufficient, not all the HCl reacts. The volume of hydrogen collected is lower than it would be if all the HCl reacted. In the calculation, this lower volume gives a smaller n(H₂), and since n(HCl) = 2n(H₂), the calculated amount of HCl is also smaller. Dividing this smaller amount by the same volume of FA 2 gives a lower calculated concentration than the true concentration.

Key Takeaways

The limiting reagent controls the amount of product. An error in the amount of a reactant propagates through the stoichiometric calculation and affects the final result.

Common Mistakes

Saying the concentration is higher.
Only stating 'lower' without explaining that not all HCl reacts.
Confusing the effect on gas volume with the effect on concentration.

Things to Be Careful About

Use the word 'lower' clearly and link it to the incomplete reaction of HCl. The mark is for both the direction and the reason.

Techniques used

relate limiting reagent to amount of gas produceddeduce effect on calculated concentration

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation16M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation16M

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