Chemistry 9701/35 — May/June 2022

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

questions

marks

120

minutes

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

Q1Presentation of Data and ObservationsManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationFree sample

Quantitative analysis

Read through the whole method before starting any practical work. Where appropriate, prepare a table for your results in the space provided.

Show the precision of the apparatus you used in the data you record.

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

Basic copper(II) carbonate contains both copper(II) carbonate, $\text{CuCO}_3$ , and copper(II) hydroxide, $\text{Cu(OH)}_2$ . The ratio of these two components can be different in samples from different sources.

This means that the formula of basic copper(II) carbonate can be written as $\text{CuCO}_3 \cdot x\text{Cu(OH)}_2$ .

Both the carbonate and the hydroxide react with acids.

\text{CuCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CuCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

\text{Cu(OH)}_2(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CuCl}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l})

You will determine the value of $x$ in a sample of basic copper(II) carbonate by reacting it with excess acid and measuring the mass of carbon dioxide given off.

FA 1 is basic copper(II) carbonate, $\text{CuCO}_3 \cdot x\text{Cu(OH)}_2$ .
FA 2 is $2.0 \text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .

(a)

Method

Use the $25\text{ cm}^3$ measuring cylinder to transfer $25.0\text{ cm}^3$ of FA 2 into a conical flask.
Weigh the flask with the acid. Record the mass.
Weigh the container with FA 1. Record the mass.
Carefully tip all of FA 1 into the acid in the conical flask. Swirl the contents of the flask and leave the flask to stand.
Weigh the container with any residual FA 1. Record the mass.
Calculate and record the mass of FA 1 added to the flask.
Calculate and record the theoretical initial mass of flask + acid + FA 1.
Swirl the flask occasionally. Weigh the flask and contents after approximately 5 minutes. Record the mass.

During this step you may wish to continue with Question 2 or Question 3.

Calculate and record the mass of carbon dioxide given off during the experiment.

Results

DifficultyMedium-Easy

Worked solution

Answer

Set up a results table with the following headings, each with the unit / g:

mass of flask + acid
mass of container + FA 1
mass of container + residue
final (constant) mass of flask + contents
mass of FA 1 added
theoretical initial mass of flask + acid + FA 1
mass of CO2 evolved

Record all four weighings to the same number of decimal places, at least 2 dp.

Calculate the three derived masses:

mass of FA 1 added = (mass of container + FA 1) − (mass of container + residue)
theoretical initial mass = mass of flask + acid + mass of FA 1 added
mass of CO2 evolved = theoretical initial mass − final mass of flask + contents

Final answer

Results table with 7 headings (each with unit / g); four weighings to ≥ 2 dp; three masses calculated by difference.

Detailed explanation

Background Concept

This experiment is a quantitative analysis by mass loss. Only copper(II) carbonate produces a gas (CO2) when it reacts with the acid; copper(II) hydroxide produces only water, which stays in the flask. The loss in mass of the flask and its contents therefore equals the mass of CO2 that escaped. The whole determination rests on accurate weighing and careful recording.

Weighing by difference is a standard laboratory technique. Instead of trying to transfer an exact mass of solid, you weigh the container holding the solid before and after tipping it in; the difference gives the mass actually added, even if some solid sticks to the container.

Understanding the Question

Part (a) asks you to record your results in a table. The four marks are awarded for the quality of the recorded data, not for a particular numerical answer: (I) unambiguous headings with correct units, (II) all four weighings recorded to the same number of decimal places (at least 2 dp), (III) the three derived masses calculated correctly, and (IV) the ratio mass of FA 1 : mass of CO2 falling within 20% of the supervisor's value.

Approach

Plan the table before you start weighing. Decide on the headings and units first, then record each measurement directly as you take it. After the experiment, calculate the three derived masses from the weighings.

Step-by-Step Reasoning

The mark scheme requires seven headings, each with the unit / g:

mass of flask + acid
mass of container + FA 1
mass of container + residue
final (constant) mass of flask + contents
mass of FA 1 added
theoretical initial mass of flask + acid + FA 1
mass of CO2 evolved

Headings must be unambiguous — 'flask + acid' rather than just 'flask' — and the unit must appear either with each entry or in the heading.

All four weighings must be recorded to the same number of decimal places, and at least 2 dp, because the precision of the balance is part of the data. Mixing 2.50 and 2.5 would lose this mark.

Three masses are then calculated:

mass of FA 1 added = (mass of container + FA 1) − (mass of container + residue)
theoretical initial mass = mass of flask + acid + mass of FA 1 added
mass of CO2 evolved = theoretical initial mass − final mass of flask + contents

Key Takeaways

Always record raw data with headings that include units.
Keep the same number of decimal places throughout a column of readings.
Mass by difference avoids the need for perfect transfer of a solid.

Common Mistakes

Missing units in headings or entries.
Recording weighings to different numbers of decimal places.
Writing ambiguous headings such as 'flask' without saying what is in it.
Using the initial mass of the container + FA 1 directly as the mass of FA 1, forgetting the residue.

Things to Be Careful About

The mass of CO2 is a mass loss, so it must be smaller than the mass of FA 1. If your calculated mass of CO2 is not sensible, check the subtraction. The mark scheme also checks the ratio mass FA 1 / mass CO2 against the supervisor's value, so consistent, careful weighing throughout is essential.

Techniques used

construct a results table with clear headings and unitsrecord weighings to a consistent number of decimal placescalculate masses by difference

(b)

(i)

Calculate the amount, in mol, of carbon dioxide given off in the reaction.

$\text{amount of CO}_2 = \text{.............................. mol}$

DifficultyEasy

Worked solution

Working

Using the example values: mass of CO2 evolved = 0.50 g.

\text{amount of CO}_2 = \frac{\text{mass of CO}_2}{M(\text{CO}_2)} = \frac{0.50}{44} = 0.0114 \text{ mol}

Answer

0.0114 mol (2–4 s.f.)

Final answer

0.0114 mol (example value; 2–4 s.f.)

Detailed explanation

Background Concept

The amount of a substance in moles is related to its mass by:

\text{amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g mol}^{-1})}

The molar mass of CO2 is 12.0 + 2 × 16.0 = 44.0 g mol⁻¹. In this experiment the mass of CO2 is the loss in mass of the flask and contents, because CO2 is the only gas that escapes.

Understanding the Question

You have already recorded the mass of CO2 given off (part (a)). This part asks you to convert that mass into an amount in moles, using the molar mass of CO2.

Approach

Take the mass of CO2 from your results and divide it by 44.0.

Step-by-Step Reasoning

Using the representative values, mass of CO2 = 0.50 g:

\text{amount of CO}_2 = \frac{0.50}{44.0} = 0.01136... = 0.0114 \text{ mol}

The answer must be given to 2–4 significant figures, matching the precision of the measurements. 0.0114 mol has 3 significant figures, which is appropriate.

Key Takeaways

The mass loss equals the mass of CO2, the only escaping gas.
Moles = mass / molar mass.

Common Mistakes

Using the mass of FA 1 instead of the mass loss.
Using the wrong molar mass for CO2 (e.g. 28 for CO).
Giving too many significant figures (e.g. 0.011363636...) when the data only justify 2–4.

Things to Be Careful About

Make sure the mass of CO2 you use is the mass loss (theoretical initial mass − final mass), not the mass of FA 1. The mark scheme awards the mark for mass loss / 44 with the answer to 2–4 s.f.

Techniques used

convert mass of CO2 to moles using the molar mass

(ii)

Calculate the amount, in mol, of copper(II) carbonate in the sample of FA 1 that you added to the flask.

$\text{amount of CuCO}_3 = \text{.............................. mol}$

Hence calculate the mass of copper(II) carbonate in the sample of FA 1 that you added to the flask.

$\text{mass of CuCO}_3 = \text{.............................. g}$

DifficultyMedium-Easy

Worked solution

Working

From the equation, 1 mol of CuCO3 produces 1 mol of CO2, so:

\text{amount of CuCO}_3 = \text{amount of CO}_2 = 0.0114 \text{ mol}

\text{mass of CuCO}_3 = 0.0114 \times 123.5 = 1.41 \text{ g}

Answer

amount of CuCO3 = 0.0114 mol; mass of CuCO3 = 1.41 g

Final answer

0.0114 mol; 1.41 g

Detailed explanation

Background Concept

The equation for the reaction of copper(II) carbonate with hydrochloric acid is:

\text{CuCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CuCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

The stoichiometry is 1 : 1 between CuCO3 and CO2: one mole of CuCO3 produces exactly one mole of CO2. So the amount of CuCO3 in the sample equals the amount of CO2 you calculated in part (b)(i).

The molar mass of CuCO3 is 63.5 + 12.0 + 3 × 16.0 = 123.5 g mol⁻¹.

Understanding the Question

This part asks you to (1) state the amount of CuCO3, using the 1 : 1 stoichiometric link to CO2, and (2) convert that amount to a mass using the molar mass.

Approach

First copy the moles of CO2 as the moles of CuCO3. Then multiply by the molar mass of CuCO3 (123.5) to get the mass in grams.

Step-by-Step Reasoning

From part (b)(i), amount of CO2 = 0.0114 mol. Therefore:

\text{amount of CuCO}_3 = 0.0114 \text{ mol}

\text{mass of CuCO}_3 = 0.0114 \times 123.5 = 1.41 \text{ g}

The answer is given to 3 significant figures, within the required 2–4 s.f. range.

Key Takeaways

The 1 : 1 stoichiometry of the carbonate reaction lets you transfer the moles of CO2 directly to moles of CuCO3.
Mass = moles × molar mass.

Common Mistakes

Forgetting the 1 : 1 ratio and applying an incorrect factor.
Using the molar mass of CuCO3·xCu(OH)2 instead of CuCO3 alone.
Using 123.5 incorrectly (e.g. 123.5 × mass instead of × moles).

Things to Be Careful About

The molar mass of CuCO3 is 123.5, not the molar mass of the whole basic carbonate. The mark scheme requires moles × 123.5 with the answer to 2–4 s.f.

Techniques used

apply 1:1 stoichiometry between CuCO3 and CO2convert moles of CuCO3 to mass using the molar mass

(iii)

Calculate the mass of copper(II) hydroxide in the sample of FA 1 that you added to the flask.

$\text{mass of Cu(OH)}_2 = \text{.............................. g}$

DifficultyEasy

Worked solution

Working

\text{mass of Cu(OH)}_2 = \text{mass of FA 1} - \text{mass of CuCO}_3 = 2.50 - 1.41 = 1.09 \text{ g}

Answer

1.09 g

Final answer

1.09 g

Detailed explanation

Background Concept

The sample of FA 1 contains only two components: CuCO3 and Cu(OH)2. Once you know the mass of CuCO3, the mass of Cu(OH)2 is simply the total mass of FA 1 minus the mass of CuCO3.

Understanding the Question

Using the mass of FA 1 added (from your results) and the mass of CuCO3 (from part (b)(ii)), calculate the mass of Cu(OH)2 by subtraction.

Approach

Subtract the mass of CuCO3 from the total mass of FA 1.

Step-by-Step Reasoning

Using the representative values: mass of FA 1 = 2.50 g, mass of CuCO3 = 1.41 g.

\text{mass of Cu(OH)}_2 = 2.50 - 1.41 = 1.09 \text{ g}

Key Takeaways

The sample is a two-component mixture, so the second component's mass is found by difference.

Common Mistakes

Subtracting the wrong way round, giving a negative mass.
Using the mass of CO2 instead of the mass of CuCO3.

Things to Be Careful About

The mass of Cu(OH)2 must be positive and smaller than the mass of FA 1. If it is not, check the value from part (b)(ii).

Techniques used

calculate mass of Cu(OH)2 by difference

(iv)

Calculate the amount, in mol, of copper(II) hydroxide in the sample of FA 1 that you added to the flask.

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

Hence calculate the value of $x$ in the formula of basic copper(II) carbonate, $\text{CuCO}_3 \cdot x\text{Cu(OH)}_2$ .

$x = \text{..............................}$

DifficultyMedium-Easy

Worked solution

Working

\text{amount of Cu(OH)}_2 = \frac{1.09}{97.5} = 0.0112 \text{ mol}

x = \frac{\text{amount of Cu(OH)}_2}{\text{amount of CuCO}_3} = \frac{0.0112}{0.0114} = 0.98

Answer

x = 1.0 (to 2 s.f.)

Final answer

x = 1.0 (to 2 s.f.)

Detailed explanation

Background Concept

The value of x in CuCO3·xCu(OH)2 is the mole ratio of Cu(OH)2 to CuCO3 in the sample. For every one mole of CuCO3, there are x moles of Cu(OH)2. So:

x = \frac{\text{amount of Cu(OH)}_2}{\text{amount of CuCO}_3}

The molar mass of Cu(OH)2 is 63.5 + 2 × (16.0 + 1.0) = 63.5 + 34.0 = 97.5 g mol⁻¹.

Understanding the Question

Convert the mass of Cu(OH)2 (from part (b)(iii)) into moles, then divide by the moles of CuCO3 (from part (b)(ii)) to find x.

Approach

First, moles of Cu(OH)2 = mass / 97.5. Then divide by the moles of CuCO3.

Step-by-Step Reasoning

\text{amount of Cu(OH)}_2 = \frac{1.09}{97.5} = 0.0112 \text{ mol}

x = \frac{0.0112}{0.0114} = 0.98 \approx 1.0

The value of x is close to 1, meaning the formula is approximately CuCO3·Cu(OH)2.

Key Takeaways

x is a mole ratio, so both components must be in moles before forming the ratio.
The result should be a small whole number or close to one; a wildly non-integer value signals an experimental error.

Common Mistakes

Dividing masses instead of moles (x = 1.09/1.41 = 0.77 is wrong).
Using the molar mass of CuCO3 for Cu(OH)2.
Forming the ratio upside down.

Things to Be Careful About

Both numerator and denominator must be amounts in mol. The mark scheme requires moles Cu(OH)2 = mass / 97.5 and then the ratio moles Cu(OH)2 / moles CuCO3.

Techniques used

convert mass of Cu(OH)2 to molesdetermine the mole ratio Cu(OH)2 : CuCO3

(c)

In this determination you assume that hydrochloric acid is in excess.
Show, by calculation, that this assumption is correct.

DifficultyMedium

Worked solution

Working

\text{moles of HCl added} = 2.0 \times \frac{25}{1000} = 0.050 \text{ mol}

\text{moles of HCl required} = 2 \times (\text{moles CuCO}_3 + \text{moles Cu(OH)}_2) = 2 \times (0.0114 + 0.0112) = 0.0452 \text{ mol}

Since 0.050 mol > 0.0452 mol, the acid is in excess.

Answer

0.050 mol of HCl was added but only 0.0452 mol is required, so the assumption that HCl is in excess is correct.

Final answer

Correct — 0.050 mol HCl added > 0.0452 mol required.

Detailed explanation

Background Concept

Both components of the basic carbonate react with acid, and each consumes two moles of HCl per mole of component:

\text{CuCO}_3 + 2\text{HCl} \rightarrow \text{CuCl}_2 + \text{CO}_2 + \text{H}_2\text{O}

\text{Cu(OH)}_2 + 2\text{HCl} \rightarrow \text{CuCl}_2 + 2\text{H}_2\text{O}

So the total moles of HCl required = 2 × (moles of CuCO3 + moles of Cu(OH)2). The amount of HCl actually added is found from its concentration and volume: amount = concentration × volume (in dm³).

Understanding the Question

You used 25.0 cm³ of 2.0 mol dm⁻³ HCl. This part asks you to show, by calculation, that this amount exceeds what the sample needs, i.e. that the acid is in excess.

Approach

Calculate the moles of HCl added. Calculate the moles of HCl required from the amounts of CuCO3 and Cu(OH)2 found in part (b). Compare the two.

Step-by-Step Reasoning

Moles of HCl added:

\text{amount of HCl} = 2.0 \times \frac{25}{1000} = 0.050 \text{ mol}

Moles of HCl required:

\text{required} = 2 \times (0.0114 + 0.0112) = 2 \times 0.0226 = 0.0452 \text{ mol}

Since 0.050 mol > 0.0452 mol, the acid is in excess.

Alternative approach: the volume of acid required is:

\text{volume required} = \frac{0.0452 \times 1000}{2.0} = 22.6 \text{ cm}^3

which is less than the 25.0 cm³ used, confirming the excess.

Key Takeaways

Verifying an assumption by calculation is a standard analysis step.
Each mole of carbonate or hydroxide consumes two moles of HCl.

Common Mistakes

Forgetting the factor of 2 (each component needs 2 HCl).
Comparing moles of HCl with moles of CuCO3 alone, ignoring Cu(OH)2.
Not converting cm³ to dm³ when calculating moles of HCl.

Things to Be Careful About

The mark scheme awards: M1 for moles HCl = 2 × 25/1000 = 0.05 and moles reacting = moles CuCO3 + moles Cu(OH)2; M2 for the comparison 0.05 > 2 × (moles CuCO3 + moles Cu(OH)2). The alternative route (volume of HCl required < 25 cm³) also scores both marks.

Techniques used

calculate moles of HCl from concentration and volumecompare moles of HCl with the stoichiometric requirement

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Q3Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations13M

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