Chemistry 9701/32 — May/June 2023

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

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.

In this experiment you will determine the relative atomic mass, $A_r$ , of metal $\mathbf{M}$ by thermal decomposition of its basic carbonate, $\mathbf{MCO}_3\cdot\mathbf{M(OH)}_2$ .

$\mathbf{FB\ 1}$ is the basic metal carbonate, $\mathbf{MCO}_3\cdot\mathbf{M(OH)}_2$ .

(a)

Method

Weigh the empty crucible with its lid. Record the mass.
Transfer all of the $\mathbf{FB\ 1}$ from the container into the crucible.
Weigh the crucible, lid and $\mathbf{FB\ 1}$ . Record the mass.
Calculate and record the mass of $\mathbf{FB\ 1}$ used.
Place the crucible and contents on a pipe-clay triangle.
Heat the crucible gently, with the lid on, for approximately 1 minute.
Heat strongly, with the lid off, for a further 4 minutes.
Replace the lid and leave the crucible to cool for at least 5 minutes.

During the cooling period, you may wish to begin work on Question 3.

When the crucible is cool, weigh the crucible with its lid and contents. Record the mass.
Place the crucible and contents on the pipe-clay triangle. Remove the lid.
Heat the crucible strongly for a further 2 minutes.
Replace the lid and leave the crucible to cool for at least 5 minutes.
When the crucible is cool, reweigh the crucible with its lid and contents. Record the mass.
Calculate and record the mass of residue obtained.

Results

Results Table 1

DifficultyMedium-Easy

Worked solution

Working

Results Table 1

Measurement	Mass / g
Mass of empty crucible and lid	25.000
Mass of crucible, lid and FB 1 (before heating)	26.200
Mass of crucible, lid and residue after 1st heating	25.800
Mass of crucible, lid and residue after 2nd heating	25.798
Mass of FB 1 used	1.200
Mass of residue (MO) obtained	0.798

Calculations:
Mass of FB 1 used = $26.200 - 25.000 = 1.200 \text{ g}$
Mass of residue = $25.798 - 25.000 = 0.798 \text{ g}$
Mass loss during heating = $26.200 - 25.798 = 0.402 \text{ g}$

Answer

See table and calculations above. The fourth weighing ( $25.798 \text{ g}$ ) is within $0.04 \text{ g}$ of the third weighing ( $25.800 \text{ g}$ ), confirming constant mass has been achieved.

Final answer

Representative results table with masses recorded to 3 decimal places, showing mass of FB 1 = 1.200 g and mass of residue = 0.798 g.

Detailed explanation

Background Concept

In quantitative thermal decomposition experiments, a sample is heated to drive off volatile products (such as CO₂ and H₂O) until only the solid residue remains. To ensure the reaction is complete, the sample is heated to constant mass: this means heating, cooling, and weighing repeatedly until the mass change between two consecutive weighings is negligible (typically within $\pm 0.04 \text{ g}$ for this type of apparatus). The mass loss corresponds to the combined mass of the gases and vapour evolved.

Understanding the Question

This part asks you to set up a results table for the thermal decomposition of the basic metal carbonate FB 1, record the four required weighings with appropriate precision, and calculate the mass of the original sample (FB 1) and the final solid residue. Since this is a practical paper, the actual masses depend on the specific sample provided, so a representative data set is used here to demonstrate the correct technique and calculations.

Approach

Table Setup: Create a table with unambiguous headings for the four weighings and a column for mass in grams. Include rows for the calculated masses of FB 1 and the residue.
Recording Data: Record the four weighings to the same number of decimal places (typically 2 or 3, depending on the balance precision). The final two weighings must be within $0.04 \text{ g}$ of each other to show constant mass.
Calculations: Subtract the empty crucible mass from the others to find the mass of FB 1 and the mass of the residue. The mass loss is the difference between the initial total mass and the final total mass.

Step-by-Step Reasoning

Weighing 1: Empty crucible + lid = $25.000 \text{ g}$ .
Weighing 2: Crucible + lid + FB 1 = $26.200 \text{ g}$ . Mass of FB 1 = $26.200 - 25.000 = 1.200 \text{ g}$ .
Weighing 3: After first heating = $25.800 \text{ g}$ .
Weighing 4: After second heating = $25.798 \text{ g}$ . Difference from weighing 3 is $0.002 \text{ g}$ , which is $< 0.04 \text{ g}$ , so constant mass is achieved.
Residue mass: $25.798 - 25.000 = 0.798 \text{ g}$ .
Mass loss: $26.200 - 25.798 = 0.402 \text{ g}$ .

Key Takeaways

Always include units in table headings, record all readings to the same decimal places, and heat to constant mass to ensure complete decomposition. The mass of the sample and the residue are derived by subtracting the empty crucible mass.

Common Mistakes

Forgetting to include units (g) in the table headings.
Recording weighings to different decimal places (e.g., mixing 25.0 and 25.000).
Failing to heat to constant mass (the last two readings must be within $0.04 \text{ g}$ ).
Calculating the mass of residue by subtracting the mass of FB 1 from the total mass loss instead of using the final weighing minus the empty crucible.

Things to Be Careful About

The mark scheme awards accuracy marks based on the ratio of mass of FB 1 to mass of residue. Ensure your calculations are carried through to at least 2 decimal places for the ratio.
Always show your working for the mass calculations as required by the instructions.

Techniques used

set up results table for thermal decompositionrecord masses to correct precisionheat to constant mass to ensure complete decomposition

(b)

(i)

When $\mathbf{FB\ 1}$ undergoes thermal decomposition, the products are the metal oxide, $\mathbf{MO}$ , carbon dioxide and water vapour.

Give the equation for the thermal decomposition of $\mathbf{FB\ 1}$ . Include state symbols.

DifficultyEasy

Worked solution

Answer

\text{MCO}_3\cdot\text{M(OH)}_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g})

Final answer

MCO3·M(OH)2(s) → 2MO(s) + CO2(g) + H2O(g)

Detailed explanation

Background Concept

Basic metal carbonates contain both the carbonate ion ( $\text{CO}_3^{2-}$ ) and the hydroxide ion ( $\text{OH}^-$ ) in their lattice. Upon strong heating, the carbonate decomposes to give the metal oxide and carbon dioxide, while the hydroxide decomposes to give the metal oxide and water vapour. The metal is the same in both parts of the formula, so the total amount of metal oxide produced is twice the amount of the original basic carbonate.

Understanding the Question

You are asked to write the balanced symbol equation for the thermal decomposition of the basic metal carbonate $\text{MCO}_3\cdot\text{M(OH)}_2$ , including state symbols. The products are given as the metal oxide (MO), carbon dioxide, and water vapour.

Approach

Write the unbalanced equation with the correct formulas and state symbols.
Balance the atoms: 1 M on the left gives 2 MO on the right (since there are two M atoms in $\text{MCO}_3\cdot\text{M(OH)}_2$ ). The $\text{CO}_3$ part gives 1 $\text{CO}_2$ , and the $\text{M(OH)}_2$ part gives 1 $\text{H}_2\text{O}$ .
Check that all atoms and charges are balanced.

Step-by-Step Reasoning

Reactant: $\text{MCO}_3\cdot\text{M(OH)}_2$ is a solid, so $(\text{s})$ .
Products: MO is a solid $(\text{s})$ , $\text{CO}_2$ is a gas $(\text{g})$ , $\text{H}_2\text{O}$ is a gas/vapour $(\text{g})$ at heating temperatures.
Balancing: The formula $\text{MCO}_3\cdot\text{M(OH)}_2$ contains 2 M atoms, 1 C atom, 5 O atoms, and 2 H atoms. On the right, $2\text{MO}$ gives 2 M and 2 O, $\text{CO}_2$ gives 1 C and 2 O, and $\text{H}_2\text{O}$ gives 2 H and 1 O. Total O on right = $2 + 2 + 1 = 5$ . Balanced.

Key Takeaways

Basic carbonates decompose to give twice the moles of metal oxide as the moles of basic carbonate, plus one mole each of $\text{CO}_2$ and $\text{H}_2\text{O}$ .

Common Mistakes

Forgetting state symbols, especially $(\text{g})$ for water vapour (liquid $(\text{l})$ is incorrect at decomposition temperatures).
Writing $\text{M}_2\text{O}$ instead of $2\text{MO}$ (the formula given is MO, so the coefficient must be 2).
Not balancing the oxygen atoms correctly.

Things to Be Careful About

Ensure the state symbol for water is $(\text{g})$ as it is produced as vapour during strong heating. The mark scheme is strict about state symbols.

Techniques used

write balanced equation for thermal decompositioninclude state symbols

(ii)

The amount, in mol, of carbon dioxide produced is given by the following formula.

\text{amount of CO}_2 = \frac{\text{mass loss during heating}}{(M_r \text{ of CO}_2 + M_r \text{ of water})}

Calculate the amount, in mol, of carbon dioxide produced in (a).

DifficultyMedium-Easy

Worked solution

Working

Mass loss during heating = $0.402 \text{ g}$ (from part (a))

\text{amount of CO}_2 = \frac{\text{mass loss}}{M_r(\text{CO}_2) + M_r(\text{H}_2\text{O})} = \frac{0.402}{44 + 18} = \frac{0.402}{62} = 0.00648 \text{ mol}

Answer

$0.00648 \text{ mol}$

Final answer

0.00648 mol

Detailed explanation

Background Concept

From the balanced equation in (b)(i), 1 mole of $\text{MCO}_3\cdot\text{M(OH)}_2$ produces 1 mole of $\text{CO}_2$ and 1 mole of $\text{H}_2\text{O}$ . Therefore, the total moles of gaseous products is $2 \times \text{moles of CO}_2$ . The total mass loss during heating is the combined mass of $\text{CO}_2$ and $\text{H}_2\text{O}$ evolved. Since they are produced in a 1:1 molar ratio, the mass loss can be attributed to $62 \text{ g}$ of gas per mole of $\text{CO}_2$ produced ( $44 \text{ g/mol}$ for $\text{CO}_2$ + $18 \text{ g/mol}$ for $\text{H}_2\text{O}$ ).

Understanding the Question

You are given a formula to calculate the amount (in mol) of $\text{CO}_2$ produced from the mass loss. You must substitute the mass loss value from your results table (part a) into this formula and calculate the answer to 2–4 significant figures.

Approach

Identify the mass loss from your calculated values in part (a).
Add the $M_r$ of $\text{CO}_2$ ( $44$ ) and $M_r$ of $\text{H}_2\text{O}$ ( $18$ ) to get $62$ .
Divide the mass loss by $62$ to get the moles of $\text{CO}_2$ .
Round to an appropriate number of significant figures (2–4).

Step-by-Step Reasoning

Mass loss = $0.402 \text{ g}$ (representative value from part a).
Denominator = $44 + 18 = 62 \text{ g mol}^{-1}$ .
Moles of $\text{CO}_2$ = $0.402 / 62 = 0.00648387... \text{ mol}$ .
Rounding to 3 significant figures: $0.00648 \text{ mol}$ .

Key Takeaways

The mass loss in this decomposition is entirely due to $\text{CO}_2$ and $\text{H}_2\text{O}$ escaping. Because they are produced in a 1:1 ratio, their combined molar mass ( $62 \text{ g/mol}$ ) can be used to find the moles of each individually.

Common Mistakes

Using the wrong mass loss value (e.g., using the mass of residue instead of the mass loss).
Forgetting to add the $M_r$ values correctly ( $44 + 18 = 62$ , not $44 - 18$ ).
Not rounding to the correct number of significant figures (must be 2–4 sf).

Things to Be Careful About

Use the exact mass loss value from your own results table. If your mass loss was different, your answer will differ, but the method is what is marked. Carry at least 3 significant figures through to the next step.

Techniques used

calculate moles from mass lossuse molar mass of combined gases

(iii)

Calculate the relative formula mass, $M_r$ , of the basic metal carbonate.
Show your working.

DifficultyMedium-Easy

Worked solution

Working

From the equation in (b)(i), 1 mol of $\text{MCO}_3\cdot\text{M(OH)}_2$ produces 1 mol of $\text{CO}_2$ .
Therefore, moles of $\text{MCO}_3\cdot\text{M(OH)}_2$ = moles of $\text{CO}_2$ = $0.00648 \text{ mol}$ .

M_r = \frac{\text{mass of FB 1 used}}{\text{moles of CO}_2} = \frac{1.200}{0.00648} = 185

Answer

$185$

Final answer

185

Detailed explanation

Background Concept

The relative formula mass ( $M_r$ ) of a compound is the mass of one mole of that compound in grams. By determining the mass of the sample used and the number of moles it contains (derived from the moles of a known product), we can calculate $M_r$ using the relationship: $M_r = \text{mass} / \text{moles}$ .

Understanding the Question

You must calculate the $M_r$ of the basic metal carbonate using the mass of FB 1 used (from part a) and the moles of $\text{CO}_2$ produced (from part b)(ii). The stoichiometry from the balanced equation shows a 1:1 molar ratio between the basic carbonate and $\text{CO}_2$ .

Approach

State that moles of basic carbonate = moles of $\text{CO}_2$ (from 1:1 ratio in equation).
Use the formula $M_r = \text{mass} / \text{moles}$ .
Substitute the mass of FB 1 and the calculated moles of $\text{CO}_2$ .
Calculate and round to 2–4 significant figures.

Step-by-Step Reasoning

Mass of FB 1 = $1.200 \text{ g}$ .
Moles of $\text{CO}_2$ = $0.00648 \text{ mol}$ .
Moles of FB 1 = $0.00648 \text{ mol}$ (1:1 ratio).
$M_r = 1.200 / 0.00648 = 185.185...$
Rounding to 3 significant figures: $185$ .

Key Takeaways

When a product is known and its moles are calculated, the moles of the reactant can be found using the stoichiometric ratio from the balanced equation. This allows calculation of the reactant's molar mass or $M_r$ .

Common Mistakes

Assuming a different molar ratio (e.g., 1:2) between the basic carbonate and $\text{CO}_2$ .
Using the mass of the residue instead of the mass of FB 1 in the numerator.
Not showing the working clearly; the mark scheme requires the correct use of the value from (b)(ii).

Things to Be Careful About

Ensure you use the unrounded value of moles from (b)(ii) if carrying forward, though for this calculation 3 sf is usually sufficient. The final answer must be to 2–4 significant figures.

Techniques used

calculate relative formula mass from moles and massuse stoichiometric ratio

(iv)

Calculate the relative atomic mass of metal $\mathbf{M}$ .

DifficultyMedium-Easy

Worked solution

Working

$M_r$ of $\text{MCO}_3\cdot\text{M(OH)}_2 = 2A_r(\text{M}) + A_r(\text{C}) + 5A_r(\text{O}) + 2A_r(\text{H})$
$M_r = 2A_r + 12 + (5 \times 16) + (2 \times 1) = 2A_r + 12 + 80 + 2 = 2A_r + 94$

Alternatively, using the mass of the non-metal parts:
$M_r = 2A_r + M_r(\text{CO}_3) + M_r(\text{OH})_2 = 2A_r + 60 + 34 = 2A_r + 94$

2A_r = M_r - 94 = 185 - 94 = 91

A_r = \frac{91}{2} = 45.5

Answer

$45.5$

Final answer

45.5

Detailed explanation

Background Concept

The relative formula mass ( $M_r$ ) of a compound is the sum of the relative atomic masses ( $A_r$ ) of all the atoms in its formula. For the basic metal carbonate $\text{MCO}_3\cdot\text{M(OH)}_2$ , the formula contains 2 atoms of metal M, 1 carbon, 5 oxygen, and 2 hydrogen atoms. The $M_r$ of the non-metal portion ( $\text{CO}_3\cdot(\text{OH})_2$ ) is $12 + (3 \times 16) + 2 \times (16 + 1) = 12 + 48 + 34 = 94$ . Alternatively, it can be seen as $M_r(\text{CO}_3) + M_r(\text{OH})_2 = 60 + 34 = 94$ .

Understanding the Question

You must calculate the relative atomic mass ( $A_r$ ) of metal M using the $M_r$ calculated in part (b)(iii). You need to set up an equation relating $M_r$ to $A_r$ and solve for $A_r$ .

Approach

Express $M_r$ in terms of $A_r$ : $M_r = 2A_r + 94$ (or $2A_r + 60 + 34$ ).
Rearrange to solve for $A_r$ : $A_r = (M_r - 94) / 2$ .
Substitute the $M_r$ value from (b)(iii) and calculate.
Round to 2–4 significant figures.

Step-by-Step Reasoning

$M_r = 185$ (from part b)(iii)).
Mass of non-metal parts = $60 (\text{for CO}_3) + 34 (\text{for (OH)}_2) = 94$ .
$2A_r = 185 - 94 = 91$ .
$A_r = 91 / 2 = 45.5$ .
The value $45.5$ is close to the $A_r$ of Scandium ( $45.0$ ), though experimental error is expected in this type of practical.

Key Takeaways

Breaking down the $M_r$ into the metal and non-metal components allows you to isolate and calculate the unknown $A_r$ . Always account for the number of metal atoms (which is 2 in this formula).

Common Mistakes

Forgetting that there are 2 atoms of M in the formula, so dividing by 1 instead of 2.
Calculating the mass of the non-metal part incorrectly (e.g., using $60 + 17 = 77$ instead of $60 + 34 = 94$ ).
Not showing the algebraic rearrangement clearly.

Things to Be Careful About

The mark scheme specifically allows the calculation as $A_r = [M_r - (60 + 34)] / 2$ . Ensure you use this exact logic or an equivalent correct algebraic form. Round the final answer to 2–4 significant figures.

Techniques used

calculate relative atomic mass from relative formula massalgebraic rearrangement

(c)

A student accidentally spilt a little of the residue before carrying out the final weighing.
Predict whether the calculated value of the relative atomic mass of $\mathbf{M}$ will be higher or lower as a result of this mistake.
Explain your answer.

The $A_r$ of $\mathbf{M}$ will be ............................. .

explanation ...............................................................................................................................

DifficultyMedium

Worked solution

Answer

The $A_r$ of M will be lower.

Explanation:
Spilling residue before the final weighing means the final mass recorded will be lower than it should be. This makes the mass loss greater than the true value. A greater mass loss leads to a greater calculated amount (mol) of CO₂ (and water). Since moles of basic carbonate = moles of CO₂, the calculated moles of basic carbonate will be greater, resulting in a lower calculated $M_r$ . A lower $M_r$ leads to a lower calculated $A_r$ for metal M.

Final answer

Lower; mass loss is greater, so moles of CO2 is greater, making calculated Mr lower and Ar lower.

Detailed explanation

Background Concept

In any multi-step calculation, an error in an initial measurement can propagate and affect the final result. It is crucial to understand the direction of the error: does it make the numerator larger or smaller, and does it make the denominator larger or smaller? Here, spilling residue affects the final mass, which is used to calculate the mass loss.

Understanding the Question

A student spills some residue before the final weighing. You must predict whether the calculated $A_r$ of M will be higher or lower, and explain why by tracing the effect of the error through the calculations.

Approach

Determine how the error affects the final mass reading.
Trace the effect on the mass loss.
Trace the effect on the moles of CO₂.
Trace the effect on the $M_r$ of the basic carbonate.
Trace the effect on the $A_r$ of M.

Step-by-Step Reasoning

Final mass: Spilling residue means the final weighed mass is lower than it should be.
Mass loss: Mass loss = initial mass - final mass. If final mass is lower, mass loss is greater.
Moles of CO₂: Moles of CO₂ = mass loss / 62. If mass loss is greater, moles of CO₂ is greater.
$M_r$ of basic carbonate: $M_r$ = mass of FB 1 / moles of CO₂. If moles of CO₂ is greater (denominator larger), $M_r$ is lower.
$A_r$ of M: $A_r = (M_r - 94) / 2$ . If $M_r$ is lower, $A_r$ is lower.

Key Takeaways

When evaluating errors, always trace the effect step-by-step through the calculation chain. A lower final mass in a decomposition experiment leads to an overestimation of mass loss, which overestimates moles of gas, which underestimates the molar mass of the reactant, and finally underestimates the atomic mass of the metal.

Common Mistakes

Predicting the wrong direction (e.g., saying $A_r$ will be higher).
Failing to explain the chain of reasoning (just stating "mass loss is greater" is not enough; you must link it to moles, then $M_r$ , then $A_r$ ).
Confusing the effect on mass loss vs. mass of residue.

Things to Be Careful About

The mark scheme requires the prediction (lower) AND the explanation. The explanation must mention that mass loss is greater (or moles of CO₂ is greater), leading to a lower $M_r$ and thus a lower $A_r$ . Any two of these linked points will earn the mark.

Techniques used

evaluate effect of experimental error on calculated valuetrace error through calculation chain

(d)

A student suggested that addition of sulfuric acid to the residue from (a) would show whether the basic metal carbonate had decomposed fully.
State whether the student is correct.
Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The student is not correct.

Explanation:
Sulfuric acid will react with any unreacted basic metal carbonate (or metal oxide/hydroxide) to produce effervescence (CO₂) or dissolve. However, it will not distinguish between complete and incomplete decomposition because the residue (MO) will also react with sulfuric acid to form a soluble salt and water (no visible gas if only MO is present, but it dissolves). More critically, if decomposition is incomplete, the unreacted basic carbonate will fizz with acid, but the student cannot easily tell if the fizzing is from residual carbonate or if the reaction is truly complete without a control. Actually, the key point is: sulfuric acid will not show if the metal hydroxide is still present, as both the carbonate and the oxide/hydroxide will react with acid (carbonate fizzes, oxide/hydroxide dissolves without fizzing, but it's not a clear visual test for completeness compared to heating to constant mass). Alternatively, if the student carries out the test and observes fizzing, it means decomposition is incomplete; if no fizzing, it is complete. But the suggestion as stated is flawed because the residue MO will also react with H₂SO₄ to form MSO₄ and H₂O, which may mask the observation or is not a definitive test without comparing to a known sample.

Correct mark scheme interpretation: The student is not correct because sulfuric acid will not show if the metal hydroxide is still present (if decomposition is incomplete, you might have MO and unreacted M(OH)₂ or MCO₃; H₂SO₄ reacts with all of them, fizzing only with carbonate, but it doesn't clearly indicate if ALL carbonate is gone without careful observation). OR the student is correct IF they observe fizzing (incomplete) or no fizzing (complete), but the mark scheme accepts either if explained correctly.

Primary Answer:
The student is not correct. Sulfuric acid will react with the metal oxide residue (MO) to form a soluble sulfate and water, which does not provide a clear visual indication of whether unreacted carbonate remains, as both will dissolve/react. (Or: Sulfuric acid will not distinguish between the oxide and the hydroxide/carbonate clearly enough).

Final answer

Not correct; sulfuric acid will react with the metal oxide residue as well, so it cannot clearly show if decomposition is complete (or: will not show if metal hydroxide is still present).

Detailed explanation

Background Concept

To determine if a thermal decomposition is complete, one typically heats to constant mass. Adding an acid to the residue is a proposed alternative test. Carbonates react with acids to produce CO₂ gas (effervescence/fizzing). Metal oxides and metal hydroxides react with acids to form salts and water (dissolution, no gas). If decomposition is incomplete, the residue contains a mixture of MO and unreacted MCO₃·M(OH)₂. If complete, it is pure MO.

Understanding the Question

A student suggests adding sulfuric acid to the residue to check if decomposition is complete. You must state whether this is correct and explain why.

Approach

Consider what happens if decomposition is complete (residue is pure MO). MO + H₂SO₄ → MSO₄ + H₂O (no fizzing, solid dissolves).
Consider what happens if decomposition is incomplete (residue is MO + MCO₃·M(OH)₂). Carbonate part will fizz (CO₂ produced).
Evaluate if this is a reliable test. The mark scheme indicates the student is not correct because the acid will react with the oxide/hydroxide components, and it may not clearly show if decomposition is fully complete, or specifically, it will not show if the metal hydroxide portion is still present.

Step-by-Step Reasoning

If decomposition is complete, residue is MO. MO reacts with H₂SO₄ to give MSO₄(aq) and H₂O. No visible gas.
If incomplete, residue contains MCO₃·M(OH)₂. The carbonate part will react with H₂SO₄ to give CO₂(g) (fizzing).
However, the mark scheme states the student is not correct because sulfuric acid will not show if the metal hydroxide is still present. The hydroxide part of the basic carbonate will react with acid to form water and salt, which is not easily distinguishable from the oxide reacting. Thus, fizzing only confirms carbonate is present, but doesn't definitively prove complete decomposition if other components are ambiguous, or the test is not reliable.
Alternative acceptance: If the student carries out the test and observes fizzing (incomplete) or no fizzing (complete), they could argue it is correct. The mark scheme allows this if explained properly.

Key Takeaways

Acid tests are useful for carbonates (fizzing), but for basic salts containing both carbonate and hydroxide/oxide, the test may not be definitive because multiple components can react with acid without producing gas (oxide, hydroxide).

Common Mistakes

Simply saying "yes, it will fizz if carbonate is present" without addressing why it's not a good test for completeness.
Not stating clearly whether the student is correct or not.
Forgetting that metal oxides also react with acids.

Things to Be Careful About

The mark scheme has two acceptable routes:

Student is not correct + explanation that acid will not show if hydroxide is present, or will react with oxide too.
Student is correct + explanation that if residue fizzes, decomposition is incomplete; if no fizz, it is complete.
Ensure your explanation matches your prediction. The primary expected answer is that the student is not correct.

Techniques used

evaluate proposed test for completeness of reactionidentify limitations of chemical tests

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 Evaluation13M

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