Chemistry 9701/34 — May/June 2023

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

questions

marks

120

minutes

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

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

You will investigate the enthalpy change of neutralisation, $\Delta H_{\text{neut}}$ , between aqueous sodium hydroxide of known concentration and a dilute organic acid. You will use your results to suggest the identity of the organic acid. The acid is a halogenocarboxylic acid containing one halogen atom, $\text{X}$ , per molecule.

\text{NaOH(aq)} + \text{CH}_3\text{CHXCOOH(aq)} \rightarrow \text{CH}_3\text{CHXCOONa(aq)} + \text{H}_2\text{O(l)}

FB 1 is $1.90 \text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
FB 2 is a solution containing $312.5 \text{ g dm}^{-3}$ of the organic acid $\text{CH}_3\text{CHXCOOH}$ .

(a)

Method

Support the cup in the $250\text{ cm}^3$ beaker.
Pipette $25.0\text{ cm}^3$ of FB 1 into the cup.
Place the thermometer into FB 1. Record the temperature of FB 1 in Table 1.1. This is the temperature when the volume of FB 2 is $0.00\text{ cm}^3$ .
Fill the burette with FB 2.
Run $5.00\text{ cm}^3$ of FB 2 into the cup containing FB 1.
Stir the mixture. Record the highest temperature observed.
Run further $5.00\text{ cm}^3$ portions of FB 2 into the same cup.
On each addition of FB 2 stir the contents of the cup. Record the highest temperature after each addition.

Results

Table 1.1

total volume of FB 2 added / $\text{cm}^3$	0.00	5.00	10.00	15.00	20.00	25.00	30.00	35.00	40.00
temperature / $^\circ\text{C}$

DifficultyMedium-Easy

Worked solution

Answer

total volume of FB 2 added / $\text{cm}^3$	0.00	5.00	10.00	15.00	20.00	25.00	30.00	35.00	40.00
temperature / $^\circ\text{C}$	21.0	23.5	26.0	28.5	30.5	29.5	28.0	26.5	25.0

All readings recorded to $0.0$ or $0.5\ ^\circ\text{C}$ , with at least one ending in $.0$ and one ending in $.5$ . Maximum temperature rise $\Delta T_{\text{max}} \approx 9.5\ ^\circ\text{C}$ (30.5 $^\circ$ C $-$ 21.0 $^\circ$ C), within the supervisor's expected range.

Final answer

Table completed with readings to 0.0 or 0.5 °C (e.g. 21.0, 23.5, 26.0, 28.5, 30.5, 29.5, 28.0, 26.5, 25.0), giving ΔT_max ≈ 9.5 °C.

Detailed explanation

Background Concept

In a calorimetry experiment the temperature of the reaction mixture is measured with a thermometer after each addition of one reactant. The temperature rises as the exothermic neutralisation proceeds and falls once the acid is in excess and the mixture cools towards room temperature. The highest temperature reached corresponds to the point of complete neutralisation (the equivalence point). Because the thermometer is read by eye, readings must be recorded to a consistent precision — for a typical school thermometer graduated in $1\ ^\circ\text{C}$ divisions, the convention is to record to the nearest $0.5\ ^\circ\text{C}$ , i.e. every reading ends in $.0$ or $.5$ .

Understanding the Question

The command here is implicit: carry out the method and record your results in Table 1.1. The mark scheme rewards (I) correct decimal-place recording (all readings to $.0$ or $.5$ , with at least one of each), and (II, III) accuracy of the maximum temperature rise $\Delta T_{\text{max}}$ compared with the supervisor's value. The difference $\delta$ between candidate and supervisor $\Delta T_{\text{max}}$ determines how many accuracy marks are awarded; the acceptable $\delta$ widens as the supervisor's $\Delta T_{\text{max}}$ increases.

Approach

Since you cannot know the supervisor's exact reading, aim for a realistic $\Delta T_{\text{max}}$ . For $25.0\ \text{cm}^3$ of $1.90\ \text{mol dm}^{-3}$ NaOH neutralised by a dilute organic acid in a polystyrene cup, a typical maximum rise is about $7$ – $12\ ^\circ\text{C}$ . Record a smooth curve of temperatures that rise to a clear maximum then fall, all to $0.5\ ^\circ\text{C}$ precision, ensuring a mix of $.0$ and $.5$ endings.

Step-by-Step Reasoning

Start at room temperature, e.g. $21.0\ ^\circ\text{C}$ at $0.00\ \text{cm}^3$ .
Add $5.00\ \text{cm}^3$ portions; the temperature climbs ( $23.5$ , $26.0$ , $28.5$ ) as more acid reacts.
Reach the maximum near the equivalence volume, here about $25.0\ \text{cm}^3$ giving $30.5\ ^\circ\text{C}$ , so $\Delta T_{\text{max}} = 30.5 - 21.0 = 9.5\ ^\circ\text{C}$ .
Beyond the maximum the temperature falls ( $29.5$ , $28.0$ , $25.0$ ) as excess cold acid is added and heat is lost.
Check decimal places: $21.0, 23.5, 26.0, 28.5, 30.5, 29.5, 28.0, 26.5, 25.0$ — all end in $.0$ or $.5$ and both types are present, satisfying point I.

The exact numbers are candidate-dependent; what matters is the precision convention and a $\Delta T_{\text{max}}$ close to the supervisor's value.

Key Takeaways

Always record thermometer readings to a fixed precision (here $0.5\ ^\circ\text{C}$ ) and keep that precision throughout the table.
A neutralisation curve rises to a maximum then falls; the maximum marks the equivalence point.
Accuracy marks depend on how close your $\Delta T_{\text{max}}$ is to the supervisor's, not on the individual readings.

Common Mistakes

Recording some readings to whole degrees ( $22$ , $25$ ) and others to $0.5$ — the mark scheme requires ALL readings to end in $.0$ or $.5$ .
Failing to include at least one $.0$ and one $.5$ ending.
Producing a $\Delta T_{\text{max}}$ far from the supervisor's value, losing both accuracy marks.
Not observing a clear maximum followed by a decrease.

Things to Be Careful About

Stir after each addition and record the HIGHEST temperature reached, not the temperature at the moment of pouring.
Keep the same thermometer and the same cup throughout to avoid systematic shifts.
The mark scheme's $\delta$ thresholds: for a supervisor $\Delta T_{\text{max}}$ of $7.0$ – $15.0\ ^\circ\text{C}$ , $\delta \le 1.5\ ^\circ\text{C}$ gives one mark and $\delta \le 1.0\ ^\circ\text{C}$ gives two.

Techniques used

record thermometer readings to 0.0 or 0.5 °Censure at least one reading ends in .0 and one in .5obtain a maximum temperature rise consistent with the supervisor's value

(b)

(i)

Plot a graph of temperature ( $y$ -axis) against volume of FB 2 added ( $x$ -axis) on the grid. Select a scale on the $y$ -axis to include a temperature of $2\ ^\circ\text{C}$ above your maximum thermometer reading. Label any points you consider to be anomalous.

Draw two lines of best fit, the first for the increase in temperature and the second for after the maximum temperature has been reached. Extrapolate the two lines so they intersect. This intersection corresponds to the volume of FB 2 required to form a neutral solution.

Fig. 1.1

DifficultyMedium

Worked solution

Answer

Plot temperature ( $^\circ\text{C}$ ) on the $y$ -axis against total volume of FB 2 added ( $\text{cm}^3$ ) on the $x$ -axis, both axes labelled with quantity and unit and a linear numerical scale. Choose the $y$ -scale so it extends to $2\ ^\circ\text{C}$ above the highest recorded temperature (e.g. if max is $30.5\ ^\circ\text{C}$ , the scale runs to about $32\ ^\circ\text{C}$ ), with the plotted points occupying at least a $5\times5$ big-square region.

Plot all recorded points accurately. Draw a line of best fit through the rising points and a second line of best fit through the falling points (straight lines or smooth curves), ignoring any point labelled anomalous. Extrapolate both lines until they intersect; the intersection gives the volume of FB 2 at maximum temperature.

Final answer

Graph of temperature against volume of FB 2 with labelled linear axes (y-scale 2 °C above maximum), all points plotted, two lines of best fit extrapolated to intersect at the maximum-temperature volume.

Detailed explanation

Background Concept

The temperature–volume graph for a neutralisation calorimetry experiment has a characteristic 'tent' shape: temperature increases linearly as acid is added and reacts, reaches a sharp maximum at the equivalence point, then decreases linearly as excess cold acid dilutes the mixture and heat is lost. Extrapolating the two straight portions to their intersection locates the equivalence volume more precisely than reading the single highest point, because it averages out the scatter of individual readings.

Understanding the Question

You must produce a graph on the supplied grid (Fig. 1.1). The mark scheme breaks this into four points: M1 axes labelled with unambiguous names/units and numerical scales; M2 linear scales with the $y$ -axis extending $2\ ^\circ\text{C}$ above the highest table temperature and the data occupying at least $5\times5$ big squares; M3 all points plotted correctly; M4 two lines of best fit extrapolated to intersect.

Approach

Set the $x$ -axis from $0$ to $40\ \text{cm}^3$ and the $y$ -axis from a little below the lowest temperature to $2\ ^\circ\text{C}$ above the highest, using a scale that spreads the points over most of the grid. Plot each (volume, temperature) pair, then fit one line through the rising branch and another through the falling branch, extending both to cross.

Step-by-Step Reasoning

Label the $y$ -axis 'temperature / $^\circ\text{C}$ ' and the $x$ -axis 'total volume of FB 2 added / $\text{cm}^3$ ' (M1).
Choose linear scales: $x$ from $0$ to $40$ , $y$ from e.g. $20$ to $32$ so the top exceeds the maximum ( $30.5$ ) by $2\ ^\circ\text{C}$ and the points fill the grid (M2).
Plot the nine points from Table 1.1 accurately (M3).
Draw a rising line of best fit through the first four or five points and a falling line through the last points; extrapolate both to intersect near $25\ \text{cm}^3$ (M4).
Any point far off the trend is labelled anomalous and ignored by the lines.

Key Takeaways

Two extrapolated lines of best fit give a more reliable equivalence volume than the single maximum reading.
Axis labelling (quantity AND unit) and scale choice are themselves markable, not just the curve.

Common Mistakes

A $y$ -scale that stops at the maximum temperature instead of $2\ ^\circ\text{C}$ above it — loses M2.
Plotting points inaccurately or joining them with straight line segments rather than drawing two best-fit lines — loses M3/M4.
Forgetting units on the axes — loses M1.
Drawing only one line of best fit through all the points.

Things to Be Careful About

The two lines must be extrapolated (extended beyond the plotted points) until they meet; the intersection is what you read in (b)(ii).
Keep scales linear and evenly spaced; do not use a broken or non-uniform axis.

Techniques used

plot temperature against volume with labelled axes and linear scaleschoose a y-scale extending 2 °C above the maximum readingdraw two lines of best fit and extrapolate to their intersection

(ii)

Use your graph to determine the volume of FB 2 required to neutralise $25.0\text{ cm}^3$ of FB 1.

$25.0\text{ cm}^3$ of FB 1 required .................................... $\text{cm}^3$ of FB 2.

DifficultyEasy

Worked solution

Answer

$25.0\text{ cm}^3$ of FB 1 required $22.5\text{ cm}^3$ of FB 2 (volume read at the intersection of the two extrapolated lines of best fit, recorded to 1 or 2 decimal places).

(Value is candidate-dependent; read from your own graph.)

Final answer

Volume of FB 2 at the intersection, e.g. 22.5 cm^3 (recorded to 1 or 2 decimal places).

Detailed explanation

Background Concept

The intersection of the two extrapolated lines on the temperature–volume graph corresponds to the volume of acid that exactly neutralises the base, because that is where the rising (reaction) branch meets the falling (excess-acid cooling) branch — the theoretical maximum-temperature point.

Understanding the Question

The command is 'use your graph to determine the volume'. You read the $x$ -coordinate of the intersection point from part (b)(i). The mark scheme requires the value to be recorded to 1 or 2 decimal places.

Approach

Drop a vertical line from the intersection point down to the $x$ -axis and read the volume, estimating to $0.1$ or $0.05\ \text{cm}^3$ .

Step-by-Step Reasoning

Locate the intersection of the two extrapolated lines of best fit.
Read the corresponding volume of FB 2 on the $x$ -axis.
Record it to 1 or 2 decimal places, e.g. $22.5\ \text{cm}^3$ . This value is then carried forward into parts (c)(i), (c)(iv) and (c)(v).

Key Takeaways

The equivalence volume is read from the intersection of extrapolated lines, not from the single highest plotted point.
Record the read-off to the precision the mark scheme demands (1–2 d.p.).

Common Mistakes

Reading to only a whole number ( $22$ or $23$ ) — the scheme requires 1 or 2 decimal places.
Reading the $y$ -value (temperature) instead of the $x$ -value (volume).
Using the volume at the highest plotted point rather than the extrapolated intersection.

Things to Be Careful About

This value is carried forward (ecf) into the calculations in (c); a consistent read-off preserves the later method marks even if the number is slightly off.

Techniques used

read the volume of FB 2 at the intersection of the extrapolated lines

(c)

Calculations

(i)

Calculate the energy change, in J, when the volume of FB 2 recorded in (b)(ii) neutralises $25.0\text{ cm}^3$ of FB 1.

energy change = .................................... J

DifficultyMedium-Easy

Worked solution

Working

Total volume of solution $= 25.0 + 22.5 = 47.5\ \text{cm}^3$ , so mass of solution $m = 47.5\ \text{g}$ (assuming $\rho = 1.00\ \text{g cm}^{-3}$ ).

Temperature rise $\Delta T$ from the graph (e.g. $30.5 - 21.0 = 9.5\ ^\circ\text{C}$ ).

Q = m c \Delta T = 47.5 \times 4.18 \times 9.5

Q = 1884\ \text{J}

Answer

energy change $= 1.88 \times 10^{3}\ \text{J}$ (to 2–3 s.f.)

Final answer

1884 J (≈ 1.88 × 10^3 J), using Q = (25.0 + 22.5) × 4.18 × 9.5

Detailed explanation

Background Concept

The heat released by a reaction in solution is calculated from $Q = mc\Delta T$ , where $m$ is the mass of the solution (taken as the total volume in $\text{cm}^3$ because the density is assumed $1.00\ \text{g cm}^{-3}$ ), $c = 4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ is the specific heat capacity of the solution, and $\Delta T$ is the temperature rise. The temperature rise must be obtained from the extrapolated maximum on the graph, not the raw highest reading, to correct for heat loss.

Understanding the Question

You are asked for the energy change in J when the volume from (b)(ii) neutralises $25.0\ \text{cm}^3$ of FB 1. The mark scheme's expression is $Q = (25 + \text{volume in (b)(ii)}) \times 4.18 \times \text{temp rise}$ , with the answer to 2 or more significant figures.

Approach

Add the two volumes to get the total mass of solution, multiply by $4.18$ and by the temperature rise read from the graph.

Step-by-Step Reasoning

Total volume $= 25.0 + 22.5 = 47.5\ \text{cm}^3 \Rightarrow m = 47.5\ \text{g}$ .
Temperature rise $\Delta T$ from the intersection of the extrapolated lines, e.g. $9.5\ ^\circ\text{C}$ .
$Q = 47.5 \times 4.18 \times 9.5 = 1884\ \text{J}$ .
Report to 2–3 s.f.: $1.88 \times 10^3\ \text{J}$ .

This value is carried forward into (c)(iii).

Key Takeaways

The mass used is the TOTAL solution mass (base + acid), not just the pipetted volume.
Use the extrapolated $\Delta T$ , which is larger than the observed maximum because it corrects for heat loss.

Common Mistakes

Using only $25.0\ \text{g}$ instead of $25.0 + V$ — the mark scheme explicitly requires the total volume.
Using the raw highest thermometer reading as $\Delta T$ instead of the extrapolated value.
Giving the answer to only 1 significant figure.

Things to Be Careful About

Keep $Q$ in joules here; the conversion to kJ happens in (c)(iii).
Carry the same $\Delta T$ and volume consistently into the enthalpy calculation.

Techniques used

calculate total volume of solutionapply q = mcΔT with c = 4.18 J g^-1 K^-1use the maximum temperature rise from the graph

(ii)

Calculate the amount, in mol, of sodium hydroxide, FB 1, pipetted into the cup.

amount of $\text{NaOH}$ = ................................ mol

DifficultyEasy

Worked solution

Working

n(\text{NaOH}) = c \times V = 1.90 \times \frac{25.0}{1000}

n = 4.75 \times 10^{-2}\ \text{mol}

Answer

amount of $\text{NaOH} = 0.0475\ \text{mol}$

Final answer

0.0475 mol

Detailed explanation

Background Concept

The amount of solute in a solution is $n = cV$ , where $c$ is the concentration in $\text{mol dm}^{-3}$ and $V$ the volume in $\text{dm}^3$ . A volume in $\text{cm}^3$ must be divided by $1000$ to convert to $\text{dm}^3$ .

Understanding the Question

Calculate the amount of FB 1 ( $1.90\ \text{mol dm}^{-3}$ NaOH) in the $25.0\ \text{cm}^3$ pipetted into the cup. The mark scheme gives $= 1.9 \times 25 / 1000 = 4.75 \times 10^{-2}$ , to 2–4 significant figures.

Approach

Multiply the concentration by the volume converted to $\text{dm}^3$ .

Step-by-Step Reasoning

$V = 25.0\ \text{cm}^3 = 0.0250\ \text{dm}^3$ .
$n = 1.90 \times 0.0250 = 0.0475\ \text{mol}$ .
Report to 2–4 s.f.: $4.75 \times 10^{-2}\ \text{mol}$ .

This value is used in (c)(iii) and (c)(iv).

Key Takeaways

Always convert $\text{cm}^3$ to $\text{dm}^3$ before using $n = cV$ .
The pipetted volume of base is fixed at $25.0\ \text{cm}^3$ regardless of the titre.

Common Mistakes

Forgetting the $/1000$ and giving $47.5\ \text{mol}$ .
Giving only 1 significant figure.

Things to Be Careful About

Carry $0.0475\ \text{mol}$ forward unchanged into the enthalpy and $M_r$ calculations.

Techniques used

calculate moles from concentration and volumeconvert cm^3 to dm^3

(iii)

Calculate the enthalpy change of neutralisation, $\Delta H_{\text{neut}}$ , in $\text{kJ mol}^{-1}$ , for $1.00\text{ mol}$ of sodium hydroxide reacting with FB 2.

\Delta H_{\text{neut}} = \text{......... ................. kJ mol}^{-1}

(sign, value)

DifficultyMedium-Easy

Worked solution

Working

\Delta H_{\text{neut}} = -\frac{Q}{n} = -\frac{1884}{0.0475 \times 1000}\ \text{kJ mol}^{-1}

\Delta H_{\text{neut}} = -39.7\ \text{kJ mol}^{-1}

Answer

$\Delta H_{\text{neut}} = -39.7\ \text{kJ mol}^{-1}$ (sign and value)

Final answer

-39.7 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of neutralisation per mole of base is $\Delta H = -Q/n$ , where $Q$ (in J) is the heat released and $n$ the moles of base. The negative sign is essential because neutralisation is exothermic. Dividing $Q$ in joules by $n$ and by $1000$ converts to $\text{kJ mol}^{-1}$ .

Understanding the Question

Use (c)(i) and (c)(ii) to find $\Delta H_{\text{neut}}$ for $1.00\ \text{mol}$ of NaOH reacting with FB 2. The mark scheme's expression is $-\text{(c)(i)}/[\text{(c)(ii)} \times 1000]$ , with the negative sign shown and 2–4 s.f.

Approach

Divide the energy change by the moles of NaOH, convert J to kJ, and place a minus sign.

Step-by-Step Reasoning

$Q = 1884\ \text{J}$ (carried forward from c(i)).
$n = 0.0475\ \text{mol}$ (from c(ii)).
$\Delta H = -1884 / (0.0475 \times 1000) = -1884 / 47.5 = -39.7\ \text{kJ mol}^{-1}$ .
The negative sign denotes the exothermic reaction.

Key Takeaways

Enthalpy changes of neutralisation are negative (exothermic); omitting the sign loses the mark.
The $\times 1000$ in the denominator converts J to kJ.

Common Mistakes

Forgetting the negative sign.
Dividing by $n$ without converting J to kJ, giving $-39700$ .
Using the moles of acid instead of the moles of base.

Things to Be Careful About

ecf is allowed: an error in (c)(i) or (c)(ii) carried through correctly still earns the method mark, but the sign must be present.

Techniques used

divide energy change by amount of substanceconvert J to kJapply the negative sign for an exothermic reaction

(iv)

Use your answers to (b)(ii) and (c)(ii) and the information given on page 2 to calculate the relative formula mass, $M_r$ , of the organic acid $\text{CH}_3\text{CHXCOOH}$ .
Show your working.

$M_r$ of $\text{CH}_3\text{CHXCOOH}$ = .......................................

DifficultyMedium

Worked solution

Working

The reaction is 1:1, so $n(\text{acid}) = n(\text{NaOH}) = 0.0475\ \text{mol}$ .

Mass of acid in the titre volume $= 312.5 \times \dfrac{22.5}{1000} = 7.031\ \text{g}$ .

M_r = \frac{\text{mass}}{\text{amount}} = \frac{312.5 \times 22.5 / 1000}{0.0475}

M_r = 148\ \text{(to 3 s.f.)}

Answer

$M_r$ of $\text{CH}_3\text{CHXCOOH} = 148$

Final answer

148

Detailed explanation

Background Concept

A mass concentration of $312.5\ \text{g dm}^{-3}$ means each $\text{dm}^3$ of FB 2 contains $312.5\ \text{g}$ of acid. The mass of acid in the titre volume $V$ (in $\text{cm}^3$ ) is $312.5 \times V/1000$ grams. Because the acid is monoprotic and reacts 1:1 with NaOH, the moles of acid neutralised equal the moles of NaOH, $0.0475\ \text{mol}$ . Hence $M_r = \text{mass}/\text{moles}$ .

Understanding the Question

Use (b)(ii) (titre volume) and (c)(ii) (moles of NaOH) with the page-2 concentration to calculate $M_r$ . The mark scheme's expression is $M_r = [312.5 \times \text{(b)(ii)}] / [\text{(c)(ii)} \times 1000]$ .

Approach

Compute the mass of acid in the titre, then divide by the moles of acid (= moles of NaOH).

Step-by-Step Reasoning

Mass of acid in titre $= 312.5 \times 22.5/1000 = 7.031\ \text{g}$ .
Moles of acid $= 0.0475\ \text{mol}$ (1:1 with NaOH).
$M_r = 7.031 / 0.0475 = 148$ .

Equivalently $M_r = (312.5 \times 22.5)/(0.0475 \times 1000) = 148$ .

Key Takeaways

The 1:1 stoichiometry lets you equate moles of acid to moles of base at the equivalence point.
$M_r$ is found from mass divided by amount, using the titre volume to get the mass.

Common Mistakes

Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ in the mass term.
Using the wrong volume (e.g. $25.0$ instead of the titre).
Not recognising the 1:1 ratio, leading to a doubled or halved $M_r$ .

Things to Be Careful About

This $M_r$ is carried forward into (c)(v) and (c)(vi); ecf applies, so a consistent value preserves the later marks.

Techniques used

relate mass concentration to moles via volumecalculate Mr from mass and amountuse the stoichiometric 1:1 ratio

(v)

The acid is known to be one of the following: $\text{CH}_3\text{CHFCOOH}$ , $\text{CH}_3\text{CHClCOOH}$ , $\text{CH}_3\text{CHBrCOOH}$ or $\text{CH}_3\text{CHICOOH}$ .
Use your answer to (c)(iv) to identify the acid used to make solution FB 2.

The acid in FB 2 is ....................................................... .

DifficultyMedium-Easy

Worked solution

Working

The fragment $\text{CH}_3\text{CHCOOH}$ has mass $= 12 + 3(1) + 12 + 1 + 12 + 2(16) + 1 = 73$ .

A_r(\text{X}) = M_r - 73 = 148 - 73 = 75

The closest halogen $A_r$ is chlorine ( $35.5$ )... checking the given ranges: $\text{CH}_3\text{CHClCOOH}$ has $M_r = 108.5$ , $\text{CH}_3\text{CHBrCOOH} = 152.9$ . The calculated $M_r = 148$ lies in the bromine range ( $130.7 < 148 \le 176.4$ ).

Answer

The acid in FB 2 is $\text{CH}_3\text{CHBrCOOH}$ (2-bromopropanoic acid).

Final answer

CH3CHBrCOOH

Detailed explanation

Background Concept

Each candidate acid has the form $\text{CH}_3\text{CHXCOOH}$ , so its $M_r$ equals the constant fragment mass plus the $A_r$ of the halogen X. The fragment $\text{CH}_3\text{CHCOOH}$ (C $_3$ H $_5$ O $_2$ ) has mass $3(12) + 5(1) + 2(16) = 73$ . Therefore $A_r(\text{X}) = M_r - 73$ . The four halogens have $A_r$ : F $= 19$ , Cl $= 35.5$ , Br $= 80$ , I $= 127$ , giving expected $M_r$ values $92.0, 108.5, 152.9, 199.9$ . The mark scheme provides ranges to assign the calculated $M_r$ to the correct acid.

Understanding the Question

Use the $M_r$ from (c)(iv) to identify which of the four acids is in FB 2. The mark scheme: $A_r = \text{(c)(iv)} - 73$ , then choose the acid whose $M_r$ range contains the calculated value.

Approach

Subtract $73$ from the calculated $M_r$ to get $A_r(\text{X})$ , then match to the nearest halogen, or place the $M_r$ in the given ranges.

Step-by-Step Reasoning

$M_r = 148$ (carried forward from c(iv)).
$A_r(\text{X}) = 148 - 73 = 75$ .
The halogen $A_r$ values are F $19$ , Cl $35.5$ , Br $80$ , I $127$ ; $75$ is closest to Br ( $80$ ).
Using the ranges: $130.7 < 148 \le 176.4$ , so the acid is $\text{CH}_3\text{CHBrCOOH}$ (expected $M_r = 152.9$ ).

Key Takeaways

Identifying an unknown by $M_r$ relies on subtracting the known fragment mass and matching the remainder to an element's $A_r$ .
Experimental $M_r$ values carry error, so ranges (not exact equality) are used.

Common Mistakes

Forgetting to subtract $73$ and comparing $M_r$ directly to an $A_r$ .
Choosing the wrong halogen because of a large experimental error in $M_r$ — pick the nearest range.
Mis-summing the fragment mass.

Things to Be Careful About

ecf: an incorrect $M_r$ from (c)(iv) still earns this mark if the right acid is chosen for that value.

Techniques used

subtract the CH3CHCOOH fragment mass from Mrcompare the resulting Ar of X with halogen atomic massesselect the matching halogenocarboxylic acid

(vi)

Calculate the percentage error in the relative formula mass, $M_r$ , you calculated in (c)(iv).

percentage error in $M_r$ = ................................... %

DifficultyMedium-Easy

Worked solution

Working

Actual $M_r$ of $\text{CH}_3\text{CHBrCOOH} = 152.9$ .

\%\ \text{error} = \frac{|\text{actual} - \text{experimental}|}{\text{actual}} \times 100 = \frac{|152.9 - 148|}{152.9} \times 100

= \frac{4.9}{152.9} \times 100 = 3.2\%

Answer

percentage error in $M_r = 3.2\%$

(Use the actual $M_r$ matching the acid identified in (c)(v).)

Final answer

3.2 %

Detailed explanation

Background Concept

Percentage error compares the difference between an experimental value and the accepted (actual) value, expressed as a fraction of the accepted value: $\%\text{error} = |\text{actual} - \text{experimental}|/\text{actual} \times 100$ . The 'actual' $M_r$ is the theoretical relative formula mass of the acid identified in (c)(v).

Understanding the Question

Calculate the percentage error in the $M_r$ from (c)(iv). The mark scheme's expression is $[\text{actual } M_r - \text{(c)(iv)}] \times 100 / \text{actual } M_r$ , with the actual values $92.0, 108.5, 152.9, 199.9$ for F, Cl, Br, I acids respectively.

Approach

Take the actual $M_r$ of the identified acid, subtract the experimental $M_r$ , divide by the actual, multiply by $100$ .

Step-by-Step Reasoning

Identified acid = $\text{CH}_3\text{CHBrCOOH}$ , actual $M_r = 152.9$ .
Experimental $M_r = 148$ (from c(iv)).
Difference $= 152.9 - 148 = 4.9$ .
$\%\text{error} = 4.9 / 152.9 \times 100 = 3.2\%$ .

Key Takeaways

Percentage error is always relative to the ACTUAL value, not the experimental one.
The actual value used must correspond to the acid identified in (c)(v).

Common Mistakes

Dividing by the experimental value instead of the actual value.
Using the wrong actual $M_r$ (not matching the identified acid).
Reporting the absolute difference without the $\times 100$ .

Things to Be Careful About

ecf applies: use the actual $M_r$ of whatever acid you named in (c)(v), and your (c)(iv) value, even if both are wrong, to keep the method mark.

Techniques used

compute percentage error from experimental and actual valuesuse the actual Mr of the identified acid

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