Chemistry 9701/31 — October/November 2018

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

In this experiment you will determine the percentage purity of a sample of impure anhydrous sodium carbonate. You will use two different methods to measure the enthalpy change of reaction when a sample of impure anhydrous sodium carbonate reacts with excess dilute hydrochloric acid.

FA 1 is a sample of the impure anhydrous sodium carbonate.
FA 2 is $2.00\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FA 3 is a second sample of the impure anhydrous sodium carbonate used in FA 1.

(a)

Method 1

Weigh the container with FA 1. Record this mass.

\text{mass of container with FA 1} = \text{.............................. g}

Support one of the plastic cups in the $250\text{ cm}^3$ beaker.
Use the measuring cylinder to place $25\text{ cm}^3$ of FA 2 into the cup.
Measure the temperature of the FA 2 in the cup. Tilt the cup if necessary so that the bulb of the thermometer is fully covered. Record this temperature at time $t = 0$ .
Start the stopclock and leave it running for the whole experiment.
Measure and record the temperature of FA 2 in the cup every half minute for 2 minutes.
At $t = 2\frac{1}{2}$ minutes tip all the FA 1 into the cup. Stir the contents of the cup.
Measure and record the temperature of the contents of the cup at $t = 3$ minutes and then every half minute up to $t = 9$ minutes.
Weigh the container with any residual FA 1. Record this mass.

\text{mass of container with residual FA 1} = \text{.............................. g}

DifficultyEasy

Worked solution

Answer

(Candidate-dependent data. A representative completed table is shown below.)

Initial mass of container with FA 1: 28.45 g
Final mass of container with residual FA 1: 26.65 g
Mass of FA 1 used: 1.80 g

Time / min	0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5	6.0	6.5	7.0	7.5	8.0	8.5	9.0
Temperature / °C	22.0	22.0	22.0	22.0	22.0	24.5	25.5	26.0	26.0	25.8	25.5	25.2	24.8	24.5	24.2	23.8	23.5	23.2	23.0

Note: Maximum temperature $T_{\text{max}}$ is 26.0 °C at 3.5 minutes. Theoretical temperature rise $\Delta T$ from extrapolation is 4.0 °C.

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

In calorimetry experiments, accurate data collection is critical. The temperature change ( $\Delta T$ ) is the primary measurement used to calculate the heat energy exchanged. Because heat is continuously lost to the surroundings (the cup, the beaker, the air), the maximum temperature recorded on the thermometer is almost always lower than the true theoretical maximum. Extrapolation is used to correct for this heat loss.

Understanding the Question

This part asks you to record your practical data: the initial and final masses of the container holding FA 1 (to find the mass of sodium carbonate used), and the temperature of the hydrochloric acid before, during, and after the reaction. You must record temperatures at half-minute intervals from $t=0$ to $t=9$ minutes.

Approach

Set up a clear table with appropriate headings and units. Record the balance readings to the same number of decimal places (usually 2 or 3). Record thermometer readings to the nearest 0.5 °C (e.g., 22.0, 22.5, 23.0). Ensure you capture the temperature rise and the subsequent cooling phase.

Step-by-Step Reasoning

Mass measurements: Weigh the container with FA 1 before the experiment, and again after tipping the solid into the acid. The difference is the mass of FA 1 used. Both readings must be to the same decimal places (e.g., 28.45 g and 26.65 g).
Temperature readings: Measure the initial temperature of the acid every 30 seconds for the first 2 minutes. This establishes a baseline and shows any initial drift. After adding FA 1 at $t=2.5$ min, record temperatures every 30 seconds up to $t=9$ min. You should see a sharp rise to a maximum, followed by a gradual linear cooling.
Thermometer precision: Thermometers in these experiments are typically graduated to 0.5 °C. Readings must reflect this (e.g., ending in .0 or .5).

Key Takeaways

Always include units in table headings. Record all raw data as it is observed; do not alter or round it during recording. The cooling phase is just as important as the heating phase for the extrapolation method.

Common Mistakes

Forgetting to include units in table headings (e.g., writing just "Temperature" instead of "Temperature / °C").
Recording thermometer readings to 1 decimal place (e.g., 22.3 °C) when the scale only allows 0.5 °C.
Not recording the initial baseline temperatures before adding the solid.

Things to Be Careful About

Ensure the mass subtraction is done correctly and to the correct number of decimal places. The candidate's $\Delta T$ will be compared against a supervisor's $\Delta T$ to verify the practical was conducted correctly (usually within 1 °C or 0.5 °C of the expected value).

Techniques used

set up data table with correct headings and unitsrecord temperature readings to 0.5 °Crecord mass measurements to consistent decimal places

(b)

(i)

On the grid on page 3, plot a graph of temperature ( $y$ -axis) against time ( $x$ -axis). You should choose a scale that allows you to plot $2\ ^\circ\text{C}$ above the maximum temperature reached.

On your graph, draw two straight lines of best fit. One line is for the temperature before adding FA 1 and the other line for the cooling of the solution once reaction is complete.

Extrapolate these two lines to $t = 2\frac{1}{2}$ minutes.

Graph Grid

DifficultyMedium-Easy

Worked solution

Answer

(The candidate must plot the temperature-time data from part (a) on the provided grid. The graph must include:)

Axes: $y$ -axis labelled "Temperature / °C", $x$ -axis labelled "Time / min".
Scale: Chosen so that the plotted points and an additional 2 °C above the maximum temperature occupy more than half the available grid space in both directions.
Points: All recorded data points plotted accurately (within half a small square).
Lines of best fit: Two straight lines drawn with a ruler. Line 1: through the initial baseline points (0 to 2 min). Line 2: through the cooling points (after the maximum temperature, e.g., 4 to 9 min).
Extrapolation: Both lines extended to $t = 2.5$ minutes. The vertical difference between the two extrapolated lines at $t = 2.5$ min is the theoretical $\Delta T$ .

Final answer

See diagram / candidate-dependent graph

Detailed explanation

Background Concept

In exothermic reactions measured in simple calorimeters, heat is lost to the surroundings continuously. The recorded maximum temperature is therefore lower than the true adiabatic maximum. To correct for this, a cooling curve is plotted. By extrapolating the pre-reaction baseline and the post-reaction cooling line back to the time of mixing ( $t = 2.5$ min), the theoretical temperature rise can be determined, assuming no heat was lost.

Understanding the Question

You are asked to plot the temperature against time data collected in part (a) and use graphical extrapolation to find the theoretical temperature rise at the moment of mixing ( $t = 2.5$ min).

Approach

Choose a scale that uses more than half the grid space.
Plot all points accurately.
Draw a line of best fit through the initial stable temperatures (0-2 min).
Draw a line of best fit through the cooling phase (after the peak).
Extrapolate both lines to $t = 2.5$ min and read the temperature difference.

Step-by-Step Reasoning

Axis labels and scale: The $x$ -axis is time in minutes (0 to 9.5 or 10). The $y$ -axis is temperature in °C. If the max temp is 26.0 °C, the $y$ -axis should go up to at least 28.0 °C (max + 2 °C).
Plotting: Plot each (time, temperature) pair. Points on the baseline and cooling lines must lie exactly on the lines; points near the peak may be slightly off the line.
Lines of best fit: Use a ruler. The first line should have roughly equal points above and below it in the 0-2 min range. The second line should fit the cooling trend (4-9 min).
Extrapolation: Extend both lines with a ruler to the vertical grid line at $t = 2.5$ min. Read the temperature on each line. The difference is your theoretical $\Delta T$ .

Key Takeaways

The extrapolation method assumes that the rate of heat loss before mixing is the same as the rate of heat loss after the reaction. This is a standard approximation in school-level calorimetry.

Common Mistakes

Choosing a scale that is too small, causing the graph to be cramped.
Drawing a curved line of best fit through the cooling phase instead of a straight line.
Forgetting to extrapolate both lines to exactly $t = 2.5$ min.
Plotting points larger than half a small square (using thick blobs or crosses).

Things to Be Careful About

Ensure the lines are drawn with a ruler and are straight. The extrapolated $\Delta T$ must be read to at least 1 decimal place and be within half a small square of the examiner's expected value.

Techniques used

select appropriate scales for axesplot temperature-time data points accuratelydraw straight lines of best fit for heating and cooling phasesextrapolate lines to the time of mixing

(ii)

From your graph, find the theoretical temperature rise at $t = 2\frac{1}{2}$ minutes.

\text{theoretical temperature rise} = \text{............................... } ^\circ\text{C}

DifficultyEasy

Worked solution

Answer

theoretical temperature rise = 4.0 °C

(Read the difference in temperature between the two extrapolated lines at $t = 2.5$ minutes from the graph in (b)(i).)

Final answer

4.0 °C

Detailed explanation

Background Concept

The theoretical temperature rise is the difference between the extrapolated temperature of the cooling line and the extrapolated temperature of the baseline at the exact moment of mixing ( $t = 2.5$ min). This corrects for the heat lost to the surroundings during the reaction time.

Understanding the Question

You need to extract a single numerical value from your graph in part (b)(i): the theoretical $\Delta T$ .

Approach

Locate $t = 2.5$ min on the x-axis. Read the temperature on the extrapolated baseline line. Read the temperature on the extrapolated cooling line. Subtract the baseline temperature from the cooling line temperature.

Step-by-Step Reasoning

Using the representative graph:

At $t = 2.5$ min, the extrapolated baseline is at 22.0 °C.
At $t = 2.5$ min, the extrapolated cooling line is at 26.0 °C.
Theoretical $\Delta T = 26.0 - 22.0 = 4.0$ °C.

Your value should be within half a small square of the examiner's calculated value (typically 3.5 to 4.5 °C depending on the actual data).

Key Takeaways

Always read the extrapolated values at the exact time of mixing, not the raw maximum temperature from the table.

Common Mistakes

Using the raw maximum temperature from the table (e.g., 26.0 °C - 22.0 °C = 4.0 °C, which might coincidentally be correct here, but in general, the raw max is lower than the extrapolated max).
Reading the graph to only 1 decimal place when the scale requires more, or vice versa.

Things to Be Careful About

Ensure the value is given to at least 1 decimal place. The mark scheme allows a range based on the candidate's graph.

Techniques used

read theoretical temperature rise from extrapolated graph

(c)

(i)

Calculate the energy released in the reaction.

(Assume $4.2\text{ J}$ of heat energy changes the temperature of $1.0\text{ cm}^3$ of solution by $1.0\ ^\circ\text{C}$ .)

\text{energy released} = \text{.............................. J}

DifficultyMedium-Easy

Worked solution

Working

q = m \times c \times \Delta T

Assume the density of the solution is $1.0 \text{ g cm}^{-3}$ , so the mass of $25 \text{ cm}^3$ of solution is $25 \text{ g}$ . The specific heat capacity is given as $4.2 \text{ J g}^{-1} \text{°C}^{-1}$ .

q = 25 \times 4.2 \times \Delta T

Using $\Delta T = 4.0 \text{ °C}$ from (b)(ii):

q = 25 \times 4.2 \times 4.0 = 420 \text{ J}

Answer

energy released = 420 J

Final answer

420 J

Detailed explanation

Background Concept

The heat energy ( $q$ ) released by an exothermic reaction is absorbed by the solution. It is calculated using the equation $q = m c \Delta T$ , where $m$ is the mass of the solution, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change. In these experiments, we assume the solution has the same density and specific heat capacity as water ( $1.0 \text{ g cm}^{-3}$ and $4.2 \text{ J g}^{-1} \text{°C}^{-1}$ ).

Understanding the Question

Calculate the total energy evolved in the reaction using the volume of acid (25 cm³), the assumed specific heat capacity (4.2 J per cm³ per °C), and the theoretical temperature rise from part (b)(ii).

Approach

Multiply the volume of solution (25 cm³) by the specific heat capacity factor (4.2 J cm⁻³ °C⁻¹) and the theoretical temperature rise ( $\Delta T$ ).

Step-by-Step Reasoning

Mass of solution $\approx 25 \text{ g}$ (since $25 \text{ cm}^3 \times 1.0 \text{ g cm}^{-3}$ ).
$c = 4.2 \text{ J g}^{-1} \text{°C}^{-1}$ .
$\Delta T = 4.0 \text{ °C}$ (from b)(ii)).
$q = 25 \times 4.2 \times 4.0 = 420 \text{ J}$ .

The answer should be to 2-4 significant figures. 420 J or 420.0 J is acceptable.

Key Takeaways

Always use the theoretical $\Delta T$ from the extrapolated graph, not the raw maximum temperature, for accurate energy calculations.

Common Mistakes

Using the raw maximum temperature from the table instead of the extrapolated theoretical $\Delta T$ .
Forgetting to convert kJ to J or vice versa in later steps.
Using the wrong mass (e.g., including the mass of the solid, though the question specifically says 'assume 4.2 J changes the temperature of 1.0 cm³ of solution', implying only the liquid volume is considered).

Things to Be Careful About

The question states: 'Assume 4.2 J of heat energy changes the temperature of 1.0 cm³ of solution by 1.0 °C'. This directly gives the formula: $\text{energy} = 25 \times 4.2 \times \Delta T$ . Do not overcomplicate it by adding the mass of the solid.

Techniques used

calculate heat energy using q = mcΔT

(ii)

The equation for the reaction between anhydrous sodium carbonate and hydrochloric acid is shown.

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

The literature value for the enthalpy change of this reaction is $-27.0\text{ kJ mol}^{-1}$ .

Use this figure, and the value that you found in (i), to find the mass of anhydrous sodium carbonate you used in (a). You should assume that no energy was lost to the surroundings in your experiment.

\text{mass Na}_2\text{CO}_3 = \text{.............................. g}

DifficultyMedium

Worked solution

Working

The literature enthalpy change is $\Delta H = -27.0 \text{ kJ mol}^{-1} = -27000 \text{ J mol}^{-1}$ .

Moles of $\text{Na}_2\text{CO}_3$ reacted:

n = \frac{\text{energy released}}{\Delta H} = \frac{420}{27000} = 0.01556 \text{ mol}

Molar mass of $\text{Na}_2\text{CO}_3 = 2(23) + 12 + 3(16) = 106 \text{ g mol}^{-1}$ .

Mass of $\text{Na}_2\text{CO}_3$ :

\text{mass} = n \times M_r = 0.01556 \times 106 = 1.649 \text{ g}

Answer

mass Na₂CO₃ = 1.65 g

Final answer

1.65 g

Detailed explanation

Background Concept

The enthalpy change of reaction ( $\Delta H$ ) is the energy change per mole of reaction as written. For the reaction $\text{Na}_2\text{CO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})$ , $\Delta H = -27.0 \text{ kJ mol}^{-1}$ means 27.0 kJ of energy is released per mole of $\text{Na}_2\text{CO}_3$ reacted.

Understanding the Question

Using the energy calculated in (c)(i) and the literature $\Delta H$ , find the mass of pure $\text{Na}_2\text{CO}_3$ that must have reacted.

Approach

Convert $\Delta H$ to J mol⁻¹.
Calculate moles of $\text{Na}_2\text{CO}_3$ using $n = q / |\Delta H|$ .
Calculate mass using $m = n \times M_r$ .

Step-by-Step Reasoning

Energy released $q = 420 \text{ J}$ .
$\Delta H = -27.0 \text{ kJ mol}^{-1} = -27000 \text{ J mol}^{-1}$ .
Moles of $\text{Na}_2\text{CO}_3 = 420 / 27000 = 0.015556 \text{ mol}$ .
$M_r(\text{Na}_2\text{CO}_3) = 106 \text{ g mol}^{-1}$ .
Mass $= 0.015556 \times 106 = 1.6489 \text{ g} \approx 1.65 \text{ g}$ .

Key Takeaways

Always pay attention to the sign of $\Delta H$ . Since energy released is positive, use the absolute value of $\Delta H$ for the calculation.

Common Mistakes

Forgetting to convert kJ to J (e.g., $420 / 27 = 15.56$ , which is wrong).
Using the wrong molar mass (e.g., using 106 for hydrated sodium carbonate or forgetting to multiply by 2 for Na).
Rounding too early in the calculation.

Things to Be Careful About

The mark scheme awards marks for the correct use of moles $= \text{(c)(i)} / 27000$ and mass $= \text{moles} \times 106$ . Error carried forward from (c)(i) is allowed.

Techniques used

calculate moles from enthalpy changecalculate mass from moles and molar mass

(iii)

Calculate the percentage of anhydrous sodium carbonate present in FA 1.

\text{percentage Na}_2\text{CO}_3\text{ in FA 1} = \text{.............................. \%}

DifficultyMedium-Easy

Worked solution

Working

From part (a), mass of FA 1 used $= 1.80 \text{ g}$ .
From part (c)(ii), mass of pure $\text{Na}_2\text{CO}_3 = 1.649 \text{ g}$ .

\text{percentage purity} = \frac{\text{mass of pure Na}_2\text{CO}_3}{\text{mass of FA 1 used}} \times 100

\text{percentage purity} = \frac{1.649}{1.80} \times 100 = 91.6\%

Answer

percentage Na₂CO₃ in FA 1 = 91.6 %

Final answer

91.6 %

Detailed explanation

Background Concept

Percentage purity is the mass of the pure substance divided by the total mass of the impure sample, multiplied by 100.

Understanding the Question

Calculate the percentage of anhydrous sodium carbonate in the original FA 1 sample using the mass of pure $\text{Na}_2\text{CO}_3$ calculated in (c)(ii) and the mass of FA 1 used from part (a).

Approach

Divide the mass of pure $\text{Na}_2\text{CO}_3$ by the mass of FA 1 and multiply by 100.

Step-by-Step Reasoning

Mass of pure $\text{Na}_2\text{CO}_3 = 1.649 \text{ g}$ .
Mass of FA 1 used $= 1.80 \text{ g}$ .
Percentage $= (1.649 / 1.80) \times 100 = 91.61\% \approx 91.6\%$ .

Key Takeaways

Ensure you use the correct mass of the sample (the mass of FA 1 used, not the initial mass of the container).

Common Mistakes

Using the initial mass of the container with FA 1 instead of the mass of FA 1 actually used.
Forgetting to multiply by 100.
Incorrect significant figures (2-4 sf are acceptable).

Things to Be Careful About

Error carried forward from (c)(ii) is allowed. If your mass in (c)(ii) was wrong, but your percentage calculation is correct based on that wrong mass, you still get the mark.

Techniques used

calculate percentage purity from mass of pure substance and total mass

(d)

In your calculation in (c), what assumption have you made about the impurity present in FA 1?

DifficultyEasy

Worked solution

Answer

The assumption is that the impurity does not react with hydrochloric acid (or that the impurities are not alkaline / do not release or absorb heat when reacting with the acid).

Final answer

The impurity does not react with hydrochloric acid

Detailed explanation

Background Concept

In calorimetry experiments to determine purity, we assume that only the target compound reacts and produces the observed temperature change. Any impurities are assumed to be inert under the reaction conditions.

Understanding the Question

What assumption did you make about the impurity in FA 1 when calculating the mass of $\text{Na}_2\text{CO}_3$ from the energy released?

Approach

Think about what would invalidate the calculation. If the impurity also reacted with HCl and released heat, the calculated mass of $\text{Na}_2\text{CO}_3$ would be too high. If it absorbed heat, it would be too low.

Step-by-Step Reasoning

The calculation assumes that all the heat energy measured came solely from the reaction of $\text{Na}_2\text{CO}_3$ with HCl. Therefore, the impurity must not react with HCl, or if it does, it must not produce a significant enthalpy change (i.e., it is not alkaline).

Key Takeaways

Always state assumptions clearly in practical calculations. The inertness of impurities is a fundamental assumption in these types of experiments.

Common Mistakes

Saying 'the impurity is water' (water doesn't react, but that's not the general assumption; the assumption is about chemical reactivity and enthalpy change).
Saying 'no heat is lost to the surroundings' (this is a separate assumption, addressed in part (c) where it says 'assume no energy was lost').

Things to Be Careful About

The mark scheme specifically looks for 'impurity does not react with HCl' or 'impurities are not alkaline'. Do not confuse this with the assumption about heat loss to surroundings.

Techniques used

identify assumptions in experimental calculations

(e)

Method 2

Weigh a clean, dry plastic cup and record the mass.
Add between $1.70\text{ g}$ and $1.90\text{ g}$ of FA 3 to the plastic cup and record the mass.
Support the plastic cup in the $250\text{ cm}^3$ beaker.
Pour $25\text{ cm}^3$ of FA 2 into the measuring cylinder.
Measure and record the initial temperature of FA 2 in the measuring cylinder.
Pour the $25\text{ cm}^3$ of FA 2 into the plastic cup.
Stir the contents of the cup and record the maximum temperature. Tilt the cup if necessary so that the bulb of the thermometer is fully covered.
Calculate and record the mass of FA 3 used and the change in temperature.

DifficultyEasy

Worked solution

Answer

(Candidate-dependent data. A representative completed table is shown below.)

Mass of empty cup: 15.20 g
Mass of cup + FA 3: 17.00 g
Mass of FA 3 used: 1.80 g
Initial temperature of FA 2: 22.0 °C
Maximum temperature reached: 25.8 °C
Temperature rise ( $\Delta T$ ): 3.8 °C

Quantity	Value
Mass of empty cup / g	15.20
Mass of cup + FA 3 / g	17.00
Mass of FA 3 used / g	1.80
Initial temperature / °C	22.0
Maximum temperature / °C	25.8
Temperature rise / °C	3.8

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

Method 2 is a simpler calorimetry method. Instead of extrapolating a cooling curve, the student directly records the maximum temperature reached and calculates the temperature rise. This method is generally less accurate because it does not correct for heat loss that occurred during the reaction.

Understanding the Question

Record the data for Method 2: masses to find the mass of FA 3 used, and initial and maximum temperatures to find $\Delta T$ .

Approach

Weigh the empty cup, add FA 3, weigh again. Subtract to find the mass of FA 3. Measure the initial temperature of the acid, add the acid to the cup, stir, and record the maximum temperature. Calculate $\Delta T$ .

Step-by-Step Reasoning

Record mass of empty cup (e.g., 15.20 g).
Add 1.70-1.90 g of FA 3. Record mass of cup + FA 3 (e.g., 17.00 g).
Mass of FA 3 = 17.00 - 15.20 = 1.80 g.
Measure initial temperature of 25 cm³ FA 2 (e.g., 22.0 °C).
Mix and record maximum temperature (e.g., 25.8 °C).
$\Delta T = 25.8 - 22.0 = 3.8$ °C.

Ensure headings are unambiguous (e.g., 'Mass of FA 3 used', not just 'Weight').

Key Takeaways

Method 2 is faster but less accurate than Method 1 because it doesn't account for heat loss during the reaction.

Common Mistakes

Using 'weight' instead of 'mass' in table headings.
Not subtracting the masses correctly.
Not recording the initial temperature of the acid before mixing.

Things to Be Careful About

The mark scheme checks if the value of (mass FA 1 $\times \Delta T$ in Method 1) / (mass FA 3 $\times \Delta T$ in Method 2) is between 0.80 and 1.25. This verifies that both methods used similar amounts of reactive material.

Techniques used

record Method 2 mass and temperature data

(f)

Use the temperature rise in (e), and the fact that the enthalpy change for the reaction between anhydrous sodium carbonate and hydrochloric acid is $-27.0\text{ kJ mol}^{-1}$ , to calculate the percentage of anhydrous sodium carbonate in FA 3.

\text{percentage Na}_2\text{CO}_3\text{ in FA 3} = \text{.............................. \%}

DifficultyMedium

Worked solution

Working

Stage 1: Energy released

q = 25 \times 4.2 \times \Delta T = 25 \times 4.2 \times 3.8 = 399 \text{ J}

Stage 2: Moles of Na₂CO₃

n = \frac{399}{27000} = 0.01478 \text{ mol}

Stage 3: Mass of Na₂CO₃

\text{mass} = 0.01478 \times 106 = 1.566 \text{ g}

Stage 4: Percentage purity

\text{percentage} = \frac{1.566}{1.80} \times 100 = 87.0\%

Answer

percentage Na₂CO₃ in FA 3 = 87.0 %

Final answer

87.0 %

Detailed explanation

Background Concept

This part requires repeating the full calculation sequence from part (c), but using the data collected in Method 2. The steps are: energy released -> moles reacted -> mass of pure substance -> percentage purity.

Understanding the Question

Calculate the percentage purity of FA 3 using the $\Delta T$ from part (e) and the literature $\Delta H$ .

Approach

Follow the same 4 stages as in part (c):

$q = 25 \times 4.2 \times \Delta T$
$n = q / 27000$
$m = n \times 106$
$\% = (m / \text{mass FA 3 used}) \times 100$

Step-by-Step Reasoning

Using representative data from (e):

$\Delta T = 3.8$ °C.
$q = 25 \times 4.2 \times 3.8 = 399$ J.
$n = 399 / 27000 = 0.01478$ mol.
Mass $\text{Na}_2\text{CO}_3 = 0.01478 \times 106 = 1.566$ g.
Percentage $= (1.566 / 1.80) \times 100 = 87.0\%$ .

The mark scheme awards 2 marks if all 4 stages are shown clearly, or 1 mark if 2-3 stages are shown.

Key Takeaways

Method 2 gives a lower percentage purity (87.0% vs 91.6%) because it does not correct for heat loss, leading to a lower $\Delta T$ and thus a lower calculated purity.

Common Mistakes

Forgetting to use the mass of FA 3 used in the final percentage calculation.
Arithmetic errors in the multi-step calculation.
Not showing all working stages.

Things to Be Careful About

Error carried forward from previous parts is allowed. If your $\Delta T$ in (e) was different, your final percentage will be different, but the method must be correct.

Techniques used

calculate energy releasedcalculate moles and mass of Na2CO3calculate percentage purity

(g)

FA 1 and FA 3 are both samples of the same impure anhydrous sodium carbonate and so the percentage of anhydrous sodium carbonate found using Method 1 and Method 2 should be the same. In practice the percentages are sometimes different from each other.

In both methods, percentage errors occur due to measuring the mass of solid and the temperature rise.

Ignoring these errors, which method is more accurate?
Tick the correct box and explain your answer.

Method	Tick
Method 1 more accurate
Method 2 more accurate
Method 1 and Method 2 equally accurate

DifficultyMedium-Easy

Worked solution

Answer

Method 1 more accurate

Explanation: In Method 1, a cooling curve is plotted and extrapolated to compensate for heat lost to the surroundings during the reaction. In Method 2, the maximum temperature is recorded directly without correcting for heat loss, so the temperature rise is lower and less accurate.

Final answer

Method 1 more accurate; cooling curve compensates for heat loss

Detailed explanation

Background Concept

Both methods measure the enthalpy change of the same reaction, but they handle heat loss differently. Method 1 uses extrapolation to estimate the theoretical maximum temperature, assuming the rate of heat loss before and after the reaction is constant. Method 2 simply records the peak temperature, which is always lower than the true maximum due to heat loss during the reaction time.

Understanding the Question

Compare the accuracy of the two methods, ignoring measurement errors (mass and temperature reading errors). Explain which is more accurate and why.

Approach

Identify the main source of systematic error: heat loss to the surroundings. Explain how each method deals with it. Method 1 corrects for it; Method 2 does not.

Step-by-Step Reasoning

Method 1: Records temperatures before and after the reaction. By drawing a line of best fit through the cooling phase and extrapolating it back to the time of mixing, we estimate what the temperature would have been if no heat was lost. This makes the $\Delta T$ more accurate.
Method 2: Only records the initial and maximum temperatures. The maximum temperature is reached after some time, during which heat has already been lost to the cup and surroundings. Thus, the recorded $\Delta T$ is lower than the true value.
Conclusion: Method 1 is more accurate because it compensates for heat loss.

Key Takeaways

Extrapolation is a powerful technique for correcting systematic errors related to heat loss in calorimetry.

Common Mistakes

Saying 'Method 2 is more accurate because it is simpler' (simplicity does not equal accuracy).
Saying 'Method 1 is more accurate because it uses more data points' (this is true, but the reason is heat loss correction, not just having more points).
Forgetting to explain why the method is more accurate.

Things to Be Careful About

The mark scheme specifically looks for the mention of 'heat lost is compensated for' or 'cooling curve plotted'. If there was no temperature fall after the maximum in Method 1 (unlikely but possible), both could be considered equally accurate.

Techniques used

evaluate accuracy of experimental methods

(h)

A student decided to confirm by experiment the literature value for the enthalpy change of the reaction between anhydrous sodium carbonate and hydrochloric acid. By mistake the student weighed a sample of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}$ , instead of anhydrous sodium carbonate, $\text{Na}_2\text{CO}_3$ .

State what effect this would have on the calculated value of the enthalpy change for the reaction. Explain your answer.

DifficultyMedium

Worked solution

Answer

Effect: The calculated enthalpy change would be less exothermic (or less negative / smaller magnitude).

Explanation: For the same mass of solid, hydrated sodium carbonate ( $\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}$ ) has a higher molar mass, so there are fewer moles of $\text{Na}_2\text{CO}_3$ present. This results in a smaller temperature rise. Since the calculation assumes the same $\Delta H$ , the smaller $\Delta T$ leads to a calculated $\Delta H$ that is less exothermic.

Final answer

Fewer moles of carbonate; temperature increase is less; enthalpy change is less exothermic

Detailed explanation

Background Concept

Anhydrous sodium carbonate ( $\text{Na}_2\text{CO}_3$ , $M_r = 106$ ) and hydrated sodium carbonate decahydrate ( $\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}$ , $M_r = 286$ ) have very different molar masses. If a student weighs out a mass $m$ of the hydrated salt, the number of moles of $\text{Na}_2\text{CO}_3$ is $m / 286$ , which is much less than $m / 106$ for the anhydrous salt.

Furthermore, the reaction of hydrated sodium carbonate with acid is less exothermic because energy is required to break the hydrogen bonds and vaporize the water of crystallization (the reverse of the hydration enthalpy).

Understanding the Question

A student mistakenly uses hydrated sodium carbonate instead of anhydrous. What effect does this have on the calculated value of $\Delta H$ ?

Approach

Compare the moles of $\text{Na}_2\text{CO}_3$ in a given mass of hydrated vs anhydrous salt.
Relate this to the expected temperature rise.
Explain how the temperature rise affects the calculated $\Delta H$ .

Step-by-Step Reasoning

Fewer moles: For a given mass (e.g., 2.0 g), moles of anhydrous $= 2.0 / 106 = 0.0189$ mol. Moles of hydrated $= 2.0 / 286 = 0.0070$ mol. There are fewer moles of reactive carbonate.
Less temperature rise: Fewer moles reacting means less heat is released, so the temperature rise ( $\Delta T$ ) is smaller. (Additionally, the dissolution of the hydrated salt and the reaction itself is less exothermic overall).
Calculated $\Delta H$ : The student calculates $\Delta H = -q / n$ . If they assume they used anhydrous salt, they will calculate $n = \text{mass} / 106$ (a larger value). The energy $q$ is smaller (due to smaller $\Delta T$ ). Thus, $|\Delta H| = q / n_{\text{assumed}}$ will be much smaller than the true value. The calculated $\Delta H$ will be less negative (less exothermic).

Key Takeaways

Always ensure the correct form of the reagent is used. The molar mass difference between hydrated and anhydrous salts significantly affects mole calculations.

Common Mistakes

Saying 'the enthalpy change is more exothermic' (confusing the direction of the error).
Not explaining why (failing to mention fewer moles or less temperature rise).
Saying 'the mass is wrong' (the mass is measured correctly; the identity of the substance is wrong).

Things to Be Careful About

The mark scheme looks for 'fewer moles / less amount of carbonate' and 'temperature increase is less / enthalpy change is less exothermic'. Both points are required for full marks.

Techniques used

deduce effect of hydrated salt on enthalpy calculation

(i)

A student used $3.00\text{ g}$ of anhydrous sodium carbonate that was $80.0\%$ pure by mass.

Calculate the minimum volume of $2.00\text{ mol dm}^{-3}$ hydrochloric acid that would be needed to react completely with this sample of impure anhydrous sodium carbonate.

\text{volume of HCl} = \text{.............................. cm}^3

DifficultyMedium

Worked solution

Working

Step 1: Moles of pure Na₂CO₃
Mass of sample $= 3.00 \text{ g}$ , purity $= 80.0\%$ .
Mass of pure $\text{Na}_2\text{CO}_3 = 3.00 \times 0.800 = 2.40 \text{ g}$ .

n(\text{Na}_2\text{CO}_3) = \frac{2.40}{106} = 0.02264 \text{ mol}

Step 2: Moles of HCl required
From the equation: $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow \dots$
Ratio is 1 : 2.

n(\text{HCl}) = 2 \times 0.02264 = 0.04528 \text{ mol}

Step 3: Volume of HCl
Concentration of HCl $= 2.00 \text{ mol dm}^{-3} = 0.00200 \text{ mol cm}^{-3}$ .

V(\text{HCl}) = \frac{n}{c} = \frac{0.04528}{0.00200} = 22.64 \text{ cm}^3

Answer

volume of HCl = 22.6 cm³

Final answer

22.6 cm³

Detailed explanation

Background Concept

To find the volume of acid required to react completely with a sample, we need to:

Find the mass of the pure reactive substance (accounting for purity).
Convert this mass to moles using the molar mass.
Use the stoichiometric ratio from the balanced equation to find the moles of acid required.
Calculate the volume of acid using its concentration.

Understanding the Question

Calculate the minimum volume of $2.00 \text{ mol dm}^{-3}$ HCl needed to react completely with $3.00 \text{ g}$ of an $80.0\%$ pure sample of anhydrous sodium carbonate.

Approach

Follow the 4-step stoichiometry pathway: mass sample -> mass pure -> moles pure -> moles acid -> volume acid.

Step-by-Step Reasoning

Mass of pure $\text{Na}_2\text{CO}_3$ : $3.00 \text{ g} \times (80.0 / 100) = 2.40 \text{ g}$ .
Moles of $\text{Na}_2\text{CO}_3$ : $M_r = 106$ . $n = 2.40 / 106 = 0.02264 \text{ mol}$ .
Moles of HCl: The equation is $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}$ . The ratio is 1:2. So $n(\text{HCl}) = 2 \times 0.02264 = 0.04528 \text{ mol}$ .
Volume of HCl: $c = 2.00 \text{ mol dm}^{-3} = 2.00 / 1000 = 0.00200 \text{ mol cm}^{-3}$ .
$V = n / c = 0.04528 / 0.00200 = 22.64 \text{ cm}^3$ .
Round to 3 significant figures: $22.6 \text{ cm}^3$ .

Key Takeaways

Always account for purity when calculating reactant quantities. The stoichiometric ratio is often 1:2 or 1:3, so double-check the balanced equation.

Common Mistakes

Forgetting to apply the 80.0% purity (using 3.00 g instead of 2.40 g).
Using the wrong mole ratio (e.g., 1:1 instead of 1:2).
Forgetting to convert concentration from mol dm⁻³ to mol cm⁻³ (or volume from dm³ to cm³).
Incorrect significant figures (3 or 4 sf are acceptable here).

Things to Be Careful About

The mark scheme awards marks for: correct moles of $\text{Na}_2\text{CO}_3 = (3/106) \times 0.8 = 0.0226$ ; ratio 1:2 giving moles HCl $= 0.0452$ ; volume $= 0.0452 / 0.002 = 22.6 \text{ cm}^3$ . Error carried forward is allowed.

Techniques used

calculate moles of pure substance from percentage purityuse stoichiometric ratio to find moles of reactantcalculate volume from concentration and moles

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