Chemistry 9701/31 — October/November 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

You will determine the value of $x$ in hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ . $x$ is not an integer.

You will carry out two methods to determine the value of $x$ . Each method involves sodium carbonate reacting with excess hydrochloric acid to release carbon dioxide.

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

(a)

Experiment 1

You will measure the volume of carbon dioxide released when hydrated sodium carbonate reacts with excess hydrochloric acid.

FA 1 is $0.500\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FA 2 is hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

Method

Weigh the container with FA 2. Record the mass.
Fill the tub with water to a depth of approximately $5\text{ cm}$ .
Fill the $250\text{ cm}^3$ measuring cylinder completely with water. Holding a piece of paper towel firmly over the top, invert the measuring cylinder and place it in the water in the tub.
Remove the paper towel and clamp the inverted measuring cylinder so the open end is in the water just above the base of the tub.
Using the $50\text{ cm}^3$ measuring cylinder, transfer $50.0\text{ cm}^3$ of FA 1 into the flask labelled Z. Check that the bung fits tightly into the neck of flask Z, clamp flask Z and place the end of the delivery tube into the inverted $250\text{ cm}^3$ measuring cylinder.
Remove the bung from the neck of the flask. Tip all the FA 2 from the container into the acid in the flask and replace the bung immediately. Remove the flask from the clamp and swirl it to mix the contents. You may need to shake the flask quite vigorously until the gas formed starts to collect in the measuring cylinder.
Return the flask to the clamp. Leave for several minutes, shaking the flask occasionally.
Weigh the container with any residual FA 2. Record the mass.
Calculate the mass of FA 2 added to the flask. Record the mass.
When no more gas is collected, measure the final volume of gas in the measuring cylinder. Record the volume.

Results

Record your results in a suitable table in the space provided.

DifficultyMedium-Easy

Worked solution

Answer

Record a table with columns:

(Mass of) container + FA 2 / g
(Mass of) container + residual FA 2 / g
(Mass of) FA 2 used / g
(Volume of) carbon dioxide / cm³

Record all balance readings to the same number of decimal places (2 or 3 dp) and the gas volume as an integer. Calculate the mass of FA 2 used by subtracting the residual mass from the initial mass.

Final answer

Table with headings and units; balance readings to same dp; volume integer; mass of FA2 used calculated.

Detailed explanation

Background Concept

In Experiment 1, hydrated sodium carbonate reacts with excess hydrochloric acid and the carbon dioxide evolved is collected by displacement of water in an inverted measuring cylinder. At room temperature and pressure, 1 mol of gas occupies about 24 dm³ (24000 cm³). The mass of FA 2 used is found by weighing the container before and after tipping the solid into the acid. A clear results table must allow these readings to be recorded with units and consistent precision.

Understanding the Question

The question asks you to record the data from Experiment 1 in a suitable table. The mark scheme rewards correct headings with units, consistent precision in balance readings, an integer gas volume, and correct calculation of the mass of FA 2 added.

Approach

Plan columns for each measured quantity: mass before, mass after, mass used, and volume of gas. Include units in the headings. Record balance readings to the same number of decimal places and the volume as a whole number of cm³. Calculate the mass used by subtraction.

Step-by-Step Reasoning

Weigh the container with FA 2: record, for example, 10.944 g.
After tipping the solid into the acid, weigh the container with residual FA 2: record, for example, 10.000 g.
Mass used = 10.944 − 10.000 = 0.944 g.
Record the volume of carbon dioxide collected, for example, 150 cm³.
Table headings: “Mass of container + FA 2 / g”, “Mass of container + residual FA 2 / g”, “Mass of FA 2 used / g”, “Volume of carbon dioxide / cm³”.
All balance readings should have the same number of decimal places; the gas volume should be an integer.

Key Takeaways

A results table needs clear headings with units, consistent precision, and correct derived values. The mass of solid used is found by difference.

Common Mistakes

Missing units or putting units in every cell instead of in the heading.
Balance readings with different numbers of decimal places.
Recording the volume with decimals when an integer is expected.
Forgetting to record the residual mass, so the mass used cannot be calculated.

Things to Be Careful About

Use the same balance precision for all mass readings. Read the gas volume at eye level after no more gas is produced. Ensure no gas escapes when replacing the bung.

Techniques used

record balance readings to consistent precisionrecord gas volumecalculate mass of FA2 used by difference

(b)

Calculations

(i)

Calculate the amount, in mol, of carbon dioxide collected in the measuring cylinder at room conditions.

\text{amount of CO}_2 = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

Hence deduce the amount, in mol, of sodium carbonate present in the FA 2 you added in your experiment.

\text{amount of Na}_2\text{CO}_3 = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

DifficultyMedium-Easy

Worked solution

Working

Using my results, for example volume of CO₂ = 150 cm³:

\text{amount CO}_2 = \frac{150}{24000} = 0.00625\text{ mol}

From the equation, 1 mol Na₂CO₃·xH₂O gives 1 mol CO₂, so:

\text{amount Na}_2\text{CO}_3 = 0.00625\text{ mol}

Answer

amount CO₂ = 0.00625 mol; amount Na₂CO₃ = 0.00625 mol

Final answer

0.00625 mol CO2; 0.00625 mol Na2CO3 (example)

Detailed explanation

Background Concept

At room temperature and pressure, one mole of any gas occupies approximately 24 dm³, which is 24000 cm³. The equation shows that 1 mol of hydrated sodium carbonate produces 1 mol of carbon dioxide, so the amount of sodium carbonate is equal to the amount of carbon dioxide collected.

Understanding the Question

You are asked to use the measured volume of carbon dioxide to calculate the amount of CO₂ in mol, then deduce the amount of Na₂CO₃ in the FA 2 added. The stoichiometric ratio from the equation is 1:1.

Approach

Divide the gas volume in cm³ by 24000 to obtain moles. Then use the balanced equation to relate moles of CO₂ to moles of Na₂CO₃.

Step-by-Step Reasoning

Using the example volume 150 cm³:

\text{amount CO}_2 = \frac{150}{24000} = 0.00625\text{ mol}

The equation shows 1 mol Na₂CO₃·xH₂O gives 1 mol CO₂, so the amount of Na₂CO₃ is also 0.00625 mol. The answer should be quoted to 2–4 significant figures.

Key Takeaways

The molar gas volume at room conditions is 24 dm³ mol⁻¹. A gas volume in cm³ is converted to moles by dividing by 24000.

Common Mistakes

Using 24000 dm³ instead of 24000 cm³.
Forgetting to convert cm³ to dm³.
Applying a 2:1 stoichiometric ratio instead of 1:1.

Things to Be Careful About

Use the measured volume of gas, not the mass of solid. Quote the final answer to an appropriate number of significant figures.

Techniques used

convert gas volume to moles using molar gas volumeapply 1:1 stoichiometry

(ii)

Use your answer to (b)(i) and the mass of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ , you used in Experiment 1 to calculate the relative formula mass, $M_r$ , of the $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

M_r\text{ of Na}_2\text{CO}_3\cdot x\text{H}_2\text{O} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

DifficultyMedium-Easy

Worked solution

Working

Using mass of FA 2 used = 0.944 g and amount of Na₂CO₃ = 0.00625 mol:

M_\text{r} = \frac{\text{mass}}{\text{amount}} = \frac{0.944}{0.00625} = 151.0

Answer

M $_\text{r}$ of Na₂CO₃·xH₂O = 151.0

Final answer

151.0

Detailed explanation

Background Concept

The relative formula mass, M_r, is the mass of one mole of a substance. It is calculated from M_r = mass / amount. Here the mass is the mass of hydrated sodium carbonate used, and the amount is the amount of Na₂CO₃ found in part (b)(i).

Understanding the Question

You are asked to use the mass of FA 2 used in Experiment 1 and the amount of Na₂CO₃ from (b)(i) to calculate the M_r of the hydrated salt.

Approach

Divide the mass of FA 2 used by the amount of Na₂CO₃. Because one mole of the hydrated salt contains one mole of Na₂CO₃, this gives the M_r of the hydrated salt.

Step-by-Step Reasoning

Using the example values:

M_\text{r} = \frac{0.944}{0.00625} = 151.0

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

Key Takeaways

M_r = mass / amount is a fundamental relationship used whenever a relative formula mass is found from experimental data.

Common Mistakes

Using the mass of the container instead of the mass of FA 2 used.
Using the amount of CO₂ incorrectly.
Quoting too many significant figures.

Things to Be Careful About

The mass must be the mass of hydrated sodium carbonate actually added, not the initial container mass. M_r has no units.

Techniques used

calculate relative formula mass from mass and amount

(iii)

Use your answer to (b)(ii) to calculate the value of $x$ in the $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

Show your working.

x = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

DifficultyMedium-Easy

Worked solution

Working

M $_\text{r}$ of Na₂CO₃ = 106.0

M_\text{r}\text{ of }x\text{H}_2\text{O} = 151.0 - 106.0 = 45.0

x = \frac{45.0}{18.0} = 2.5

Answer

x = 2.5

Final answer

2.5

Detailed explanation

Background Concept

The M_r of anhydrous sodium carbonate, Na₂CO₃, is 106.0. Each water of crystallisation contributes 18.0 to the M_r. Therefore M_r(Na₂CO₃·xH₂O) = 106.0 + 18.0x.

Understanding the Question

You are asked to use the M_r from (b)(ii) to find the value of x, the number of moles of water of crystallisation per mole of sodium carbonate.

Approach

Subtract the M_r of anhydrous Na₂CO₃ from the M_r of the hydrated salt to find the total M_r due to water. Then divide by 18.0 to find x.

Step-by-Step Reasoning

Using the example M_r = 151.0:

M_\text{r}\text{ of }x\text{H}_2\text{O} = 151.0 - 106.0 = 45.0

x = \frac{45.0}{18.0} = 2.5

The final answer should be quoted to 2–4 significant figures. Here x is not an integer, as stated in the question.

Key Takeaways

Water of crystallisation is found from the difference between hydrated and anhydrous M_r, divided by 18.0.

Common Mistakes

Dividing by 18 before subtracting 106.
Using 106 instead of 106.0.
Quoting x as an integer when the calculation gives a non-integer.

Things to Be Careful About

Use the M_r from your own result in (b)(ii), not a textbook value. Show the subtraction and division clearly.

Techniques used

subtract anhydrous M_r from hydrated M_rdivide by M_r of water to find x

(c)

A student suggests that it would be better to use hot water in the tub.

(i)

State whether using hot water would be an improvement. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Yes, it would be an improvement. Carbon dioxide is less soluble in hot water, so more of the CO₂ produced would be collected and measured.

Final answer

Yes – CO2 is less soluble in hot water, so more gas is collected.

Detailed explanation

Background Concept

Gases are less soluble in water at higher temperatures. In this experiment, some carbon dioxide dissolves in the water in the tub and is therefore not collected in the measuring cylinder. Using hot water reduces this loss.

Understanding the Question

The question asks whether using hot water in the tub would improve the experiment. You must state whether it is an improvement and explain why.

Approach

Consider the effect of temperature on the solubility of carbon dioxide in water. If less CO₂ dissolves, more gas is collected, making the measurement more reliable.

Step-by-Step Reasoning

Carbon dioxide is soluble in cold water.
Hot water holds less dissolved gas.
Therefore more of the CO₂ produced escapes into the measuring cylinder.
The measured volume is closer to the true volume of gas produced, so the experiment is improved.

Key Takeaways

Gas solubility decreases as temperature increases. This is a common practical improvement when collecting gases over water.

Common Mistakes

Saying hot water makes the gas expand, so the volume is larger. The intended point is solubility, not thermal expansion.
Saying it makes no difference.

Things to Be Careful About

Link the improvement explicitly to reduced solubility of carbon dioxide.

Techniques used

evaluate effect of temperature on gas solubility

(ii)

State the effect, if any, that using hot water would have on the value of $x$ calculated.

DifficultyMedium

Worked solution

Answer

The measured amount of CO₂, and hence Na₂CO₃, would be larger, so the calculated M $_\text{r}$ would be smaller. Therefore the calculated value of x would be smaller.

Final answer

x would be smaller.

Detailed explanation

Background Concept

A systematic error that increases the measured volume of gas increases the calculated amount of Na₂CO₃. Since M_r = mass / amount, a larger amount gives a smaller M_r. Because x = (M_r − 106) / 18, a smaller M_r gives a smaller x.

Understanding the Question

You are asked to state the effect, if any, that using hot water would have on the value of x calculated from Experiment 1.

Approach

Trace the effect step by step: more gas collected → larger amount of CO₂ → larger amount of Na₂CO₃ → smaller M_r → smaller x.

Step-by-Step Reasoning

Hot water reduces CO₂ solubility, so more CO₂ is collected.
Amount of CO₂ = volume / 24000, so the calculated amount is larger.
Amount of Na₂CO₃ is equal to amount of CO₂, so it is also larger.
M_r = mass of FA 2 / amount, so M_r is smaller.
x = (M_r − 106) / 18, so x is smaller.

Key Takeaways

Systematic errors propagate through calculations. Understanding the direction of each step is essential for predicting the effect on the final answer.

Common Mistakes

Saying x would increase.
Confusing M_r with x.
Saying there is no effect.

Things to Be Careful About

The question asks for the effect on x, not on the quality of the experiment. State clearly that x is smaller.

Techniques used

predict effect of larger measured gas volume on calculated x

(d)

Experiment 2

You will carry out a titration to measure the volume of hydrochloric acid that neutralises an aqueous solution of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

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

FA 3 is $0.100\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FA 4 is an aqueous solution containing $14.30\text{ g dm}^{-3}$ of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .
FA 5 is bromophenol blue indicator.

Method

Fill the burette with FA 3.
Pipette $25.0\text{ cm}^3$ of FA 4 into a conical flask.
Add a few drops of FA 5.
Carry out a rough titration and record your burette readings in the space below.

\text{The rough titre is } \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ cm}^3.

Carry out as many titrations as you think necessary to obtain consistent results.
Make sure your recorded results show the precision of your practical work.
Record, in a suitable form below, all of your burette readings and the volume of FA 3 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Record the rough titre, then at least two accurate titrations. For each accurate titration show:

initial (burette) reading / cm³
final (burette) reading / cm³
titre / cm³ (final − initial)

All burette readings should be to the nearest 0.05 cm³. Accurate titres should agree within 0.10 cm³, for example 47.30, 47.35 and 47.25 cm³.

Final answer

Rough titre plus accurate titration table with initial/final readings and titres to 0.05 cm3; concordant within 0.10 cm3.

Detailed explanation

Background Concept

In a titration, the volume of solution added from the burette is found by subtracting the initial reading from the final reading. Accurate work requires readings to the nearest 0.05 cm³ and repeat titrations that agree closely. The indicator, bromophenol blue, signals the end-point of the neutralisation.

Understanding the Question

You are asked to record the rough titre and then all accurate titration results. The mark scheme requires two burette readings and a titre for the rough titration, and initial and final readings for at least two accurate titrations. Headings and units must be shown.

Approach

Record the rough titre first. Then carry out accurate titrations, refilling the burette as needed, and record initial reading, final reading and titre for each. Repeat until at least two titres are within 0.10 cm³ of each other.

Step-by-Step Reasoning

Rough titre, for example 47.4 cm³, gives an approximate end-point.
Accurate titration 1: initial 0.00 cm³, final 47.30 cm³, titre 47.30 cm³.
Accurate titration 2: initial 0.00 cm³, final 47.35 cm³, titre 47.35 cm³.
Accurate titration 3: initial 0.00 cm³, final 47.25 cm³, titre 47.25 cm³.
All readings are to the nearest 0.05 cm³.
The titres agree within 0.10 cm³.

Key Takeaways

Good titration records include headings with units, readings to 0.05 cm³, and concordant results. The titre is always final reading minus initial reading.

Common Mistakes

Recording only the titre without initial and final readings.
Using inconsistent decimal places.
Not repeating the titration enough to obtain concordant results.

Things to Be Careful About

Read the burette at eye level and record the meniscus correctly. Refill the burette when necessary and record the new initial reading.

Techniques used

record rough and accurate burette readingscalculate titresassess concordance of titres

(e)

From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value.

25.0\text{ cm}^3\text{ of FA 4 required } \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ cm}^3\text{ of FA 3.}

DifficultyMedium-Easy

Worked solution

Working

Using concordant titres 47.30, 47.35 and 47.25 cm³:

\text{mean titre} = \frac{47.30 + 47.35 + 47.25}{3} = 47.30\text{ cm}^3

Answer

25.0 cm³ of FA 4 required 47.30 cm³ of FA 3.

Final answer

47.30 cm3 (example)

Detailed explanation

Background Concept

The mean titre is calculated only from concordant results, usually those within a total spread of 0.20 cm³. The mean should be quoted to 2 decimal places, the same precision as the individual burette readings.

Understanding the Question

You are asked to calculate a suitable mean titre from your accurate titration results and show how you obtained it. The mark scheme requires that the selected titres agree closely and that the working is shown.

Approach

Select two or more accurate titres that are within 0.20 cm³ of each other. Add them and divide by the number of titres used. Quote the mean to 2 decimal places.

Step-by-Step Reasoning

Using the example titres 47.30, 47.35 and 47.25 cm³:

\text{mean titre} = \frac{47.30 + 47.35 + 47.25}{3} = \frac{141.90}{3} = 47.30\text{ cm}^3

The mean is rounded to the nearest 0.01 cm³.

Key Takeaways

Only concordant titres should be averaged. Show the selected values and the calculation clearly.

Common Mistakes

Averaging an outlier that is not concordant.
Quoting the mean to too many decimal places.
Not showing which titres were selected.

Things to Be Careful About

The mean must be quoted to 2 decimal places. If one titre is clearly different, do not include it in the mean.

Techniques used

select concordant titrescalculate mean titre to 2 decimal places

(f)

Calculations

(i)

Give your answers to (f)(ii), (f)(iii) and (f)(iv) to an appropriate number of significant figures.

DifficultyEasy

Worked solution

Answer

Answers to (f)(ii), (f)(iii) and (f)(iv) are quoted to 3–4 significant figures.

Final answer

3–4 significant figures

Detailed explanation

Background Concept

Calculated values should be quoted to a number of significant figures consistent with the precision of the measurements. The mark scheme for this question requires 3–4 significant figures for the answers in (f)(ii), (f)(iii) and (f)(iv).

Understanding the Question

This part is a reminder to give the calculated answers to an appropriate number of significant figures. It is a marking point in its own right.

Approach

After completing the calculations in (f)(ii), (f)(iii) and (f)(iv), check that each final value has 3–4 significant figures.

Step-by-Step Reasoning

0.004730 mol has 4 significant figures.
0.002365 mol has 4 significant figures.
0.09460 mol has 4 significant figures.
x = 2.51 has 3 significant figures.
These are all acceptable.

Key Takeaways

Always match the significant figures of a final answer to the precision expected by the question or mark scheme.

Common Mistakes

Quoting too many significant figures, such as 0.0047300.
Quoting too few, such as 0.005.

Things to Be Careful About

Leading zeros are not significant. For example, 0.004730 has four significant figures because the leading zeros only locate the decimal point.

Techniques used

quote calculated values to 3–4 significant figures

(ii)

Calculate the amount, in mol, of hydrochloric acid present in the volume of FA 3 you calculated in (e).

\text{amount of HCl} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

DifficultyMedium-Easy

Worked solution

Working

Using mean titre 47.30 cm³:

\text{amount HCl} = 0.100 \times \frac{47.30}{1000} = 0.004730\text{ mol}

Answer

amount of HCl = 0.004730 mol

Final answer

0.004730 mol

Detailed explanation

Background Concept

The amount of solute in a solution is given by n = cV, where c is concentration in mol dm⁻³ and V is volume in dm³. A volume in cm³ must be divided by 1000 to convert to dm³.

Understanding the Question

You are asked to calculate the amount of hydrochloric acid in the mean titre of FA 3 found in part (e).

Approach

Use n = cV. FA 3 is 0.100 mol dm⁻³ HCl. Convert the mean titre from cm³ to dm³ by dividing by 1000.

Step-by-Step Reasoning

Using the example mean titre 47.30 cm³:

\text{amount HCl} = 0.100 \times \frac{47.30}{1000} = 0.004730\text{ mol}

The answer is quoted to 4 significant figures.

Key Takeaways

n = cV is the core relationship for titration calculations. Always convert cm³ to dm³ before multiplying by concentration.

Common Mistakes

Forgetting to divide the volume by 1000.
Using the concentration of FA 4 instead of FA 3.
Quoting too many significant figures.

Things to Be Careful About

Use the mean titre from part (e), not a single titre. Include the unit mol in the final answer.

Techniques used

calculate moles from concentration and volume

(iii)

Use the equation for the neutralisation to deduce the amount, in mol, of sodium carbonate present in $25.0\text{ cm}^3$ of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

\text{amount of Na}_2\text{CO}_3 = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

Hence calculate the amount, in mol, of sodium carbonate in $1.00\text{ dm}^3$ of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

\text{amount of Na}_2\text{CO}_3\text{ in } 1.00\text{ dm}^3 = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

DifficultyMedium-Easy

Worked solution

Working

From the equation, 2 mol HCl react with 1 mol Na₂CO₃·xH₂O:

\text{amount Na}_2\text{CO}_3\text{ in }25.0\text{ cm}^3 = \frac{0.004730}{2} = 0.002365\text{ mol}

\text{amount Na}_2\text{CO}_3\text{ in }1.00\text{ dm}^3 = 0.002365 \times 40 = 0.09460\text{ mol}

Answer

0.002365 mol in 25.0 cm³; 0.09460 mol in 1.00 dm³

Final answer

0.002365 mol; 0.09460 mol

Detailed explanation

Background Concept

The balanced equation shows that 2 mol HCl react with 1 mol Na₂CO₃·xH₂O. Therefore the amount of sodium carbonate is half the amount of HCl. To scale from 25.0 cm³ to 1.00 dm³, multiply by 40 because 1.00 dm³ = 1000 cm³ and 1000/25 = 40.

Understanding the Question

You are asked to use the amount of HCl from (f)(ii) to find the amount of Na₂CO₃ in 25.0 cm³ of FA 4, then calculate the amount in 1.00 dm³.

Approach

Divide the amount of HCl by 2 to get the amount of Na₂CO₃ in the 25.0 cm³ sample. Then multiply by 40 to find the amount in 1.00 dm³.

Step-by-Step Reasoning

Using the example amount of HCl = 0.004730 mol:

\text{amount Na}_2\text{CO}_3\text{ in }25.0\text{ cm}^3 = \frac{0.004730}{2} = 0.002365\text{ mol}

\text{amount Na}_2\text{CO}_3\text{ in }1.00\text{ dm}^3 = 0.002365 \times 40 = 0.09460\text{ mol}

Key Takeaways

Stoichiometry from the balanced equation is essential. Scaling a 25.0 cm³ portion to 1.00 dm³ requires multiplying by 40.

Common Mistakes

Using a 1:1 ratio instead of 2:1.
Multiplying by 4 instead of 40.
Quoting the amount in 25.0 cm³ as the amount in 1.00 dm³.

Things to Be Careful About

Keep the two answers clearly labelled: one for 25.0 cm³ and one for 1.00 dm³.

Techniques used

use 2:1 stoichiometryscale from 25.0 cm3 to 1.00 dm3

(iv)

Calculate the value of $x$ in the sample of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

Show your working.

x = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

DifficultyMedium

Worked solution

Working

FA 4 contains 14.30 g dm⁻³ of Na₂CO₃·xH₂O.

M_\text{r} = \frac{14.30}{0.09460} = 151.2

M_\text{r}\text{ of }x\text{H}_2\text{O} = 151.2 - 106.0 = 45.2

x = \frac{45.2}{18.0} = 2.51

Answer

x = 2.51

Final answer

2.51

Detailed explanation

Background Concept

The concentration of FA 4 is given in g dm⁻³. The amount of Na₂CO₃·xH₂O in 1.00 dm³ was found in (f)(iii). Since concentration in mol dm⁻³ = amount in 1 dm³, the M_r is:

M_\text{r} = \frac{\text{concentration in g dm}^{-3}}{\text{concentration in mol dm}^{-3}}

Once M_r is known, x is found by subtracting the M_r of anhydrous Na₂CO₃ (106.0) and dividing by 18.0.

Understanding the Question

You are asked to calculate x using the mass concentration of FA 4 and the amount of Na₂CO₃ per dm³ from (f)(iii).

Approach

First find M_r of the hydrated salt using M_r = 14.30 / amount per dm³. Then subtract 106.0 and divide by 18.0.

Step-by-Step Reasoning

Using the example amount per dm³ = 0.09460 mol:

M_\text{r} = \frac{14.30}{0.09460} = 151.2

M_\text{r}\text{ of }x\text{H}_2\text{O} = 151.2 - 106.0 = 45.2

x = \frac{45.2}{18.0} = 2.51

The answer is quoted to 3 significant figures.

Key Takeaways

The M_r of a dissolved substance can be found from its mass concentration divided by its molar concentration. Water of crystallisation is then found from the difference in M_r.

Common Mistakes

Using the amount in 25.0 cm³ instead of the amount in 1.00 dm³.
Forgetting to subtract 106.0 before dividing by 18.0.
Quoting x as an integer when the calculation gives a non-integer.

Things to Be Careful About

Use the amount in 1.00 dm³ from (f)(iii), not the amount in 25.0 cm³. Show all working clearly.

Techniques used

calculate M_r from mass concentration and molar concentrationdeduce x from M_r

(g)

The aqueous solution of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ , FA 4, was prepared by weighing and dissolving the solid to make $1.00\text{ dm}^3$ of solution.

\begin{aligned} \text{Mass of container } + \text{ Na}_2\text{CO}_3\cdot x\text{H}_2\text{O} &= 32.509\text{ g} \\ \text{Mass of empty container} &= 18.209\text{ g} \\ \text{Mass of } \text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O} &= 14.300\text{ g} \end{aligned}

(i)

State the maximum uncertainty in a single balance reading for the balance used.

\text{maximum uncertainty} = \pm \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ g}

Calculate the maximum percentage uncertainty in this mass of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ .

Show your working.

\text{maximum percentage uncertainty} = \pm \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\%

DifficultyMedium

Worked solution

Working

The balance reads to 0.001 g, so the maximum uncertainty in a single reading is ±0.001 g.

The mass is a difference of two readings:

\text{absolute uncertainty} = 2 \times 0.001 = \pm 0.002\text{ g}

\text{percentage uncertainty} = \frac{0.002}{14.300} \times 100 = 0.0140\%

Answer

maximum uncertainty = ±0.001 g; maximum percentage uncertainty = ±0.014%

Final answer

±0.001 g; ±0.014%

Detailed explanation

Background Concept

The balance readings are given to three decimal places, so the uncertainty in a single reading is ±0.001 g. When a mass is found by difference, the absolute uncertainties of the two readings add, giving 2 × 0.001 = ±0.002 g. Percentage uncertainty is calculated as (absolute uncertainty / measured value) × 100.

Understanding the Question

You are asked to state the maximum uncertainty in a single balance reading and then calculate the maximum percentage uncertainty in the mass of Na₂CO₃·xH₂O, which is 14.300 g.

Approach

Use the balance precision to state the single-reading uncertainty. Double it for the mass by difference. Then divide by the mass and multiply by 100.

Step-by-Step Reasoning

The mass of FA 4 is found from two balance readings:

\text{mass} = 32.509 - 18.209 = 14.300\text{ g}

Single-reading uncertainty = ±0.001 g.

Absolute uncertainty in the mass = 2 × 0.001 = ±0.002 g.

\text{percentage uncertainty} = \frac{0.002}{14.300} \times 100 = 0.013986\% \approx 0.014\%

Key Takeaways

A mass obtained by difference has double the uncertainty of a single balance reading. Percentage uncertainty is absolute uncertainty divided by the measured value, multiplied by 100.

Common Mistakes

Using ±0.001 g as the uncertainty in the mass without doubling it.
Dividing by 32.509 instead of 14.300.
Forgetting to multiply by 100.

Things to Be Careful About

The uncertainty in the mass is ±0.002 g, not ±0.001 g. Quote the percentage uncertainty to 2–3 significant figures.

Techniques used

state balance uncertaintypropagate uncertainty through mass differencecalculate percentage uncertainty

(ii)

Using the method in Experiment 2 a student calculated the relative formula mass, $M_r$ , of $\text{Na}_2\text{CO}_3\cdot x\text{H}_2\text{O}$ to be $242.2$ . Assume that the uncertainty in the mass of FA 4 is the only source of error in the experiment.

Calculate the maximum value for the relative formula mass of FA 4.

\text{maximum value for the relative formula mass of FA 4} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

DifficultyMedium

Worked solution

Working

Using the percentage uncertainty from (g)(i), ±0.014%:

\text{maximum } M_\text{r} = 242.2 \times \frac{100 + 0.0140}{100}

\text{maximum } M_\text{r} = 242.2 \times 1.00014 = 242.23

Answer

maximum value for the relative formula mass of FA 4 = 242.23

Final answer

242.23

Detailed explanation

Background Concept

If the only source of error is the uncertainty in the mass of FA 4, the percentage uncertainty in the mass is the same as the percentage uncertainty in the calculated M_r. To find the maximum possible M_r, add the percentage uncertainty to the calculated value.

Understanding the Question

You are given a student’s calculated M_r of 242.2 and asked to find the maximum value, assuming the only error is the uncertainty in the mass of FA 4.

Approach

Multiply the calculated M_r by (100 + percentage uncertainty)/100. This increases the M_r by the percentage uncertainty.

Step-by-Step Reasoning

Using the percentage uncertainty from (g)(i), 0.0140%:

\text{maximum } M_\text{r} = 242.2 \times \frac{100 + 0.0140}{100}

\text{maximum } M_\text{r} = 242.2 \times 1.00014 = 242.23

The maximum value is slightly greater than 242.2.

Key Takeaways

Percentage uncertainty in a directly proportional quantity is transferred directly to the final calculated value. To find the maximum value, add the percentage uncertainty.

Common Mistakes

Subtracting the percentage uncertainty instead of adding it.
Using the absolute uncertainty in grams directly without converting to a percentage.
Quoting the answer as 242.2, losing the effect of the uncertainty.

Things to Be Careful About

Use the percentage uncertainty from your answer to (g)(i). The maximum value must be larger than 242.2.

Techniques used

apply percentage uncertainty to M_rcalculate maximum value

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