Chemistry 5070/31 — May/June 2024
Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme
Topics Experimental Contexts · Analysis, Conclusions and Evaluation · Planning Experiments and Investigations · Observations and Measurements · Qualitative Analysis · Use of Techniques, Apparatus and Materials
You are going to investigate the reactions of three metals, , and , with aqueous copper(II) sulfate.
Read all the instructions carefully before starting the experiments.
Instructions
You are going to do four experiments.
Experiment 1
- Use a measuring cylinder to add of aqueous copper(II) sulfate to a beaker.
- Use a thermometer to measure the initial temperature of the aqueous copper(II) sulfate in the beaker.
- Record this temperature to the nearest in Table 1.1.
- Add all the sample of metal to the beaker. Carefully stir the mixture.
- The temperature of the mixture will increase. Continue stirring until there is no further temperature increase.
- Measure the highest temperature of the mixture. Record this temperature to the nearest in Table 1.1.
- Leave the beaker to stand while you complete Experiments 2 and 3. You will need it for Experiment 4.
Experiment 2
Use a second beaker to repeat Experiment 1 using instead of .
Experiment 3
Use a third beaker to repeat Experiment 1 using instead of .
Determine the temperature increase in each reaction and write your answers in Table 1.1.
Table 1.1
| experiment | initial temperature / | highest temperature / | temperature increase / |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 |
Experiment 4
Slowly add of dilute sulfuric acid to the contents of the beaker from Experiment 1.
Answer
Complete Table 1.1 with:
- initial and highest temperatures recorded to one decimal place, ending in .0 or .5
- every highest temperature greater than the corresponding initial temperature
- temperature increase = highest temperature initial temperature, correctly calculated
- Experiment 2 has the greatest temperature increase
- Experiment 3 has the lowest temperature increase
Completed Table 1.1 with readings to 1 decimal place (.0 or .5), highest temperatures above initial, increases calculated, Experiment 2 greatest and Experiment 3 lowest.
Walkthrough
This part is about carrying out the temperature measurements and recording them properly. Read the initial and highest temperatures to the nearest 0.5 °C and write them in the table. Then subtract the initial temperature from the highest temperature for each experiment to get the temperature increase. The mark scheme requires all values to one decimal place ending in .0 or .5, and all highest temperatures must be greater than the initial temperatures. Since the reaction is exothermic, the temperature rises. Experiment 2 should give the greatest temperature increase and Experiment 3 the lowest, because the temperature increase is a measure of how much reaction occurred and therefore of the reactivity of the metal.
Key Takeaways
- Temperature readings for a thermometer should be recorded to the precision required, here nearest 0.5 °C.
- Temperature increase = highest temperature − initial temperature.
- A larger temperature increase means a more exothermic reaction and usually a more reactive metal.
Common Mistakes
- Recording temperatures as whole numbers instead of to one decimal place ending in .0 or .5.
- Writing a highest temperature that is lower than the initial temperature, which is impossible for an exothermic reaction.
- Arithmetic errors in subtracting the initial temperature.
- Not making sure Experiment 2 has the greatest increase and Experiment 3 the lowest, which is needed for later parts.
Things to Be Careful About
- Use the same thermometer and read it at eye level to avoid parallax error.
- Stir continuously until the temperature stops rising, so you record the true highest temperature.
- Use the units given in the table header () and do not add units to every cell.
Describe the initial appearance of the aqueous copper(II) sulfate used in Experiment 1.
______
Answer
Blue solution.
Blue solution
Walkthrough
Aqueous copper(II) sulfate contains hydrated copper(II) ions, , which give the solution its blue colour. The question only asks for the initial appearance, so a single observation is enough.
Key Takeaways
- Copper(II) sulfate solution is blue.
- Qualitative observations must describe colour and state (solution).
Common Mistakes
- Writing 'copper sulfate' without saying 'blue'.
- Confusing the solid (blue crystals) with the solution.
Things to Be Careful About
- The mark is for 'blue'; 'solution' is not required but helps.
- Do not describe the solution as colourless.
Answer
Colourless solution and brown/pink solid.
Colourless solution and brown/pink solid
Walkthrough
In Experiment 1, metal X displaces copper from copper(II) sulfate, so copper(II) ions are removed from solution and copper metal is deposited. The blue solution becomes colourless when all the copper(II) ions have reacted. In Experiment 4, dilute sulfuric acid is added to the beaker. If X is in excess, some solid X remains; the acid reacts with it, but the appearance after the reaction is still a colourless solution (metal sulfate) with brown/pink solid (copper, and possibly unreacted X). The mark scheme awards one mark for 'colourless solution' and one for 'brown/pink solid'.
Key Takeaways
- Displacement of copper from copper(II) sulfate removes the blue colour.
- Copper metal is brown/pink.
- The final mixture contains a colourless solution and a brown/pink solid.
Common Mistakes
- Saying 'blue solution remains' – the blue colour disappears because copper(II) ions are used up.
- Only giving one observation when two marks are available.
- Calling the solid 'red' instead of brown/pink.
Things to Be Careful About
- The acid itself does not change the colour; it reacts with excess metal, but the solution stays colourless.
- Include both the solution and the solid in the description.
Answer
No blue colour remains, so all the copper(II) ions have reacted and some X is left unreacted. (Alternatively, effervescence when acid is added shows unreacted X is present.)
No blue colour remains / unreacted X remains / effervescence with acid, showing X is in excess.
Walkthrough
In Experiment 1, X reacts with copper(II) sulfate. If X is in excess, there is more than enough X to react with all the copper(II) ions. Evidence: the blue colour of the solution disappears, meaning no copper(II) ions remain; solid X is left over; and when acid is added in Experiment 4, effervescence occurs because the acid reacts with the remaining X, releasing hydrogen. Any one of these observations shows X is in excess.
Key Takeaways
- A reactant is in excess when some of it remains after the reaction.
- The disappearance of the blue colour shows the limiting reactant (copper(II) sulfate) has been used up.
- Effervescence with acid is a test for a reactive metal.
Common Mistakes
- Saying 'the temperature increased' – this does not show excess.
- Saying 'the solution became colourless, so X is used up' – actually X is the reactant left over.
- Not linking the observation to the idea of excess.
Things to Be Careful About
- The blue colour is due to copper(II) ions, not to X.
- If the solution is colourless, copper(II) ions are all used up, so X must be in excess.
Use your results to arrange , , and in decreasing order of reactivity.
Explain how the results give this order of reactivity.
most reactive ______
______
least reactive ______
explanation ______
Answer
Most reactive: Y
X
Least reactive: Z
The greater the temperature increase, the more reactive the metal.
Y, X, Z; greater temperature increase means more reactive.
Walkthrough
The temperature increase in each experiment comes from the heat released when the metal displaces copper from copper(II) sulfate. A more reactive metal reacts more vigorously and releases more energy, so it gives a larger temperature rise. Using the results, put the metal with the greatest temperature increase first (Y), then the next (X), then the smallest (Z). The explanation must state the link between temperature increase and reactivity.
Key Takeaways
- Displacement reactions are exothermic.
- Temperature increase can be used to compare reactivity of metals.
- The order of reactivity matches the order of temperature increases.
Common Mistakes
- Ordering by initial or highest temperature instead of by temperature increase.
- Forgetting to state that greater temperature increase means greater reactivity.
- Writing the order from least to most reactive when the question asks for decreasing order.
Things to Be Careful About
- Use the temperature increase, not the highest temperature.
- The mark scheme expects the order Y, X, Z based on the intended results.
- Include both the order and the explanation for full marks.
A student repeats the experiment using a fourth metal.
This metal is the second most reactive of the four metals.
Suggest a temperature increase for this experiment.
______
Answer
Any temperature increase between the highest and second highest values from the student's results.
A value between the highest and second highest temperature increases from the student's results
Walkthrough
A fourth metal that is second most reactive of the four must give a temperature increase between the most reactive metal (highest increase) and the next most reactive (second highest increase). So the suggested value should lie between those two values. For example, if the highest increase is 20.0 °C and the second highest is 15.0 °C, any value such as 17.5 °C would be acceptable.
Key Takeaways
- Reactivity order corresponds to temperature increase order.
- A metal between two others in reactivity should give a temperature increase between their values.
Common Mistakes
- Giving a value outside the range.
- Giving a value equal to one of the existing values instead of between them.
- Not using the student's own results.
Things to Be Careful About
- The question says 'suggest', so any value in the correct range is accepted.
- Make sure the value is between the highest and second highest, not between the highest and lowest.
The temperature increases calculated are less than the true values for these experiments.
Suggest a reason for this.
Describe an improvement to the method which makes the results closer to the true values.
reason ______
improvement ______
Answer
Reason: heat loss to the surroundings.
Improvement: insulate the beaker / use a lid / use a polystyrene cup.
Heat loss to surroundings; improve by insulating the beaker or using a lid/polystyrene cup.
Walkthrough
The temperature increase is measured by taking the highest temperature reached. Some heat is always lost to the surroundings through the sides of the beaker and the surface, so the measured highest temperature is lower than the true value. To reduce this, insulate the beaker, put a lid on it, or use a polystyrene cup, which traps heat and keeps the reaction warmer for longer.
Key Takeaways
- Heat loss is the main source of error in temperature-change experiments.
- Insulation, a lid, or a polystyrene cup reduces heat loss.
- Improvements should target the identified source of error.
Common Mistakes
- Saying 'the thermometer was wrong' without a reason.
- Suggesting an improvement that does not reduce heat loss, such as stirring faster.
- Only giving a reason or only an improvement when two marks are available.
Things to Be Careful About
- The reason must be about heat loss, not about the reaction itself.
- The improvement must be practical and directly reduce heat loss.
- 'Use a lid' and 'insulate' are both acceptable.
State and explain the effect of using half the concentration of aqueous copper(II) sulfate on the temperature increase in Experiment 2.
effect ______
explanation ______
Answer
Effect: the temperature increase will be lower, about half.
Explanation: half the concentration means half the number of copper(II) ions in the same volume, so fewer displacement reactions occur and less heat is released.
Lower temperature increase, about half; because fewer copper(II) ions are present to react.
Walkthrough
Concentration tells you how many particles are in a given volume. If the concentration of copper(II) sulfate is halved, the same volume contains half the number of copper(II) ions. Each ion that reacts with metal Y releases heat, so with half the ions there are half as many reactions and roughly half the heat released. Therefore the temperature increase is approximately halved. The mark scheme wants three ideas: lower temperature change, approximately halved, and fewer copper ions to react.
Key Takeaways
- Temperature increase depends on the amount of reacting particles, not just the metal.
- Halving concentration halves the number of ions in the same volume.
- The energy released, and hence temperature rise, is proportional to the number of reacting particles.
Common Mistakes
- Saying the temperature increase stays the same.
- Saying the temperature increase doubles.
- Forgetting to mention the number of copper(II) ions.
- Not saying 'approximately' when predicting the halving.
Things to Be Careful About
- The question asks for both the effect and the explanation.
- Use 'approximately' because heat loss and other factors make the exact halving unlikely.
- Link the lower temperature to fewer copper(II) ions, not to the metal Y.
You are provided with solution .
You will do a series of experiments.
You should:
- record your observations and conclusions for each of these experiments
- test and name any gases evolved.
To prepare for the experiment in (d), place depth of in a test-tube. Place a wooden splint into the test-tube and leave it while doing the experiments in (a), (b) and (c).
To depth of in a test-tube, add a few drops of dilute nitric acid.
Add depth of aqueous barium nitrate.
observations ______
conclusions ______
Answer
Observations: solution (remains) colourless / no change / no precipitate.
Conclusions: does not contain sulfate ions () or carbonate ions ().
No precipitate; no sulfate or carbonate ions present
Walkthrough
Barium nitrate is used to test for sulfate ions. If sulfate ions were present, a white precipitate of barium sulfate, , would form. Carbonate ions would also give a white precipitate of barium carbonate, . Dilute nitric acid is added first to remove carbonate ions, so any white precipitate would be due to sulfate. Here no precipitate is seen, so the solution does not contain sulfate ions (and, because the acid would have destroyed carbonate, it also does not contain carbonate ions).
Key Takeaways
- Barium nitrate gives a white precipitate with sulfate ions.
- Carbonate ions also give a white precipitate with barium ions, so nitric acid is added first.
- No precipitate means neither sulfate nor carbonate is present.
Common Mistakes
- Writing 'no reaction' without giving the conclusion.
- Concluding that the solution contains no ions at all.
- Forgetting that carbonate ions would also precipitate with barium ions.
Things to Be Careful About
- Record observations and conclusions separately.
- Use 'solution remains colourless' or 'no precipitate' rather than 'nothing happened'.
- Use dilute nitric acid, not dilute sulfuric acid, because sulfate ions would interfere.
To depth of in a test-tube, add a few drops of dilute nitric acid.
Add depth of aqueous silver nitrate.
observations ______
conclusions ______
Answer
Observations: white precipitate.
Conclusions: contains chloride ions ().
White precipitate; chloride ions present
Walkthrough
Silver nitrate is the test for chloride ions. Chloride ions react with silver ions to form silver chloride, , which is a white precipitate. Dilute nitric acid is added first to remove carbonate ions, which would also give a white precipitate with silver ions. Because a white precipitate forms, the solution contains chloride ions.
Key Takeaways
- Silver nitrate with chloride ions gives a white precipitate of silver chloride.
- The precipitate is insoluble in dilute nitric acid.
- Nitric acid prevents carbonate ions from giving a false positive.
Common Mistakes
- Giving the observation as 'silver chloride' instead of 'white precipitate'.
- Forgetting to state the conclusion that chloride ions are present.
- Saying the precipitate is soluble in nitric acid.
Things to Be Careful About
- Use dilute nitric acid, not hydrochloric acid, because chloride ions would interfere.
- The observation is the colour and state; the conclusion is the ion identified.
To depth of aqueous sodium carbonate in a test-tube, add depth of aqueous silver nitrate. A white precipitate should form.
Add dilute nitric acid a drop at a time until no further change is seen.
Effervescence of a colourless gas should be observed. The gas turns limewater milky.
Answer
The white precipitate dissolves, forming a colourless solution.
Precipitate dissolves / colourless solution formed
Walkthrough
In experiment (c), silver carbonate, , is a white precipitate. When dilute nitric acid is added, the carbonate reacts with the acid to produce carbon dioxide gas (the effervescence already described) and a soluble silver salt. As the acid is added, the white precipitate dissolves and a colourless solution forms. This is the additional observation asked for.
Key Takeaways
- Carbonates react with acids to give a salt, water and carbon dioxide.
- An insoluble carbonate dissolves as it reacts with acid.
- The gas produced turns limewater milky, confirming carbon dioxide.
Common Mistakes
- Repeating the effervescence or the limewater test, which are already given in the question.
- Saying 'bubbles' without mentioning the precipitate dissolving.
- Writing 'the precipitate disappears' without saying what forms.
Things to Be Careful About
- The question asks for 'one other observation', so give a new observation, not one already stated.
- 'Colourless solution' is a useful phrase for a soluble salt solution.
Answer
To prevent carbonate ions (or other ions) from forming a precipitate with silver nitrate, so that any white precipitate can be attributed to chloride ions.
To prevent other ions (e.g. carbonate) from precipitating with silver nitrate, making the test specific for chloride
Walkthrough
Silver nitrate gives a white precipitate with chloride ions, but it also gives a white precipitate with carbonate ions. If carbonate ions were still present, the white precipitate in (b) could be silver carbonate rather than silver chloride. Adding dilute nitric acid first removes carbonate ions by converting them to carbon dioxide gas, so any white precipitate formed with silver nitrate must be due to chloride ions. This makes the test specific for chloride.
Key Takeaways
- Acidifying a test solution can remove interfering ions.
- Carbonate ions are removed by nitric acid as carbon dioxide.
- The purpose is to make the silver nitrate test specific for chloride.
Common Mistakes
- Saying 'to make the solution acidic' without explaining why.
- Mentioning sulfate ions, which are not removed by nitric acid and do not interfere with silver nitrate.
- Not linking the acid to the removal of carbonate ions.
Things to Be Careful About
- Use dilute nitric acid, not sulfuric or hydrochloric acid, because those would add interfering ions.
- The answer should mention carbonate ions specifically.
Place the end of the wooden splint which has been in into the flame of a Bunsen burner with the air hole open. Record the first flame colour seen.
first flame colour seen ______
conclusions ______
Answer
First flame colour seen: orange-red.
Conclusions: calcium ions () or lithium ions () present.
Orange-red flame; calcium or lithium ions present
Walkthrough
A flame test is used to identify metal cations. The wooden splint soaked in solution is placed in a hot Bunsen flame. The first flame colour seen is orange-red. Calcium ions give a brick-red/orange-red flame and lithium ions give a crimson flame, which are very similar. So the conclusion is that calcium ions or lithium ions are present. The phrase 'first flame colour' is important because sodium ions, if present as an impurity, give a strong yellow flame that may appear later and mask the true colour.
Key Takeaways
- Flame tests identify some metal cations by their characteristic flame colour.
- Calcium and lithium give similar orange-red/crimson colours.
- The first colour should be recorded before any sodium yellow appears.
Common Mistakes
- Writing 'yellow flame' and concluding sodium.
- Giving only calcium and not mentioning lithium as an alternative.
- Forgetting to record the 'first' flame colour.
Things to Be Careful About
- Use a clean splint to avoid contamination.
- The air hole should be open to give a blue, non-luminous flame.
- The conclusion must include both possible ions because the colours are similar.
Explain why it is difficult to make a definite conclusion from the flame colour in (d)(i).
______
Answer
Calcium and lithium ions give similar flame colours, so the flame test alone cannot distinguish between them.
Calcium and lithium give similar flame colours, so no definite conclusion
Walkthrough
The flame colour observed in (d)(i) could be produced by either calcium ions or lithium ions because the two ions give very similar flame colours. Therefore, from the flame test alone, it is not possible to say definitely which of the two ions is present. A further test, such as the sodium hydroxide test, would be needed to distinguish them.
Key Takeaways
- A flame test is not always conclusive if two ions give similar colours.
- Additional tests are needed to confirm the identity of the ion.
Common Mistakes
- Saying 'the flame colour is masked by sodium' when the mark scheme wants the overlap of calcium and lithium.
- Saying 'the test is unreliable' without explaining why.
Things to Be Careful About
- The reason is the similarity of the two flame colours, not contamination.
- Keep the answer short and specific.
To depth of in a boiling tube, add aqueous sodium hydroxide drop by drop until a change is seen.
Then add excess aqueous sodium hydroxide.
Keep the mixture for use in (f).
observations ______
conclusions ______
Answer
Observations: white precipitate forms; precipitate is insoluble in excess sodium hydroxide.
Conclusions: contains calcium ions ().
White precipitate, insoluble in excess NaOH; calcium ions present
Walkthrough
Sodium hydroxide is used to test for metal cations. When added dropwise to a solution containing calcium ions, a white precipitate of calcium hydroxide, , forms. The precipitate is then tested with excess sodium hydroxide. Calcium hydroxide is not amphoteric, so it does not dissolve in excess sodium hydroxide. This distinguishes calcium from ions such as aluminium, zinc and lead, whose hydroxides dissolve in excess sodium hydroxide. Therefore the conclusion is that calcium ions are present.
Key Takeaways
- Sodium hydroxide gives a white precipitate with calcium ions.
- Insolubility in excess sodium hydroxide is characteristic of calcium hydroxide.
- Amphoteric hydroxides (aluminium, zinc, lead) dissolve in excess sodium hydroxide.
Common Mistakes
- Saying the precipitate is soluble in excess sodium hydroxide, which would suggest aluminium, zinc or lead.
- Giving the observation but forgetting the conclusion.
- Not mentioning 'insoluble in excess'.
Things to Be Careful About
- Add sodium hydroxide dropwise first, then excess, to observe both stages.
- 'White precipitate, insoluble in excess' is the key phrase.
Answer
Observations: gas evolved turns red litmus blue.
Conclusions: ammonia gas is evolved, so ammonium ions () are present.
Gas turns red litmus blue; ammonia; ammonium ions present
Walkthrough
Ammonium ions react with hydroxide ions when warmed to release ammonia gas:
Ammonia is an alkaline gas, so it turns damp red litmus paper blue. In this experiment, the mixture from (e) contains sodium hydroxide. On gentle warming, ammonia gas is evolved. The observation is that the gas turns red litmus blue, and the conclusion is that ammonia gas is produced, so the solution contains ammonium ions.
Key Takeaways
- Ammonium ions are detected by warming with sodium hydroxide and testing the gas with red litmus.
- Ammonia turns red litmus blue because it is alkaline.
- The gas must be warmed out of solution to be detected.
Common Mistakes
- Identifying the gas as hydrogen instead of ammonia.
- Saying the gas turns blue litmus red (that would be an acidic gas).
- Forgetting to mention warming.
Things to Be Careful About
- Use damp red litmus paper.
- The observation is 'red litmus turns blue'; the conclusion is 'ammonia gas, so ammonium ions present'.
Solution is made from a mixture of two different ionic compounds.
Suggest the names of these two compounds.
______
Answer
Calcium chloride and ammonium chloride.
Calcium chloride and ammonium chloride
Walkthrough
Combine the evidence from all the experiments. Experiment (b) showed chloride ions are present. Experiments (d) and (e) showed calcium ions are present (flame colour and white precipitate insoluble in excess sodium hydroxide). Experiment (f) showed ammonium ions are present (ammonia gas evolved on warming with sodium hydroxide). The solution is made from two different ionic compounds. The two cations are calcium and ammonium, and the anion is chloride. Therefore the two compounds are calcium chloride, , and ammonium chloride, .
Key Takeaways
- Qualitative tests identify individual ions.
- An ionic compound is made of a cation and an anion.
- When more than one cation is present, more than one compound may be needed.
Common Mistakes
- Naming only one compound.
- Suggesting a compound containing both calcium and ammonium, which is not a simple ionic compound.
- Forgetting that both compounds must contain chloride because chloride is the only anion identified.
Things to Be Careful About
- Check that the two compounds account for all the ions identified.
- Both calcium chloride and ammonium chloride are soluble and would be present as ions in solution.
You are not expected to do any experimental work for this question
Barium carbonate decomposes when heated. The word equation for the reaction is shown.
Plan an experiment to determine the percentage loss in mass when barium carbonate is heated.
Your plan must include the use of common laboratory apparatus and a sample of barium carbonate. No other chemicals should be used.
Your plan must include:
- the apparatus needed
- the method to use and the measurements to take
- procedures to ensure that the percentage determined is as accurate as possible
- how the measurements are used to determine the percentage loss in mass.
You may draw a diagram to help answer the question.
Answer
Apparatus
- crucible and lid, or a heat-proof dish/boiling tube
- balance
- Bunsen burner, tripod and pipeclay triangle
- tongs and a heat-proof mat
Method and measurements
- Place the empty crucible on the balance and record its mass, or set the balance to zero (tare) with the empty crucible on it.
- Add a sample of barium carbonate and record the mass of the crucible plus sample.
- Heat the crucible strongly with the Bunsen burner for several minutes.
- Allow the crucible to cool, then reweigh the crucible and its contents.
Procedures to improve accuracy
- Reheat and reweigh repeatedly until the mass is constant (no further change), showing that all the barium carbonate has decomposed.
- Repeat the whole experiment and use an average, or at least repeat to check reliability and identify anomalous results.
- Let the crucible cool before weighing to avoid affecting the balance reading.
- Keep the lid partly on to prevent loss of solid, while still allowing carbon dioxide to escape.
Using the measurements
- Mass of barium carbonate used = mass of (crucible + sample) − mass of empty crucible.
- Mass of barium oxide left = mass after heating − mass of empty crucible.
- Mass loss = mass of barium carbonate used − mass of barium oxide left.
- Percentage loss = .
Since the gas escapes, the loss in mass is the loss of carbon dioxide, so:
Plan: weigh a sample in a crucible, heat strongly, cool and reweigh, reheating to constant mass; percentage loss = (initial mass of sample − final mass of sample) / initial mass of sample × 100
Walkthrough
This is a planning question, so no actual readings are taken. The answer must describe an experiment that measures the loss in mass when barium carbonate is heated. The word equation tells us that barium carbonate decomposes into barium oxide (solid) and carbon dioxide (gas). Because the carbon dioxide escapes, the mass of the container plus solid decreases. The loss in mass is therefore the mass of carbon dioxide given off.
The mark scheme divides the marks into four sections: apparatus, method and measurements, accuracy, and calculation. One mark is available for each section, and two further marks can be gained from any suitable points, making six marks in total.
Apparatus: A named container that can be heated safely is needed, such as a crucible (with lid), heat-proof dish or boiling tube. A balance is needed to measure mass. A Bunsen burner, tripod, pipeclay triangle and tongs are the usual apparatus for heating a crucible.
Method and measurements: The key measurements are the mass of the container, the mass of the container plus sample before heating, and the mass of the container plus contents after heating. If the balance is tared (set to zero) with the empty container on it, the displayed masses are directly the masses of the sample before and after heating. This is simpler to calculate with.
Accuracy: The most important accuracy point is heating to constant mass. If the sample is heated only once, some barium carbonate may remain undecomposed. Reheating and reweighing until two consecutive readings are the same proves that no further mass is being lost, so the decomposition is complete. Repeating the whole experiment and averaging the results checks reliability and helps identify anomalous results. Cooling the crucible before weighing avoids damaging the balance and prevents convection currents above the hot crucible from affecting the reading. A lid can be used partially open to prevent solid from spitting out while still letting carbon dioxide escape.
Calculation: The mass of the sample used is initial mass of sample. After heating, the mass of the sample is the final mass (mass of barium oxide). Since the only mass lost is the carbon dioxide gas, mass loss = initial mass − final mass. The percentage loss is then:
The question says no other chemicals should be used, and this plan uses only barium carbonate with common laboratory apparatus, so it satisfies that instruction.
Key Takeaways
- Thermal decomposition of a carbonate produces a metal oxide and carbon dioxide gas.
- When a gas escapes, the loss in mass is equal to the mass of gas given off.
- Heating to constant mass is the standard way to ensure a thermal decomposition is complete.
- Repeating experiments and averaging results improves reliability and helps spot anomalies.
- A percentage change is always calculated as change divided by the original value, multiplied by 100.
Common Mistakes
- Forgetting to reheat to constant mass, so the decomposition may be incomplete and the calculated percentage loss is too low.
- Using the mass of the container + sample as if it were the mass of the sample alone, without subtracting the mass of the empty container.
- Calculating percentage loss from final mass divided by initial mass, rather than loss in mass divided by initial mass.
- Closing the lid completely, which traps carbon dioxide and prevents the mass from falling correctly.
- Not allowing the crucible to cool before weighing, which can damage the balance or give inaccurate readings.
- Not repeating the experiment, so reliability cannot be judged.
Things to Be Careful About
- The calculation must use the mass of the sample before heating, not the mass of the container plus sample.
- If the balance is tared with the empty container, the displayed masses are already the masses of the sample, which avoids an extra subtraction.
- Ensure the final answer expresses the result as a percentage, so include the multiplication by 100.
- The phrase “constant mass” is a key accuracy point; heating once is not sufficient.
- No numerical answer is expected because this is a plan, not an actual experiment.