JC2 H2 Chemistry Tutor | Preparing a Calculation That Requires Selecting Relevant Data

Start with the output: the mass percentage of anhydrous sodium carbonate in a sample containing only anhydrous sodium carbonate and sodium chloride. This definition makes the denominator the original sample mass and the numerator the carbonate mass inferred from the analysis.

Specify that sodium chloride does not consume the acid under the stated titration conditions, and that the endpoint represents complete carbonate neutralisation. Those conditions establish why the acid amount can be related to carbonate rather than to an unspecified mixture of reactive substances.

Keep the exercise paper-based. The question supplies a valid mean titre rather than asking students to choose readings. That leaves room to study relevance and scaling. A separate practical-data review can address endpoint technique, repeated readings and uncertainty.

The tutor should solve the task before adding background information. If the main calculation is already ambiguous, extra numbers will amplify a wording problem. A useful selection task has a clear valid route and enough evidence for students to justify it.

Write the scope of the answer in the heading of the tutor key: mass percentage in the original dry two-component sample. That small step prevents an unnoticed switch to concentration in the prepared solution, percentage in one aliquot or mass fraction of a product mixture. The target quantity should stay consistent throughout the worked solution.

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A 1.060 g sample is quantitatively dissolved and the solution is made up to 250.0 cm³. A 25.00 cm³ aliquot requires a mean titre of 10.00 cm³ of 0.1000 mol/dm³ hydrochloric acid for complete neutralisation of its carbonate.

Supply the relative formula mass of anhydrous Na₂CO₃ as 106.0 for this exercise. Also state that the prepared solution is stored in a bottle of capacity 500 cm³, whose empty mass is 72.40 g. The storage details are background information.

Another supplied line gives a gas molar volume of 24.0 dm³/mol under specified conditions. Carbon dioxide is formed in the reaction, but no gas-volume measurement is needed to answer the original mass-percentage question.

Label the numbers with their actual roles and units. A storage bottle’s capacity is different from the volume of solution prepared. The question must not rely on deliberately confusing typography or an unstated assumption about whether the bottle is full.

Keep the background data realistic but limited. A storage detail and an available gas constant are enough to create a selection decision. Adding many unrelated molar masses or temperatures can test reading endurance more than chemical judgment and may make an otherwise focused diagnostic task unnecessarily tiring.

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The full chemical equation for the stated neutralisation is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. One mole of sodium carbonate consumes two moles of hydrochloric acid under the specified complete-neutralisation condition.

The amount of acid used for the aliquot is 0.1000 × 0.01000 = 0.001000 mol. The amount of sodium carbonate in that aliquot is therefore 0.001000 ÷ 2 = 0.0005000 mol.

Write aliquot beside that amount. It describes the carbonate in the 25.00 cm³ portion that was titrated, rather than the entire original sample. A chemically correct mole ratio can still lead to the wrong final answer when the portion is not tracked.

Sodium chloride is present in the sample and is also a product in the equation. That does not mean its initial amount can be read from the product coefficient. The titration determines the reactive carbonate amount; the original chloride component is inferred only if additional relationships, such as the stated two-component composition, are used.

Ask why acid-solution volume must be combined with concentration before using the reaction ratio. A measured volume of solution is not itself an amount of HCl. This question reveals whether the student understands n = cV or is moving directly between numbers because the problem resembles a familiar titration layout.

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The full prepared solution is ten aliquot volumes: 250.0/25.00 = 10.00. Assuming a homogeneous solution and quantitative preparation, the amount of carbonate in the original sample is 0.0005000 × 10.00 = 0.005000 mol.

The carbonate mass is 0.005000 × 106.0 = 0.5300 g. Divide by the original mixture mass and multiply by 100: 0.5300/1.060 × 100 = 50.00%. The answer describes mass percentage in the original solid sample.

Because the sample contains only the two stated components, the remaining 0.5300 g is sodium chloride under the exercise assumptions. This subtraction is a possible extension; it is not required to obtain the carbonate percentage.

Check the result before rounding away useful information. The carbonate mass is positive and below the sample mass, and the percentage lies between zero and one hundred. These checks cannot prove the full method correct, but they can expose some scaling or unit mistakes.

An alternative valid route first finds carbonate concentration in the prepared solution: 0.0005000 mol in 0.02500 dm³ gives 0.02000 mol/dm³. Multiplying by the full 0.2500 dm³ gives 0.005000 mol, agreeing with the aliquot-factor route. Comparing the routes can make the scaling less mysterious.

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The 500 cm³ bottle capacity does not determine the dilution factor. The actual prepared solution volume is 250.0 cm³. Using capacity in place of solution volume would double the inferred carbonate amount and produce an incorrect result.

The empty bottle mass is not needed because the original solid sample mass is already supplied directly. It would become relevant if the sample mass had to be found from a difference between a filled and an empty container.

The gas molar volume would be useful if the question asked for the carbon dioxide volume corresponding to a known amount. In the original task, carbonate is determined from acid consumption, so no gas-volume conversion is needed.

The planning table below makes these relationships explicit. Relevance belongs to a quantity’s role in the reasoning, rather than to its position in the paragraph or whether it looks like a familiar Chemistry constant.

Ask students to change one line so that the bottle mass becomes useful. For example, replace the directly given sample mass with masses of the same container before and after adding the dry sample. The student must then calculate a mass difference. The purpose is to explain data relevance, not merely to identify disposable numbers.

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A fictional answer of 5.000% may arise when the student calculates the carbonate mass in one aliquot and divides it by the full sample mass. Ask which portion each number represents. The repair concerns scaling, even though the equation and acid amount may be correct.

An answer of 100.0% can result from omitting the two-to-one acid-to-carbonate ratio. The inferred full-sample carbonate amount then doubles. A plausible percentage alone is insufficient evidence of a valid route; inspect the chemical relationship.

A student who uses 500/25 as the factor has selected a container property instead of the prepared solution volume. Ask what material was actually sampled by the pipette. The distinction is between capacity and contained solution, rather than between a good and bad number.

If the student begins by converting 10.00 cm³ with the gas molar volume, identify the physical meaning of the measured volume. It is a volume of acid solution, not a gas volume. Unit matching must include the quantity being measured, not just the appearance of cm³.

Keep unit errors separate from portion errors. A student may correctly identify the aliquot but enter 10.00 as dm³, producing an amount a thousand times too large. Another may convert every volume correctly and still omit the factor of ten. Their next practice questions should address different relationships.

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Ask the student to write the requested output with its definition: carbonate mass divided by original mixture mass. Then work backwards. Carbonate mass needs carbonate amount and molar mass; carbonate amount needs the neutralisation relationship and the aliquot factor.

This chain identifies which supplied values have a role before numerical substitution begins. It also exposes a missing value. If acid concentration were absent, the titre alone would not determine acid amount without some other valid information.

Invite the student to mark a brief reason beside an unused value. This discourages arbitrary crossing out. The explanation can be short: bottle capacity is not prepared volume, or gas volume is not requested.

Do not require students to write lengthy justifications for every number in all future assessments. The routine is a learning scaffold. As selection becomes reliable, a concise plan or annotation can preserve the reasoning without adding unnecessary writing.

A concise plan could read: acid amount → carbonate in aliquot → carbonate in full solution → carbonate mass → mass percentage. Each arrow represents a justified relationship. Ask the student to name the factor or equation at one arrow, so the plan remains chemical reasoning rather than a sequence memorised by shape.

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Keep the dataset and now ask for the carbon dioxide volume corresponding to complete reaction of the carbonate in one aliquot at the supplied gas conditions. The equation gives 0.0005000 mol of CO₂, so its volume is 0.0005000 × 24.0 = 0.0120 dm³, or 12.0 cm³.

The gas molar volume has become relevant because the requested output changed. The original sample mass and full-solution scaling are unnecessary for this aliquot-only gas question. This comparison helps students avoid learning that a particular number is always a distractor.

For a second variation, ask for the carbon dioxide corresponding to the entire original sample. Now use 0.005000 mol and obtain 0.120 dm³, or 1.20 × 10² cm³. The aliquot factor becomes relevant again because the target portion changed.

State that these are theoretical volumes under the given assumptions, rather than collected experimental measurements. A real gas collection can be affected by conditions and losses. Keep the conceptual selection exercise distinct from claims about an actual laboratory yield.

The theoretical aliquot gas volume is quoted as about 12.0 cm³ to reflect the three significant figures in the supplied molar volume. The full-sample result is about 120 cm³, likewise at three significant figures. Carry extra calculator digits during working, then use the precision of the supplied data when presenting the final answer.

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Prepare a fresh mixture problem with changed values, then ask for a short plan before calculation. For example, a 1.272 g two-component sample made up to 200.0 cm³ has a 20.00 cm³ aliquot requiring 12.00 cm³ of 0.1000 mol/dm³ HCl.

The acid amount is 0.001200 mol; carbonate in the aliquot is 0.0006000 mol; the full solution contains 0.006000 mol. With the same supplied formula mass, carbonate mass is 0.6360 g and its mass percentage is 50.00%. The equal final percentage does not make the intermediate quantities interchangeable.

Change one item of background information so students must judge its role again. Avoid an enormous list of irrelevant constants. The difficulty should come from connecting chemical quantities, rather than from endurance in reading a deliberately cluttered question.

In group work, collect an individual relevance plan before sharing methods. Parents can be shown the particular improvement: the student now tracks the aliquot and identifies actual solution volume. That is clearer evidence than simply reporting that a harder calculation was completed.

For the fresh dataset, remove the original values from view before the first attempt. If the student is stuck, ask which sample portion the acid reacted with. A targeted prompt can reveal the missing link without supplying the whole sequence. Record that assistance before judging whether the new calculation was independent.

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Check every unit conversion, reaction coefficient, aliquot factor and numerical result in the tutor key. Confirm that the specified endpoint matches the acid-to-carbonate ratio and that the sample composition supports the interpretation.

Keep separate notes for selection, stoichiometry, scaling and arithmetic. A student may need repair in only one part. Reassigning the full calculation repeatedly can obscure a small missing connection that could be taught directly.

Use repair when quantity meanings are confused, stabilise with another valid aliquot chain, and extend by changing the requested output. This provides a controlled increase in demand while preserving the chemistry that the student has already understood.

The preparation is complete when the student set, relevance map, verified calculations and likely-error responses agree. The tutor can then use an incomplete answer to choose a precise teaching move, while the student learns to make every calculation serve the question.

When discussing progress with a family, show a short example of the student explaining why one value was unused. That explanation makes a hidden planning skill visible. It is stronger evidence than saying the student has learned to ignore distractors, because the same value may be needed when a later question changes the output.

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