Chemistry 9701/53 — October/November 2022

Background Concept

This question tests the use of a balanced chemical equation to relate the amount of a product formed to the amount of a reactant consumed. The key relationship is:

$\text{amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g mol}^{-1})}$

The balanced equation given is:

$\text{MgCl}_2(\text{aq}) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Mg(NO}_3)_2(\text{aq}) + 2\text{AgCl}(\text{s})$

This shows that one mole of magnesium chloride produces two moles of silver chloride. The stoichiometric ratio is therefore 1:2.

Understanding the Question

The student weighed the dried silver chloride precipitate and obtained 3.63 g. You are asked to calculate the amount of magnesium chloride originally present in the sample. The only source of chloride ions is the magnesium chloride, so all the silver chloride formed comes from the MgCl2.

Approach

1. Calculate the molar mass of silver chloride, AgCl.
2. Convert the mass of AgCl (3.63 g) into moles using the molar mass.
3. Use the stoichiometric ratio from the balanced equation (1 MgCl2 : 2 AgCl) to find the moles of MgCl2.

Step-by-Step Reasoning

• Molar mass of AgCl: Ag = 107.9, Cl = 35.5, so Mr = 107.9 + 35.5 = 143.4 g mol⁻¹.
• Moles of AgCl = 3.63 / 143.4 = 0.02531381 mol.
• The equation shows 1 mol MgCl2 produces 2 mol AgCl, so moles of MgCl2 = 0.02531381 / 2 = 0.0126569 mol.
• The answer should be given to at least 2 significant figures, so 0.0127 mol is acceptable (0.01266 or 0.0127 both fine).

Key Takeaways

• Always use the balanced equation to find the mole ratio between the species you know and the one you want.
• The molar mass of a compound is the sum of the relative atomic masses of its constituent atoms.

Common Mistakes

• Forgetting to divide by 2: the equation shows 2 mol AgCl are formed per 1 mol MgCl2. Using 0.0253 mol directly as the moles of MgCl2 is a common error.
• Using an incorrect molar mass for AgCl, e.g. forgetting the chlorine atom or using 143.5 instead of 143.4 (the mark scheme uses 143.4, but 143.5 is often accepted; stick to the value given in the mark scheme).

Things to Be Careful About

• Give the final answer to at least 2 significant figures.
• Include the unit "mol" in the final answer.
• Use the exact value from the calculation in subsequent parts, not a rounded intermediate value, to avoid rounding errors.

Background Concept

Percentage by mass is defined as:

$\text{percentage by mass} = \frac{\text{mass of component}}{\text{total mass of sample}} \times 100\%$

To find the mass of magnesium chloride, we use the relationship:

$\text{mass} = \text{amount (mol)} \times \text{molar mass}$

The molar mass of MgCl2 is 24.3 + 2(35.5) = 95.3 g mol⁻¹.

Understanding the Question

You are asked to calculate the percentage by mass of magnesium chloride in the original 1.52 g sample. You have already found the amount of MgCl2 in part (a)(i). The question also provides a fallback value of 0.0102 mol in case you could not answer (a)(i), but you should use your own answer if you have it.

Approach

1. Convert the amount of MgCl2 (from part (a)(i)) to mass using the molar mass.
2. Divide this mass by the total mass of the sample (1.52 g) and multiply by 100.

Step-by-Step Reasoning

• Using the answer from (a)(i), 0.0126569 mol:
  • Mass of MgCl2 = 0.0126569 × 95.3 = 1.2062 g.
• Percentage by mass = (1.2062 / 1.52) × 100 = 79.355%, which rounds to 79.4%.
• The mark scheme gives M1 for the mass calculation and M2 for the percentage calculation. Both must be shown to gain both marks.
• If you used the fallback value of 0.0102 mol, you would get a different (incorrect) answer, but you would still gain method marks for the correct procedure.

Key Takeaways

• The molar mass of MgCl2 must be calculated correctly: 24.3 + 2 × 35.5 = 95.3 g mol⁻¹.
• The percentage is always (part / whole) × 100.

Common Mistakes

• Using the wrong molar mass for MgCl2, e.g. using 95.5 or forgetting the two chlorine atoms.
• Dividing the mass of AgCl by the sample mass instead of the mass of MgCl2.
• Forgetting to multiply by 100.

Things to Be Careful About

• Use the unrounded value from part (a)(i) to avoid rounding errors.
• Give the final answer to at least 2 significant figures (79.4% is fine).
• Include the % sign in the final answer.

Background Concept

The rate at which a solid dissolves in a solvent depends on three main factors:

• Agitation (stirring): brings fresh solvent into contact with the solid surface.
• Temperature: increasing temperature increases the kinetic energy of particles and speeds up dissolving.
• Surface area: a finer powder has a larger surface area, so more solvent can contact the solid at once.

Understanding the Question

The student dissolves the solid mixture in distilled water in step 2. You are asked to suggest what could be done to make the solid dissolve as quickly as possible.

Approach

Recall the factors that affect the rate of dissolving and pick one that is easy to apply in this context.

Step-by-Step Reasoning

• Stirring the mixture is the most obvious and practical answer: it continually brings fresh water into contact with the solid surface, speeding up dissolution.
• Alternatively, warming the water or grinding the solid into a finer powder would also work.
• The mark scheme accepts any one of these three: stir, increase temperature, or increase the state of division (grind to powder).

Key Takeaways

• Dissolving rate is controlled by agitation, temperature, and surface area.
• In a practical context, stirring is the simplest way to speed up dissolving.

Common Mistakes

• Saying "add more water" — this does not speed up the dissolving of a given mass of solid; it only changes the concentration.
• Saying "wait longer" — this is not a method to speed it up.

Things to Be Careful About

• The answer must be a practical action the student could take, not a theoretical statement.

Background Concept

When a precipitate is formed, it is often contaminated with soluble species from the solution. In this reaction, the precipitate is silver chloride, AgCl, which is insoluble in water. The solution contains magnesium nitrate, Mg(NO3)2, and any excess silver nitrate, AgNO3, that was added. These soluble salts would remain with the precipitate if it were not washed.

Understanding the Question

The student filters the precipitate and then washes it with distilled water. You are asked to explain why this washing is done.

Approach

Identify what impurities are present in the precipitate and why removing them matters for the accuracy of the final mass.

Step-by-Step Reasoning

• The precipitate of AgCl is collected on the filter paper. The filter cake also contains some of the solution, which holds dissolved Mg(NO3)2 and excess AgNO3.
• If these are not removed, they would add to the mass of the dried precipitate, making the calculated mass of AgCl too high.
• Washing with distilled water removes these soluble impurities, leaving only pure AgCl.
• The mark scheme specifically accepts "to remove magnesium nitrate / (excess) silver nitrate".

Key Takeaways

• Washing a precipitate removes soluble contaminants that would otherwise inflate the measured mass.
• The washing must be done before drying so the impurities are removed while still wet.

Common Mistakes

• Saying "to remove water" — that is what drying does, not washing.
• Saying "to cool the precipitate" — not the purpose.
• Being vague: "to clean it" is not enough; the mark scheme wants the specific impurities named or at least "soluble impurities".

Things to Be Careful About

• The answer should mention that the impurities are soluble (or name them specifically) to gain the mark.

Background Concept

Silver chloride is thermally unstable at high temperatures. Strong heating (as with a Bunsen burner) can cause it to decompose, for example:

$2\text{AgCl}(\text{s}) \rightarrow 2\text{Ag}(\text{s}) + \text{Cl}_2(\text{g})$

This would change the mass of the solid and give an incorrect result. An oven provides a gentler, controlled temperature that is sufficient to evaporate water but not to decompose the precipitate.

Understanding the Question

The student dries the precipitate in an oven. You are asked to suggest why an oven is used instead of direct heating with a Bunsen burner.

Approach

Consider what a Bunsen burner would do to silver chloride that an oven would not.

Step-by-Step Reasoning

• A Bunsen burner produces a very high temperature flame.
• Silver chloride can decompose when heated strongly, producing silver and chlorine gas.
• This would reduce the mass of the solid (chlorine is lost as a gas), leading to an incorrect low mass of AgCl and hence an incorrect percentage of MgCl2.
• An oven provides a controlled, lower temperature that dries the precipitate without decomposition.

Key Takeaways

• Some compounds decompose on strong heating; drying must be done at a temperature below their decomposition temperature.
• An oven is a safer, controlled method of drying than a naked flame.

Common Mistakes

• Saying "to avoid burning the filter paper" — the precipitate is usually transferred to a crucible before drying, so this is not the main reason.
• Saying "to dry it more evenly" — not the key point; decomposition is the crucial reason.
• Not mentioning decomposition specifically.

Things to Be Careful About

• The mark scheme wants the idea of avoiding thermal decomposition; the word "decomposition" is important.

Background Concept

When a balance is accurate to two decimal places, it means the reading can be relied upon to the nearest 0.01 g, but the uncertainty in a single reading is half of that, i.e. ±0.005 g. When you measure a mass by difference (weighing the beaker, then weighing the beaker plus sample), you take two readings, so the uncertainties add: total uncertainty = 2 × 0.005 = 0.01 g.

Percentage error is calculated as:

$\text{percentage error} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100\%$

Understanding the Question

The beaker was weighed, then the sample was added and the mass increased by 1.52 g. The balance is accurate to two decimal places. You are asked to calculate the percentage error in measuring this mass, and to show your working.

Approach

1. Determine the uncertainty in each reading (±0.005 g).
2. Recognise that two readings are involved, so total uncertainty = 0.01 g.
3. Divide by the measured mass (1.52 g) and multiply by 100.

Step-by-Step Reasoning

• The balance reads to 2 decimal places, so each reading has an uncertainty of ±0.005 g.
• The mass of the sample is found by difference: (mass of beaker + sample) − (mass of beaker). Both readings have the same uncertainty, so the total absolute uncertainty is 0.005 + 0.005 = 0.01 g.
• Percentage error = (0.01 / 1.52) × 100 = 0.6579%, which rounds to 0.66%.
• The mark scheme requires the working to be shown along with the correct answer.

Key Takeaways

• The uncertainty of a balance reading to 2 decimal places is ±0.005 g, not ±0.01 g.
• When a mass is measured by difference (two readings), the uncertainties add.
• Percentage error is the absolute uncertainty divided by the measured value, times 100.

Common Mistakes

• Using ±0.01 g as the uncertainty for a single reading (that is the uncertainty for the difference, not for one reading).
• Forgetting to double the uncertainty for two readings.
• Not showing the working — the mark scheme specifically requires it.

Things to Be Careful About

• The final answer should be given to at least 1 significant figure (0.66% is fine).
• Include the % sign.
• The calculation must be shown clearly to gain the mark.

Background Concept

Percentage error decreases as the measured quantity increases, because the absolute uncertainty stays the same while the denominator (the measured value) becomes larger. This is a fundamental idea in measurement: relative error = absolute error / measured value.

Understanding the Question

The question asks how the percentage error could be reduced, with the restriction "other than by changing the balance". So you cannot suggest a more accurate balance; you must suggest something else.

Approach

Think about what appears in the denominator of the percentage error formula: the measured mass. Increasing it reduces the percentage error.

Step-by-Step Reasoning

• The percentage error is (0.01 / mass) × 100. If the mass is larger, the fraction is smaller.
• Therefore, using a larger mass of the solid mixture (e.g. 3 g instead of 1.5 g) would reduce the percentage error.
• The mark scheme specifically accepts "use a larger mass of solid (mixture)".

Key Takeaways

• Relative error decreases when you measure a larger quantity with the same absolute uncertainty.
• This is why weighing a larger sample is often better than weighing a tiny one.

Common Mistakes

• Suggesting a more accurate balance — explicitly disallowed by the question.
• Suggesting repeating the experiment — that reduces random error, but not the systematic error from the balance's precision.
• Being vague, e.g. "be more careful" — not a specific method.

Things to Be Careful About

• The answer must respect the constraint "other than by changing the balance".

Background Concept

To be certain that a solid is completely dry, you cannot simply assume that a fixed drying time is enough. The standard technique is to dry, cool, weigh, and then repeat the drying and weighing until two consecutive masses are the same (constant mass). If the mass stops changing, all the water has been removed.

Understanding the Question

The student dries the precipitate in an oven in step 5. You are asked what could be done to ensure the precipitate is completely dried.

Approach

Recall the standard method for confirming complete drying: repeated drying and weighing until constant mass.

Step-by-Step Reasoning

• After an initial period of drying, the student should allow the precipitate to cool (in a desiccator, for example) and weigh it.
• The precipitate is then returned to the oven for more drying, cooled, and weighed again.
• If the second mass is the same as the first, the precipitate is completely dry. If not, the process is repeated.
• The mark scheme accepts "continue drying and reweigh until mass remains constant".

Key Takeaways

• Constant mass is the criterion for complete drying.
• This technique is used in gravimetric analysis to ensure accuracy.

Common Mistakes

• Saying "dry it for longer" without mentioning reweighing — you cannot know it is dry without weighing.
• Saying "cool it before weighing" — that is part of the technique but not the complete answer.
• Saying "use a desiccator" — that is an alternative drying method, not a way to check complete drying.

Things to Be Careful About

• The answer must include the idea of reweighing and checking that the mass is constant.