Chemistry 9701/51 — October/November 2022

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

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Topics Analysis, Conclusions and Evaluation · Planning

Q1Analysis, Conclusions and EvaluationPlanningFree sample

A student attempts to determine the percentage by mass of magnesium chloride in the solid mixture containing magnesium chloride, $\text{MgCl}_2$ , and anhydrous magnesium nitrate, $\text{Mg(NO}_3)_2$ , using the following method.

step 1 Accurately weigh about $1.5\text{ g}$ of the solid mixture and record the mass.

step 2 Dissolve the solid mixture in distilled water.

step 3 Add an excess of silver nitrate solution.

step 4 Filter the solid mixture and wash the precipitate collected with distilled water.

step 5 Dry the precipitate in an oven.

step 6 Weigh the precipitate and record the mass.

In this process only the chloride ions from the magnesium chloride form a precipitate with the silver nitrate solution.

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

One student in the class obtains the following results.

mass of solid mixture = $1.52\text{ g}$

mass of $\text{AgCl}$ solid after drying = $3.63\text{ g}$

(a)

(i)

Calculate the amount, in mol, of magnesium chloride present in the sample.

DifficultyMedium-Easy

Worked solution

Working

$\text{moles of AgCl} = \frac{3.63}{143.4} = 0.02531381 \text{ mol}$

From the equation, 1 mol MgCl $_2$ produces 2 mol AgCl:

$\text{moles of MgCl}_2 = \frac{0.02531381}{2} = 0.0126569 \text{ mol}$

Answer

0.0127 mol (to at least 2 SF)

Final answer

0.0127 mol

Detailed explanation

Background Concept

This is a gravimetric analysis problem. A known mass of precipitate (AgCl) is formed, and we work backwards through the balanced equation to find the amount of the original substance (MgCl $_2$ ). The key stoichiometric relationship is that 1 mol of MgCl $_2$ produces 2 mol of AgCl.

Understanding the Question

We are told that 3.63 g of AgCl was obtained. We need the amount, in mol, of MgCl $_2$ in the original sample. The path is: mass of AgCl → moles of AgCl → moles of MgCl $_2$ (using the 1:2 ratio).

Approach

Convert the mass of AgCl to moles using its molar mass (143.4 g mol $^{-1}$ ).
Use the stoichiometric ratio from the balanced equation to convert moles of AgCl to moles of MgCl $_2$ .

Step-by-Step Reasoning

Moles of AgCl = 3.63 / 143.4 = 0.02531381 mol.

The equation MgCl $_2$ + 2AgNO $_3$ → Mg(NO $_3$ ) $_2$ + 2AgCl shows 1 mol MgCl $_2$ : 2 mol AgCl.

So moles of MgCl $_2$ = 0.02531381 / 2 = 0.0126569 mol.

Key Takeaways

In gravimetric analysis, always start from the measured precipitate and work back through the stoichiometric ratio.

Common Mistakes

Forgetting to divide by 2 — the 1:2 ratio is the most common error. Also, using the wrong molar mass for AgCl.

Things to Be Careful About

Molar mass of AgCl = 107.9 + 35.5 = 143.4 g mol $^{-1}$ . The answer must be given to at least 2 significant figures.

Techniques used

calculate moles from mass and molar massapply stoichiometric ratio from a balanced equation

(ii)

Use your answer to (i) to calculate the percentage by mass of magnesium chloride in the sample. (If you were unable to answer (i) use $0.0102\text{ mol}$ . This is not the correct answer.)

DifficultyMedium-Easy

Worked solution

Working

$\text{mass of MgCl}_2 = 0.0126569 \times 95.3 = 1.2062 \text{ g}$

$\text{percentage by mass} = \frac{1.2062}{1.52} \times 100 = 79.355\%$

Answer

79.4% (to at least 2 SF)

Final answer

79.4%

Detailed explanation

Background Concept

Percentage by mass is the mass of the component divided by the total mass of the sample, multiplied by 100.

Understanding the Question

We have the moles of MgCl $_2$ from (a)(i). We need to convert this to a mass, then express it as a percentage of the 1.52 g sample.

Approach

Multiply the moles of MgCl $_2$ by its molar mass (95.3 g mol $^{-1}$ ) to get the mass.
Divide by the total sample mass (1.52 g) and multiply by 100.

Step-by-Step Reasoning

Mass of MgCl $_2$ = 0.0126569 × 95.3 = 1.2062 g.

Percentage by mass = (1.2062 / 1.52) × 100 = 79.355% ≈ 79.4%.

Key Takeaways

Percentage by mass always relates the component mass to the total sample mass.

Common Mistakes

Using the mass of AgCl instead of the mass of MgCl $_2$ . Forgetting to divide by the total sample mass.

Things to Be Careful About

Molar mass of MgCl $_2$ = 24.3 + 2(35.5) = 95.3 g mol $^{-1}$ . Give the answer to at least 2 significant figures.

Techniques used

convert moles to mass using molar masscalculate percentage by mass

(b)

(i)

Suggest what the student could do in step 2 to ensure the solid dissolves as quickly as possible.

DifficultyEasy

Worked solution

Answer

Stir the mixture (when the solid is mixed with water).

(Also acceptable: increase the temperature of the water, or increase the state of division of the solid mixture by grinding.)

Final answer

Stir the mixture.

Detailed explanation

Background Concept

The rate at which a solid dissolves in a solvent can be increased by stirring (bringing fresh solvent into contact with the solid), heating (increasing kinetic energy), or increasing the surface area (grinding the solid).

Understanding the Question

Step 2 is dissolving the solid mixture in distilled water. We need a practical suggestion to make this happen as quickly as possible.

Approach

Recall the standard physical factors that speed up dissolving.

Step-by-Step Reasoning

Stirring, heating, or grinding all increase the rate of dissolution without changing the chemistry of the analysis.

Key Takeaways

Three standard answers: stir, heat, or increase surface area (grind).

Common Mistakes

Suggesting a chemical change (e.g., adding acid) instead of a physical method.

Things to Be Careful About

The suggestion must be a physical method that does not interfere with the subsequent analysis.

Techniques used

suggest a method to increase the rate of dissolution

(ii)

Explain why the precipitate was washed with distilled water before it was dried.

DifficultyEasy

Worked solution

Answer

To remove magnesium nitrate / excess silver nitrate from the precipitate before drying.

Final answer

To remove magnesium nitrate / excess silver nitrate from the precipitate.

Detailed explanation

Background Concept

After filtration, the precipitate (AgCl) is contaminated with soluble substances from the solution — the magnesium nitrate produced in the reaction and any excess silver nitrate.

Understanding the Question

Washing the precipitate removes these soluble impurities so that the final dried mass is only silver chloride.

Approach

Identify what contaminants are present on the precipitate and why they must be removed.

Step-by-Step Reasoning

The precipitate is washed with distilled water to dissolve and remove the magnesium nitrate and excess silver nitrate that would otherwise add to the measured mass of the dried precipitate.

Key Takeaways

Washing removes soluble impurities; the answer should name the specific impurities.

Common Mistakes

Saying "to remove impurities" without specifying which ones.

Things to Be Careful About

Be specific — name magnesium nitrate and excess silver nitrate.

Techniques used

explain the purpose of washing a precipitate

(iii)

Suggest why the precipitate is dried in an oven and not by direct heating with a Bunsen burner.

DifficultyEasy

Worked solution

Answer

To avoid (thermal) decomposition of the silver chloride / precipitate / solid.

Final answer

To avoid thermal decomposition of the silver chloride.

Detailed explanation

Background Concept

Silver chloride is thermally unstable — strong heating causes it to decompose. An oven provides a gentler, lower-temperature drying method.

Understanding the Question

The precipitate is dried in an oven rather than by direct Bunsen heating. We need to explain why.

Approach

Recall that AgCl decomposes on strong heating.

Step-by-Step Reasoning

Direct heating with a Bunsen burner would cause thermal decomposition of the silver chloride, changing its mass and giving an incorrect result. Drying in an oven avoids this.

Key Takeaways

The key term is "thermal decomposition".

Common Mistakes

Not mentioning decomposition — e.g., saying "to avoid losing mass" without the reason.

Things to Be Careful About

The answer must state that decomposition is avoided.

Techniques used

explain the choice of drying method for a thermally unstable precipitate

(c)

(i)

In step 1, a small beaker was weighed, using a balance accurate to two decimal places, and its mass recorded. The sample was placed in the beaker and the mass of the beaker increased by $1.52\text{ g}$ .

Calculate the percentage error in measuring the mass of this sample.

Show your working.

DifficultyMedium-Easy

Worked solution

Working

The balance is accurate to two decimal places, so each reading has an error of ±0.005 g.
Two readings are taken (beaker, then beaker + sample), so the total error is:

$2 \times 0.005 = 0.01 \text{ g}$

$\text{percentage error} = \frac{0.01}{1.52} \times 100 = 0.6579\%$

Answer

0.66% (to at least 1 SF)

Final answer

0.66%

Detailed explanation

Background Concept

Percentage error = (absolute error / measured value) × 100%. A balance accurate to two decimal places has an error of ±0.005 g per reading.

Understanding the Question

The mass of the sample (1.52 g) is found by difference: the beaker is weighed, then the beaker with the sample. Each weighing has an error of ±0.005 g, so the total error is 2 × 0.005 = 0.01 g.

Approach

Determine the absolute error (double the single-reading error).
Divide by the measured mass and multiply by 100.

Step-by-Step Reasoning

Absolute error = 2 × 0.005 = 0.01 g.

Percentage error = (0.01 / 1.52) × 100 = 0.6579% ≈ 0.66%.

Key Takeaways

When a mass is obtained by difference (two readings), the error is doubled.

Common Mistakes

Using only one reading's error (0.005 g) instead of doubling it.

Things to Be Careful About

The answer must be given to at least 1 significant figure, and working must be shown.

Techniques used

calculate percentage error from balance precisionaccount for error doubling in a mass-by-difference measurement

(ii)

Other than by changing the balance, state how this percentage error could be reduced.

DifficultyEasy

Worked solution

Answer

Use a larger mass of the solid (mixture).

Final answer

Use a larger mass of the solid mixture.

Detailed explanation

Background Concept

Percentage error decreases as the measured quantity increases, because the same absolute error becomes a smaller fraction of a larger value.

Understanding the Question

We need to reduce the percentage error in measuring the mass of the sample, without changing the balance.

Approach

Recall that a larger measured value gives a smaller percentage error.

Step-by-Step Reasoning

Using a larger mass of the solid mixture means the same absolute error (0.01 g) is divided by a larger number, giving a smaller percentage error.

Key Takeaways

Larger sample → smaller percentage error.

Common Mistakes

Suggesting a more precise balance — the question explicitly excludes changing the balance.

Things to Be Careful About

Read the constraint: "other than by changing the balance".

Techniques used

suggest a method to reduce percentage error

(iii)

State what could be done in step 5 to ensure that the precipitate was completely dried.

DifficultyEasy

Worked solution

Answer

Continue drying and reweigh until the mass remains constant.

Final answer

Continue drying and reweighing until the mass remains constant.

Detailed explanation

Background Concept

Complete drying is confirmed by repeated drying and weighing: when the mass stops changing, all water has been removed.

Understanding the Question

Step 5 is drying the precipitate. We need to state what could be done to ensure it is completely dry.

Approach

Recall the constant-mass technique.

Step-by-Step Reasoning

Continue drying in the oven and reweighing until the mass remains constant. This confirms that no more water is being lost.

Key Takeaways

The key idea is "reweigh until constant mass".

Common Mistakes

Saying "dry for longer" without mentioning the reweighing check.

Things to Be Careful About

The answer must include both drying and reweighing until constant mass.

Techniques used

state the constant-mass technique to confirm complete drying

(d)

Another student in the class did not dry their silver chloride.

State how this would affect the value of the percentage by mass of magnesium chloride in the sample. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The measured mass / amount of AgCl (precipitate) would be greater than the true value, so the percentage by mass of MgCl $_2$ in the sample would be greater than the true value.

Final answer

The percentage by mass of MgCl2 would be greater than the true value.

Detailed explanation

Background Concept

If the precipitate is not dried, it retains water, so its measured mass is too high.

Understanding the Question

A student did not dry their silver chloride. We need to state how this affects the calculated percentage by mass of MgCl $_2$ and explain why.

Approach

Trace the effect through the calculation: wet precipitate → higher mass of AgCl → higher calculated moles of MgCl $_2$ → higher percentage by mass.

Step-by-Step Reasoning

The undried precipitate contains water, so its measured mass is greater than the true value. This makes the calculated moles of AgCl (and therefore moles of MgCl $_2$ ) too high, so the percentage by mass of MgCl $_2$ is greater than the true value.

Key Takeaways

An error that increases the precipitate mass inflates the calculated percentage of MgCl $_2$ .

Common Mistakes

Saying the percentage would decrease, or giving the effect without the explanation.

Things to Be Careful About

Both the direction (greater) and the reason (water adds to the mass) must be stated.

Techniques used

deduce the effect of an error on a calculated result

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