Chemistry 5070/42 — May/June 2021

Walkthrough

The experiment requires strong heating to drive off water of crystallisation. Glass beakers are not designed to withstand the high temperatures of a Bunsen burner flame directly and may crack or melt. A crucible is made of ceramic (e.g., porcelain) which has a high melting point and is chemically inert. Additionally, ceramic is a better thermal conductor than glass, ensuring the contents heat through efficiently.

Key Takeaways

Always select apparatus appropriate for the thermal conditions of the experiment. Crucibles are used for strong heating of solids.

Common Mistakes

• Saying "glass is too hot" (glass doesn't get too hot, it breaks). Must mention breaking or melting.
• Forgetting the ORA (Opposite Reason Answer): if you say glass breaks, you can also say crucible conducts heat better.

Things to Be Careful About

The question asks to suggest why a crucible is used instead of a glass beaker. Focus on the limitations of glass (breaking/melting) or the advantages of a crucible (thermal conductivity, heat resistance).

Walkthrough

Heating hydrated salts drives off water. If heating is stopped too early, some water remains, making the final mass too high and the calculated water lost too low. By repeating the heating, cooling, and weighing process, the student checks if the mass is still decreasing. When two consecutive readings are the same (constant mass), no more water is being lost, meaning the dehydration is complete.

Key Takeaways

"Heating to constant mass" is a standard technique to ensure a reaction or physical change is complete.

Common Mistakes

• Saying "to make the experiment accurate" (too vague). Must specify what is being made accurate (complete removal of water).
• Saying "to get a better average" (this is for repeated readings of the same measurement, not for driving a reaction to completion).

Things to Be Careful About

The mark scheme accepts "to make sure ALL the water driven off" or "no water remains". Be specific about the water.

Walkthrough

The mass of water lost is the difference between the initial mass of the hydrated salt and the final mass of the anhydrous salt.

• Row 1, Column 1: $0.52 - 0.25 = 0.27$ g.
• Row 3, Column 4: $2.04 - 1.00 = 1.04$ g.

Key Takeaways

Mass lost = Initial mass - Final mass. Always check units and decimal places.

Common Mistakes

• Forgetting to copy the table headers correctly.
• Calculation errors in subtraction.

Things to Be Careful About

Ensure the values are placed in the correct empty cells in the table.

Walkthrough

The question asks to plot "final mass of $\text{MgSO}_4$ against the mass of $\text{H}_2\text{O}$ lost". This means:

• x-axis (horizontal): mass of $\text{H}_2\text{O}$ lost (independent variable, though here it's derived).
• y-axis (vertical): final mass of $\text{MgSO}_4$ (dependent variable).

The grid provided has x-axis from 0 to 1.60 and y-axis from 0 to 1.60. Major markings are every 0.20. There are 10 small squares between 0.00 and 0.20, so each small square is 0.02.

Plotting the points:

1. x=0.27 (1.5 small squares past 0.20), y=0.25 (1.25 small squares past 0.20).
2. x=0.50 (on the line), y=0.48 (1 small square below 0.50).
3. x=0.67 (3.5 small squares past 0.60), y=0.83 (1.5 small squares above 0.80).
4. x=1.04 (2 small squares past 1.00), y=1.00 (on the line).
5. x=1.30 (on the line), y=1.23 (1.5 small squares above 1.20).
6. x=1.53 (1.5 small squares past 1.50), y=1.46 (3 small squares below 1.50).

Key Takeaways

Always check which variable goes on which axis. Plotting is a key skill in Paper 4.

Common Mistakes

• Swapping x and y axes.
• Misreading the grid scale (e.g., thinking each small square is 0.01 or 0.05).

Things to Be Careful About

Use crosses (x) or dots for points. The mark scheme shows crosses. Ensure points are within half a small square of the correct position.

Walkthrough

Looking at the data, there is a roughly linear relationship. For every ~0.50 g of water lost, ~0.48-0.50 g of MgSO4 remains.

• Point 1: 0.27 -> 0.25 (ratio ~0.92)
• Point 2: 0.50 -> 0.48 (ratio ~0.96)
• Point 3: 0.67 -> 0.83 (ratio ~1.24) -> Anomaly
• Point 4: 1.04 -> 1.00 (ratio ~0.96)
• Point 5: 1.30 -> 1.23 (ratio ~0.95)
• Point 6: 1.53 -> 1.46 (ratio ~0.95)

The third point (0.67, 0.83) has a much higher mass of anhydrous salt than expected for the amount of water lost. It is the outlier.

Key Takeaways

An anomalous result is a data point that does not fit the general pattern or trend of the data.

Common Mistakes

• Circling a point that is just slightly off but part of the trend.
• Not circling clearly.

Things to Be Careful About

The circle must be around the specific data point (the cross), not just near it.

Walkthrough

For the anomalous point (0.67, 0.83):

• Mass of water lost = 0.67 g (calculated from 1.50 - 0.83).
• Mass of final solid = 0.83 g.
• Expected final mass for 0.67 g water lost (based on other points, ratio ~0.95) would be around $0.67 \times 0.95 \approx 0.64$ g.
• The actual final mass (0.83 g) is much higher than expected (0.64 g). This means the final solid is heavier than it should be. Since the only thing left after heating should be anhydrous MgSO4, the extra mass must be unevaporated water.
• Therefore, the heating was insufficient to remove all the water.

Another possible error: If the crucible was left in the open air to cool, anhydrous MgSO4 is hygroscopic and can reabsorb water from the atmosphere, increasing the final mass.

Key Takeaways

An anomalous result where the final mass is too high usually indicates incomplete reaction (not all water removed) or reabsorption of moisture.

Common Mistakes

• Saying "the balance was wrong" (too vague, needs a specific chemical/physical reason).
• Saying "too much water was lost" (this would make the final mass lower, not higher).

Things to Be Careful About

The reason must explain why the final mass of MgSO4 is too high (or water lost is too low). "Sample not heated to constant mass" is a key phrase.

Walkthrough

A line of best fit should represent the trend of the data. Since the relationship is proportional (if 0 g water is lost, 0 g anhydrous salt is produced - well, actually if you start with 0 hydrated salt you get 0, so it passes through origin), the line should go through (0,0).

The line should pass as close as possible to the majority of the points (the 5 good ones), with roughly equal numbers of points above and below the line on either side of the anomaly. The anomalous point (0.67, 0.83) is ignored for the line.

Looking at the points: (0.27, 0.25), (0.50, 0.48), (1.04, 1.00), (1.30, 1.23), (1.53, 1.46). These are very close to a straight line $y = 0.95x$ (roughly). The line should go from (0,0) to roughly (1.53, 1.46).

Key Takeaways

• Line of best fit must be a straight line (for this data).
• Must pass through origin if the relationship is proportional (start with 0, get 0).
• Exclude anomalous points.

Common Mistakes

• Drawing a curve.
• Forcing the line through the anomalous point.
• Not starting the line at the origin (0,0).

Things to Be Careful About

Use a ruler. The line should be thin and clear.

Walkthrough

The question gives a new final mass of 1.20 g. We need to find the corresponding mass of water lost using the graph.

1. Locate 1.20 on the vertical axis (y-axis).
2. Move horizontally to the right until you hit the line of best fit.
3. From that intersection, move vertically down to the horizontal axis (x-axis).
4. Read the value. The line passes through (1.30, 1.23) and (1.04, 1.00). At y=1.20, x is slightly less than 1.30. 1.26 is a reasonable reading.

Note: Reading from a graph always has some uncertainty. The mark scheme accepts 1.26 (likely 1.24 - 1.28 range).

Key Takeaways

Reading values from a graph requires precision and using the line of best fit, not the raw data points (unless reading a specific point).

Common Mistakes

• Reading the wrong axis.
• Not using the line of best fit (reading off the anomalous point or a specific data point incorrectly).

Things to Be Careful About

The mark scheme gives 1.26. Ensure your reading is consistent with the line drawn.

Walkthrough

Use the formula: $\text{moles} = \frac{\text{mass}}{M_r}$.
Mass of water = 1.26 g.
Relative molecular mass ($M_r$) of water ($\text{H}_2\text{O}$) = $1 + 1 + 16 = 18$.

Key Takeaways

Always calculate $M_r$ correctly. Water is 18, not 17 or 19.

Common Mistakes

• Using the wrong mass (e.g., using 1.20 g of MgSO4 instead of the water mass).
• Calculation error: $1.26 / 18$. $18 \times 7 = 126$, so $1.26 / 18 = 0.07$.

Things to Be Careful About

The mark scheme awards 1 mark for $M_r=18$ and 1 mark for the correct mole calculation (0.07). Show the working.

Walkthrough

We have the moles of both components in the hydrated salt.

• Moles of anhydrous $\text{MgSO}_4$ = 0.01 mol.
• Moles of water $\text{H}_2\text{O}$ = 0.07 mol.

The chemical formula $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$ tells us that for every 1 mole of $\text{MgSO}_4$, there are $x$ moles of water.

Ratio = $\frac{0.07}{0.01} = 7$.

Therefore, $x = 7$. The formula is $\text{MgSO}_4 \cdot 7\text{H}_2\text{O}$ (magnesium sulfate heptahydrate, also known as Epsom salts).

Key Takeaways

To find the number of water molecules in a hydrate, calculate the mole ratio of water to the anhydrous salt.

Common Mistakes

• Inverting the ratio (calculating $0.01/0.07$).
• Forgetting that the ratio is $1:x$, not $x:1$.

Things to Be Careful About

The question asks for the value of $x$, which is a whole number. 7 is a reasonable integer.