Chemistry 9701/52 — May/June 2020

Background Concept

When collecting a gas for quantitative analysis, the system (flasks, delivery tubes, collection vessel) initially contains atmospheric air. The first portion of gas generated will displace this air rather than being pure product.

Understanding the Question

The question asks why the initial gas collected should be discarded. This is a standard practical technique when collecting gases for accurate mass or volume measurements.

Approach

Consider what is in the apparatus before the reaction starts. The reaction generates gas X, which will push the existing contents out of the apparatus first.

Step-by-Step Reasoning

Before the reaction begins, the round-bottom flask, delivery tube, and collection vessel are filled with air. As the reaction starts and gas X is produced, the first gas to exit the apparatus is a mixture of the newly formed gas X and the displaced air. For a precise determination of $M_r$, the volume and mass must correspond to pure gas X only. Therefore, the first sample is discarded to ensure the subsequent sample collected is pure gas X.

Key Takeaways

Always discard the initial gas collected in quantitative gas experiments to remove air from the system.

Common Mistakes

Stating "the gas is impure" without explaining why (i.e., mixing with air from the apparatus).

Things to Be Careful About

Ensure the explanation clearly identifies that the impurity is air from the apparatus, not just a vague "impurity".

Background Concept

Gas collection methods must be matched to the properties of the gas. Water displacement (Method A) is suitable only for gases that are insoluble or very slightly soluble in water. Gas syringes (Method B) are suitable for any gas, provided the syringe is compatible and leak-free.

Understanding the Question

The question asks for two reasons why Method A (water displacement) is unsuitable for gas X, which is produced from concentrated sulfuric acid and copper (likely $\text{SO}_2$, a soluble acidic gas).

Approach

Consider the physical and chemical properties of gas X and the mechanics of Method A. Solubility and pressure changes during heating/cooling are key factors.

Step-by-Step Reasoning

Reason 1: Gas X (e.g., $\text{SO}_2$) is soluble in water. If collected over water, a significant portion will dissolve, leading to an inaccurate (lower) measured volume. The mark scheme accepts "gas is soluble so should not be collected over water".
Reason 2: Method A involves heating the reaction flask. If the heat is removed or the reaction stops, the gas in the flask cools and contracts, reducing the pressure. This can cause water from the trough to be sucked back up the delivery tube into the hot flask (suck-back), potentially cracking the glass or causing a hazard. Method B uses a gas syringe, avoiding this risk.

Key Takeaways

Always consider solubility and the risk of suck-back when choosing a gas collection method.

Common Mistakes

Suggesting that gas X reacts with water as the primary reason (while true, solubility is the direct mark). Forgetting to mention the physical danger of suck-back.

Things to Be Careful About

The question asks for reasons other than the first sample containing air (part a(i)). Ensure both reasons are distinct and directly address the apparatus or gas properties.

Background Concept

To determine the mass of a gas absorbed by a solid, a gravimetric analysis approach is used. The solid is weighed before and after exposure to the gas. The increase in mass is attributed to the absorbed gas.

Understanding the Question

The question asks for the specific measurements needed to find the mass of gas X absorbed by the soda lime in the U-tube.

Approach

Identify the initial and final states of the absorbing medium (soda lime in the U-tube) that need to be measured.

Step-by-Step Reasoning

The mass of gas X absorbed is the difference between the final mass and the initial mass of the absorbing system. Therefore, the student must measure:

1. The mass of the U-tube plus the soda lime before the gas is passed through.
2. The mass of the U-tube plus the soda lime after all the gas has been passed through and absorbed.
   Subtracting the initial mass from the final mass gives the mass of gas X.

Key Takeaways

Gravimetric absorption requires pre- and post-experiment mass measurements of the absorbent.

Common Mistakes

Only stating "mass before and after" without specifying that it is the U-tube + soda lime. Forgetting that the U-tube itself has mass.

Things to Be Careful About

Ensure both "before" and "after" measurements are explicitly stated, as the mark scheme requires both (AND).

Background Concept

Soda lime is a mixture of $\text{NaOH}$ and $\text{Ca(OH)}_2$, both of which are strong bases. Acidic gases (like $\text{SO}_2$, $\text{CO}_2$, $\text{HCl}$) react with bases to form salts and water.

Understanding the Question

The question asks why soda lime is specifically chosen to absorb gas X, which is identified as an acidic gas in the stem.

Approach

Connect the chemical nature of gas X (acidic) with the chemical nature of soda lime (alkaline) to explain the absorption mechanism.

Step-by-Step Reasoning

Gas X is described as an acidic gas. Soda lime is alkaline. When an acidic gas comes into contact with an alkaline solid, an acid-base neutralisation reaction occurs (e.g., $\text{SO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_3 + \text{H}_2\text{O}$). This chemical reaction ensures the gas is permanently and effectively removed from the gas stream and retained in the U-tube, allowing its mass to be measured accurately.

Key Takeaways

Match the chemical properties of the gas (acidic) with the absorbent (alkaline) for effective absorption.

Common Mistakes

Suggesting soda lime is used because it is "dry" or "porous" without mentioning the chemical neutralisation reaction.

Things to Be Careful About

The mark scheme specifically looks for "alkaline" or "neutralisation reaction". Ensure these key terms are used.

Background Concept

The ideal gas equation $PV = nRT$ relates pressure ($P$), volume ($V$), number of moles ($n$), the gas constant ($R$), and absolute temperature ($T$). For the equation to work with $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ (which uses SI units), $P$ must be in pascals (Pa), $V$ in cubic metres ($\text{m}^3$), and $T$ in kelvin (K).

Understanding the Question

Calculate the number of moles of gas X using the given $P$, $V$, and $T$ values and the ideal gas law.

Approach

1. Convert all given values to SI units.
2. Rearrange the equation to solve for $n$.
3. Substitute and calculate.

Step-by-Step Reasoning

Step 1: Unit conversions

• Temperature: $T = 21^\circ\text{C} + 273 = 294 \text{ K}$
• Volume: $V = 100 \text{ cm}^3$. Since $1 \text{ m}^3 = 10^6 \text{ cm}^3$, $V = 100 \times 10^{-6} \text{ m}^3 = 1.00 \times 10^{-4} \text{ m}^3$
• Pressure: $P = 9.8 \times 10^4 \text{ Pa}$ (already in SI units)
• Gas constant: $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$

Step 2: Rearrange and substitute
$n = \frac{PV}{RT} = \frac{(9.8 \times 10^4) \times (100 \times 10^{-6})}{8.31 \times 294}$
$n = \frac{9.8}{8.31 \times 294} = \frac{9.8}{2443.14} = 4.0104... \times 10^{-3} \text{ mol}$

Step 3: Final answer
Rounding to 3 significant figures: $n = 4.01 \times 10^{-3} \text{ mol}$.

Key Takeaways

Always check units when using the ideal gas equation. $R = 8.31$ requires $\text{m}^3$, Pa, and K.

Common Mistakes

Forgetting to convert $\text{cm}^3$ to $\text{m}^3$ (using $100$ instead of $100 \times 10^{-6}$). Forgetting to convert $^\circ\text{C}$ to K. Arithmetic errors in the calculation.

Things to Be Careful About

The mark scheme awards method marks for correct $T$ and $V$ conversions, and accuracy marks for the final value. Show your working clearly.

Background Concept

Relative molecular mass ($M_r$) is numerically equal to the molar mass in $\text{g mol}^{-1}$. The relationship is:
$\text{mass} = n \times M_r \quad \text{or} \quad M_r = \frac{\text{mass}}{n}$

Understanding the Question

The student has the mass of the gas sample ($0.251 \text{ g}$) and has just calculated the number of moles ($4.01 \times 10^{-3} \text{ mol}$). Explain how to find $M_r$.

Approach

Use the formula $M_r = \text{mass} / n$.

Step-by-Step Reasoning

The student already has the mass of gas X ($0.251 \text{ g}$) from the U-tube experiment and the number of moles ($n$) from the ideal gas calculation in (d)(i). To find the relative molecular mass, simply divide the mass by the number of moles:
$M_r = \frac{0.251 \text{ g}}{4.01 \times 10^{-3} \text{ mol}} = 62.6 \text{ g mol}^{-1}$
The question asks to "explain how", so stating "divide mass by number of moles" is sufficient, but showing the calculation confirms understanding.

Key Takeaways

$M_r = \text{mass} / \text{moles}$.

Common Mistakes

Using the wrong formula (e.g., multiplying mass and moles). Forgetting to use the moles from (d)(i) instead of calculating new moles.

Things to Be Careful About

The question asks to "explain how", so a verbal explanation (divide mass by moles) is acceptable and often preferred over just a number, though showing the number is good practice.

Background Concept

Error analysis in calculations requires tracing how an error in a measured quantity propagates through the formula to affect the final result.

Understanding the Question

Not all gas X is absorbed by the soda lime. This means the mass increase of the U-tube is less than it should be. How does this affect the calculated $M_r$?

Approach

1. Determine the effect on the measured mass.
2. Use the formula $M_r = \text{mass} / n$ to determine the effect on $M_r$.

Step-by-Step Reasoning

Step 1: Effect on mass
If not all gas X is absorbed, the mass of gas X recorded (final mass - initial mass of U-tube) will be less than the actual mass of gas X that was in the syringe ($0.251 \text{ g}$).

Step 2: Effect on $M_r$
The formula for $M_r$ is:
$M_r = \frac{\text{mass of X}}{n}$
Since the number of moles $n$ is calculated from the gas volume, pressure, and temperature (which are assumed correct), $n$ is unchanged. However, the numerator (mass) is smaller than it should be. Dividing a smaller number by the same $n$ gives a smaller result.
Therefore, the calculated $M_r$ will be lower than the true value.

Key Takeaways

If the measured mass is too low and $n$ is constant, $M_r$ will be too low.

Common Mistakes

Saying "no effect" or "higher $M_r$" without tracing the logic through the formula.

Things to Be Careful About

Clearly state the direction of the error (lower) and briefly explain why (smaller mass / same moles).

Background Concept

The ideal gas equation can be rearranged to relate pressure and temperature directly:
$PV = nRT \implies \frac{P}{T} = \frac{nR}{V}$
This shows that $\frac{P}{T}$ is directly proportional to the number of moles $n$ (assuming $V$ and $R$ are constant).

Understanding the Question

In Experiment 2, the same mass of gas is used, but $T$ and $P$ are not recorded. The student calculates a higher $M_r$ for Experiment 2 than for Experiment 1. We need to state and explain how $\frac{P}{T}$ changes from Exp 1 to Exp 2.

Approach

1. Relate $M_r$ to $n$ using the constant mass.
2. Relate $n$ to $\frac{P}{T}$ using the ideal gas equation.
3. Combine to find the relationship between $M_r$ and $\frac{P}{T}$.

Step-by-Step Reasoning

Step 1: Effect on $n$
We know $M_r = \frac{\text{mass}}{n}$. Rearranging gives $n = \frac{\text{mass}}{M_r}$.
The mass is the same in both experiments. If the calculated $M_r$ for Experiment 2 is higher, then the calculated number of moles $n$ must be less.

Step 2: Relate $n$ to $\frac{P}{T}$
From $PV = nRT$, we can rearrange to:
$\frac{P}{T} = \frac{nR}{V}$
Since $R$ is a constant and $V$ ($100 \text{ cm}^3$) is the same in both experiments, $\frac{P}{T}$ is directly proportional to $n$.

Step 3: Conclusion
Since $n$ is less in Experiment 2, $\frac{P}{T}$ must also be less in Experiment 2 compared to Experiment 1.

Key Takeaways

$M_r \propto \frac{1}{n}$ and $n \propto \frac{P}{T}$, so $M_r \propto \frac{T}{P}$. A higher calculated $M_r$ implies a lower $\frac{P}{T}$.

Common Mistakes

Stating $\frac{P}{T}$ is higher without explaining the link through $n$. Failing to mention that $n$ is less.

Things to Be Careful About

The question asks to "State and explain". You must explicitly state "P/T is less" and then provide the logical explanation involving $n$.