Chemistry 9701/21 — October/November 2010

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Atoms, Molecules and Stoichiometry · Introduction to Organic Chemistry · Group 17 · Equilibria · Reaction Kinetics · Nitrogen and Sulfur · +4 more

Q1Atoms, Molecules and StoichiometryFree sample

In 1814, Sir Humphrey Davy and Michael Faraday collected samples of a flammable gas, A, from the ground near Florence in Italy.
They analysed A which they found to be a hydrocarbon. Further experiments were then carried out to determine the molecular formula of A.

(a)

What is meant by the term molecular formula?

DifficultyEasy

Worked solution

Answer

The molecular formula shows the actual number of atoms of each element present in one molecule of a compound.

Final answer

The actual number of atoms of each element in one molecule of a compound.

Detailed explanation

Background Concept

Compounds can be represented by formulae at different levels of detail. The empirical formula gives the simplest whole-number ratio of atoms of each element; the molecular formula gives the actual numbers of each type of atom in one molecule. For example, glucose has empirical formula CH₂O but molecular formula C₆H₁₂O₆.

Understanding the Question

The command word is 'What is meant by' — a definition worth 2 marks. The mark scheme splits it into two halves: the actual number of atoms of each element (1), and in one molecule of a compound (1).

Approach

State the definition precisely, making sure both halves appear: 'actual number' and 'one molecule'.

Step-by-Step Reasoning

Saying only 'the formula showing the atoms in a compound' is too vague — the key word is actual number, which distinguishes it from the empirical (simplest ratio) formula.
The second mark requires specifying it applies to one molecule of the compound.

Key Takeaways

Always give full definitions with the precise distinguishing words; here 'actual' and 'one molecule' each carry a mark.

Common Mistakes

Confusing molecular formula with empirical formula.
Omitting 'in one molecule' and losing the second mark.

Things to Be Careful About

Write the definition as a complete sentence; a fragment like 'actual number of atoms' alone scores only 1 of the 2 marks.

Techniques used

define molecular formula

(b)

Davy and Faraday deduced the formula of A by exploding it with an excess of oxygen and analysing the products of combustion.

Complete and balance the following equation for the complete combustion of a hydrocarbon with the formula $\text{C}_x\text{H}_y$ .

\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow \dots\dots\dots\dots\dots\dots\dots\dots + \dots\dots\dots\dots\dots\dots\dots\dots

DifficultyMedium-Easy

Worked solution

Answer

\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}

Final answer

xCO2 + (y/2)H2O

Detailed explanation

Background Concept

Complete combustion of any hydrocarbon in excess oxygen produces carbon dioxide and water only. Balancing with general subscripts x and y requires algebra: each carbon atom gives one CO₂, and each pair of hydrogen atoms gives one H₂O.

Understanding the Question

You must complete the right-hand side of the general combustion equation so that C, H and O atoms balance.

Approach

Balance carbon first (x CO₂), then hydrogen (y/2 H₂O), then check oxygen: right side has 2x + y/2 oxygen atoms = (x + y/4)O₂ molecules, matching the given coefficient.

Step-by-Step Reasoning

Carbon: x carbons on the left → x CO₂ on the right.
Hydrogen: y hydrogens on the left → y/2 H₂O on the right.
Oxygen check: x CO₂ contains 2x O atoms; y/2 H₂O contains y/2 O atoms; total 2x + y/2 = 2(x + y/4), which is exactly the oxygen in (x + y/4)O₂ — the equation balances.

Key Takeaways

The general combustion equation CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O is a powerful tool for gas-volume problems like parts (c) and (d).

Common Mistakes

Writing H₂O coefficient as y instead of y/2 (forgetting each water contains 2 H).
Confusing the O₂ coefficient with the products.

Things to Be Careful About

Fractional coefficients are perfectly acceptable when balancing with general formulae; do not try to 'clear' the fractions here.

Techniques used

balance a combustion equation algebraically

(c)

When $10\text{ cm}^3$ of A was mixed at room temperature with $50\text{ cm}^3$ of oxygen (an excess) and exploded, $40\text{ cm}^3$ of gas remained after cooling the apparatus to room temperature and pressure.

When this $40\text{ cm}^3$ of gas was shaken with an excess of aqueous potassium hydroxide, $\text{KOH}$ , $30\text{ cm}^3$ of gas still remained.

(i)

What is the identity of the $30\text{ cm}^3$ of gas that remained at the end of the experiment?

DifficultyMedium-Easy

Worked solution

Answer

Oxygen, $\text{O}_2$ — the excess oxygen that did not react; KOH absorbs only the carbon dioxide.

Final answer

Oxygen (O2)

Detailed explanation

Background Concept

Aqueous KOH (like NaOH) is an alkali that absorbs acidic oxides, notably CO₂: CO₂ + 2KOH → K₂CO₃ + H₂O. Oxygen is not absorbed by KOH.

Understanding the Question

After combustion and cooling, 40 cm³ of gas remains. Shaking with excess KOH removes 10 cm³, leaving 30 cm³. You must identify what the 30 cm³ is.

Approach

The KOH removes the CO₂ (acidic oxide). The gas left behind must be the unreacted excess oxygen, since the original mixture had 50 cm³ O₂ — an excess over what was needed.

Step-by-Step Reasoning

The combustion products are CO₂ and H₂O. On cooling, water condenses to liquid, so it is not part of the gas volume.
KOH absorbs the CO₂ (10 cm³ absorbed).
The remaining 30 cm³ is the oxygen that was in excess and did not react.

Key Takeaways

In gas-volume combustion problems, KOH/NaOH absorption is the standard way to measure the CO₂ produced.

Common Mistakes

Saying nitrogen (there is no nitrogen — the mixture was hydrocarbon + pure oxygen).
Saying CO₂ — that is the gas absorbed, not the residue.

Things to Be Careful About

Remember water vapour condenses on cooling to room temperature, so it never appears in the residual gas volume.

Techniques used

deduce residual gas identity from KOH absorption

(ii)

The combustion of A produced a gas that reacted with the $\text{KOH(aq)}$ .

What is the identity of this gas?

DifficultyEasy

Worked solution

Answer

Carbon dioxide, $\text{CO}_2$ — the acidic combustion product absorbed by the alkali.

Final answer

Carbon dioxide (CO2)

Detailed explanation

Background Concept

CO₂ is an acidic oxide that reacts with aqueous alkalis such as KOH, so it is removed from a gas mixture on shaking with KOH(aq). Oxygen and hydrocarbons are not absorbed.

Understanding the Question

The 10 cm³ of gas that disappeared on shaking with KOH must be identified.

Approach

Link 'gas absorbed by KOH' directly to 'carbon dioxide', the only acidic gaseous product of complete hydrocarbon combustion.

Step-by-Step Reasoning

Complete combustion of a hydrocarbon gives only CO₂ and H₂O.
H₂O has condensed to liquid; O₂ is in excess and not absorbed by KOH.
Therefore the gas absorbed by KOH is CO₂.

Key Takeaways

KOH(aq) absorption is the classical method for quantifying CO₂ in gas analysis.

Common Mistakes

Naming SO₂ or other acidic oxides — no sulfur is present in a hydrocarbon.

Things to Be Careful About

State 'carbon dioxide', not just 'an acidic gas'.

Techniques used

identify the gas absorbed by aqueous KOH

(iii)

What volume of the gas you have identified in (ii) was produced by the combustion of A?

DifficultyMedium-Easy

Worked solution

Answer

Volume of $\text{CO}_2 = 40 - 30 = 10\text{ cm}^3$

Final answer

10 cm^3

Detailed explanation

Background Concept

Gay-Lussac's law of combining volumes: at constant temperature and pressure, gas volumes react in simple ratios equal to their mole ratios. Volume changes on absorption directly give the volume of the absorbed gas.

Understanding the Question

You need the volume of CO₂ produced, knowing the total gas was 40 cm³ before KOH and 30 cm³ after.

Approach

The volume lost on absorption equals the volume of the gas absorbed (CO₂).

Step-by-Step Reasoning

Before KOH: 40 cm³ (excess O₂ + CO₂).
After KOH: 30 cm³ (excess O₂ only).
Volume of CO₂ = 40 − 30 = 10 cm³.

Key Takeaways

Volume of absorbed gas = volume before − volume after.

Common Mistakes

Reporting 30 cm³ (the remaining gas) instead of the 10 cm³ absorbed.

Things to Be Careful About

Read the question carefully: it asks for the CO₂ produced, i.e. the gas that reacted with the KOH.

Techniques used

deduce gas volume by difference

(iv)

What volume of oxygen was used up in the combustion of A?

DifficultyMedium

Worked solution

Working

Total volume before explosion: $10 + 50 = 60\text{ cm}^3$

After cooling, gas remaining: $40\text{ cm}^3$ (excess $\text{O}_2$ + $\text{CO}_2$ ; water has condensed)

Volume decrease = volume of $\text{O}_2$ consumed + volume of $\text{CO}_2$ formed − 0 (CO₂ is still gas)

\text{O}_2\text{ used} = (60 - 40) + 10 = 30 - 10 = 20\text{ cm}^3

Alternatively: excess $\text{O}_2$ remaining = $30\text{ cm}^3$ , so $\text{O}_2$ used $= 50 - 30 = 20\text{ cm}^3$ .

Answer

$20\text{ cm}^3$ of oxygen was used.

Final answer

20 cm^3

Detailed explanation

Background Concept

In combustion gas-volume problems, water condenses on cooling so it contributes no gas volume. The volume contraction on explosion equals the O₂ consumed minus the CO₂ produced (since CO₂ replaces some gas volume). Careful bookkeeping of volumes is essential.

Understanding the Question

Given: 10 cm³ hydrocarbon + 50 cm³ O₂ exploded; 40 cm³ gas remains; 10 cm³ of that is CO₂ (from iii). Find the volume of O₂ consumed.

Approach

Two equivalent routes:

Excess route: O₂ remaining after combustion = 40 − 10 (CO₂) = 30 cm³. O₂ used = 50 − 30 = 20 cm³.
Contraction route: total before = 60 cm³; after = 40 cm³; contraction = 20 cm³. Since CO₂ formed (10 cm³) stays as gas, contraction = O₂ used − CO₂ formed, so O₂ used = 20 + 10 = 30? No — careful: contraction = O₂ used − CO₂ formed only if water condenses. Here contraction = 20, so O₂ used = 20 + 10 = 30 cm³? That contradicts route 1. Recheck: contraction = O₂ consumed − CO₂ produced (water removed from gas phase). So O₂ used = contraction + CO₂ = 20 + 10 = 30 cm³. But route 1 gives 20. The discrepancy arises because the contraction route must account for the water: CₓHᵧ + O₂ → CO₂ + H₂O(l). Gas before: 60 cm³. Gas after: excess O₂ + CO₂ = 40 cm³. O₂ consumed = 60 − 40 − 10 = ... Actually: gas after = (50 − O₂ used) + CO₂. So 40 = 50 − O₂ used + 10 → O₂ used = 20 cm³. The contraction equation: contraction = O₂ used − CO₂ formed = 20 − 10 = 10 cm³, but observed contraction is 60 − 40 = 20 cm³. The extra 10 cm³ is the hydrocarbon itself disappearing from the gas phase! So contraction = O₂ used − CO₂ + hydrocarbon consumed = 20 − 10 + 10 = 20 ✓. The mark scheme answer is 20 cm³.

Step-by-Step Reasoning

O₂ remaining after combustion: the 40 cm³ consists of excess O₂ plus 10 cm³ CO₂, so excess O₂ = 30 cm³.
O₂ used = initial 50 cm³ − remaining 30 cm³ = 20 cm³.
Check with the general equation (part d): 10 cm³ CH₄ needs 2 × 10 = 20 cm³ O₂ ✓.

Key Takeaways

Track each gas separately: remaining O₂ = total residual gas − CO₂; O₂ used = initial O₂ − remaining O₂.

Common Mistakes

Forgetting that the residual gas contains both excess O₂ and CO₂, and subtracting incorrectly.
Ignoring that the hydrocarbon gas itself disappears from the gas volume, which complicates contraction-based methods.

Things to Be Careful About

Water condenses and must be excluded from all gas volumes at room temperature. The mark scheme answer is 20 cm³ — verify consistency with part (d) before moving on.

Techniques used

calculate oxygen consumed by differenceapply Avogadro's law to gas volumes

(d)

Use your equation in (b) and your results from (c)(iii) and (c)(iv) to calculate the molecular formula of A.
Show all of your working.

DifficultyMedium

Worked solution

Working

From the general equation:

\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}

Volumes (equal to mole ratios by Avogadro's law):

10\text{ cm}^3 \quad 20\text{ cm}^3 \quad 10\text{ cm}^3

$10\text{ cm}^3$ of $\text{C}_x\text{H}_y$ gives $10\text{ cm}^3$ of $\text{CO}_2$ , so $x = 1$ .
$10\text{ cm}^3$ of $\text{C}_x\text{H}_y$ reacts with $20\text{ cm}^3$ of $\text{O}_2$ , so $\left(x + \frac{y}{4}\right) = 2$ ; with $x = 1$ , $\frac{y}{4} = 1$ and $y = 4$ .

Answer

Molecular formula of A is $\text{CH}_4$ (methane).

Final answer

CH4

Detailed explanation

Background Concept

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of moles. Therefore, for gaseous reactions, volume ratios equal mole ratios, so the coefficients in the balanced equation can be read directly from experimental gas volumes.

Understanding the Question

Using the general combustion equation from (b) and the volumes deduced in (c) — 10 cm³ hydrocarbon, 20 cm³ O₂ consumed, 10 cm³ CO₂ produced — determine x and y and hence the molecular formula of A.

Approach

Write the volume of each species under the equation. The ratio 10 : 20 : 10 simplifies to 1 : 2 : 1, giving two equations: x = 1 (from CO₂) and x + y/4 = 2 (from O₂). Solve for y.

Step-by-Step Reasoning

Place volumes under the equation: CₓHᵧ (10), O₂ (20 used), CO₂ (10).
Mole ratio CₓHᵧ : CO₂ = 10 : 10 = 1 : 1. Since each molecule of hydrocarbon produces x molecules of CO₂, x = 1. (1 mark)
Mole ratio CₓHᵧ : O₂ = 10 : 20 = 1 : 2. From the equation, 1 mol hydrocarbon uses (x + y/4) mol O₂, so x + y/4 = 2. With x = 1, y/4 = 1, so y = 4. (1 mark)
Molecular formula = C₁H₄ = CH₄ — methane, famously the 'fire-damp' Davy studied in mines. (1 mark)

Key Takeaways

Gas-volume analysis + Avogadro's law is a classic method for finding molecular formulae of gaseous hydrocarbons. The general equation CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O converts volumes directly into x and y.

Common Mistakes

Using the initial 50 cm³ of O₂ instead of the 20 cm³ consumed — this gives the wrong y.
Forgetting to show working, which loses method marks even if CH₄ is stated.
Confusing the water volume (which would be y/2 × 10 = 5 cm³ as vapour, but water condenses so it is not measured).

Things to Be Careful About

Show all three working steps explicitly (x = 1; x + y/4 = 2, y = 4; formula CH₄) — each carries a separate mark. State the formula, not just 'methane'.

Techniques used

apply Avogadro's law to equate volume and mole ratiossolve simultaneous equations for x and y

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