Chemistry 9701/21 — May/June 2011

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

questions

marks

minutes

Topics Atoms, Molecules and Stoichiometry · Hydrocarbons · Chemical Bonding · Carbonyl Compounds · Introduction to Organic Chemistry · States of Matter · +7 more

Q1Introduction to Organic ChemistryAtoms, Molecules and StoichiometryStates of MatterFree sample

Some intercontinental jet airliners use kerosene as fuel. The formula of kerosene may be taken as C₁₄H₃₀.

(a)

To which homologous series of compounds does kerosene belong?

DifficultyEasy

Worked solution

Answer

Alkanes (paraffins).

Final answer

Alkanes (paraffins)

Detailed explanation

Background Concept

A homologous series is a family of compounds with the same general formula, similar chemical properties and successive members differing by CH2. The alkanes are the saturated hydrocarbons with general formula CnH2n+2; they contain only C–C and C–H single bonds.

Understanding the Question

'State which homologous series' — a one-mark recall item. The formula C14H30 fits CnH2n+2 (n = 14, 2n + 2 = 30), so it is an alkane.

Approach

Check the formula against the general formula of the alkanes, then name the series.

Step-by-Step Reasoning

C14H30: hydrogen count = 2(14) + 2 = 30, so the compound is a saturated, unbranched-chain-type hydrocarbon — an alkane. The mark scheme explicitly rejects the answer 'hydrocarbon': although true, it is not the name of a homologous series and does not answer the question.

Key Takeaways

Match a molecular formula to CnH2n+2 to identify alkanes; always name the series, not the broader class.

Common Mistakes

Writing 'hydrocarbon' — this is rejected because it names a class of compounds, not a homologous series.

Things to Be Careful About

Command words like 'which homologous series' demand the family name (alkane), not a description of bonding.

Techniques used

identify the homologous series from the molecular formula

(b)

When kerosene burns in an excess of air, carbon dioxide and water form.
Balance the following equation for the complete combustion of kerosene.

......C₁₄H₃₀(l) + ......O₂(g) → ......CO₂(g) + ......H₂O(g)

DifficultyMedium-Easy

Worked solution

Answer

2\text{C}_{14}\text{H}_{30}(\text{l}) + 43\text{O}_2(\text{g}) \rightarrow 28\text{CO}_2(\text{g}) + 30\text{H}_2\text{O}(\text{g})

(Equivalently: $\text{C}_{14}\text{H}_{30}(\text{l}) + \frac{43}{2}\text{O}_2(\text{g}) \rightarrow 14\text{CO}_2(\text{g}) + 15\text{H}_2\text{O}(\text{g})$ )

Final answer

2C14H30(l) + 43O2(g) -> 28CO2(g) + 30H2O(g)

Detailed explanation

Background Concept

Complete combustion of a hydrocarbon produces carbon dioxide and water only. Balancing relies on conservation of atoms: count C first, then H, then balance O last (because O2 is the only source of oxygen on the left).

Understanding the Question

'Balance the following equation' — fill in the coefficients. States are given: kerosene is (l), products are gases.

Approach

Balance C and H using one molecule of C14H30, then halve the odd oxygen count, or double everything to clear the fraction.

Step-by-Step Reasoning

One C14H30 gives 14 C, so 14 CO2; 30 H, so 15 H2O. Oxygen needed on the right: (14 × 2) + 15 = 43 atoms = 43/2 O2. To avoid fractions, multiply all coefficients by 2:

2\text{C}_{14}\text{H}_{30} + 43\text{O}_2 \rightarrow 28\text{CO}_2 + 30\text{H}_2\text{O}

Check: C: 28 = 28; H: 60 = 60; O: 86 = 56 + 30 = 86. ✓

Key Takeaways

Always balance C and H first and O last; fractions in O2 are acceptable but doubling clears them.

Common Mistakes

Using whole-number O2 and getting an unbalanced equation; forgetting to double all coefficients when clearing the fraction.

Things to Be Careful About

Both the doubled form and the fractional form are accepted; keep the state symbols as printed.

Techniques used

balance a combustion equation by atom counting

(c)

In this section, give your answers to one decimal place.

The flight path from Beijing to Paris is approximately 8195 km.
A typical intercontinental jet airliner burns 10.8 kg of kerosene for each kilometre covered.

(i)

Calculate the mass, in tonnes, of C₁₄H₃₀ burnt on a flight from Beijing to Paris.
[1 tonne = 1 000 kg]

DifficultyEasy

Worked solution

Working

\text{mass} = 8195 \text{ km} \times 10.8 \text{ kg km}^{-1} = 88506 \text{ kg}

\text{mass in tonnes} = \frac{88506}{1000} = 88.5 \text{ tonnes}

Answer

88.5 tonnes of C14H30

Final answer

88.5 tonnes

Detailed explanation

Background Concept

Fuel consumption is given per kilometre; multiplying by the total distance gives the total mass burnt. Unit conversion: 1 tonne = 1000 kg.

Understanding the Question

Given: distance 8195 km, consumption 10.8 kg per km. Asked: total mass in tonnes, to one decimal place.

Approach

Multiply distance by consumption rate, then divide by 1000 to convert kg to tonnes.

Step-by-Step Reasoning

8195 × 10.8 = 88506 kg. Dividing by 1000 gives 88.506 tonnes, which to one decimal place is 88.5 tonnes.

Key Takeaways

Watch the units: km × kg km⁻¹ = kg; then convert to tonnes.

Common Mistakes

Forgetting the tonne conversion, or rounding too early and losing the one-decimal-place answer.

Things to Be Careful About

The instruction says one decimal place — 88.5, not 88.506 or 89.

Techniques used

calculate mass from a rate of consumption and distance

(ii)

Use your equation in (b) to calculate the mass, in tonnes, of CO₂ produced during this flight.

DifficultyMedium-Easy

Worked solution

Working

M_r(\text{C}_{14}\text{H}_{30}) = (14 \times 12) + (30 \times 1) = 198

From the equation $2\text{C}_{14}\text{H}_{30} \rightarrow 28\text{CO}_2$ :

2 \times 198 \text{ t of C}_{14}\text{H}_{30} \rightarrow 28 \times 44 \text{ t of CO}_2

88.5 \text{ t of C}_{14}\text{H}_{30} \rightarrow \frac{28 \times 44 \times 88.5}{2 \times 198} = 275.3 \text{ t of CO}_2

Answer

275.3 tonnes of CO2

Final answer

275.3 tonnes

Detailed explanation

Background Concept

In a balanced equation, the coefficients give the ratio of moles. Since moles = mass/Mr, mass ratios scale as (coefficient × Mr). Tonnes can be used directly in place of grams because the tonne factor cancels in the ratio.

Understanding the Question

Given: 88.5 tonnes of C14H30 burnt (from (c)(i)) and the balanced equation from (b). Asked: mass of CO2 produced, one decimal place.

Approach

Find Mr of the fuel and of CO2, set up the stoichiometric mass ratio from the equation, and scale to 88.5 tonnes.

Step-by-Step Reasoning

M_r(C14H30) = 14(12) + 30(1) = 198; M_r(CO2) = 12 + 2(16) = 44.
The equation 2C14H30 + 43O2 → 28CO2 + 30H2O shows 2 mol fuel produce 28 mol CO2, i.e. 2 × 198 = 396 tonnes of fuel produce 28 × 44 = 1232 tonnes of CO2.
Scaling: 88.5 × 1232/396 = 275.3 tonnes of CO2. (Using the unrounded 88.506 gives 275.4, which the mark scheme also allows.)

Key Takeaways

Mass–mass calculations: convert via (mass/Mr) × mole ratio × Mr. Using tonne units throughout is valid because the conversion cancels.

Common Mistakes

Using the wrong mole ratio (e.g. 1:14 instead of 2:28 — actually equivalent, but errors arise with the O2 ratio); using M_r = 198 incorrectly; forgetting the coefficient 2 on the fuel.

Things to Be Careful About

Error carried forward applies if a wrong Mr was used, provided the method is correct. Round only at the end; both 275.3 and 275.4 (from 88.506) are credited.

Techniques used

calculate Mr from a formulause stoichiometric mass ratios from a balanced equation

(d)

Bicycles may be carried on commercial airliners. When carried on airliners, bicycles are placed in the luggage hold. This is a part of the aircraft which, in flight, will have different temperatures and air pressures from those at sea level.

This question concerns the change in pressure in an inflated bicycle tyre from when it is at sea level to when it is in the hold of an airliner in flight.

At sea level and a temperature of 20 °C an inflated bicycle tyre contains $710 \text{ cm}^3$ of air at an internal pressure of $6 \times 10^5$ Pa.

Use the general gas equation $PV = nRT$ to calculate the amount, in moles, of air in the tyre at sea level.

DifficultyMedium-Easy

Worked solution

Working

n = \frac{PV}{RT} = \frac{6 \times 10^5 \times 710 \times 10^{-6}}{8.31 \times 293}

n = \frac{426}{2434.83} = 0.175 \text{ mol}

Answer

n = 0.175 mol of air

Final answer

0.175 mol

Detailed explanation

Background Concept

The ideal gas equation PV = nRT relates pressure (Pa), volume (m³), amount (mol) and temperature (K), with R = 8.31 J K⁻¹ mol⁻¹. Consistent SI units are essential: 1 cm³ = 1 × 10⁻⁶ m³, and T(K) = T(°C) + 273.

Understanding the Question

Given: V = 710 cm³, P = 6 × 10⁵ Pa, T = 20 °C. Asked: n, the amount of air in the tyre.

Approach

Rearrange PV = nRT for n, convert V to m³ and T to kelvin, then substitute.

Step-by-Step Reasoning

V = 710 × 10⁻⁶ m³ = 7.10 × 10⁻⁴ m³; T = 20 + 273 = 293 K.

n = \frac{PV}{RT} = \frac{(6 \times 10^5)(710 \times 10^{-6})}{8.31 \times 293} = \frac{426}{2434.83} = 0.175 \text{ mol}

Key Takeaways

The most common failure in gas-law calculations is unit inconsistency — always convert to Pa, m³ and K before substituting.

Common Mistakes

Using 710 instead of 710 × 10⁻⁶ (volume in cm³); using 20 instead of 293 for temperature; quoting an inappropriate number of significant figures.

Things to Be Careful About

R = 8.31 J K⁻¹ mol⁻¹ with SI units; the answer 0.175 mol is carried forward to part (e).

Techniques used

use the ideal gas equationconvert volume units and temperature to kelvin

(e)

The same bicycle, with its tyres inflated at sea level as described in (d) above, is placed in the luggage hold of an airliner. At a height of 10 000 m, the temperature in the luggage hold is 5 °C and the air pressure is $2.8 \times 10^4$ Pa.

Assuming the volume of the tyre does not change, use your answer to (d) to calculate the pressure inside the tyre at a height of 10 000 m.

DifficultyMedium-Easy

Worked solution

Working

At 10 000 m: T = 5 + 273 = 278 K, V unchanged, n = 0.175 mol.

P = \frac{nRT}{V} = \frac{0.175 \times 8.31 \times 278}{710 \times 10^{-6}}

P = 5.7 \times 10^5 \text{ Pa}

Answer

P = 5.7 × 10^5 Pa

Final answer

5.7 × 10^5 Pa

Detailed explanation

Background Concept

For a fixed amount of gas in a fixed volume, PV = nRT shows pressure is directly proportional to temperature (in kelvin): P1/T1 = P2/T2. Cooling the gas lowers its pressure; the external pressure drop does not change the pressure inside the sealed, rigid tyre (assumed constant volume).

Understanding the Question

Given: n = 0.175 mol (from (d)), V = 710 cm³ unchanged, T = 5 °C = 278 K. Asked: the internal pressure of the tyre in flight. Note the external pressure (2.8 × 10⁴ Pa) is irrelevant to the internal pressure calculation — it is context/distractor information.

Approach

Rearrange PV = nRT for P and substitute n, R, T (in K) and V (in m³).

Step-by-Step Reasoning

P = \frac{nRT}{V} = \frac{0.175 \times 8.31 \times 278}{710 \times 10^{-6}} = \frac{404.1}{7.10 \times 10^{-4}} \approx 569411 \text{ Pa} \approx 5.7 \times 10^5 \text{ Pa}

The pressure falls only slightly (from 6.0 × 10⁵ to 5.7 × 10⁵ Pa) because the temperature drop from 293 K to 278 K is modest. Error carried forward applies if the (d) value differs, provided the method is correct.

Key Takeaways

In sealed rigid containers, n and V are fixed, so P ∝ T (kelvin). Ignore irrelevant given data (external pressure here).

Common Mistakes

Using the external pressure 2.8 × 10⁴ Pa in the calculation; using 5 instead of 278 K; using volume in cm³ instead of m³.

Things to Be Careful About

Temperature must be in kelvin; express the answer to an appropriate precision (5.7 × 10⁵ Pa, two significant figures).

Techniques used

apply the ideal gas equation at new conditionsassume constant volume and fixed amount of gas

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