Chemistry 9701/21 — May/June 2013

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

questions

marks

minutes

Topics Introduction to Organic Chemistry · Carbonyl Compounds · Atoms, Molecules and Stoichiometry · Nitrogen and Sulfur · Equilibria · Chemical Energetics · +7 more

Q1Atoms, Molecules and StoichiometryNitrogen and SulfurFree sample

A sample of a fertiliser was known to contain ammonium sulfate, (NH₄)₂SO₄, and sand only.

A 2.96 g sample of the solid fertiliser was heated with 40.0 cm³ of NaOH(aq), an excess, and all of the ammonia produced was boiled away.

After cooling, the remaining NaOH(aq) was exactly neutralised by 29.5 cm³ of 2.00 mol dm⁻³ HCl.

In a separate experiment, 40.0 cm³ of the original NaOH(aq) was exactly neutralised by 39.2 cm³ of the 2.00 mol dm⁻³ HCl.

(a)

(i)

Write balanced equations for the following reactions.

NaOH with HCl

(NH₄)₂SO₄ with NaOH

DifficultyMedium-Easy

Worked solution

Answer

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

$(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow 2\text{NH}_3 + \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Final answer

NaOH + HCl -> NaCl + H2O ; (NH4)2SO4 + 2NaOH -> 2NH3 + Na2SO4 + 2H2O

Detailed explanation

Background Concept

This question tests two classic acid-base reactions. The first is a simple neutralisation: a strong base (NaOH) reacts with a strong acid (HCl) to give a salt and water. The second is a displacement reaction: a strong base displaces the weaker base ammonia from its salt. Ammonium sulfate contains the ammonium ion, $\text{NH}_4^+$ , which is the conjugate acid of ammonia. When a strong base such as NaOH is added and the mixture is heated, the equilibrium $\text{NH}_4^+ + \text{OH}^- \rightleftharpoons \text{NH}_3 + \text{H}_2\text{O}$ is driven to the right and the volatile ammonia is boiled off. This is exactly why the fertiliser is heated with excess NaOH in the experiment.

Understanding the Question

The command word is "write balanced equations". Two reactions are named: NaOH with HCl, and $(\text{NH}_4)_2\text{SO}_4$ with NaOH. You must supply the products and balance the atoms on both sides. The mark scheme allows ionic equations as an alternative, but a full balanced equation is safest.

Approach

For each reaction, identify the type, predict the products, then balance atom by atom. For the neutralisation, the metal ion from the base pairs with the anion from the acid. For the ammonium salt, the base removes a proton from each ammonium ion, releasing one ammonia molecule per $\text{NH}_4^+$ .

Step-by-Step Reasoning

NaOH + HCl: the $\text{Na}^+$ pairs with $\text{Cl}^-$ to give NaCl, and $\text{H}^+$ pairs with $\text{OH}^-$ to give $\text{H}_2\text{O}$ . Already balanced: one Na, one O, one H on each side (plus the H from HCl and the H from OH combining).
$(\text{NH}_4)_2\text{SO}_4$ + NaOH: the sulfate keeps its $\text{Na}^+$ counter-ion, forming $\text{Na}_2\text{SO}_4$ . Each of the two ammonium ions releases one $\text{NH}_3$ , giving $2\text{NH}_3$ . The two protons removed are captured by two $\text{OH}^-$ to form $2\text{H}_2\text{O}$ . Therefore two NaOH are needed: $(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow 2\text{NH}_3 + \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ .
Check balancing: left has 2 N, 8 H (from ammonium) + 2 H (from NaOH) = 10 H, 1 S, 2 Na, 6 O. Right has 2 N (in $2\text{NH}_3$ ), 6 H (in $2\text{NH}_3$ ) + 4 H (in $2\text{H}_2\text{O}$ ) = 10 H, 1 S, 2 Na, 4 O (sulfate) + 2 O (water) = 6 O. Balanced.

Key Takeaways

A strong base displaces ammonia from an ammonium salt, and the stoichiometry is 2 mol NaOH per 1 mol $(\text{NH}_4)_2\text{SO}_4$ because there are two ammonium ions per formula unit. Neutralisation always gives salt + water.

Common Mistakes

Forgetting the coefficient 2 in front of NaOH, or writing only one $\text{NH}_3$ per formula unit. Writing $\text{NH}_4\text{OH}$ as a product rather than $\text{NH}_3 + \text{H}_2\text{O}$ . Omitting the balancing of the sodium sulfate.

Things to Be Careful About

The mark scheme allows ionic equations, but if you write full equations make sure every atom balances. State symbols are not demanded here but adding them is good practice. Do not forget that the ammonium ion has four hydrogens, one of which becomes part of the water.

Techniques used

write balanced equations with state symbolsrecognise acid-base neutralisationapply displacement of a weak base by a strong base

(ii)

Calculate the amount, in moles, of NaOH present in the 40.0 cm³ of the original NaOH(aq) that was neutralised by 39.2 cm³ of 2.00 mol dm⁻³ HCl.

DifficultyEasy

Worked solution

Working

$n(\text{HCl}) = \frac{39.2 \times 2.00}{1000} = 0.0784 \text{ mol}$

From $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ , the ratio is 1:1, so $n(\text{NaOH}) = n(\text{HCl})$ .

Answer

$n(\text{NaOH}) = 0.0784 \text{ mol}$

Final answer

0.0784 mol

Detailed explanation

Background Concept

The amount in moles of a solute is found from $n = c \times V$ , where $c$ is the concentration in $\text{mol dm}^{-3}$ and $V$ is the volume in $\text{dm}^3$ . Since the volume is given in $\text{cm}^3$ , it must be divided by 1000.

Understanding the Question

This is the standardisation titration: 40.0 $\text{cm}^3$ of the original NaOH is exactly neutralised by 39.2 $\text{cm}^3$ of 2.00 $\text{mol dm}^{-3}$ HCl. You are asked for the moles of NaOH in that 40.0 $\text{cm}^3$ .

Approach

Convert the HCl volume to $\text{dm}^3$ , multiply by its concentration to get moles of HCl, then use the 1:1 stoichiometry of the neutralisation to get moles of NaOH.

Step-by-Step Reasoning

Volume of HCl $= 39.2 \text{ cm}^3 = 0.0392 \text{ dm}^3$ .
$n(\text{HCl}) = 2.00 \times 0.0392 = 0.0784 \text{ mol}$ .
The balanced equation shows one mole of NaOH reacts with one mole of HCl, so $n(\text{NaOH}) = 0.0784 \text{ mol}$ .

Key Takeaways

Always convert $\text{cm}^3$ to $\text{dm}^3$ before multiplying by concentration. The 1:1 ratio comes straight from the balanced equation.

Common Mistakes

Forgetting to divide by 1000, giving a value 1000 times too large. Using the wrong volume (e.g. the 29.5 $\text{cm}^3$ from the other titration).

Things to Be Careful About

Keep the value unrounded (0.0784) for use in the subtraction in part (iv). Rounding too early loses accuracy. Units are moles.

Techniques used

calculate moles from concentration and volumeapply 1:1 stoichiometry from a balanced equation

(iii)

Calculate the amount, in moles, of NaOH present in the 40.0 cm³ of NaOH(aq) that remained after boiling the (NH₄)₂SO₄.

DifficultyEasy

Worked solution

Working

$n(\text{HCl}) = \frac{29.5 \times 2.00}{1000} = 0.0590 \text{ mol}$

1:1 ratio, so $n(\text{NaOH}) = n(\text{HCl})$ .

Answer

$n(\text{NaOH}) = 0.0590 \text{ mol}$

Final answer

0.0590 mol

Detailed explanation

Background Concept

This is the back-titration reading. After the fertiliser has been boiled with the NaOH, the NaOH that is left over (the excess that did not react with the ammonium sulfate) is titrated with HCl. The moles of HCl used here equal the moles of unreacted NaOH.

Understanding the Question

29.5 $\text{cm}^3$ of 2.00 $\text{mol dm}^{-3}$ HCl neutralised the NaOH remaining in the 40.0 $\text{cm}^3$ after boiling. Find the moles of that remaining NaOH.

Approach

Convert volume to $\text{dm}^3$ , multiply by concentration, apply the 1:1 ratio.

Step-by-Step Reasoning

Volume $= 29.5 \text{ cm}^3 = 0.0295 \text{ dm}^3$ .
$n(\text{HCl}) = 2.00 \times 0.0295 = 0.0590 \text{ mol}$ .
1:1 with NaOH, so $n(\text{NaOH remaining}) = 0.0590 \text{ mol}$ .

Key Takeaways

The smaller titre here reflects the fact that some NaOH was consumed by the ammonium sulfate, so less remained to be titrated.

Common Mistakes

Swapping the two titres (using 39.2 here). Forgetting the /1000 conversion.

Things to Be Careful About

Keep three significant figures (0.0590) so the subtraction in (iv) is accurate. This value represents the excess NaOH, not the total.

Techniques used

calculate moles from concentration and volumeapply 1:1 stoichiometry from a balanced equation

(iv)

Use your answers to (ii) and (iii) to calculate the amount, in moles, of NaOH that reacted with the (NH₄)₂SO₄.

DifficultyMedium-Easy

Worked solution

Working

$n(\text{NaOH reacted}) = n(\text{total}) - n(\text{excess}) = 0.0784 - 0.0590$

Answer

$n(\text{NaOH}) = 0.0194 \text{ mol}$

Final answer

0.0194 mol

Detailed explanation

Background Concept

This is the heart of a back-titration. The original 40.0 $\text{cm}^3$ of NaOH contained a known total amount of base (from part ii). After reaction with the fertiliser, only the excess base remained, and that was measured by titration (part iii). The base that actually reacted with the ammonium sulfate is the difference.

Understanding the Question

Use your answers to (ii) and (iii) to find how much NaOH reacted with the $(\text{NH}_4)_2\text{SO}_4$ .

Approach

Subtract the moles of excess NaOH from the moles of NaOH originally present.

Step-by-Step Reasoning

Total NaOH originally present $= 0.0784 \text{ mol}$ (part ii).
NaOH remaining after boiling $= 0.0590 \text{ mol}$ (part iii).
NaOH that reacted $= 0.0784 - 0.0590 = 0.0194 \text{ mol}$ .

Key Takeaways

Back-titration logic: amount reacted = amount added − amount left over. This is a standard technique when the reacting species is volatile or slow to react directly.

Common Mistakes

Adding the two values instead of subtracting. Using the wrong pair of numbers. Rounding intermediate values.

Things to Be Careful About

Keep the subtraction exact to avoid losing the answer mark. The result must be smaller than the total, which is a quick sanity check.

Techniques used

subtract moles to find the amount that reactedapply back-titration logic

(v)

Use your answers to (i) and (iv) to calculate the amount, in moles, of (NH₄)₂SO₄ that reacted with the NaOH.

DifficultyMedium-Easy

Worked solution

Working

From $(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow 2\text{NH}_3 + \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ , 2 mol NaOH react with 1 mol $(\text{NH}_4)_2\text{SO}_4$ .

$n[(\text{NH}_4)_2\text{SO}_4] = \frac{0.0194}{2} = 9.7 \times 10^{-3} \text{ mol}$

Answer

$n[(\text{NH}_4)_2\text{SO}_4] = 9.7 \times 10^{-3} \text{ mol}$

Final answer

9.7 x 10^-3 mol

Detailed explanation

Background Concept

The balanced equation from part (i) gives the mole ratio between NaOH and the ammonium sulfate. Because each formula unit of $(\text{NH}_4)_2\text{SO}_4$ contains two ammonium ions, two moles of NaOH are needed per mole of the salt.

Understanding the Question

Use the NaOH that reacted (part iv) and the equation from (i) to find the moles of $(\text{NH}_4)_2\text{SO}_4$ .

Approach

Divide the moles of NaOH by 2, since the ratio is 2:1.

Step-by-Step Reasoning

From the equation, $n(\text{NaOH}) : n[(\text{NH}_4)_2\text{SO}_4] = 2 : 1$ .
$n[(\text{NH}_4)_2\text{SO}_4] = 0.0194 / 2 = 0.0097 = 9.7 \times 10^{-3} \text{ mol}$ .

Key Takeaways

The stoichiometric coefficient from the balanced equation is the conversion factor between the two species.

Common Mistakes

Multiplying by 2 instead of dividing. Using a 1:1 ratio by forgetting the two ammonium ions per formula unit.

Things to Be Careful About

This value is carried forward into the mass calculation, so keep it unrounded. Note the answer is in the $10^{-3}$ range, consistent with a small mass of fertiliser.

Techniques used

apply the stoichiometric ratio from the balanced equationconvert moles of one reactant to moles of another

(vi)

Hence calculate the mass of (NH₄)₂SO₄ that reacted.

DifficultyMedium-Easy

Worked solution

Working

$M_r[(\text{NH}_4)_2\text{SO}_4] = 2(14.0) + 8(1.0) + 32.1 + 4(16.0) = 132.1$

$\text{mass} = 9.7 \times 10^{-3} \times 132.1 = 1.2814 \text{ g}$

Answer

$\text{mass of } (\text{NH}_4)_2\text{SO}_4 = 1.28 \text{ g}$

Final answer

1.28 g

Detailed explanation

Background Concept

Mass, moles and molar mass are related by $m = n \times M_r$ . The relative formula mass is the sum of the relative atomic masses of every atom in the formula unit.

Understanding the Question

Use the moles of $(\text{NH}_4)_2\text{SO}_4$ from part (v) to find the mass that reacted.

Approach

First compute $M_r$ from the formula, then multiply by the moles.

Step-by-Step Reasoning

$(\text{NH}_4)_2\text{SO}_4$ contains: N $2 \times 14.0 = 28.0$ ; H $8 \times 1.0 = 8.0$ ; S $32.1$ ; O $4 \times 16.0 = 64.0$ . Total $M_r = 132.1$ .
$\text{mass} = 9.7 \times 10^{-3} \times 132.1 = 1.2814 \text{ g}$ .

Key Takeaways

Always compute $M_r$ carefully, counting every atom — here 8 hydrogens and 4 oxygens.

Common Mistakes

Miscounting hydrogens (using 4 instead of 8) or oxygens. Using the atomic mass of the whole sulfate incorrectly. Forgetting to multiply by the moles.

Things to Be Careful About

Carry the unrounded 1.2814 g into part (vii) to keep the percentage accurate. The mass (1.28 g) is less than the 2.96 g sample, which is sensible since the sample also contains sand.

Techniques used

calculate relative formula massconvert moles to mass using m = n x M

(vii)

Use your answer to (vi) to calculate the percentage, by mass, of (NH₄)₂SO₄ present in the fertiliser.

Write your answer to a suitable number of significant figures.

DifficultyMedium-Easy

Worked solution

Working

$\% (\text{NH}_4)_2\text{SO}_4 = \frac{1.2814}{2.96} \times 100 = 43.304...\%$

Answer

$\%(\text{NH}_4)_2\text{SO}_4 = 43.3\% \text{ (3 s.f.)}$

Final answer

43.3%

Detailed explanation

Background Concept

Percentage by mass of a component in a mixture is (mass of component / total mass of sample) $\times$ 100. The total sample mass here is 2.96 g, which includes the sand as well as the ammonium sulfate.

Understanding the Question

Use the mass of $(\text{NH}_4)_2\text{SO}_4$ from part (vi) and the total sample mass to find the percentage, then give the answer to a suitable number of significant figures.

Approach

Divide the mass of ammonium sulfate by the sample mass, multiply by 100, and round to a sensible number of significant figures.

Step-by-Step Reasoning

$\% = (1.2814 / 2.96) \times 100$ .
$= 43.30405405...\%$ .
The data are given to three significant figures (2.96 g, 2.00 mol dm⁻³), so the answer should be quoted to three significant figures: 43.3%.

Key Takeaways

Percentage by mass compares the component to the whole sample. Choose significant figures to match the least precise given data.

Common Mistakes

Dividing by the mass of ammonium sulfate instead of the sample mass. Giving too many significant figures (e.g. 43.304) or too few (43%).

Things to Be Careful About

The mark scheme awards one mark for the correct expression and one for the answer given to 3 s.f. Error carried forward is allowed, so an earlier slip does not necessarily lose this mark if the method is right.

Techniques used

calculate percentage by massapply significant figures convention

(b)

The uncontrolled use of nitrogenous fertilisers can cause environmental damage to lakes and streams. This is known as eutrophication.

What are the processes that occur when excessive amounts of nitrogenous fertilisers get into lakes and streams?

DifficultyMedium-Easy

Worked solution

Answer

The nitrate fertiliser acts as a nutrient, causing excessive growth of aquatic plants/algae (an algal bloom).
When these plants and algae die, they are decomposed by bacteria, which use up the dissolved oxygen in the water, so fish and other aquatic life die.

Final answer

Excessive growth of algae/plants (algal bloom); on death, decomposition uses up dissolved oxygen, killing fish/aquatic life.

Detailed explanation

Background Concept

Eutrophication is the enrichment of a water body with nutrients, especially nitrates and phosphates, leading to excessive plant and algal growth and ultimately to oxygen depletion. Nitrogenous fertilisers are soluble in water, so rain washes nitrate ions from fields into lakes and streams (leaching and run-off).

Understanding the Question

The question asks for the processes that occur once excessive fertiliser enters a lake or stream. It is a recall/explanation question worth two marks, so two linked stages are needed.

Approach

Describe the cause-and-effect chain: nutrient input → rapid plant/algal growth → death and decomposition → oxygen depletion → death of aquatic animals.

Step-by-Step Reasoning

The nitrate acts as a plant nutrient. With an abundant supply, algae and aquatic plants grow rapidly, forming an algal bloom that covers the surface.
The bloom blocks light, and when the plants and algae die, they sink and are broken down by aerobic decomposer bacteria. This decomposition consumes the dissolved oxygen in the water.
With dissolved oxygen depleted, fish and other aerobic aquatic organisms suffocate and die.

Key Takeaways

Eutrophication is a chain of events; the key chemical idea is that nitrate is a limiting nutrient whose excess triggers biological growth, and decomposition then removes dissolved oxygen.

Common Mistakes

Giving only "algae grow" without the oxygen-depletion consequence, or only "fish die" without the cause. Confusing eutrophication with acid rain or direct toxicity of nitrate.

Things to Be Careful About

The mark scheme credits "excessive growth of aquatic plants/algae or algal bloom" and "when plants and algae die O2 is used up or fish/aquatic life die". Both halves must be present to score full marks.

Techniques used

describe the sequence of environmental processeslink nutrient enrichment to oxygen depletion

(c)

Large quantities of ammonia are manufactured by the Haber process.
Not all of this ammonia is used to make fertilisers.
State one large-scale use for ammonia, other than in the production of nitrogenous fertilisers.

DifficultyEasy

Worked solution

Answer

Manufacture of nitric acid (by the Ostwald process).

Final answer

Manufacture of nitric acid

Detailed explanation

Background Concept

Ammonia is one of the most important industrial chemicals, produced on a huge scale by the Haber process. Although most of it goes into nitrogenous fertilisers, a significant fraction is used for other purposes.

Understanding the Question

The question asks for one large-scale use of ammonia other than fertiliser production. One mark is available.

Approach

Recall the major industrial outlets for ammonia: nitric acid manufacture (Ostwald process), explosives (e.g. TNT, ammonium nitrate), nylon and other polyamides, and smaller uses such as cleaning agents and refrigerants.

Step-by-Step Reasoning

Ammonia is oxidised catalytically to nitrogen monoxide, which is further oxidised and dissolved in water to make nitric acid — the Ostwald process. This is a major non-fertiliser use.
Other valid answers include explosives, nylon manufacture, cleaning agents and refrigerants.

Key Takeaways

Ammonia is a key feedstock for nitric acid and nitrogen-containing industrial products.

Common Mistakes

Answering "detergent" — the mark scheme explicitly rejects this. Also avoid repeating "fertiliser", which the question excludes.

Things to Be Careful About

The mark scheme accepts nitric acid manufacture, explosives, nylon, cleaning agent or refrigerant, but not detergent. Give one clear use.

Techniques used

recall an industrial use of ammoniadistinguish non-fertiliser uses

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