JAN 2020 4CH1 2C QP

Question 1

This question is about elements, compounds and mixtures.

(a) Name the element that burns with a lilac flame.

(b) Name the technique used to separate the mixture of colours in black ink.

(c) The box gives the names of some substances.

Choose substances from the box to answer these questions.

(i) Identify the compound.
(ii) Identify the mixture.
(iii) Identify the non-metal element that is a solid at room temperature.

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 1(b): Elements, compounds and mixtures — parts (a), (b), (c)(i), (c)(ii), (c)(iii)
• 2(h): Chemical tests — part (a)
• 2.44–2.48: Tests for gases and ions (flame tests) — part (a)
• 1.10: Separation techniques — part (b)

▶️ Answer/Explanation

Solution

(a) potassium (ALLOW K)

(b) (paper) chromatography

(c)(i) sodium chloride (ALLOW NaCl)

(c)(ii) air

(c)(iii) sulfur (ALLOW S)

Notes:
• Part (a) tests knowledge of flame test colours for Group 1 ions (\(K^+\) gives a lilac flame).
• Part (b) requires recall of the separation technique for soluble coloured substances.
• Part (c) assesses the definitions of compound (NaCl), mixture (air), and the physical state of a non-metal element (sulfur is solid at room temperature).

Question 2

Crude oil is a mixture of hydrocarbons.

(a) Name the process used to separate crude oil into fractions.

(b) Give one use of the kerosene fraction.

(c) One of the hydrocarbons in the refinery gas fraction is an alkane with the structural formula \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \)

(i) Give the name of this alkane.
(ii) Calculate the relative molecular mass (\( M_r \)) of this alkane.

(d) One of the alkanes in the gasoline fraction has the displayed formula:

(i) Determine the molecular formula of this alkane.
(ii) Give the general formula for the alkanes.

(e) Catalytic cracking is used to convert long-chain alkanes into shorter-chain alkanes.

(i) Name the catalyst used in catalytic cracking.
(ii) Explain why it is necessary to convert long-chain alkanes into shorter-chain alkanes.

(f) Catalytic cracking also produces alkenes.

\( C_{11}H_{24} \) can undergo cracking to give pentane (\( C_5H_{12} \)) and two different alkenes.

Complete the equation for this cracking reaction.

\[C_{11}H_{24} \rightarrow C_5H_{12} + \ldots + \ldots\]

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 4(b): Crude oil — Fractional distillation (point 4.8)
• 4(b): Crude oil — Uses of fractions (point 4.9)
• 4(c): Alkanes — Naming, formulas, general formula (points 4.21, 4.19)
• 1(e): Chemical formulae, equations and calculations — Relative formula masses (point 1.26)
• 4(b): Crude oil — Catalytic cracking (point 4.17)
• 4(b): Crude oil — Need for cracking (point 4.18)
• 4(d): Alkenes — Products of cracking

▶️ Answer/Explanation

Solution

Part (a): fractional distillation

ACCEPT fractionation / fractionating

Part (b): aircraft fuel / jet fuel / paraffin / fuel for lamps or heaters

ACCEPT heating oil / cooking fuel

Part (c)(i): butane

Part (c)(ii): \( M_r = 4 \times 12 + 10 \times 1 = 58 \)

Part (d)(i): \( C_7H_{16} \)

Part (d)(ii): \( C_nH_{2n+2} \)

ACCEPT different letters instead of n

Part (e)(i): alumina / silica / zeolite(s) / aluminosilicate(s)

ACCEPT aluminium oxide (\( \text{Al}_2\text{O}_3 \)) / silicon dioxide (\( \text{SiO}_2 \))

Part (e)(ii): An explanation that links two points:

There is greater demand for short-chain alkanes (e.g., for petrol/gasoline) than for long-chain alkanes.
Fractional distillation of crude oil produces more long-chain alkanes than needed / not enough short-chain alkanes to meet demand.

Part (f): \( C_{11}H_{24} \rightarrow C_5H_{12} + C_2H_4 + C_4H_8 \)

ACCEPT alkenes in either order: \( C_{11}H_{24} \rightarrow C_5H_{12} + C_4H_8 + C_2H_4 \)

ACCEPT 1 mark for alternative products: \( C_6H_{12} \) OR \( 2C_3H_6 \)

Question 3

This question is about copper and its compounds.

(a) Copper is a metal used for electrical wiring.
Explain why copper is a good conductor of electricity.

(b) This apparatus is used to investigate the electrolysis of copper(II) sulfate solution with graphite electrodes.

Copper forms at the negative electrode and oxygen forms at the positive electrode.

(i) State what would be observed at each electrode.

(ii) The ionic half-equation for the reaction at the negative electrode is

\[Cu^{2+} + 2e^- \rightarrow Cu\]

State why this is a reduction reaction.

(iii) Explain why the copper(II) sulfate solution becomes paler blue during the electrolysis.

(c) When hydrated copper(II) sulfate crystals are heated, anhydrous copper(II) sulfate forms.

A mass of 12.5 g of hydrated copper(II) sulfate crystals is heated in a crucible until all the water of crystallisation is removed.

A mass of 8.0 g of anhydrous copper(II) sulfate forms.

Show by calculation that the formula of hydrated copper(II) sulfate is \(CuSO_4 \cdot 5H_2O\)
\[[M_r \text{ of } CuSO_4 = 159.5 \quad M_r \text{ of } H_2O = 18]\]

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 1(i): Electrolysis — parts (b)(i), (b)(ii), (b)(iii)
• 1(h): Metallic bonding — part (a)
• 1(e): Chemical formulae, equations and calculations — part (c)
• 2(g): Acids, bases and salt preparations — part (c) context

▶️ Answer/Explanation

Solution

(a) An explanation that links the following two points:
M1: Electrons are delocalised.
M2: (Electrons) can move / can flow / are mobile.
IGNORE ‘sea of electrons’ / ‘free electrons’. REJECT ‘cations/atoms move’ for both marks. M2 is dependent on M1 or a mention of electrons. ‘Electrons move’ alone scores 1 mark.

(b)(i)
Negative electrode: Brown / pink / pink-brown solid formed.
Positive electrode: Bubbles / fizzing / effervescence.
ACCEPT brown/pink coating/deposit on the electrode. ALLOW red-brown. REJECT orange. ALLOW 1 mark if both observations are correct but at the incorrect electrodes.

(b)(ii) Copper ion(s) / \(Cu^{2+}\) gains electrons.
ACCEPT ‘oxidation state of copper goes down’ / ‘goes from +2 to 0’. IGNORE references to loss of oxygen. ALLOW ‘electrons are gained’. REJECT ‘copper/Cu gains electrons’.

(b)(iii) An explanation that links the following two points:
M1: The (blue) colour is caused by copper ions / \(Cu^{2+}\).
M2: Copper ions / \(Cu^{2+}\) are being discharged / removed from the solution.
ACCEPT ‘concentration of copper ions / \(Cu^{2+}\) decreases’. ALLOW ‘copper ions / \(Cu^{2+}\) form copper’.

(c) Show by calculation:
• Mass of water removed = \(12.5 – 8.0 = 4.5 \text{ g}\)
• Moles of \(CuSO_4\) = \(\frac{8.0}{159.5} = 0.0502 \text{ mol}\) (accept 0.05 mol)
• Moles of \(H_2O\) = \(\frac{4.5}{18} = 0.25 \text{ mol}\)
• Ratio \(\frac{\text{mol } H_2O}{\text{mol } CuSO_4} = \frac{0.25}{0.05} = 5\)
Therefore, the formula is \(CuSO_4 \cdot 5H_2O\).
ACCEPT alternative methods which show that the answer is 5.

Question 4

A student investigates the reaction between sodium hydroxide solution and dilute sulfuric acid. He does a titration to find the concentration of the sulfuric acid.

This is his plan for the titration. There are some mistakes and omissions in his plan.

rinse a conical flask with the sodium hydroxide solution
use a measuring cylinder to measure out 25 cm\(^3\) of the sodium hydroxide solution and add it to the conical flask
add a few drops of methyl orange indicator to the conical flask
rinse a burette with water and then fill it with the sulfuric acid
add the acid from the burette to the conical flask until the indicator changes colour at the end-point of the titration
record the final burette reading

(a) Give the colour change of the methyl orange indicator at the end-point.

(b) Describe four changes that the student could make to improve his plan.

(c) The student then does the titration correctly. He finds that \(16.70 \text{ cm}^3\) of the dilute sulfuric acid neutralises \(25.0 \text{ cm}^3\) of sodium hydroxide solution of concentration \(0.200 \text{ mol/dm}^3\).

The equation for the reaction is:

\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

Calculate the concentration, in \(\text{mol/dm}^3\), of the sulfuric acid.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 2(f): Acids, alkalis and titrations — parts (a), (b) plan changes
• 2(g): Acids, bases and salt preparations — parts (a), (b) plan changes
• 1(e): Chemical formulae, equations and calculations — part (c)
• 1.34C: Calculations involving amount, volume, and concentration — part (c)

▶️ Answer/Explanation

Solution

(a) from yellow to orange
Note: ACCEPT ‘from red to orange’ or ‘from pink to orange’. 1 mark is awarded for two correct colours in the wrong order.

(b) Description that makes reference to any four of the following points:

He should not rinse the flask with sodium hydroxide solution (or he should rinse it with water).
He should use a pipette (and pipette filler) to measure out the sodium hydroxide solution instead of a measuring cylinder.
He should not rinse the burette with water (or he should rinse it with the sulfuric acid).
He should record the initial burette reading.
He should place a white tile (or white paper) under the flask.
He should swirl the flask whilst adding the acid.
He should add the acid dropwise near the end-point.
He should repeat the titration to obtain an average/concordant results.

(c) Calculation of the concentration of sulfuric acid:

Step 1: Calculate moles of NaOH
\( n(\text{NaOH}) = \text{volume} \times \text{concentration} \)
\( = 0.0250 \, \text{dm}^3 \times 0.200 \, \text{mol dm}^{-3} \)
\( = 0.00500 \, \text{mol} \)

Step 2: Use the stoichiometry to find moles of H\(_2\)SO\(_4\)
From the equation: \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
Molar ratio \(\text{NaOH} : \text{H}_2\text{SO}_4 = 2 : 1\)
\( n(\text{H}_2\text{SO}_4) = \frac{0.00500}{2} = 0.00250 \, \text{mol} \)

Step 3: Calculate concentration of H\(_2\)SO\(_4\)
\( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \)
\( = \frac{0.00250 \, \text{mol}}{0.01670 \, \text{dm}^3} \)
\( = 0.1497 \, \text{mol dm}^{-3} \)

Final Answer: \( 0.150 \, \text{mol/dm}^3 \) (to 3 significant figures).

Question 5

Oxygen can be prepared from hydrogen peroxide using a catalyst.

(a) Which is a correct statement about oxygen?

A) it burns with a squeaky pop
B) it relights a glowing splint
C) it turns blue litmus red
D) it turns limewater milky

(b) Explain how a catalyst increases the rate of a reaction.

(c) The equation for the preparation of oxygen from hydrogen peroxide is:

\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]

This equation can also be written using displayed formulae to show all the covalent bonds in the molecules.

\[ 2H—O—O—H \rightarrow 2H—O—H + O=O \]

The table gives the bond energies for these bonds.

Bond	H—O	O—O	O=O
Bond energy in kJ/mol	463	143	498

(i) Use the values in the table to calculate the enthalpy change, \(\Delta H\), for the reaction. Include a sign in your answer.

(ii) Complete the energy level diagram to show the position of the products and the enthalpy change, \(\Delta H\), for the reaction.

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 2(h): Chemical tests — part (a)
• 3(b): Rates of reaction — part (b)
• 3(a): Energetics — parts (c)(i), (c)(ii)
• 1(g): Covalent bonding — part (c) context

▶️ Answer/Explanation

Solution

(a) B (it relights a glowing splint)
A is incorrect as this is the test for hydrogen.
C is incorrect as oxygen is not an acidic gas.
D is incorrect as this is the test for carbon dioxide.

(b) An explanation that links the following two points:
• provides an alternative pathway (M1)
• with a lower activation energy (M2)
ACCEPT: more collisions with energy greater than the activation energy.
ALLOW: lowers the energy needed to start the reaction.

(c)(i) \(\Delta H = -212 \text{ kJ/mol}\)
Working:
Bonds broken (energy in):
In \(2H—O—O—H\): 4 × H—O bonds + 2 × O—O bonds = (4 × 463) + (2 × 143) = 1852 + 286 = 2138 kJ
Bonds formed (energy out):
In \(2H—O—H\) + \(O=O\): 4 × H—O bonds + 1 × O=O bond = (4 × 463) + 498 = 1852 + 498 = 2350 kJ
\(\Delta H\) = Energy in – Energy out = 2138 – 2350 = -212 kJ
IGNORE any signs in M1 and M2. -212 with or without working scores 3 marks.

(c)(ii)

The energy level diagram should show:
• The reactants (\(2H_2O_2\)) at a higher energy level.
• The products (\(2H_2O + O_2\)) at a lower energy level.
• A downward arrow labelled \(\Delta H\) from the reactants’ level to the products’ level, indicating an exothermic reaction.
• Correct labelling of both levels.
ACCEPT: double-headed arrow or arrow pointing from reactants to products.
REJECT: arrow pointing from products to reactants.
IGNORE: any attempts at including activation energy.

Question 6

Ethanol, C₂H₅OH, can be manufactured from ethene and steam using a phosphoric acid catalyst.

(a) (i) State the temperature and pressure used in this manufacturing process.
(ii) Draw the displayed formula of ethanol.

(b) Ethanol burns in a plentiful supply of air to form carbon dioxide and water.

(i) Give the chemical equation for this reaction.
(ii) When the air supply is limited, incomplete combustion occurs and carbon monoxide forms. State why carbon monoxide is poisonous to humans.

(c) When ethanol reacts with ethanoic acid, an ester forms. Give the name of this ester.

(d) Butanedioic acid and ethanediol react together to form a polyester and water.

(i) Give the name of this type of polymerisation.

(ii) Complete the equation. Show only one repeat unit of the polyester.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 4(e) Alcohols — parts (a), (b)
• 4(g) Esters — part (c)
• 4(h) Synthetic polymers — parts (d)(i), (d)(ii)
• 1(e) Chemical formulae, equations and calculations — part (b)(i)

▶️ Answer/Explanation

Solution

(a)(i)
Temperature: 300 °C (accept any temperature in the range 250–350 °C)
Pressure: 60–70 atm (accept any pressure in the range 60–70 atm)

(a)(ii)
Displayed formula of ethanol (C₂H₅OH):
(Structure: H₃C—CH₂—O—H, with all bonds shown)

(b)(i)
Balanced chemical equation for complete combustion:
\[ \ce{C2H5OH + 3O2 -> 2CO2 + 3H2O} \]

(b)(ii)
Carbon monoxide is poisonous because it prevents blood from carrying oxygen / reduces the capacity of haemoglobin in red blood cells to transport oxygen (by forming carboxyhaemoglobin).

(c)
Name of ester: ethyl ethanoate (accept ethyl acetate).

(d)(i)
Type of polymerisation: condensation polymerisation.

(d)(ii)
Equation showing one repeat unit of the polyester (polyester from butanedioic acid and ethanediol):
\[ \ce{->[{-}O-C-CH2CH2-C-O-CH2CH2-O-} + 2H2O] \]
OR with correct ester link structure:
\[ \ce{->[\text{repeat unit}][\underset{\text{O}}{\parallel}\text{C}-\text{CH}_2\text{CH}_2-\text{C}-\text{O}-\text{CH}_2\text{CH}_2-\text{O}-] + 2H2O} \]

Question 7

This question is about some Group 2 elements and their compounds.

(a) Calcium reacts with water to produce calcium hydroxide and hydrogen gas.

(i) Give the word equation for this reaction.
(ii) State two observations that would be made during this reaction.

(b) (i) Describe how a pure, dry sample of the insoluble salt, barium sulfate, could be made from the two solids sodium sulfate and barium chloride.
(ii) Give an ionic equation for the reaction that occurs. Include state symbols in your equation.

(c) When magnesium nitrate is heated, magnesium oxide, nitrogen dioxide and oxygen form.

The equation for the reaction is:

\[ 2\text{Mg(NO}_3\text{)}_2\text{(s)} \rightarrow 2\text{MgO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)} \]

(i) What is the name for this type of reaction?

A) addition
B) combustion
C) decomposition
D) neutralisation

(ii) Calculate the total volume, in \(\text{dm}^3\), of gas produced at rtp when 7.7 g of magnesium nitrate completely reacts.

[Assume that the molar volume of a gas at rtp is 24 \(\text{dm}^3\)]
\[ [M_r \text{ of Mg(NO}_3\text{)}_2 = 148] \]

Give your answer to two significant figures.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 1(e): Chemical formulae, equations and calculations — parts (a)(i), (c)(ii)
• 2(a): Group 1 (alkali metals) — part (a)(i), (a)(ii) [by analogy with Group 2 reactivity]
• 2(d): Reactivity series — part (a)(ii) [reactivity with water]
• 2(g): Acids, bases and salt preparations — part (b)(i)
• 3(a): Energetics — part (a)(ii) [exothermic reaction]
• 1(c): Atomic structure / 1(e): Chemical formulae — part (b)(ii) [ionic equations]
• 3(a): Energetics / 3(c): Reversible reactions and equilibria — part (c)(i) [decomposition]

▶️ Answer/Explanation

Solution

(a)(i) calcium + water → calcium hydroxide + hydrogen

(a)(ii) Any two from:
• Effervescence/fizzing/bubbles
• Calcium/metal/solid disappears/becomes smaller/dissolves
• Test tube/beaker feels warm/hot (temperature increases)

(b)(i) A description that makes reference to the following five points:
1. Dissolve each of the solids in water/make a solution of each.
2. Mix/add the two solutions together.
3. Filter the mixture.
4. Wash the precipitate/barium sulfate/residue with water.
5. Suitable method of drying the solid (e.g., dry in a warm oven, between filter papers, leave to dry).

(b)(ii) \[ \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} \]

(c)(i) C (decomposition)
A is incorrect as it is not an addition reaction.
B is incorrect as oxygen is not a reactant.
D is incorrect as nothing is neutralised here.

(c)(ii)
Example calculation:
1. \( n(\text{magnesium nitrate}) = \frac{7.7}{148} = 0.052 \text{ mol} \)
2. From equation: \( 2 \text{ mol Mg(NO}_3)_2 \) produces \( 5 \text{ mol gas} \)
\( n(\text{gas}) = 0.052 \times \frac{5}{2} = 0.13 \text{ mol} \)
3. Volume of gas \( = 0.13 \times 24 = 3.12 \text{ dm}^3 \)
4. To two significant figures: \( 3.1 \text{ dm}^3 \)

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