Chemistry 5070 / 21 — May/June 2025

(b)

in aqueous solution, reacts with an excess of aqueous ammonia to give a dark blue solution

______

(c)

reacts with warm aqueous sodium hydroxide and aluminium foil to give a gas that turns damp red litmus paper blue

______

(d)

has an aqueous solution that is used to test for an oxidising agent

______

(e)

has an aqueous solution that reacts with copper metal.

______

Q29MMedium-EasyAtoms, Elements and CompoundsElectrochemistry

A concentrated aqueous solution of copper(II) bromide is electrolysed using graphite electrodes.

(a)

Graphite has good electrical conductivity.

3M

(i)

Explain why graphite has good electrical conductivity.

Use ideas about structure and bonding.

______

DifficultyMedium-Easy

Worked solution

Answer

Each carbon atom forms three covalent bonds using three of its outer-shell electrons. The fourth outer-shell electron from each carbon is delocalised (mobile) between the layers. These mobile electrons can move through the structure and carry electrical charge, so graphite conducts electricity.

Final answer

Each carbon uses three outer-shell electrons in covalent bonds; the fourth electron is delocalised and mobile

Detailed explanation

Walkthrough

Recall that graphite is a giant covalent substance with layers. In each layer, every carbon atom is covalently bonded to three other carbons, so it uses three of its four outer-shell electrons. The fourth electron is not held by one atom; it is delocalised over the layer and can move. When a voltage is applied these mobile electrons drift, so graphite conducts electricity. Here, carbon has four outer-shell electrons (configuration 2,4); using three for bonds leaves one delocalised electron per carbon.

Key Takeaways

Graphite is special among covalent substances because it can conduct electricity.
The delocalised fourth electron from each carbon carries the current.
Conduction in graphite is electronic, not ionic.

Common Mistakes

Saying that 'all' electrons are free – only one electron per carbon is delocalised.
Confusing graphite with diamond, in which all four electrons are used in covalent bonds and there are no delocalised electrons.
Writing that graphite conducts because it contains ions.

Things to Be Careful About

The two marks are earned by two ideas: (1) each carbon uses three outer-shell electrons in covalent bonds, and (2) there are mobile (delocalised) electrons between the layers. Use the words 'delocalised' or 'mobile'. Do not mention layer sliding or high melting point, as these are not needed for this mark.

Techniques used

describe the layered giant covalent structure of graphitestate that each carbon uses three outer-shell electrons in covalent bondsidentify the fourth electron as delocalised and mobile

(ii)

State one other property of graphite that makes it suitable for use as an electrode during electrolysis.

______

(b)

Predict the products of the electrolysis of concentrated aqueous copper(II) bromide with graphite electrodes.

product at anode = ______
product at cathode = ______

DifficultyMedium-Easy

Worked solution

Answer

product at anode = bromine ( $\text{Br}_2$ )

product at cathode = copper ( $\text{Cu}$ )

Final answer

anode: bromine (Br2); cathode: copper (Cu)

Detailed explanation

Walkthrough

Copper(II) bromide solution contains $\text{Cu}^{2+}$ , $\text{Br}^-$ , $\text{H}^+$ and $\text{OH}^-$ ions. At the anode, bromide ions are discharged in preference to hydroxide ions when the halide solution is concentrated. The half-equation is $2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-$ , so bromine gas is produced. At the cathode, copper(II) ions are below hydrogen in the reactivity series, so they are discharged in preference to hydrogen ions: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ . Copper metal is deposited on the cathode. Graphite is inert, so it is not a product.

Key Takeaways

In aqueous electrolysis, the ion that is discharged is the one that is easier to oxidise or reduce (or the one present in high concentration).
In a concentrated halide solution, halide ions are discharged at the anode rather than hydroxide ions.
A metal ion less reactive than hydrogen is discharged at the cathode in preference to hydrogen ions.

Common Mistakes

Writing oxygen at the anode because the solution contains water – in concentrated copper(II) bromide, bromide ions are discharged first.
Writing hydrogen at the cathode – copper(II) ions are less reactive than hydrogen, so copper is deposited.
Forgetting that graphite is inert and so does not react or dissolve.

Things to Be Careful About

Name the products as requested: bromine at the anode and copper at the cathode. It is helpful to know the half-equations even though the question only asks for the product names, because they show why these ions are discharged.

Techniques used

identify the ions present in concentrated copper(II) bromide solutionrecall that bromide ions are discharged at the anode in concentrated halide solutionrecall that copper(II) ions are discharged at the cathode in preference to hydrogen ions

(c)

Dilute sulfuric acid is electrolysed using graphite electrodes to form oxygen and hydrogen.

Construct the ionic half-equation for the reaction at each electrode.

reaction at anode = ______
reaction at cathode = ______

DifficultyMedium-Easy

Worked solution

Answer

anode: $4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-$

cathode: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$

Final answer

anode: 4OH- -> O2 + 2H2O + 4e-; cathode: 2H+ + 2e- -> H2

Detailed explanation

Walkthrough

In dilute sulfuric acid the relevant ions from water are $\text{H}^+$ and $\text{OH}^-$ . At the anode, hydroxide ions are oxidised: four $\text{OH}^-$ ions give one $\text{O}_2$ molecule and two water molecules, releasing four electrons. The equation is balanced for oxygen (4 O atoms on each side), for hydrogen (4 H atoms on each side) and for charge (a total of 4− on both sides). At the cathode, hydrogen ions are reduced: two $\text{H}^+$ ions take two electrons to form one $\text{H}_2$ molecule. This equation balances atoms and charge as well.

Key Takeaways

Half-equations must be balanced for atoms and for charge.
Oxidation at the anode releases electrons; reduction at the cathode uses electrons.
In dilute sulfuric acid, the anode reaction oxidises hydroxide ions from water to oxygen, and the cathode reaction reduces hydrogen ions to hydrogen gas.

Common Mistakes

Unbalanced equations, e.g. one $\text{OH}^-$ giving $\text{O}_2$ , or ignoring water molecules.
Omitting the electrons from the half-equation.
Reversing the equations or putting the anode reaction at the cathode.
Writing the alternative 2H2O -> O2 + 4H+ + 4e- at the anode; the mark scheme allows 4OH- -> O2 + 2H2O + 4e- or 4OH- - 4e- -> O2 + 2H2O.

Things to Be Careful About

Check that each half-equation has equal numbers of atoms and equal total charge on both sides. The cathode equation must be $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$ because hydrogen is diatomic.

Techniques used

write the oxidation half-equation for hydroxide ions at the anodebalance atoms and charge in the half-equationwrite the reduction half-equation for hydrogen ions at the cathode

(d)

Hydrogen and oxygen are used in a fuel cell to produce electricity.

2M

(i)

Name the only chemical product formed in a fuel cell.

______

(ii)

Describe one disadvantage of using hydrogen–oxygen fuel cells in vehicles compared to gasoline or petrol engines.

______

Q39MMediumOrganic ChemistryChemical EnergeticsAtoms, Elements and Compounds

The equation for the reaction between methane and chlorine is shown in Fig. 3.1.

Fig. 3.1

(a)

State one condition for this reaction.

______

(b)

Explain why this reaction is an example of substitution.

______

(c)

Table 3.1 shows some bond energies.

Table 3.1

bond	bond energy in $\text{kJ/mol}$
C—H	410
C—Cl	340
Cl—Cl	242
H—Cl	431

Show by calculation that the enthalpy change of the reaction between methane and chlorine, $\Delta H$ , is $-119\text{ kJ/mol}$ .

DifficultyMedium-Easy

Worked solution

Working

Bonds broken:

$1 \times (\text{C—H}) = 410\text{ kJ / mol}$
$1 \times (\text{Cl—Cl}) = 242\text{ kJ / mol}$

\text{Energy absorbed (bond breaking)} = 410 + 242 = +652\text{ kJ / mol}

Bonds formed:

$1 \times (\text{C—Cl}) = 340\text{ kJ / mol}$
$1 \times (\text{H—Cl}) = 431\text{ kJ / mol}$

\text{Energy released (bond forming)} = 340 + 431 = 771\text{ kJ / mol}

Enthalpy change of reaction:

\begin{aligned} \Delta H &= \text{Energy absorbed} - \text{Energy released} \\ &= 652 - 771 \\ &= -119\text{ kJ / mol} \end{aligned}

(Alternatively, breaking all $4 \times \text{C—H}$ bonds gives $4(410) + 242 = 1882\text{ kJ / mol}$ ; forming $3 \times \text{C—H}$ , $1 \times \text{C—Cl}$ , $1 \times \text{H—Cl}$ gives $3(410) + 340 + 431 = 2001\text{ kJ / mol}$ ; $\Delta H = 1882 - 2001 = -119\text{ kJ / mol}$ )

Answer

$-119\text{ kJ / mol}$

Final answer

-119 kJ / mol

Detailed explanation

Walkthrough

To calculate the overall enthalpy change ( $\Delta H$ ) from bond energies:

Identify bonds broken (endothermic process, requires energy input):
Only one $\text{C—H}$ bond and one $\text{Cl—Cl}$ bond are broken in the reactants. $\text{Energy for bond breaking} = 410 + 242 = 652\text{ kJ / mol}$
Identify bonds formed (exothermic process, releases energy):
One $\text{C—Cl}$ bond and one $\text{H—Cl}$ bond are formed in the products. $\text{Energy for bond forming} = 340 + 431 = 771\text{ kJ / mol}$
Calculate overall $\Delta H$ : $\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed}) = 652 - 771 = -119\text{ kJ / mol}$

Because bond making releases more energy than bond breaking requires, the reaction is exothermic ( $\Delta H < 0$ ).

Key Takeaways

Bond breaking is endothermic ( $+ \Delta H$ ).
Bond forming is exothermic ( $- \Delta H$ ).
$\Delta H = \text{Total energy in to break bonds} - \text{Total energy out when forming bonds}$ .

Common Mistakes

Inverting the subtraction ( $771 - 652 = +119$ ) and giving an endothermic value.
Forgetting to include the negative sign in the final answer.

Things to Be Careful About

You can either calculate the net bonds changed (1 $\text{C—H}$ broken, 1 $\text{C—Cl}$ formed) or break all 4 $\text{C—H}$ bonds and form 3 $\text{C—H}$ bonds back. Both methods give the same correct result.

Techniques used

calculate energy required for bond breakingcalculate energy released during bond formingapply the enthalpy change formula

(d)

Complete the reaction pathway diagram in Fig. 3.2 for the reaction between methane and chlorine.

Label the:

reactants
products
enthalpy change of the reaction, $\Delta H$
activation energy, $E_\text{a}$

Fig. 3.2

DifficultyMedium

Worked solution

Answer

Reactants line drawn on the left at a higher energy level than the products line on the right.
A curved energy hump drawn rising above the reactants and falling to the products.
Activation energy, $E_\text{a}$ , shown as an upward arrow from the reactant energy level to the peak of the curve.
Enthalpy change, $\Delta H$ , shown as a downward arrow from the reactant energy level to the product energy level.

Final answer

Exothermic reaction pathway diagram with reactants above products, activation energy Ea from reactants to peak, and enthalpy change ΔH from reactants to products

Detailed explanation

Walkthrough

Since $\Delta H = -119\text{ kJ / mol}$ , the reaction is exothermic.

Energy levels of reactants and products: In an exothermic reaction, the products have less chemical energy than the reactants. Therefore, the horizontal line for the reactants ( $\text{CH}_4 + \text{Cl}_2$ ) is drawn at a higher level than the horizontal line for the products ( $\text{CH}_3\text{Cl} + \text{HCl}$ ).
Reaction pathway curve: The curve rises from the reactant level up to a peak (representing the transition state) and then slopes down to the product level.
Activation energy ( $E_\text{a}$ ): This is the minimum energy particles must have to react. It is represented by a vertical arrow starting precisely at the reactant energy level and pointing up to the maximum height of the curve.
Enthalpy change ( $\Delta H$ ): This is the difference between the reactant and product energy levels. For an exothermic reaction, it is represented by a downward vertical arrow pointing from the reactant level down to the product level.

Key Takeaways

For exothermic reactions, reactants are higher in energy than products.
$E_\text{a}$ is always measured from the reactants level up to the peak of the hump.
$\Delta H$ is measured from the reactants level to the products level (pointing down for exothermic).

Common Mistakes

Drawing $E_\text{a}$ starting from the products level or from the bottom axis (it must start from the reactants level).
Drawing $\Delta H$ pointing upwards or from the peak of the curve.
Drawing an endothermic profile (products higher than reactants) despite the negative $\Delta H$ calculated in part (c).

Things to Be Careful About

Ensure arrowheads clearly indicate the correct direction: $E_\text{a}$ points up, $\Delta H$ points down.

Techniques used

draw an exothermic reaction pathway diagramlabel activation energy with an upward arrowlabel enthalpy change with a downward arrow

(e)

Draw a dot-and-cross diagram to show the electronic configuration in a molecule of methane.

Show only the outer shell electrons.

Q412MMedium-EasyOrganic ChemistryStoichiometryAcids, Bases and Salts

Ethanol, $\text{C}_2\text{H}_5\text{OH}$ , is a member of the homologous series of alcohols.

(a)

Give the general formula of the homologous series of alcohols.

______

(b)

Members of a homologous series have the same general formula and share similar chemical properties.

State two other general characteristics of a homologous series.

______
______

(c)

The equation for the reaction between ethanol and sodium is shown.

2\text{Na} + 2\text{C}_2\text{H}_5\text{OH} \rightarrow \text{H}_2 + 2\text{C}_2\text{H}_5\text{ONa}

A sample of $1.35\text{ g}$ of sodium is added to excess ethanol.

4M

(i)

Calculate the volume of hydrogen formed measured at room temperature and pressure.

Give your answer to two significant figures.

volume = ______ $\text{dm}^3$

DifficultyMedium-Easy

Worked solution

Working

Calculate the moles of sodium ( $A_r = 23$ ):

\text{moles of Na} = \frac{1.35}{23} = 0.05870\text{ mol}

From the balanced equation, $2\text{Na} \rightarrow \text{H}_2$ , so the mole ratio of $\text{Na} : \text{H}_2$ is $2 : 1$ :

\text{moles of }\text{H}_2 = \frac{0.05870}{2} = 0.02935\text{ mol}

Calculate the volume of $\text{H}_2$ at r.t.p. ( $24\text{ dm}^3\text{ / mol}$ ):

\text{volume of }\text{H}_2 = 0.02935 \times 24 = 0.7044\text{ dm}^3

Rounding to two significant figures gives $0.70\text{ dm}^3$ .

Answer

$0.70\text{ dm}^3$

Final answer

0.70 dm3

Detailed explanation

Walkthrough

Find the moles of sodium added: $n(\text{Na}) = \frac{\text{mass}}{A_r} = \frac{1.35\text{ g}}{23\text{ g / mol}} = 0.0587\text{ mol}$
Use the stoichiometry from the equation:
According to the equation $2\text{Na} + 2\text{C}_2\text{H}_5\text{OH} \rightarrow \text{H}_2 + 2\text{C}_2\text{H}_5\text{ONa}$ , $2\text{ moles}$ of $\text{Na}$ produce $1\text{ mole}$ of $\text{H}_2$ . $n(\text{H}_2) = \frac{0.0587}{2} = 0.02935\text{ mol}$
Calculate the volume of gas produced at r.t.p.:
Molar gas volume at r.t.p. is $24\text{ dm}^3\text{ / mol}$ . $V(\text{H}_2) = n \times 24\text{ dm}^3\text{ / mol} = 0.02935 \times 24 = 0.7044\text{ dm}^3$
Apply significant figures:
The question specifies two significant figures, so $0.7044\text{ dm}^3$ rounds to $0.70\text{ dm}^3$ .

Key Takeaways

Molar gas volume at r.t.p. is always $24\text{ dm}^3\text{ / mol}$ .
Always follow stoichiometric ratios from the balanced chemical equation.
Trailing zeroes after the decimal point are significant (e.g., $0.70$ has 2 s.f.).

Common Mistakes

Forgetting to divide the moles of $\text{Na}$ by 2 to get the moles of $\text{H}_2$ .
Forgetting to round the final answer to the requested 2 significant figures, or writing 0.7 instead of 0.70.

Things to Be Careful About

Ensure you use $A_r(\text{Na}) = 23$ from the Periodic Table.

Techniques used

convert mass to moles using Arapply stoichiometric mole ratioscalculate gas volume at r.t.p.round to specified significant figures

(ii)

Water is added to the reaction mixture to make an aqueous solution.

A few drops of litmus are then added. The litmus changes colour to blue.

Suggest the name of the ion present in the aqueous solution responsible for the colour change.

______

(d)

State two uses for ethanol.

______
______

(e)

Describe the manufacture of ethanol from ethene.

Include the other reactant and the conditions for the manufacture.

______

Q512MMedium-EasyOrganic ChemistryChemistry of the EnvironmentChemical Reactions

Vehicles that use petrol as a fuel produce several air pollutants.

(a)

Petrol is a mixture of hydrocarbons which includes octane, $\text{C}_8\text{H}_{18}$ .

Explain why octane is a hydrocarbon.

______

(b)

Two of the air pollutants produced are carbon monoxide and nitrogen monoxide.

4M

(i)

Explain how carbon monoxide, $\text{CO}$ , is formed in a petrol engine.

Include a symbol equation.

______

(ii)

Explain how nitrogen monoxide, $\text{NO}$ , is formed in a petrol engine.

Include a symbol equation.

______

(c)

A catalytic converter removes most of the $\text{CO}$ and $\text{NO}$ formed in a petrol engine.

2\text{CO}(\text{g}) + 2\text{NO}(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + \text{N}_2(\text{g})

7M

(i)

Explain why this reaction involves both oxidation and reduction.

______

(ii)

The reaction is catalysed using platinum metal.

Explain how a catalyst increases the rate of a reaction.

______

(iii)

State and explain the effect of increasing the temperature on the rate of this reaction.

______

(iv)

State and explain the effect of decreasing the pressure on the rate of this reaction.

______

Q616MMediumStates of MatterStoichiometryChemical ReactionsAcids, Bases and Salts

Chlorine, $\text{Cl}_2$ , is in Group VII of the Periodic Table.

(a)

The melting point of chlorine is $-101\,^\circ\text{C}$ and the boiling point is $-35\,^\circ\text{C}$ .

4M

(i)

Explain why chlorine is a liquid at $-50\,^\circ\text{C}$ .

______

(ii)

Describe the arrangement and motion of chlorine molecules at $-50\,^\circ\text{C}$ .

______

(b)

A sample of chlorine gas contains $1.204 \times 10^{20}$ molecules.

One mole of chlorine gas contains $6.02 \times 10^{23}$ molecules.

Calculate the mass of this sample of chlorine gas.

mass of chlorine = ______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Working

Calculate the number of moles of chlorine molecules, $\text{Cl}_2$ :

\text{moles of } \text{Cl}_2 = \frac{1.204 \times 10^{20}}{6.02 \times 10^{23}} = 2.00 \times 10^{-4}\text{ mol} = 0.000200\text{ mol}

Calculate the relative molecular mass ( $M_r$ ) of $\text{Cl}_2$ :

M_r(\text{Cl}_2) = 2 \times 35.5 = 71.0

Calculate the mass of the sample:

\text{mass} = \text{moles} \times M_r = 2.00 \times 10^{-4} \times 71.0 = 0.0142\text{ g}

Answer

$0.0142\text{ g}$

Final answer

0.0142 g

Detailed explanation

Walkthrough

First find the moles of chlorine gas using Avogadro's constant ( $6.02 \times 10^{23}$ particles/mol): $\text{moles} = \frac{\text{number of molecules}}{6.02 \times 10^{23}} = \frac{1.204 \times 10^{20}}{6.02 \times 10^{23}} = 2.00 \times 10^{-4}\text{ mol}$
Chlorine gas exists as diatomic molecules, $\text{Cl}_2$ . The relative atomic mass of chlorine is $35.5$ , so the relative molecular mass $M_r$ of $\text{Cl}_2$ is $2 \times 35.5 = 71.0$ .
Use $\text{mass} = \text{moles} \times M_r$ : $\text{mass} = 2.00 \times 10^{-4} \times 71.0 = 0.0142\text{ g}$

Key Takeaways

$\text{moles} = \frac{\text{number of particles}}{\text{Avogadro constant}}$
$\text{mass} = \text{moles} \times M_r$
Chlorine is diatomic ( $ext{Cl}_2$ ), so its molar mass is $71\text{ g / mol}$ , not $35.5\text{ g / mol}$ .

Common Mistakes

Using $35.5$ instead of $71$ for the molar mass of chlorine gas molecules.
Errors in handling powers of 10 on a calculator.

Things to Be Careful About

Ensure you check whether the question refers to chlorine atoms or chlorine molecules.

Techniques used

calculate moles from number of particles and Avogadro constantcalculate molar mass of a diatomic gasconvert moles to mass

(c)

The ionic equation for the reaction of chlorine with cold dilute aqueous sodium hydroxide is shown.

\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{Cl}^- + \text{H}_2\text{O}

3M

(i)

State the oxidation number of chlorine in $\text{Cl}_2$ and in $\text{Cl}^-$ .

$\text{Cl}_2$ = ______
$\text{Cl}^-$ = ______

(ii)

During the reaction chlorine is reduced.

Explain why, using ideas about electrons.

______

(d)

Chlorine reacts with cold water to form an equilibrium mixture containing the acids $\text{HCl}(\text{aq})$ and $\text{HOCl}(\text{aq})$ .

The forward reaction releases thermal energy into the surroundings.

\text{Cl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{HCl}(\text{aq}) + \text{HOCl}(\text{aq})

4M

(i)

The temperature of the equilibrium mixture is increased.

State and explain what happens to the acidity of the equilibrium mixture.

statement = ______
explanation = ______

(ii)

$\text{HCl}(\text{aq})$ is a strong acid and $\text{HOCl}(\text{aq})$ is a weak acid.

Describe the difference between a strong acid and a weak acid.

______

(e)

A chloride of iron contains $34.5\%$ iron by mass.

Calculate the empirical formula of this chloride.

Show your working.

empirical formula = ______

DifficultyMedium-Easy

Worked solution

Working

Percentage of chlorine:

100\% - 34.5\% = 65.5\%

Calculate the number of moles of each element in $100\text{ g}$ of the compound ( $A_r(\text{Fe}) = 56$ , $A_r(\text{Cl}) = 35.5$ ):

\text{moles of Fe} = \frac{34.5}{56} = 0.6161\text{ mol}

\text{moles of Cl} = \frac{65.5}{35.5} = 1.8451\text{ mol}

Divide each by the smallest mole value ( $0.6161$ ):

\text{ratio of Fe} = \frac{0.6161}{0.6161} = 1

\text{ratio of Cl} = \frac{1.8451}{0.6161} = 2.995 \approx 3

Simplest whole number ratio is $\text{Fe} : \text{Cl} = 1 : 3$ .

Answer

$\text{FeCl}_3$

Final answer

FeCl3

Detailed explanation

Walkthrough

Find percentage of chlorine: Since the compound contains only iron and chlorine, $\%\text{Cl} = 100\% - 34.5\% = 65.5\%$ .
Convert mass percentages to moles: Divide the mass of each element by its relative atomic mass ( $A_r$ $A_{r}$ ):
- $\text{Moles of Fe} = \frac{34.5}{56} = 0.616\text{ mol}$
- $\text{Moles of Cl} = \frac{65.5}{35.5} = 1.85\text{ mol}$
Find the simplest whole-number ratio: Divide both mole amounts by the smaller value ( $0.616$ $0.616$ ):
- $\text{Fe} = \frac{0.616}{0.616} = 1$
- $\text{Cl} = \frac{1.85}{0.616} = 3.00$
The empirical formula is therefore $\text{FeCl}_3$ .

Key Takeaways

To find empirical formula:
1. Find masses/percentages of all elements.
2. Divide each by its $A_r$ to find moles.
3. Divide all mole values by the smallest mole value to find the simplest ratio.

Common Mistakes

Dividing the mass of chlorine by $71$ ( $M_r$ of $\text{Cl}_2$ ) instead of its atomic mass $35.5$ ( $A_r$ of $\text{Cl}$ ). Empirical formula calculations always use atomic masses ( $A_r$ ).

Things to Be Careful About

Round only at the final step to avoid rounding errors.

Techniques used

determine mass percentage of remaining elementcalculate moles of each elementfind simplest whole number mole ratio

Q76MMedium-EasyMetalsStoichiometryAcids, Bases and Salts

Aluminium is used in the manufacture of aircraft and food containers.

Aluminium is resistant to corrosion by water and oxygen.

(a)

Give one other reason why aluminium is used in the manufacture of aircraft.

______

(b)

Explain why aluminium is resistant to corrosion by water and oxygen.

______

(c)

Aluminium metal reacts with hot dilute sulfuric acid to form hydrogen and aqueous aluminium sulfate as the only products.

Construct the symbol equation for this reaction.

Include state symbols.

______

DifficultyMedium-Easy

Worked solution

Answer

2\text{Al}(\text{s}) + 3\text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{Al}_2(\text{SO}_4)_3(\text{aq}) + 3\text{H}_2(\text{g})

Final answer

2Al(s) + 3H2SO4(aq) -> Al2(SO4)3(aq) + 3H2(g)

Detailed explanation

Walkthrough

Identify the formulae of reactants and products:
- Aluminium: $\text{Al}$
- Sulfuric acid: $\text{H}_2\text{SO}_4$
- Aluminium sulfate: formed from $\text{Al}^{3+}$ and $\text{SO}_4^{2-}$ ions, giving $\text{Al}_2(\text{SO}_4)_3$
- Hydrogen gas: $\text{H}_2$
Balance the equation:
- Two $\text{Al}$ atoms are needed on the left: $2\text{Al}$
- Three sulfate groups are needed on the right, so three $\text{H}_2\text{SO}_4$ molecules are required: $3\text{H}_2\text{SO}_4$
- The six hydrogen atoms form three $\text{H}_2$ molecules: $3\text{H}_2$
Add state symbols:
- Aluminium is solid: (s)
- Dilute sulfuric acid is aqueous: (aq)
- Aluminium sulfate solution is aqueous: (aq)
- Hydrogen is a gas: (g)

2\text{Al}(\text{s}) + 3\text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{Al}_2(\text{SO}_4)_3(\text{aq}) + 3\text{H}_2(\text{g})

Key Takeaways

A metal reacting with an acid produces a salt and hydrogen gas.
Ensure charges balance when writing formulae for polyatomic salts: $\text{Al}^{3+}$ and $\text{SO}_4^{2-} \rightarrow \text{Al}_2(\text{SO}_4)_3$ .

Common Mistakes

Writing incorrect formula for aluminium sulfate, such as $\text{AlSO}_4$ .
Forgetting or writing incorrect state symbols (e.g. writing (l) instead of (aq) for acid or aqueous solution).

Things to Be Careful About

Double check the balancing numbers: $2, 3 \rightarrow 1, 3$ .

Techniques used

deduce chemical formulae for ionic compoundsbalance the symbol equationassign correct state symbols

(d)

Aluminium oxide reacts with sulfuric acid and with the alkali aqueous sodium hydroxide.

State the name of the type of oxide that reacts with both acids and alkalis.

______

Q811MMediumOrganic Chemistry

Fig. 8.1 shows the displayed formula of methylbut-2-enoate.

Fig. 8.1

(a)

Methylbut-2-enoate is an unsaturated ester.

3M

(i)

Explain why methylbut-2-enoate is unsaturated.

______

(ii)

Describe a chemical test to show that methylbut-2-enoate is unsaturated.

______

(b)

Methylbut-2-enoate is made by the reaction of an alcohol and a carboxylic acid in the presence of a catalyst.

3M

(i)

State the name of the type of catalyst used in this reaction.

______

(ii)

State the name of the alcohol used in this reaction.

______

(iii)

Draw the displayed formula of the carboxylic acid used in this reaction.

DifficultyMedium-Easy

Worked solution

Answer

The carboxylic acid is but-2-enoic acid. Its displayed formula is:

\begin{array}{c} \text{H} \quad \text{H} \quad \text{H} \quad \text{O} \\ | \quad \quad | \quad \quad | \quad \quad || \\ \text{H}-\text{C}-\text{C}=\text{C}-\text{C}-\text{O}-\text{H} \\ | \quad \quad | \quad \quad | \\ \text{H} \quad \quad \text{H} \quad \quad \text{H} \end{array}

(All single and double bonds explicitly drawn between every atom.)

Final answer

Displayed formula of but-2-enoic acid: H3C-CH=CH-COOH with all bonds shown

Detailed explanation

Walkthrough

The ester is methylbut-2-enoate. As established in part (b)(ii), the "methyl" part comes from methanol, and the "but-2-enoate" part comes from the carboxylic acid. Therefore, the acid is but-2-enoic acid.

But-2-enoic acid has:

A 4-carbon chain (but-).
A carboxyl group (-COOH) at carbon 1.
A carbon-carbon double bond starting at carbon 2 (between C2 and C3).

To draw the displayed formula, every atom must be shown, and every bond must be drawn as a line:

C1 is bonded to =O (double bond), -OH (single bond), and C2 (single bond).
C2 is bonded to C1 (single), C3 (double), and one H (single).
C3 is bonded to C2 (double), C4 (single), and one H (single).
C4 is bonded to C3 (single) and three H atoms (single bonds).

Key Takeaways

When converting an ester name to the parent carboxylic acid, replace the '-oate' suffix with '-oic acid' and the alkyl group with 'H' (to form the -OH of the carboxyl group). A displayed formula requires every single bond and atom to be explicitly drawn.

Common Mistakes

Drawing a structural formula (e.g., CH3CH=CHCOOH) instead of a displayed formula. The question asks for a displayed formula, so all C-H and C-C bonds must be shown as lines.
Placing the double bond in the wrong position. "but-2-enoic" means the double bond starts at C2, so it is between C2 and C3, not C1 and C2 (which would be but-2-enoic? No, C1 is the carboxyl carbon, so C2=C3 is correct).
Forgetting the hydrogen on the -OH group of the carboxyl group.

Things to Be Careful About

Count the carbons carefully. The carboxyl carbon is C1. The double bond is between C2 and C3. The methyl group is on C4. Ensure the valency of every carbon is exactly 4.

Techniques used

draw displayed formula of carboxylic acid from ester name

(c)

Methylbut-2-enoate is a monomer used to make an addition polymer.

4M

(i)

Draw the structure of this addition polymer.

Include at least two repeat units.

DifficultyMedium

Worked solution

Answer

The addition polymer has a 4-carbon backbone (derived from the C=C bond opening). Each repeat unit consists of two carbon atoms in the main chain. One carbon has a methyl group (-CH3) and a hydrogen; the other has a hydrogen and the ester group (-C(=O)OCH3). The structure must show at least two repeat units with continuation bonds extending from both ends.

Final answer

Polymer structure with two repeat units: -[CH(CH3)-CH(COOCH3)]- with continuation bonds

Detailed explanation

Walkthrough

Methylbut-2-enoate has the structure CH3-CH=CH-C(=O)OCH3. The carbon-carbon double bond is between C2 and C3 (numbering from the carbonyl carbon as C1).

In addition polymerisation, the C=C double bond breaks to form single bonds with adjacent monomer units. The atoms attached to the C=C carbons become substituents on the polymer backbone.

C2 is attached to: H and -C(=O)OCH3 (the ester group).
C3 is attached to: H and -CH3 (the methyl group).

The repeat unit is therefore: -[CH(COOCH3)-CH(CH3)]-. (The order of substituents on the two carbons can be swapped as long as they are on the correct carbons).

To draw at least two repeat units:

Draw a horizontal backbone of 4 carbon atoms (two repeat units of 2 carbons each).
On the first C (from left), attach -CH3 (top) and -H (bottom).
On the second C, attach -H (top) and -C(=O)OCH3 (bottom). The ester group should show the C=O double bond and C-O-CH3 single bonds.
Repeat this pattern for the next two carbons (third and fourth).
Add continuation bonds (lines extending outwards) from the first and last carbons of the drawn chain to indicate the polymer continues.

Key Takeaways

In addition polymerisation of an alkene, the C=C double bond opens up to form the main chain. The substituents on the original C=C carbons become side chains on the polymer backbone. Always show continuation bonds and at least two repeat units as requested.

Common Mistakes

Drawing the C=O double bond inside the backbone instead of as a side chain. The C=O is part of the ester substituent, not the backbone.
Forgetting the continuation bonds at the ends of the polymer chain.
Drawing only one repeat unit when the question asks for at least two.
Incorrectly placing the methyl and ester groups on the wrong carbons of the repeat unit.

Things to Be Careful About

The ester group (-COOCH3) must be drawn correctly with the C=O and C-O bonds. Do not break the ester group when drawing the polymer; it remains intact as a side chain. The backbone is formed solely from the carbons that were part of the C=C double bond.

Techniques used

draw addition polymer repeat unit from monomer

(ii)

A sample of $80\text{ g}$ of methylbut-2-enoate is reacted to make the addition polymer.

There is a $100\%$ yield.

State the mass of addition polymer made.

Explain your answer.

mass of addition polymer = ______ $\text{g}$
explanation = ______

(d)

Fig. 8.2 shows compound B.

Fig. 8.2

Explain why methylbut-2-enoate and compound B are a pair of structural isomers.

______