Chemistry 5070/22 — May/June 2024

Walkthrough

The Contact process is used to manufacture sulfur trioxide for sulfuric acid. A catalyst is needed to speed up this reaction without being used up. The substance from the list that is the catalyst in the Contact process is vanadium(V) oxide.

Key Takeaways

• A catalyst speeds up a reaction without being used up.
• The Contact process uses vanadium(V) oxide as its catalyst.
• Catalysts are specific to particular industrial processes.

Common Mistakes

• Choosing iron: iron is the catalyst for the Haber process, not the Contact process.
• Writing only 'vanadium' instead of 'vanadium(V) oxide'.

Things to Be Careful About

• The oxidation state (V) is part of the name; do not confuse this with other vanadium oxides.
• Do not confuse the Contact process with the Haber process.

Walkthrough

In the blast furnace, coke is burned to form carbon dioxide, which is then reduced to carbon monoxide. Carbon monoxide is the reducing agent because it removes oxygen from iron(III) oxide, leaving iron. Therefore the substance from the list is carbon monoxide.

Key Takeaways

• A reducing agent removes oxygen from another substance.
• Carbon monoxide is the active reducing agent in the blast furnace.
• The extraction of iron is a reduction process.

Common Mistakes

• Choosing iron because it is the product.
• Choosing carbon or coke instead of carbon monoxide.
• Saying carbon dioxide is the reducing agent.

Things to Be Careful About

• Carbon monoxide is toxic gas, but it is essential in the blast furnace.
• The reducing agent is carbon monoxide, not carbon dioxide.

Walkthrough

Anhydrous copper(II) sulfate has no water of crystallisation, so it is a white solid. When water is added, it becomes hydrated copper(II) sulfate, which is blue. This colour change is a standard test for water.

Key Takeaways

• Anhydrous means without water.
• Hydrated copper(II) sulfate is blue; anhydrous copper(II) sulfate is white.
• This reaction is used as a chemical test for water.

Common Mistakes

• Writing 'copper sulfate' without 'anhydrous'.
• Reversing the colour change: anhydrous is white, hydrated is blue.

Things to Be Careful About

• The word 'anhydrous' is essential because the clue says the white solid changes to blue.
• Heating blue hydrated copper(II) sulfate removes water and turns it white again.

Walkthrough

During treatment of the domestic water supply, chlorine is added to kill microbes. It acts as a disinfectant. Therefore the correct substance from the list is chlorine.

Key Takeaways

• Chlorine is used to kill microbes in water treatment.
• Water treatment includes filtration and disinfection.
• Chlorine is added in carefully controlled amounts.

Common Mistakes

• Choosing calcium carbonate or iron, which are not used to kill microbes.
• Thinking chlorine is only used in bleach.

Things to Be Careful About

• Chlorine is a toxic gas, but the amount added to water is small and safe.
• Do not choose another chemical that is not a disinfectant.

Walkthrough

Ethanoic acid has the molecular formula $\text{CH}_3\text{COOH}$, which is $\text{C}_2\text{H}_4\text{O}_2$. The empirical formula is the simplest whole-number ratio, so divide each subscript by 2: $\text{C}_1\text{H}_2\text{O}_1$, written as $\text{CH}_2\text{O}$. Methanoic acid is $\text{CH}_2\text{O}_2$, not $\text{CH}_2\text{O}$, and methanol is $\text{CH}_4\text{O}$. So the correct answer is ethanoic acid.

Key Takeaways

• The empirical formula is the simplest whole-number ratio of atoms in a compound.
• To find it, divide the molecular formula subscripts by the highest common factor.
• Different compounds can have the same empirical formula.

Common Mistakes

• Choosing methanoic acid because its name contains 'meth' and it has one carbon atom.
• Forgetting to reduce $\text{C}_2\text{H}_4\text{O}_2$ to $\text{CH}_2\text{O}$.
• Thinking empirical formula is always the same as molecular formula.

Things to Be Careful About

• Write the formula of ethanoic acid correctly before reducing it.
• Check each option's formula before deciding.
• The empirical formula does not show the full number of atoms in the molecule.

Walkthrough

The question states the reactants are aluminium carbide ($\text{Al}_4\text{C}_3$) and water ($\text{H}_2\text{O}$), and the products are methane ($\text{CH}_4$) and aluminium hydroxide. Aluminium hydroxide is formed from the $\text{Al}^{3+}$ ion and the $\text{OH}^-$ ion, giving the formula $\text{Al(OH)}_3$.

To balance the equation:

1. Aluminium: There are 4 Al atoms in $\text{Al}_4\text{C}_3$, so we need 4 $\text{Al(OH)}_3$ on the right.
2. Carbon: There are 3 C atoms in $\text{Al}_4\text{C}_3$, so we need 3 $\text{CH}_4$ on the right.
3. Hydrogen: On the right, we have $(4 \times 3) + (3 \times 4) = 12 + 12 = 24$ hydrogen atoms. Therefore, we need 12 $\text{H}_2\text{O}$ on the left to provide 24 H atoms.
4. Oxygen: On the left, 12 $\text{H}_2\text{O}$ gives 12 O atoms. On the right, 4 $\text{Al(OH)}_3$ gives $4 \times 3 = 12$ O atoms. The equation is balanced.

Key Takeaways

When constructing symbol equations, ensure you know the correct formulae for common ions (like $\text{Al}^{3+}$ and $\text{OH}^-$). Balancing often requires adjusting coefficients for water or oxygen last.

Common Mistakes

• Writing the wrong formula for aluminium hydroxide (e.g., $\text{AlOH}_3$ instead of $\text{Al(OH)}_3$).
• Forgetting to multiply the subscripts when balancing (e.g., counting 3 H in $\text{Al(OH)}_3$ instead of 3).

Things to Be Careful About

The mark scheme awards one mark for the correct formula of aluminium hydroxide and one for the balanced equation. Ensure the equation is fully balanced with integer coefficients.

Walkthrough

The concentration is given in $\text{mol / dm}^3$, so the volume must be in $\text{dm}^3$. Convert $35.0\text{ cm}^3$ to $\text{dm}^3$ by dividing by 1000: $35.0/1000 = 0.0350\text{ dm}^3$.

Then find the amount of hydrogen peroxide:

The balanced equation shows $2\text{ mol } \text{H}_2\text{O}_2$ gives $1\text{ mol } \text{O}_2$, so divide by 2 to get $0.004655\text{ mol } \text{O}_2$.

At r.t.p. one mole of any gas occupies $24\text{ dm}^3$. Multiply the amount of oxygen by 24:

Round to two significant figures: $0.11\text{ dm}^3$.

Key Takeaways

• Moles of solute = concentration $\times$ volume in $\text{dm}^3$.
• The balanced equation gives the mole ratio between reactants and products.
• Volume of gas at r.t.p. = moles $\times$ $24\text{ dm}^3\text{ / mol}$.
• Round only at the end to the required number of significant figures.

Common Mistakes

• Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ before using concentration.
• Using the mole ratio the wrong way round and multiplying by 2 instead of dividing.
• Quoting $0.11172\text{ dm}^3$ instead of rounding to two significant figures.
• Writing the final answer without the unit $\text{dm}^3$.

Things to Be Careful About

• The volume in concentration must be in $\text{dm}^3$.
• "Completely decomposed" means all the hydrogen peroxide reacts, so there is no limiting reactant to worry about.
• At r.t.p. the molar gas volume is $24\text{ dm}^3\text{ / mol}$, not $22.4\text{ dm}^3\text{ / mol}$ (that is s.t.p.).
• Two significant figures: the leading zero after the decimal point is not significant, so $0.11$ has two significant figures.

Walkthrough

For a reaction to happen, particles must collide with enough energy (at least the activation energy). Lowering the temperature removes energy from the particles, so they move more slowly and have less kinetic energy. This means they collide less often, and a smaller proportion of collisions have enough energy to react. The rate therefore decreases.

The mark scheme wants two ideas: (1) particles move slower / less kinetic energy, and (2) fewer successful collisions / fewer collisions with energy equal to or greater than activation energy.

Key Takeaways

• Temperature affects the average kinetic energy of particles.
• Rate depends on collision frequency and the fraction of collisions with energy equal to or greater than activation energy.
• Decreasing temperature lowers both, so rate decreases.

Common Mistakes

• Saying "the activation energy increases" — activation energy is a fixed property of the reaction; temperature changes how many particles have enough energy.
• Only saying "fewer collisions" without mentioning energy or success — the mark scheme accepts "fewer collisions" but the explanation is stronger with "successful" or "with energy equal to or greater than activation energy".
• Missing the direction: the question asks for the effect of decreasing temperature, so the answer must say rate decreases.

Things to Be Careful About

• Use "successful collisions" or "effective collisions" rather than just "collisions" to be safe.
• The mark scheme allows "particles move slower" or "particles have less kinetic energy"; either wording scores.
• Do not say "less frequent collisions" alone if the mark scheme wants both kinetic energy and collision success; here two marks are available.

Walkthrough

Concentration is the amount of solute in a given volume. Increasing the concentration of hydrogen peroxide puts more reactant particles in the same volume, so they are closer together. This makes collisions between particles more frequent. More collisions per second means more successful collisions per second, so the rate increases.

The mark scheme wants: (1) particles more crowded / more particles per unit volume, and (2) more collisions per second / greater collision frequency.

Key Takeaways

• Concentration affects how crowded the particles are.
• Higher concentration leads to higher collision frequency and a faster reaction.
• Unlike temperature, concentration does not change the average kinetic energy of particles.

Common Mistakes

• Saying "particles move faster" — increasing concentration does not make particles move faster; it makes them closer together.
• Saying "more successful collisions" without first saying "more collisions" or "greater frequency" — the mark scheme specifically wants collision frequency.
• Forgetting to state that the rate increases.

Things to Be Careful About

• The mark scheme accepts "more particles per unit volume" or "particles are more crowded".
• "More collisions per second" and "greater collision frequency" are equivalent; use one.
• Do not confuse concentration with temperature effects.

Walkthrough

Calcium bromide is an ionic compound. In the solid it forms a giant lattice of alternating $\text{Ca}^{2+}$ and $\text{Br}^-$ ions. To melt it, the lattice must be broken apart. The ions are held together by strong electrostatic forces of attraction between opposite charges. Melting therefore needs a large amount of energy, giving a high melting point.

Key Takeaways

• Ionic compounds have giant lattices, not molecules.
• Their high melting points are explained by strong electrostatic attraction between positive and negative ions.

Common Mistakes

• Saying strong covalent bonds or strong forces between molecules, because ionic compounds contain ions, not molecules.
• Not mentioning that the attraction is between positive and negative ions; the mark is for a strong attraction between oppositely charged ions.

Things to Be Careful About

Use the words strong electrostatic attraction. The mark scheme accepts strong attraction between positive and negative ions, so state the charges of the ions if you can.

Walkthrough

Calcium is a Group II metal with the electron configuration 2,8,8,2. It needs to lose its two outer electrons to reach a stable 2,8,8 noble-gas configuration, so each calcium atom loses 2 electrons and becomes $\text{Ca}^{2+}$.

Bromine is a Group VII non-metal with configuration 2,8,8,7. Each bromine atom needs one electron to complete its outer shell. A bromine molecule is $\text{Br}_2$, so two bromine atoms together need two electrons. The electrons lost by calcium are gained by the bromine molecule, producing two $\text{Br}^-$ ions.

This is electron transfer: the calcium atom gives electrons, and the bromine molecule accepts them. The ions formed are then held in an ionic lattice by electrostatic attraction.

Key Takeaways

• Metals lose electrons to form positive ions; non-metals gain electrons to form negative ions.
• Electron transfer creates ions with noble-gas electron configurations.
• A bromine molecule needs two electrons because it contains two bromine atoms.

Common Mistakes

• Saying calcium gains electrons.
• Saying bromine atoms lose electrons.
• Forgetting that the question asks about a bromine molecule, not a single bromine atom; the molecule gains two electrons.

Things to Be Careful About

Use the phrase loses electrons for calcium and gains electrons for bromine. Both statements are needed for full marks. The final ions are $\text{Ca}^{2+}$ and $\text{Br}^-$.

Walkthrough

Dilute aqueous calcium bromide contains $\text{Ca}^{2+}$ and $\text{Br}^-$ ions from the salt, and $\text{H}^+$ and $\text{OH}^-$ ions from water.

At the cathode, positive ions are attracted. Between $\text{Ca}^{2+}$ and $\text{H}^+$, hydrogen is discharged because calcium is more reactive than hydrogen. So hydrogen gas, $\text{H}_2$, is produced.

At the anode, negative ions are attracted. Bromide ions can be oxidised to bromine, but in a dilute solution hydroxide ions/water are discharged in preference to bromide ions. The product is oxygen gas, $\text{O}_2$, formed from water.

Key Takeaways

• In aqueous electrolysis, water supplies $\text{H}^+$ and $\text{OH}^-$.
• At the cathode, hydrogen is discharged instead of a more reactive metal such as calcium.
• At the anode, dilute halide solutions give oxygen rather than the halogen.

Common Mistakes

• Writing bromine at the anode; this is only favoured in concentrated solutions.
• Writing calcium at the cathode; calcium is too reactive, so hydrogen is discharged.

Things to Be Careful About

Read the word dilute carefully, because it changes the anode product. The mark scheme accepts oxygen or $\text{O}_2$ for the anode and hydrogen or $\text{H}_2$ for the cathode.

Walkthrough

Look at the electron transfer in the ionic equation. Two bromide ions each lose one electron, producing a bromine molecule. Loss of electrons is oxidation, so bromide ions are oxidised.

Chlorine molecules gain those two electrons, producing two chloride ions. Gain of electrons is reduction, so chlorine is reduced.

The two changes must happen together: the electrons lost by bromide are gained by chlorine. That is why the reaction involves both oxidation and reduction.

The half-equations make this clear:

Adding these cancels the electrons and gives the original equation.

Key Takeaways

• Oxidation is loss of electrons.
• Reduction is gain of electrons.
• A reaction with simultaneous oxidation and reduction is called redox.

Common Mistakes

• Saying bromine is reduced or chlorine is oxidised.
• Only writing oxidation and reduction occur, without saying which species gains or loses electrons.

Things to Be Careful About

The mark scheme gives one mark for bromide losing electrons/oxidised and one for chlorine gaining electrons/reduced. Use the words loses electrons and gains electrons clearly, and name the species.

Walkthrough

When the temperature is decreased, the system responds to oppose the change. Since the forward reaction is endothermic (absorbs heat), the reverse reaction is exothermic (releases heat). Lowering the temperature favours the exothermic direction, so the equilibrium shifts to the left, producing more zinc carbonate and releasing thermal energy.

Key Takeaways

• Le Chatelier's principle: a system at equilibrium responds to a change by counteracting it.
• For an endothermic forward reaction, decreasing temperature shifts equilibrium to the left (exothermic direction).

Common Mistakes

• Saying the equilibrium moves to the right because the forward reaction is endothermic — this would be correct for an increase in temperature, not a decrease.
• Omitting the energy explanation; the mark requires both the direction and the reason.

Things to Be Careful About

• Use the correct terminology: 'position of equilibrium moves to the left' or 'reactant side'.
• The explanation must explicitly mention 'release thermal energy' or 'favours the exothermic reaction'.

Walkthrough

Increasing pressure favours the side with fewer moles of gas. On the left, ZnCO₃ is a solid, so there are 0 moles of gas. On the right, there is 1 mole of CO₂ gas. Therefore, the equilibrium shifts to the left, reducing the number of gas moles and thus opposing the pressure increase.

Key Takeaways

• Le Chatelier's principle: increasing pressure shifts equilibrium to the side with fewer gas moles.
• Solids do not count as gas moles; only gaseous species matter.

Common Mistakes

• Forgetting that ZnCO₃ and ZnO are solids, so they contribute zero gas moles.
• Saying the equilibrium moves to the right because there is only one gas on the right — the correct comparison is left (0) vs right (1).

Things to Be Careful About

• Explicitly state 'fewer moles of gas on the left-hand side'.
• The mark is for the direction and the gas mole comparison.

Walkthrough

First, find the relative formula masses: ZnCO₃ = 65 + 12 + 48 = 125, and ZnO = 65 + 16 = 81. Then convert the given mass of zinc carbonate to moles: 4.23 g ÷ 125 g/mol = 0.03384 mol. The balanced equation shows a 1:1 mole ratio, so the same number of moles of ZnO is produced. Finally, multiply by the Mr of ZnO: 0.03384 × 81 = 2.74104 g. Round to 3 significant figures (since 4.23 has 3 sig figs) to get 2.74 g.

Key Takeaways

• Moles = mass / Mr.
• Use the balanced equation to find mole ratios.
• Mass = moles × Mr.
• Round the final answer to the appropriate number of significant figures.

Common Mistakes

• Using the wrong Mr (e.g., forgetting the oxygen atoms in ZnCO₃).
• Incorrect mole ratio (e.g., assuming 2 moles of ZnO per mole of ZnCO₃).
• Not rounding to the correct significant figures.

Things to Be Careful About

• Check the formula of zinc carbonate: ZnCO₃, not ZnCO₄.
• The final answer should be in grams with the unit.
• The mark scheme allows intermediate values; show them clearly.

Walkthrough

Zinc oxide is a classic amphoteric oxide: it reacts with acids (e.g., HCl) to form a salt and water, and with bases (e.g., NaOH) to form a zincate salt. Carbon dioxide is an acidic oxide: it dissolves in water to form carbonic acid and reacts with alkalis to form carbonates. Therefore, CO₂ does not react with acids.

Key Takeaways

• Amphoteric oxides react with both acids and bases.
• Acidic oxides react with bases but not with acids.
• Examples: ZnO, Al₂O₃ are amphoteric; CO₂, SO₂ are acidic.

Common Mistakes

• Calling zinc oxide basic only, ignoring its reaction with NaOH.
• Saying carbon dioxide is neutral; it is acidic.

Things to Be Careful About

• Use the correct terms: 'amphoteric' and 'acidic'.
• The explanation should link the property to the observed reactions.

Walkthrough

Zinc carbonate is a solid. Nitric acid is aqueous. The products are zinc nitrate (aqueous), water (liquid), and carbon dioxide (gas). Write the unbalanced equation: ZnCO₃ + HNO₃ → Zn(NO₃)₂ + H₂O + CO₂. Balance: there are 2 nitrate groups on the right, so need 2 HNO₃ on the left. Check atoms: Zn:1, C:1, O:3+6=9 on left; on right: O in Zn(NO₃)₂=6, H₂O=1, CO₂=2, total 9. H:2 on both sides. Balanced.

Key Takeaways

• Carbonates react with acids to give salt, water, and carbon dioxide.
• Nitric acid produces nitrate salts.
• State symbols: (s) solid, (aq) aqueous, (l) liquid, (g) gas.

Common Mistakes

• Forgetting the 2 in front of HNO₃.
• Writing ZnNO₃ instead of Zn(NO₃)₂.
• Missing state symbols.

Things to Be Careful About

• The nitrate ion is NO₃⁻, so zinc nitrate is Zn(NO₃)₂.
• State symbols are essential for the second mark.

Walkthrough

Methane is a greenhouse gas. When its concentration in the atmosphere rises, it traps more heat radiated from the Earth's surface, strengthening the greenhouse effect and causing global warming and climate change. The mark is awarded for naming this effect.

Key Takeaways

Methane is a potent greenhouse gas; higher levels enhance the natural greenhouse effect, leading to global warming and climate change.

Common Mistakes

• Saying "air pollution" or "acid rain" — methane does not cause acid rain.
• Giving a vague answer like "it is bad for the environment" without naming the greenhouse effect or global warming.

Things to Be Careful About

The mark scheme accepts any one of: increased global warming, enhanced greenhouse effect, or climate change. Any of these is sufficient.

Walkthrough

Sulfur dioxide is produced when sulfur-containing fossil fuels burn. To reduce its emission:

1. Flue gas desulfurisation — the waste gases (flue gases) are passed through calcium oxide or calcium carbonate, which reacts with and removes sulfur dioxide before it reaches the atmosphere.
2. Use low-sulfur fuels — choosing fuels with less sulfur means less sulfur dioxide is formed in the first place.
3. Burn less fossil fuel / use renewables — reducing the amount of fossil fuel burned directly reduces the total sulfur dioxide produced.
   Any two of these score the two marks.

Key Takeaways

Strategies to reduce sulfur dioxide emissions target either removing it after combustion or preventing its formation by using cleaner fuels or less fuel overall.

Common Mistakes

• Saying "use a catalytic converter" — that removes nitrogen oxides, not sulfur dioxide.
• Giving only one strategy when two are required.
• Saying "filter the gases" without naming a reactive substance — filtering does not remove a gas.

Things to Be Careful About

The mark scheme allows any two from the three listed. "Reacting with calcium oxide" and "reacting with calcium carbonate" are two separate acceptable points, but do not count as two different strategies — they are both forms of flue gas desulfurisation.

Walkthrough

The sample is a gas at room temperature and pressure. When its temperature increases at constant pressure, the particles gain kinetic energy and move faster. They collide with the walls harder and more often. If the pressure is kept the same, the gas must be allowed to expand, so the volume increases. In kinetic particle theory terms, the particles spread further apart and the distance between them increases. The mark scheme gives one mark for "volume increases" and one mark for "particles spread out / distance between particles increases."

Key Takeaways

• Heating a gas increases the kinetic energy of its particles.
• At constant pressure, a hotter gas occupies a larger volume.
• The volume change is explained by particles spreading out, not by particles getting bigger.

Common Mistakes

• Saying the pressure increases instead of the volume. The question states pressure is unchanged.
• Saying the particles themselves expand. Particles do not change size; the space between them increases.
• Missing the phrase "spread out" or "distance between particles increases", which is the second marking point.

Things to Be Careful About

• The mark scheme wants both the observation (volume increases) and the particle explanation.
• Use kinetic particle theory language: kinetic energy, move faster, collide with walls, spread out.
• No calculation is needed; this is a qualitative explanation.

Walkthrough

Sulfur is a gas above 445 °C and a solid at room temperature, so cooling the gas to room temperature means it changes from gas to solid. As particles cool, they lose kinetic energy. Their separation decreases: they move closer together. Their arrangement changes from random/disordered in the gas to ordered/regular in the solid. Their motion changes from moving freely from place to place to vibrating about fixed positions. The mark scheme awards one mark for each of these three changes.

Key Takeaways

• Cooling removes kinetic energy from particles.
• Gas to solid involves decreasing separation, increasing order, and loss of free movement.
• Particles in a solid are not still; they vibrate about fixed positions.

Common Mistakes

• Reversing the changes, e.g. saying particles move further apart.
• Saying particles stop moving in a solid. They vibrate.
• Describing arrangement as "ordered to random" instead of "random to ordered".

Things to Be Careful About

• Match each of the three headings: separation, arrangement, motion.
• Use "closer together", "random to ordered", "moving freely to vibrating".
• Room temperature is below sulfur's melting point, so the final state is solid, not liquid.

Walkthrough

Diffusion is explained by the continuous random motion of particles. In a region where there is a higher concentration of particles, more particles are moving randomly into the lower-concentration region than in the opposite direction. The net result is that particles move from high concentration to low concentration. This is the one-mark definition required.

Key Takeaways

• Diffusion is a net movement down a concentration gradient.
• It happens because particles are in constant random motion.
• It applies to gases and liquids.

Common Mistakes

• Saying particles move from low to high concentration, which is the opposite.
• Saying particles move until they stop; they continue moving but concentrations become equal.
• Omitting "random motion" if asked to explain in terms of kinetic particle theory.

Things to Be Careful About

• The mark scheme accepts the simple idea: particles move from high concentration to low concentration.
• If the question says "in terms of kinetic particle theory", include random motion.

Walkthrough

The notation $^{33}_{16}\text{S}^{2-}$ tells us the mass number is 33 and the proton number (atomic number) is 16. The proton number equals the number of protons, so there are 16 protons. In a neutral sulfur atom there would also be 16 electrons. However, the 2- charge means the ion has gained two extra electrons, so it has 18 electrons. The number of neutrons is found by subtracting the proton number from the mass number: $33 - 16 = 17$. The completed table is electrons 18, neutrons 17, protons 16.

Key Takeaways

• Mass number = protons + neutrons.
• Proton number = number of protons = number of electrons in a neutral atom.
• A negative ion has gained electrons; a 2- ion has two extra electrons.

Common Mistakes

• Writing 16 electrons for the sulfide ion, forgetting the 2- charge.
• Writing 33 neutrons, using the mass number as the neutron number.
• Confusing the top and bottom numbers: top is mass number, bottom is proton number.

Things to Be Careful About

• The charge is written as a superscript after the symbol: $\text{S}^{2-}$.
• Neutrons = mass number - proton number.
• Protons never change when an ion forms; only electrons change.

Walkthrough

The question asks for the displayed formula of compound A, which is identified in the diagram as butan-1-ol. A displayed formula must show every atom and every covalent bond. Butan-1-ol has a four-carbon chain (butane backbone) with an $-\text{OH}$ group on the first carbon. We draw four carbon atoms in a row, single-bonded. Carbon 1 is bonded to two hydrogens and the oxygen. Carbons 2 and 3 are bonded to two hydrogens each. Carbon 4 is bonded to three hydrogens. The oxygen is bonded to carbon 1 and one hydrogen.

Key Takeaways

A displayed formula is different from a structural or formula representation: it shows every single bond. For alcohols, the $\text{C-O-H}$ linkage must be fully drawn out, not written as $\text{-OH}$.

Common Mistakes

• Writing $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ (this is a structural/condensed formula, not displayed).
• Forgetting the hydrogen on the oxygen (writing $\text{C-O}$ instead of $\text{C-O-H}$).
• Drawing a double bond anywhere in the carbon chain (it is an alcohol, not an alkene).

Things to Be Careful About

Ensure all carbon valencies are four. The terminal methyl group ($\text{CH}_3$) needs three hydrogens. The carbon attached to oxygen ($\text{CH}_2\text{OH}$) needs two hydrogens plus the $\text{O-H}$ bond.

Walkthrough

Compound A is butan-1-ol (a 4-carbon alcohol, providing the 'butyl' group). It reacts with ethanoic acid (a 2-carbon acid, providing the 'ethanoate' group) in the presence of an acid catalyst. This is esterification. The name of the ester is formed by taking the alkyl group name from the alcohol (butyl) and the carboxylate group name from the acid (ethanoate).

Key Takeaways

Ester naming: [Alkyl from alcohol] [Carboxylate from acid]. Butan-1-ol gives 'butyl'. Ethanoic acid gives 'ethanoate'.

Common Mistakes

• Reversing the order (writing 'ethanoate butyl').
• Confusing the carbon count (butan-1-ol has 4 carbons, ethanoic acid has 2).

Things to Be Careful About

The alcohol is butan-1-ol, so the alkyl group is 'butyl', not 'butane'. The acid is ethanoic acid, so the suffix is '-oate'.

Walkthrough

Compound C is butanoic acid ($\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$). It reacts with sodium hydroxide ($\text{NaOH}$), a base. This is a neutralisation reaction.
$\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$
$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COONa} + \text{H}_2\text{O}$
The salt D is sodium butanoate. The formula is formed by replacing the H of the $-\text{COOH}$ group with $\text{Na}^+$.

Key Takeaways

Carboxylic acids react with alkalis to form carboxylate salts and water. The cation comes from the base ($\text{Na}^+$), the anion from the acid (butanoate, $\text{C}_3\text{H}_7\text{COO}^-$).

Common Mistakes

• Writing the formula as $\text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_3\text{Na}$ (confusing with carbonate).
• Naming it 'sodium butyrate' (common name).

Things to Be Careful About

The formula must show the sodium attached to the oxygen of the carboxylate group: $\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}$ or $\text{C}_3\text{H}_7\text{COONa}$.

Walkthrough

Compound C (butanoic acid) reacts with sodium carbonate ($\text{Na}_2\text{CO}_3$). Carboxylic acids react with carbonates to produce a salt, water, and carbon dioxide gas.
$2\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$
From the diagram, the products are D (sodium butanoate), gas E, and liquid F.
Therefore, gas E is carbon dioxide ($\text{CO}_2$) and liquid F is water ($\text{H}_2\text{O}$).
This is consistent with part (d) where C + NaOH gives D and F (water).

Key Takeaways

Acids + Carbonates -> Salt + Water + Carbon dioxide. This is a standard test for acids (effervescence of $\text{CO}_2$).

Common Mistakes

• Thinking the gas is hydrogen (that's acid + metal).
• Forgetting water is a product.

Things to Be Careful About

State symbols: E is a gas ($\text{CO}_2$), F is a liquid ($\text{H}_2\text{O}$). The question asks for formulas.

Walkthrough

Compound C is butanoic acid ($\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$). It reacts with ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) in the presence of an acid catalyst. This is esterification.
$\text{Acid} + \text{Alcohol} \rightleftharpoons \text{Ester} + \text{Water}$
$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$
The ester G is ethyl butanoate. The structural formula shows the alkyl chain from the acid and the alkyl group from the alcohol linked by the ester group ($-\text{COO}-$).

Key Takeaways

Esterification: Carboxylic acid + Alcohol -> Ester + Water. The $-\text{OH}$ from the acid and the $-\text{H}$ from the alcohol combine to form water. The remaining parts join at the oxygen.

Common Mistakes

• Writing the formula backwards (e.g., putting the ethyl group on the carbonyl carbon side incorrectly).
• Forgetting the ester linkage $-\text{COO}-$.

Things to Be Careful About

Structural formula can be written semi-structurally as $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_3$. Ensure the $\text{CH}_2\text{CH}_3$ (ethyl) is clearly attached to the single-bonded oxygen of the ester group.

Walkthrough

To find the empirical formula from percentage composition, assume a 100 g sample. This converts the percentages directly into masses in grams: 10.8 g of carbon, 17.1 g of fluorine, and 72.1 g of bromine. Next, divide each mass by the relative atomic mass ($A_r$) of the element to find the number of moles: C = 10.8 / 12 = 0.90 mol; F = 17.1 / 19 = 0.90 mol; Br = 72.1 / 80 = 0.90125 mol. Finally, divide all mole values by the smallest mole value (0.90) to get the simplest whole-number ratio: 1 : 1 : 1. Write the empirical formula using these whole numbers.

Key Takeaways

The empirical formula is the simplest whole-number ratio of atoms in a compound. Convert % to mass (assume 100g), then to moles using $A_r$, then find the ratio by dividing by the smallest value.

Common Mistakes

Forgetting to divide by $A_r$ and using the percentages directly as the ratio. Rounding Br to 0.9 too early and losing precision, though in this case 72.1 / 80 = 0.90125 which rounds cleanly to 0.90. Writing the formula with incorrect subscripts (e.g., $\text{C}_2\text{F}_2\text{Br}_2$) when the simplest ratio is 1:1:1.

Things to Be Careful About

Use the correct $A_r$ values from the Periodic Table: C = 12, F = 19, Br = 80. The ratio must be reduced to the simplest whole numbers. The formula is written as CFBr.