Chemistry 5070/22 — October/November 2025

Walkthrough

Structural isomers are molecules that have the same molecular formula (same number and types of atoms) but different structural arrangements of atoms.
We need to find two compounds from the list with the same molecular formula.
Compound E is cyclobutane, a cyclic alkane. The image shows a four-membered ring of CH2 groups. Molecular formula: C4H8.
Compound G is but-1-ene: CH3—CH2—CH=CH2. Molecular formula: C4H8.
Both have the formula C4H8, but E is a ring and G is a straight chain with a double bond. They are structural isomers.

Key Takeaways

Structural isomers must have the exact same molecular formula. Calculate the molecular formula for each compound by counting all atoms, then compare them.

Common Mistakes

Assuming compounds with similar names or similar structures are isomers without checking the molecular formula. For example, F (butane, C4H10) and G (but-1-ene, C4H8) have different molecular formulas and are not isomers.

Things to Be Careful About

When counting atoms in cyclic compounds like E, remember that each corner of the ring is a carbon atom with enough hydrogen atoms to make four bonds total. In cyclobutane, each carbon is bonded to two other carbons and two hydrogens (CH2), giving C4H8.

Walkthrough

Isotopes are atoms of the same element that contain the same number of protons (same atomic number, 22) but different numbers of neutrons (and thus different mass/nucleon numbers).

For $^{46}_{22}\text{Ti}$:

• Number of neutrons = $46 - 22 = 24$

For $^{47}_{22}\text{Ti}$:

• Number of neutrons = $47 - 22 = 25$

Therefore, the valid differences are:

• Different number of neutrons (or $^{47}\text{Ti}$ has one more neutron than $^{46}\text{Ti}$).
• Different nucleon number / mass number (47 compared to 46).

Key Takeaways

• Isotopes share the same proton number but differ in their neutron count and mass number.

Common Mistakes

• Stating that they have different numbers of electrons or protons (isotopes always have identical numbers of protons and electrons in neutral atoms).

Things to Be Careful About

• Ensure you specify 'neutrons' or 'mass/nucleon number', not just 'mass' alone.

Walkthrough

Chemical reactions involve the loss, gain, or sharing of electrons. Because all isotopes of an element have the exact same number of protons, neutral atoms of those isotopes also have the exact same number of electrons and identical electronic configurations (specifically, the same number of valence/outer electrons). Hence, they participate in chemical bonding in exactly the same way.

Key Takeaways

• Chemical properties depend entirely on the number and arrangement of electrons (electronic configuration).
• Physical properties (such as density or rate of diffusion) can differ between isotopes due to differences in mass.

Common Mistakes

• Saying they have the "same number of protons" without mentioning electrons or electronic configuration; chemical reactions are governed by electrons.

Things to Be Careful About

• Always link chemical reactivity directly to the electron arrangement or number of electrons.

Walkthrough

The relative atomic mass ($A_r$) of an element is the weighted average of the masses of its naturally occurring isotopes relative to $\frac{1}{12}$ of the mass of a carbon-12 atom.

To find $A_r$:

1. Multiply the mass number of each isotope by its percentage abundance:

   • $^{46}\text{Ti}$: $9 \times 46 = 414$
   • $^{47}\text{Ti}$: $7 \times 47 = 329$
   • $^{48}\text{Ti}$: $75 \times 48 = 3600$
   • $^{49}\text{Ti}$: $9 \times 49 = 441$

2. Sum these values together to find the numerator:

1. Divide by the total abundance (the sum of percentages $= 9 + 7 + 75 + 9 = 100$):

Since this is a "show that" question, all stages of the fraction must be clearly set out.

Key Takeaways

• Formula for relative atomic mass from percentage abundances:

Common Mistakes

• Forgetting to divide by 100.
• Taking a simple arithmetic average of the four mass numbers $\frac{46 + 47 + 48 + 49}{4} = 47.5$ instead of weighting by percentage abundance.

Things to Be Careful About

• In "show that" questions, full working showing substitution into the formula is required to score all marks.

Walkthrough

Hydrogen is a flammable gas. The standard test involves introducing a flame to the gas. The hydrogen ignites rapidly with a small explosion, producing a characteristic 'squeaky pop' sound.

Key Takeaways

Hydrogen test: lighted splint -> 'pop'. Oxygen test: glowing splint -> relights. Carbon dioxide test: limewater -> turns milky.

Common Mistakes

Saying 'it burns' is not specific enough; many gases burn. The key observation is the 'pop' sound. Also, do not say 'glowing splint'; that is for oxygen. It must be a 'lighted' or 'burning' splint.

Things to Be Careful About

The mark scheme specifically looks for 'lighted splint' and 'pops'. Ensure both the action (test) and the observation are clearly stated if asked, though here they are combined in 1 mark.

Walkthrough

First, convert the volume of hydrogen gas to moles using the molar gas volume at room temperature and pressure (r.t.p.), which is $24000 \text{ cm}^3\text{ / mol}$.

Next, use the stoichiometric ratio from the balanced chemical equation. The equation shows that 2 moles of Al produce 3 moles of H$_2$. Therefore, the moles of Al required is $\frac{2}{3}$ of the moles of H$_2$.

Finally, calculate the mass using the relative atomic mass of aluminium ($A_r = 27$).

The question asks for the answer to two significant figures, so $0.27459$ rounds to $0.27$.

Key Takeaways

Always check the state and conditions for gas volumes. At r.t.p., use $24000 \text{ cm}^3\text{ / mol}$ (or $24 \text{ dm}^3\text{ / mol}$). At s.t.p., use $22400 \text{ cm}^3\text{ / mol}$. Remember to use the correct mole ratio from the balanced equation, not just 1:1.

Common Mistakes

Using the wrong molar gas volume (e.g., $22400$ for r.t.p.). Forgetting to use the $2:3$ ratio and assuming moles of Al = moles of H$_2$. Forgetting to round to the correct number of significant figures at the end.

Things to Be Careful About

The volume is given in $\text{cm}^3$, so divide by $24000$, not $24$. If you convert to $\text{dm}^3$ first ($0.366 \text{ dm}^3$), then divide by $24$, you get the same result. Ensure final answers respect the requested significant figures.

Walkthrough

In Group VII, reactivity decreases down the group: fluorine is the most reactive, then chlorine, bromine, then iodine. A more reactive halogen can displace a less reactive halogen from a solution of its salt. Chlorine is above bromine in the group, so it is more reactive and displaces bromine from aqueous lithium bromide. The bromide ions lose electrons to become bromine molecules, and the chloride ions stay in solution with the lithium ions, forming lithium chloride. The two products are therefore lithium chloride and bromine.

Key Takeaways

• Halogen reactivity decreases down Group VII.
• A more reactive halogen displaces a less reactive halogen from its salt solution.
• The displacement reaction is a redox reaction: bromide ions are oxidised to bromine, chlorine is reduced to chloride ions.

Common Mistakes

• Naming 'lithium bromine' or 'chlorine bromide' – the products are lithium chloride and bromine.
• Saying bromine displaces chlorine – bromine is less reactive than chlorine.
• Forgetting that bromine is formed as an element, not as bromide ions.

Things to Be Careful About

• The question asks for two products; give both.
• Chlorine is more reactive than bromine because it is higher in Group VII.
• The aqueous solution contains $\text{Li}^+$ and $\text{Br}^-$; after displacement the ions are $\text{Li}^+$ and $\text{Cl}^-$, so lithium chloride remains in solution.

Walkthrough

Nickel is widely used as an industrial catalyst. In organic chemistry, finely divided nickel catalyses the addition of hydrogen across the carbon-carbon double bond ($\text{C}=\text{C}$) in alkenes to produce saturated alkanes (or in the manufacture of margarine from vegetable oils).

Key Takeaways

• Nickel catalyses the hydrogenation of alkenes (reaction of alkenes with hydrogen at around $150\text{ }^\circ\text{C}$ to $200\text{ }^\circ\text{C}$).
• Catalytic activity is a characteristic property of transition elements.

Common Mistakes

• Confusing the nickel catalyst with the iron catalyst used in the Haber process or vanadium(V) oxide used in the Contact process.

Things to Be Careful About

• Clearly name the reaction or state the reactants (e.g. "ethene with hydrogen" or "hydrogenation of alkenes").

Walkthrough

Transition elements exhibit characteristic chemical properties, including:

1. Variable oxidation states in their compounds (e.g. $\text{Ni}^{2+}$, $\text{Ni}^{3+}$).
2. Formation of coloured compounds/ions.
3. Catalytic activity (which was already mentioned in part (a)).

State one other chemical property such as variable oxidation states or forming coloured compounds.

Key Takeaways

• Transition metals have variable oxidation states and form coloured compounds.
• Physical properties (high density, high melting point) are not chemical properties.

Common Mistakes

• Stating a physical property (such as "high melting point" or "good conductor") instead of a chemical property.

Things to Be Careful About

• Ensure the answer focuses strictly on a chemical property.

Walkthrough

1. Find percentage of nickel: Subtract the percentages of $\text{H}$ and $\text{O}$ from $100\%$: $$100\% - (2.2\% + 34.5\%) = 63.3\% \text{ Ni}$$
2. Convert percentages to moles: Divide each percentage by the respective relative atomic mass ($A_r$) of the element:
   • $\text{Ni}: \frac{63.3}{59} = 1.073\text{ mol}$
   • $\text{H}: \frac{2.2}{1} = 2.200\text{ mol}$
   • $\text{O}: \frac{34.5}{16} = 2.156\text{ mol}$
3. Determine the simplest ratio: Divide all mole values by the smallest value ($1.073$):
   • $\text{Ni}: 1$
   • $\text{H}: \frac{2.200}{1.073} \approx 2$
   • $\text{O}: \frac{2.156}{1.073} \approx 2$
4. Write the empirical formula: $\text{NiH}_2\text{O}_2$ or $\text{Ni(OH)}_2$.

Key Takeaways

• Empirical formula is the simplest whole-number ratio of atoms of each element in a compound.
• Steps: $\% \rightarrow \text{moles } (\div A_r) \rightarrow \text{simplest ratio } (\div \text{smallest})$.

Common Mistakes

• Dividing percentage by atomic number ($Z$) instead of relative atomic mass ($A_r$).
• Rounding mole values prematurely before finding the ratio.

Things to Be Careful About

• Check that the percentages add up to $100\%$.
• Use standard Cambridge $A_r$ values from the Periodic Table: $\text{Ni} = 59$, $\text{H} = 1$, $\text{O} = 16$.

Walkthrough

1. Identify the gaseous species:
   • Left-hand side: $4\text{ mol of CO(g)}$ and $1\text{ mol of Ni(s)}$ $\rightarrow$ total gas = $4\text{ moles of gas}$.
   • Right-hand side: $1\text{ mol of Ni(CO)}_4\text{(g)}$ $\rightarrow$ total gas = $1\text{ mole of gas}$.
   • Solids do not contribute to gas pressure.
2. Apply Le Chatelier's principle:
   • An increase in pressure causes the equilibrium position to shift toward the side with fewer moles of gas to counteract the increase and reduce pressure.
   • Since the right-hand side has $1\text{ mole of gas}$ compared to $4\text{ moles of gas}$ on the left, the equilibrium shifts to the right (forward direction).

Key Takeaways

• When pressure is increased, the equilibrium shifts towards the side with fewer moles of gas.
• Only count species in the gaseous state ($ext{g}$); ignore solids ($ext{s}$) and liquids ($ext{l}$).

Common Mistakes

• Counting the solid $\text{Ni(s)}$ as part of the gas volume on the left side.
• Forgetting to explicitly compare the number of moles / volumes of gas on both sides.

Things to Be Careful About

• Ensure you specify "moles of gas" rather than just "moles".

Walkthrough

Hydrogen peroxide has the structure $\text{H}-\text{O}-\text{O}-\text{H}$:

1. Each hydrogen atom shares one pair of electrons with an oxygen atom to achieve a stable duplet (2 outer electrons).
2. The two central oxygen atoms share one pair of electrons with each other.
3. Oxygen is in Group VI and has 6 valence electrons. In $\text{H}_2\text{O}_2$, each oxygen atom forms two single covalent bonds (using 2 valence electrons) and retains two non-bonding lone pairs (4 non-bonding electrons) to complete its stable octet (8 outer electrons).

Key Takeaways

• Covalent bonds consist of shared pairs of electrons.
• Remember to include all non-bonding lone pairs on central or terminal atoms when drawing full outer-shell dot-and-cross diagrams.

Common Mistakes

• Forgetting the lone pairs on the oxygen atoms.
• Drawing double bonds instead of single bonds.

Things to Be Careful About

• Ensure distinct symbols (dots and crosses) are used consistently to show which atom provides each electron.

Walkthrough

1. Reaction with acidified $\text{KMnO}_4$: Acidified potassium manganate(VII) is a powerful oxidising agent that turns from purple to colourless when it is reduced to $\text{Mn}^{2+}$. Because hydrogen peroxide reduces it, hydrogen peroxide is acting as a reducing agent.
2. Reaction with $\text{KI}$: Aqueous potassium iodide is a reducing agent. Colourless iodide ions ($\text{I}^-$) are oxidised to brown iodine ($\text{I}_2$). Because hydrogen peroxide oxidises iodide, hydrogen peroxide is acting as an oxidising agent.
3. Therefore, hydrogen peroxide can act as both an oxidising agent and a reducing agent.

Key Takeaways

• Decolourisation of acidified aqueous potassium manganate(VII) is the standard test for a reducing agent.
• Oxidation of colourless aqueous potassium iodide to brown iodine is the standard test for an oxidising agent.

Common Mistakes

• Confusing the roles and stating that $\text{H}_2\text{O}_2$ is oxidised by iodide.
• Stating that $\text{H}_2\text{O}_2$ is acidic/alkaline rather than identifying redox properties.

Things to Be Careful About

• The question asks for the chemical properties indicated by both observations, requiring both 'reducing agent' and 'oxidising agent' to be stated.

Walkthrough

1. The given ions are barium, $\text{Ba}^{2+}$, and peroxide, $\text{O}_2^{2-}$.
2. The charges are $+2$ and $-2$, which balance in a $1:1$ ratio.
3. Combining one $\text{Ba}^{2+}$ ion and one $\text{O}_2^{2-}$ ion gives the empirical/chemical formula $\text{BaO}_2$.

Key Takeaways

• The overall charge on an ionic compound must be zero.
• A peroxide ion is polyatomic with the formula $\text{O}_2^{2-}$.

Common Mistakes

• Writing $\text{Ba}_2\text{O}_2$ or simplifying the peroxide ion incorrectly to $\text{BaO}$.

Things to Be Careful About

• Subscripts must be written clearly: $\text{BaO}_2$.

Walkthrough

The question provides structural diagrams of the two monomers used to make PET. The first monomer has two carboxylic acid groups (—COOH) at either end of a central block, making it a dicarboxylic acid. The second monomer has two hydroxyl groups (—OH) at either end, making it a diol. These are the standard names for the monomers in polyester synthesis.

Key Takeaways

Polyesters are formed from a dicarboxylic acid and a diol. Recognising the functional groups (—COOH and —OH) allows you to name the monomers correctly.

Common Mistakes

Calling the monomers 'carboxylic acid' or 'alcohol' without specifying 'di' (dicarboxylic acid and diol) is not precise enough for the mark. 'Ester' is the product linkage, not the monomer name.

Things to Be Careful About

Ensure you write 'dicarboxylic acid' and not just 'dicarboxylic'. The mark scheme accepts 'dicarboxylic (acid)'.

Walkthrough

Polyamides are condensation polymers formed from diamines and dicarboxylic acids (or amino acids). The most common example is nylon. Proteins are also natural polyamides formed from amino acids.

Key Takeaways

Nylon and proteins are the standard examples of polyamides in the 5070 syllabus.

Common Mistakes

Naming a polyester like PET or a different polymer type. Ensure you name a polyamide specifically.

Things to Be Careful About

The mark scheme accepts 'nylon' or 'protein'. 'Nylon' is the most common synthetic example expected at this level.

Walkthrough

An unsaturated organic compound is one that contains at least one carbon–carbon double bond ($\text{C}=\text{C}$) or triple bond. Looking at the displayed formula in Fig. 8.1, there is a double bond between two carbon atoms ($\text{C}=\text{C}$).

Key Takeaways

• Saturated compounds contain only single carbon–carbon bonds ($\text{C}-\text{C}$).
• Unsaturated compounds contain one or more carbon–carbon double (or triple) bonds ($\text{C}=\text{C}$).

Common Mistakes

• Stating just "contains a double bond" without specifying that it is between carbon atoms (since the molecule also contains a $\text{C}=\text{O}$ bond, which does not make a hydrocarbon chain unsaturated in this context).

Things to Be Careful About

• Always specify carbon–carbon double bond or write $\text{C}=\text{C}$.

Walkthrough

Aqueous bromine (bromine water) undergoes an addition reaction across the $\text{C}=\text{C}$ double bond of an unsaturated compound. As the bromine is consumed in the reaction to form a dibromo compound, the orange/brown colour disappears and the solution turns colourless.

Key Takeaways

• Bromine water is the standard chemical test for unsaturation ($\text{C}=\text{C}$ bonds).
• A positive result is the decolourisation of bromine water: orange (or brown/red-brown) to colourless.

Common Mistakes

• Writing "clear" instead of "colourless". A coloured solution can be clear (transparent); "colourless" means having no colour.

Things to Be Careful About

• Include both the initial colour (orange/red-brown/brown) and the final colour (colourless).

Walkthrough

An acid is a substance that ionises in water to produce hydrogen ions, $\text{H}^+$. A strong acid fully ionises (completely dissociates) in aqueous solution, whereas a weak acid only partially ionises (partially dissociates), establishing an equilibrium where most of the acid remains as un-ionised molecules.

Key Takeaways

• Weak refers to the extent of ionisation/dissociation, not the concentration (dilute vs. concentrated).
• Weak acids only partially ionise in water.

Common Mistakes

• Confusing weak with dilute. "Dilute" means low amount of solute per unit volume of solution; "weak" means incomplete ionisation.

Things to Be Careful About

• Use the word partially (or incompletely) with ionises or dissociates.

Walkthrough

Carboxylic acids exhibit typical acidic properties. When an acid reacts with a metal carbonate, the products are a salt, water, and carbon dioxide gas:

Therefore, the gas produced is carbon dioxide.

Key Takeaways

• $\text{Acid} + \text{carbonate} \rightarrow \text{salt} + \text{water} + \text{carbon dioxide}$.

Common Mistakes

• Writing "hydrogen" or "oxygen" instead of carbon dioxide.

Things to Be Careful About

• The question asks to name the gas, so write "carbon dioxide" (though $\text{CO}_2$ is often accepted, writing the name is safest).

Walkthrough

An ester is formed via an esterification (condensation) reaction between a carboxylic acid and an alcohol:

This reaction requires heating in the presence of an acid catalyst, typically concentrated sulfuric acid ($\text{H}_2\text{SO}_4$).

Key Takeaways

• Esterification: $\text{Carboxylic acid} + \text{alcohol} \rightleftharpoons \text{ester} + \text{water}$.
• Conditions: acid catalyst (commonly concentrated sulfuric acid) and heat.

Common Mistakes

• Forgetting the acid catalyst and only stating the alcohol.

Things to Be Careful About

• Clearly state both the required reactant (an alcohol) and the required catalyst (acid / concentrated sulfuric acid).

Walkthrough

The term hydrated in chemistry refers to a solid substance (typically a salt) that contains water of crystallisation chemically bonded within its crystal lattice structure.

Key Takeaways

• Hydrated means chemically combined with water.
• Anhydrous means containing no water of crystallisation.

Common Mistakes

• Confusing "hydrated" with "dissolved in water" or "aqueous solution". A hydrated substance is a solid, not a solution.
• Saying "contains moisture" or "is wet", which implies physical mixing rather than chemical combination.

Things to Be Careful About

• Ensure you mention that the water is chemically combined or refer directly to water of crystallisation.

Walkthrough

When testing for $\text{Zn}^{2+}$ cations with aqueous ammonia ($\text{NH}_3(\text{aq})$):

1. Adding a few drops produces a white precipitate of zinc hydroxide, $\text{Zn(OH)}_2$:

1. Adding excess aqueous ammonia causes the precipitate to dissolve, forming a colourless solution containing a complex ion (tetraamminezinc(II) ion, $[\text{Zn(NH}_3)_4]^{2+}$).

Both parts of the observation (the initial precipitate and what happens in excess) are required to gain full marks.

Key Takeaways

• $\text{Zn}^{2+}$ with $\text{NH}_3(\text{aq})$ gives a white ppt. that is soluble in excess forming a colourless solution.
• $\text{Al}^{3+}$ and $\text{Pb}^{2+}$ give white precipitates insoluble in excess $\text{NH}_3(\text{aq})$, which distinguishes $\text{Zn}^{2+}$ from them.

Common Mistakes

• Stating only "white precipitate" and omitting what happens in excess.
• Stating that the precipitate remains / is insoluble in excess (which would be true for $\text{Al}^{3+}$ or $\text{Mg}^{2+}$, but not $\text{Zn}^{2+}$).

Things to Be Careful About

• Always describe both stages: (1) dropwise addition, (2) addition until in excess.