§-Past paper

VICChemistry2022

VCE Chemistry 2022

Walkthrough of the 2022 VCE chemistry exam: what it assessed, strategy tips, and the common errors flagged in the official marker report.

Marks: 120
Time: 150 min
Authority: VCAA
Updated: 2026-05-29

Open the official PDF ->Official marking notes

What this paper assessed

The VCAA Units 3 & 4 Chemistry examination is worth 120 marks over 2.5 hours of writing time (plus 15 minutes reading), with the supplied data book. It covers:

Unit 3 - How can design and innovation help to optimise chemical processes? Chemical energy and fuels (combustion, thermochemistry, fuel cells), the rate and extent of chemical reactions (collision theory, catalysts, equilibrium and $K$ ), and electrochemistry (galvanic cells, electrolysis, Faraday's laws).
Unit 4 - How are carbon-based compounds designed for purpose? Structure, nomenclature and reactions of organic families, organic reaction pathways, analytical techniques (mass spectrometry, IR and NMR spectroscopy, chromatography), and the chemistry of food (proteins, carbohydrates, fats, and enzyme catalysis).

The 2022 paper rewarded quantitative reasoning shown step by step (stoichiometry, thermochemistry, electrolysis) and the ability to link spectroscopic data to molecular structure rather than just reading peaks.

Structure and timing

120 marks in 150 minutes (about 1.25 min/mark), in two sections:

Section A - Multiple choice: typically 30 questions, 30 marks. Budget about 35 minutes (~1.1 min each).
Section B - Short and extended answer: 90 marks. Budget about 110 minutes, leaving ~5 minutes to check.

Use the supplied data book for constants, the periodic table and spectroscopic data rather than recalling them. For any calculation chain, label the quantity at each step so method marks are visible even if the final number is wrong, and check significant figures against the data given.

Worked practice questions (exam-style)

Worked example

Question 1 (5 marks): Stoichiometry and limiting reagent

$5.00\ \text{g}$ of calcium carbonate reacts with $40.0\ \text{mL}$ of $1.50\ \text{mol L}^{-1}$ hydrochloric acid: $\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$ . Determine the limiting reagent and the mass of $\text{CO}_2$ produced. ( $M(\text{CaCO}_3) = 100.1\ \text{g mol}^{-1}$ , $M(\text{CO}_2) = 44.0\ \text{g mol}^{-1}$ .)

The key strategy is to find moles of each reactant first, then compare the available amounts against the stoichiometric ratio to identify which runs out.

Step 1: Calculate moles of each reactant

Convert the given mass and volume-concentration data to moles:

n(\text{CaCO}_3) = \frac{5.00}{100.1} = 0.0500\ \text{mol}, \qquad n(\text{HCl}) = 1.50 \times 0.0400 = 0.0600\ \text{mol}.

Step 2: Identify the limiting reagent

The balanced equation requires 2 mol of HCl for every 1 mol of $\text{CaCO}_3$ . To react all of the $\text{CaCO}_3$ we would need $2 \times 0.0500 = 0.100\ \text{mol}$ of HCl, but only $0.0600\ \text{mol}$ is available. HCl runs out first, so HCl is the limiting reagent.

Step 3: Use the limiting reagent to find moles of $\text{CO}_2$

From the equation, the mole ratio of HCl to $\text{CO}_2$ is 2:1. Using the moles of the limiting reagent:

n(\text{CO}_2) = \frac{0.0600}{2} = 0.0300\ \text{mol}.

Step 4: Convert to mass

Multiply moles by the molar mass of $\text{CO}_2$ :

m(\text{CO}_2) = n \times M = 0.0300 \times 44.0 = 1.32\ \text{g}.

Final answer: HCl is the limiting reagent; the mass of $\text{CO}_2$ produced is $1.32\ \text{g}$ (3 significant figures).

Worked example

Question 2 (4 marks): Equilibrium and Le Chatelier

For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ , $\Delta H = +57\ \text{kJ mol}^{-1}$ . Predict and explain the effect on the amount of $\text{NO}_2$ of (a) increasing temperature and (b) increasing the volume of the container at constant temperature.

Solution.
(a) The forward reaction is endothermic ( $\Delta H > 0$ ). Increasing temperature favours the endothermic (forward) direction to absorb the added energy, shifting the equilibrium right and increasing the amount of $\text{NO}_2$ (2 marks).
(b) Increasing volume decreases the pressure, so the system shifts toward the side with more gas moles. There is 1 mol of gas on the left and 2 mol on the right, so it shifts right, increasing $\text{NO}_2$ (2 marks). Full marks require naming the driver (opposing the imposed change), not just the direction.

Worked example

Question 3 (6 marks): Spectroscopy and structure

A compound with molecular formula $\text{C}_3\text{H}_6\text{O}$ shows a strong IR absorption near $1715\ \text{cm}^{-1}$ , no broad O-H band, and a $^{13}$ C NMR spectrum with three distinct carbon environments. Deduce the functional group and propose a structure, justifying with the data.

Work through the evidence in order: degrees of unsaturation, IR functional group, then NMR carbon count.

Step 1: Calculate degrees of unsaturation

For a molecular formula $\text{C}_n\text{H}_m\text{O}_x$ , the degrees of unsaturation are $\dfrac{2n + 2 - m}{2}$ . For $\text{C}_3\text{H}_6\text{O}$ :

\text{degrees of unsaturation} = \frac{2(3) + 2 - 6}{2} = 1.

One degree of unsaturation indicates a single double bond or ring. No nitrogen or halogen is present, and there is one oxygen, so the double bond is most likely a carbonyl C=O.

Step 2: Assign the IR evidence

The strong band near $1715\ \text{cm}^{-1}$ is characteristic of a C=O (carbonyl) stretch. The absence of a broad O-H absorption in the $2500\text{-}3300\ \text{cm}^{-1}$ region rules out a carboxylic acid. The formula has only one oxygen, so an ester (which needs two oxygens) is also ruled out, leaving a ketone or aldehyde as candidates.

Step 3: Use the $^{13}$ C NMR to select between candidates

Three distinct carbon environments are observed. Consider propanone ( $\text{CH}_3\text{COCH}_3$ ): its two methyl carbons are equivalent by symmetry, giving only two environments, so propanone is inconsistent. Propanal ( $\text{CH}_3\text{CH}_2\text{CHO}$ ) has three inequivalent carbons: the $\text{CH}_3$ group, the $\text{CH}_2$ group, and the carbonyl carbon. This matches the NMR data.

An aldehyde C-H stretch near $2700\text{-}2850\ \text{cm}^{-1}$ in the IR spectrum would provide additional confirmation.

Final answer: the compound is propanal, $\text{CH}_3\text{CH}_2\text{CHO}$ . Full marks require linking each spectroscopic feature explicitly to a structural feature.

Common errors students made

The 2022 VCAA examination report flagged students losing marks by omitting state symbols, mixing up oxidising and reducing agents, quoting Le Chatelier shifts without naming the driver, and drawing energy profile diagrams without an activation-energy marker. Add these recurring traps:

Limiting-reagent neglect - calculating yield from the reagent in excess instead of identifying the limiting reagent first.
Significant figures and units dropped at the final step or carried inconsistently.
Equivalent vs inequivalent environments - miscounting NMR peaks by ignoring molecular symmetry.
Mole-ratio errors - using a 1:1 ratio when the balanced equation is not (e.g. the 2:1 HCl-to- $\text{CO}_2$ ratio above).

How to use this paper

Sit Section A in ~35 minutes, then Section B in ~110 minutes with only the VCAA data book. Mark against the official VCAA examination report and assessment criteria (linked in the frontmatter above) to see where method marks fall. Rebuild every multi-step calculation that lost more than half its marks, writing one labelled line per quantity and identifying the limiting reagent first. Drill the IR/NMR/MS fingerprint table until you can map a peak to a functional group and a structural feature instantly, and practise stating the Le Chatelier driver in words.

Use this paper well

Sit the paper under exam conditions (150 minutes, 120 marks).
Mark yourself against the official VCAA marking notes.
Compare against the Chemistry hub to find the syllabus dot points this paper tested.

What this paper assessed

Structure and timing

Worked practice questions (exam-style)

Common errors students made

How to use this paper

Use this paper well

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