June 2026
The 2025 VCE Chemistry exam showed that calculation discipline is one of the clearest differences between capable and high-scoring students.
Many of the calculations were not conceptually unusual. Students were asked to calculate energy content, calibration factor, energy absorbed by water, fuel efficiency, limiting reagent, electroplating current, equilibrium concentrations, titre averages and masses from redox titrations.
The chemistry was familiar.
The execution was demanding.
Students needed to track units, convert volumes correctly, use balanced equations, apply mole ratios, show working, quote appropriate units and interpret what the final number meant.
This is where marks were often lost.
In Chemistry, a correct method can still collapse if the units, ratios or final interpretation are wrong.
Unit errors appeared from the beginning
Question 2 asked students to calculate the energy content per 100 g of a snack bar.
The 65 g snack bar contained:
- 45.0 g carbohydrate
- 10.0 g protein
- 10.0 g fats
Using the energy values provided in the Data Book, the energy in the 65 g snack bar was:
(16 × 45.0) + (17 × 10.0) + (37 × 10.0) = 1260 kJ
Per 100 g:
1260 × 100 ÷ 65 = 1938 kJ
The answer needed to be expressed as:
1.9 × 10⁶ J
The Examination Report noted that the most common error was a unit error.
This is an important warning. Students may have calculated the energy correctly in kilojoules, but the options were in joules. That final conversion changed the answer.
In VCE Chemistry, units are not a finishing detail. They are part of the calculation.
Calibration factor required correct substitution
Question 5 asked students to calculate the calibration factor of a solution calorimeter.
Oscar passed a current of 2.50 A through a heater with a voltage of 12.0 V for 150 s. The temperature increased from 22.3 °C to 28.4 °C, so:
ΔT = 6.1 °C
The energy supplied was:
E = VIt
The calibration factor was:
CF = VIt ÷ ΔT
CF = 12.0 × 2.50 × 150 ÷ 6.1 = 738 J °C⁻¹
This was one of the more successfully answered calculations, but it still illustrates a key skill.
Students needed to know what the calibration factor represents: the energy required to raise the calorimeter’s temperature by one degree Celsius.
That is why the unit is J °C⁻¹.
A calculation is stronger when the student understands the unit that the formula should produce.
Thermochemical equations needed states and enthalpy
Section B Question 1a asked students to write a balanced thermochemical equation for the complete combustion of propane.
The expected equation was:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) ΔH = −2220 kJ
The report noted common errors: failing to provide an enthalpy value, using incorrect states, or balancing the equation incorrectly.
This was not just an equation-writing question.
It required students to match the enthalpy value to the correct states at SLC. If the state of water is wrong, the enthalpy value no longer corresponds to the equation being written.
That is why Chemistry marking is so precise.
The equation, states and enthalpy value form one complete chemical statement.
Energy released needed the correct sign language
Question 1b asked students to calculate the amount of energy released by the complete combustion of 9.0 g of propane.
The amount of propane was:
n(C₃H₈) = 9.0 ÷ 44.0 = 0.2045 mol
Using the heat of combustion:
Energy released = 0.2045 × 2220 = 454 kJ
The report noted that some students incorrectly included the negative sign and described the energy released as −454 kJ.
This is a subtle but important communication issue.
The enthalpy change for combustion is negative because the reaction is exothermic. But when asked for the amount of energy released, the answer should be given as a positive quantity of energy released.
The sign and the wording must agree.
A reaction may have ΔH = −454 kJ, but the energy released is 454 kJ.
Energy absorbed by water required q = mcΔT
Question 1c.i asked students to calculate the energy absorbed by the water in the saucepan.
The water mass was 166.0 g. The temperature increased from 24.1 °C to 90.1 °C, so:
ΔT = 66.0 °C
Using:
q = mcΔT
q = 166.0 × 4.18 × 66.0 = 45 796 J
So:
q = 45.8 kJ
This calculation was handled reasonably well, but it contains two important habits.
First, students must calculate ΔT correctly. Second, they must manage the unit conversion from joules to kilojoules if they later compare the value with energy released by combustion.
A common Chemistry error is comparing values in different units without noticing.
That can destroy the efficiency calculation that follows.
Efficiency required comparing useful energy with total energy
Question 1c.ii asked students to determine the overall energy efficiency of heating the water.
The water absorbed 45.8 kJ.
The propane combustion released 454 kJ.
The efficiency was:
45.8 ÷ 454 × 100 = 10%
This result also makes chemical sense. In a camp stove and saucepan set-up, not all energy released by combustion is transferred to the water. Much is lost to the surroundings, saucepan, air and incomplete heat transfer.
Efficiency calculations require students to identify the useful energy output and the total energy input.
In this question, the useful energy was the energy absorbed by the water.
The energy released by the fuel was the input.
The distinction matters.
Limiting reagent required volume conversion
Question 1d was one of the clearest calculation discriminators.
Kim combusted 9.0 g of propane in a sealed container with total air volume of 0.125 m³. The air contained 21.0% v/v oxygen at SLC. Students were asked to determine the mass of the reactant in excess.
The report identified two major issues. Some students did not convert m³ to L correctly. Others did not determine the quantity of excess reagent remaining after reaction.
The conversion was essential:
0.125 m³ = 125 L
The volume of oxygen was:
0.210 × 125 = 26.25 L
At SLC:
n(O₂) = 26.25 ÷ 24.8 = 1.058 mol
From earlier:
n(C₃H₈) = 0.2045 mol
The combustion equation requires 5 mol of O₂ per 1 mol of C₃H₈, so oxygen required was:
5 × 0.2045 = 1.0225 mol
Oxygen was in excess:
n(O₂ excess) = 1.058 − 1.0225 = 0.0355 mol
Mass of oxygen remaining:
m(O₂) = 0.0355 × 32.0 = 1.1 g
This question shows how multi-step Chemistry calculations work. Every step depends on the one before it.
A unit conversion error at the start makes the limiting reagent conclusion wrong.
Excess reagent meant what was left, not what was present
Question 1d also exposed a conceptual calculation issue.
Students were not simply asked to identify which reactant was in excess. They were asked to determine the mass of the reactant in excess.
That means the amount remaining after reaction had to be calculated.
It was not enough to calculate the initial amount of oxygen. It was not enough to state that oxygen was in excess. Students needed to subtract the oxygen consumed by combustion and then convert the remaining oxygen to mass.
This is a common VCE Chemistry issue.
The question may ask for the excess reagent, the limiting reagent, the amount consumed or the amount remaining. These are different quantities.
High-scoring students read that wording carefully.
Fuel comparison needed energy per mass or mole
Question 1e asked why different masses of ethanol and propane were required to produce the same temperature change in the water.
The report noted that this question was poorly answered.
The key point was that ethanol has a lower energy content than propane, whether expressed per gram or per mole.
The report gave the comparison:
Heat of combustion of ethanol = 29.7 kJ g⁻¹
Heat of combustion of propane = 40.5 kJ g⁻¹
Because ethanol releases less energy per gram, a larger mass of ethanol was required to produce the same temperature change.
The second part of the explanation was chemical. Ethanol is already partially oxidised because of the hydroxy group in its structure, so it releases less energy on complete combustion than propane.
A vague response about “different fuels having different energy contents” was not strong enough.
The comparison needed to be direct.
Electroplating current required Faraday’s law
Question 2d asked students to calculate the current required to electroplate 7.80 g of chromium over 2.50 hours.
The chromium half-equation was:
Cr³⁺(aq) + 3e⁻ → Cr(s)
First, calculate the amount of chromium:
n(Cr) = 7.80 ÷ 52.0 = 0.150 mol
Each mole of chromium requires 3 moles of electrons:
n(e⁻) = 0.150 × 3 = 0.450 mol
Charge transferred:
Q = n(e⁻)F
Q = 0.450 × 96500 = 43 425 C
Time:
2.50 h = 2.50 × 60 × 60 = 9000 s
Current:
I = Q ÷ t = 43 425 ÷ 9000 = 4.83 A
This calculation required students to link electrochemistry and stoichiometry.
The half-equation provided the electron ratio. Faraday’s constant converted moles of electrons into charge. Time converted charge into current.
Each step had a unit.
Rhodium required amount of electrons, not voltage comparison
Question 2e asked why less current would be required to plate the same mass of rhodium compared with chromium.
This was not asking students to compare the half-cell potentials.
The report noted that a common misconception was attempting to explain the current difference using voltages.
The key was molar mass and electron requirement.
Both chromium and rhodium ions had a 3+ charge, so each metal ion required three electrons to form the metal atom. However, rhodium has a larger atomic mass than chromium. For the same mass of metal, fewer moles of rhodium atoms are needed than chromium atoms. Therefore, fewer moles of electrons are required.
For the same plating time, a smaller amount of charge needs to pass through the circuit, so the current is lower.
This is a beautiful example of VCE Chemistry reasoning.
The answer is not always in the most obvious data. Here, the half-cell potential was not the relevant quantity. The mole ratio and molar mass were.
Equilibrium calculations required reaction quotient reasoning
Question 3c asked students to determine what needed to happen for a methanol synthesis system to return to equilibrium.
The equilibrium expression was:
Kc = [CH₃OH] ÷ ([CO][H₂]²) = 1.21 M⁻²
At time t₁:
- [CO] = 2.00 M
- [H₂] = 1.25 M
- [CH₃OH] = 5.60 M
Students needed to calculate the reaction quotient:
Q = 5.60 ÷ (2.00 × 1.25²)
Q = 5.60 ÷ 3.125 = 1.792
Since Q > Kc, the system had too much product relative to reactants. To return to equilibrium, the system needed to shift left, decreasing the concentration of methanol and increasing the concentrations of carbon monoxide and hydrogen.
This question rewarded students who understood the purpose of Q.
It is not enough to plug values into an expression. Students must compare Q with Kc and state the direction of shift.
ICE-style calculations required careful stoichiometry
Question 3d asked students to calculate the initial amount of hydrogen added to a 1.00 L container when 5.00 mol of carbon monoxide and an unknown amount of hydrogen were introduced, and the equilibrium concentration of methanol was 1.30 M.
The reaction was:
CO(g) + 2H₂(g) ⇌ CH₃OH(g)
If [CH₃OH]eq = 1.30 M, then 1.30 mol of methanol was produced in the 1.00 L container.
That means:
- CO consumed = 1.30 mol
- H₂ consumed = 2 × 1.30 = 2.60 mol
- CO at equilibrium = 5.00 − 1.30 = 3.70 M
Using Kc:
1.21 = 1.30 ÷ (3.70 × [H₂]eq²)
Solving gives the equilibrium concentration of hydrogen, then the initial amount is found by adding back the 2.60 mol consumed.
This is the kind of multi-step equilibrium calculation where students must keep the reaction stoichiometry visible.
The most common errors in such questions usually come from forgetting the 2:1 ratio between hydrogen and methanol.
Yield calculations required proportional thinking
Question 24 in Section A asked about a two-step manufacturing process.
Step 1 had a yield of 45%. The overall yield was 16%. Students had to calculate the yield of Step 2.
The relationship was:
overall yield = step 1 yield × step 2 yield
So:
0.16 = 0.45 × step 2 yield
step 2 yield = 0.16 ÷ 0.45 = 0.355
That is approximately 36%.
This calculation is straightforward, but it tests whether students understand multiplicative yield.
A common mistake is to subtract percentages or treat each yield independently. In a multi-step process, losses compound.
Chemistry manufacturing calculations often require proportional reasoning rather than a single formula.
Titration calculations required dilution and stoichiometry
Later in the paper, Question 6 involved oxalate ions in spinach and titration with permanganate.
The report noted that the most common issue was students not allowing for dilution or reaction stoichiometry in their calculations.
A strong response required several steps:
- calculate moles of permanganate in the titre
- use the balanced redox stoichiometry to find moles of oxalate in the aliquot
- scale from the aliquot to the full sample solution
- calculate mass of oxalate ion
This is a familiar pattern in analytical Chemistry.
Titration questions are rarely just cV = n. They often include aliquots, dilution factors, reaction ratios and sample masses.
Students must track what portion of the sample is being analysed and how it relates to the whole.
Calculation working mattered
The exam instructions were clear: students needed to show working in numerical questions, and incorrect answers without working could not earn method marks.
This is especially important in Chemistry because many calculations involve several steps.
A student may make an arithmetic error but still show correct chemical reasoning. Without working, that reasoning cannot be credited.
Good working is not excessive. It should show:
- the formula or chemical relationship used
- substituted values
- mole ratios from equations or half-equations
- unit conversions
- final answer with units and appropriate significant figures
This is not just presentation.
It protects marks.
Why calculation errors are often avoidable
Many calculation errors in Chemistry are not caused by not knowing the topic.
They come from rushing.
Students:
- forget to convert kJ to J
- use mL instead of L
- forget that 1 m³ = 1000 L
- compare values in different units
- use the wrong mole ratio
- calculate initial amount instead of excess remaining
- include a negative sign when asked for energy released
- ignore dilution factors
- treat voltage as relevant when current depends on charge and time
- forget to quote units
These mistakes are avoidable with a disciplined calculation routine.
Read. Identify. Convert. Calculate. Check units. Interpret.
That routine matters.
What future Chemistry students should learn from 2025
The 2025 VCE Chemistry exam shows that calculation skills need to be practised as chemical reasoning, not arithmetic.
Students should be able to:
- convert between kJ and J accurately
- calculate energy per 100 g from food composition data
- calculate calibration factor using voltage, current, time and temperature change
- write thermochemical equations with states and enthalpy values
- distinguish ΔH sign from energy released
- use q = mcΔT correctly
- calculate energy efficiency
- perform limiting reagent calculations involving gas volumes
- determine excess reagent remaining
- compare fuels using energy per gram or mole
- apply Faraday’s law in electroplating
- use molar mass and ion charge to compare current requirements
- calculate reaction quotients and equilibrium shifts
- handle multi-step titration calculations with dilution and stoichiometry
These skills decide marks across the whole paper.
In Chemistry, calculation accuracy is chemical accuracy.
How ATAR STAR approaches Chemistry calculations
At ATAR STAR, Chemistry calculations are taught as structured reasoning.
Students learn to set up the chemistry before touching the calculator: identify the reaction, balance the equation, determine the mole ratio, convert units, carry significant figures and interpret the final answer in context.
The 2025 Examination Report confirms why this matters. High-scoring students did not simply know formulas. They controlled the calculation from start to finish.
They made the numbers obey the chemistry.