Problem-solving¶
Syllabus mapping
Working Scientifically outcome — Problem-solving. Covers applying mathematical and conceptual relationships systematically, including in unfamiliar or multi-step contexts, and selecting the appropriate relationship from the data sheet rather than a memorised list of "the formula for this topic."
Outcome codes: CH-11WS-06 (Year 11) and CH-12WS-06 (Year 12) — Problem-solving. Same outcome wording in both years, with a separate code for each. Wording on this page is paraphrased for teaching use — check the official syllabus before treating any line as verbatim NESA text.
You need to know: a reliable process for turning a word problem into a solved answer, and — the harder skill — how to handle a problem that needs several relationships chained together, not just one you can look up directly.
Core¶
A reliable process¶
Every numeric chemistry problem in this course can be worked through the same way:
- List what's given, with units, and what's being asked for. Write the balanced equation if a reaction is involved.
- Identify the relevant relationship(s) — \(n = \dfrac{m}{MM}\), \(C = \dfrac{n}{V}\), \(PV = nRT\), \(q = mc\Delta T\) and so on — and check the data sheet for constants and molar masses rather than relying on memory.
- Rearrange algebraically first, before substituting numbers — this catches mistakes far more easily than rearranging with numbers already in place.
- Substitute with units (converting mL to L and °C to K where needed), and calculate.
- Check the answer: right order of magnitude, right unit, right number of significant figures (see Precision and Uncertainty).
Step 5 is the one students skip under time pressure, and it catches a large fraction of careless errors — 450 g of product from 5 g of reactant should immediately look wrong before you've even checked the working.
Reading the data sheet, not memorising a topic list¶
The exam data sheet gives you constants, a periodic table with relative atomic masses, and a set of formulae — but it doesn't tell you which one applies to your specific problem. That's a skill NESA is deliberately testing. Practise with the actual data sheet (see Resources) so you know where everything is before the exam.
Multi-step problems: chaining relationships¶
The single biggest jump between a "Core" problem and an Advanced one is whether it needs one relationship or several, chained so the result of one calculation becomes an input to the next. In chemistry, almost every chain goes through moles.
Worked example — a chained problem
\(25.0\ \text{mL}\) of \(0.100\ \text{mol L}^{-1}\) silver nitrate is added to excess sodium chloride solution. What mass of silver chloride precipitates?
Step 1 — moles of AgNO₃ from the solution (see Measures of Concentration):
Step 2 — mole ratio (see Mole Concept): \(\ce{AgNO3} : \ce{AgCl} = 1 : 1\), so \(n(\ce{AgCl}) = 0.00250\ \text{mol}\)
Step 3 — mass of AgCl: \(MM = 107.9 + 35.45 = 143.3\ \text{g mol}^{-1}\)
No single step is hard — each is covered on its own page. The skill being tested is recognising that a volume and concentration gives you moles, and moles then plug straight into the mole ratio and \(n = \dfrac{m}{MM}\). Nothing in the question told you to do this explicitly.
Sanity-checking with units¶
Checking that units cancel correctly is a fast way to catch an algebra mistake. If \(n = CV\) gives you mol L⁻¹ × mL, you forgot to convert to litres — the answer will be 1000 times too big. If a molar mass comes out in "g mol" instead of g mol⁻¹, you've multiplied where you should have divided.
Advanced
Recognising which relationship applies when the problem doesn't say. The hardest problems don't name the topic — they describe a real situation (a tablet dissolving in acid, a gas collected over water, a fuel heating a can of water) and expect you to decide which relationships are relevant, sometimes from more than one focus area at once. A common trap is a hidden limiting reagent: if a question gives you amounts of both reactants, always check which one runs out before calculating the product. There's no shortcut beyond understanding what each formula represents — a formula understood as "concentration times volume gives moles" can be spotted in any disguise.
Extension
Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Not every chemical situation has a neat formula. The concentration of a reactant during a reaction keeps changing, and so does the rate, so working out "how much is left after 30 seconds" needs numerical methods: break time into many small steps and, at each step, use the current concentration to calculate the current rate, then use that rate to update the concentration — and repeat (Euler's method, easily built in a spreadsheet). Building a step-by-step spreadsheet model of a reaction and comparing it with real data from a rate prac is a strong, technical depth study angle — and a first taste of how chemists model reactions computationally.
Video/visual resources¶
- 🎥 TODO: source a worked-examples video on multi-step stoichiometry problems (solution → moles → ratio → mass or gas volume).
Check yourself¶
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\(0.50\ \text{g}\) of magnesium is added to \(50.0\ \text{mL}\) of \(0.10\ \text{mol L}^{-1}\) hydrochloric acid. Outline (don't fully solve) the steps you'd use to find the volume of hydrogen produced at STP, and explain why you can't just start from the magnesium.
Answer
First find the moles of each reactant (\(n = \dfrac{m}{MM}\) for Mg, \(n = CV\) for HCl). Then use the mole ratio (\(\ce{Mg + 2HCl -> MgCl2 + H2}\)) to find which is the limiting reagent — here the acid runs out first. Use the moles of the limiting reagent and the ratio to find moles of \(\ce{H2}\), then \(V = n \times 22.71\). You can't start from the magnesium because it's in excess — using it would give too much hydrogen.
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A student calculates that \(5.0\ \text{g}\) of calcium carbonate produces \(450\ \text{g}\) of carbon dioxide. Explain why this should immediately be treated as suspicious.
Answer
By conservation of mass, a product can't have more mass than the reactants it came from — and here the carbonate is only 5.0 g. An answer 90 times bigger points to an arithmetic or conversion error (e.g. multiplying by a molar mass instead of dividing).
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Rearrange \(PV = nRT\) to make \(T\) the subject, and check that the units work out to kelvin (with \(P\) in kPa, \(V\) in L, and \(R\) in J mol⁻¹ K⁻¹).
Answer
\[ T = \frac{PV}{nR} \]kPa × L = J (that's why these units work with \(R = 8.314\)). So the units are \(\dfrac{\text{J}}{\text{mol} \times \text{J mol}^{-1}\,\text{K}^{-1}} = \text{K}\) ✓.