Common mistakes · Chemistry · 10 min read
IB Chemistry common mistakes: 17 slips that cost marks, with the fix for each
Seventeen errors that cost marks on the current syllabus, with the wrong working next to the right working and every number checked.
On this page
- Which calculation mistakes cost the most marks?
- Where do energetics answers go wrong?
- What do students get wrong about equilibrium and acids?
- Which redox and atomic-structure slips are common?
- Which bonding mistakes keep coming back?
- Which organic chemistry mistakes lose marks?
- Why do "explain" answers score low?
- How do you make these fixes stick?
The mistakes that cost the most marks in IB Chemistry are mechanical: a mole ratio left out, a volume in cm3 that should be in dm3, a ΔH with the wrong sign, a curly arrow starting from the wrong place. These slips cost marks even when you know the topic. Below are 17 of them, grouped by topic, each with the wrong working next to the right working. Redo the calculations yourself rather than just reading them.
Which calculation mistakes cost the most marks?
1. Rounding in the middle, or to the wrong precision (Structure 1.4)
Keep full precision on your calculator and round once, at the end, to the number of significant figures of the least precise data. Rounding early can move the answer outside the accepted range.
Example. 1.23 g of NaCl (M = 58.44 g mol−1) is dissolved to make 250.0 cm3 of solution.
Loses marks n = 1.23 ÷ 58.44 ≈ 0.02 mol; c = 0.02 ÷ 0.2500 = 0.08 mol dm−3.
Earns marks n = 1.23 ÷ 58.44 = 0.021047 mol (kept in the calculator); c = 0.021047 ÷ 0.2500 = 0.0842 mol dm−3 (3 s.f., matching 1.23 g).
2. Forgetting the mole ratio (Reactivity 2.1)
Write the balanced equation before any numbers. Most wrong titration answers come from assuming 1:1.
Example. 25.0 cm3 of 0.100 mol dm−3 NaOH is neutralised by 12.5 cm3 of H2SO4. H2SO4 + 2NaOH → Na2SO4 + 2H2O.
Loses marks n(NaOH) = 2.50 × 10−3 mol = n(H2SO4); c = 2.50 × 10−3 ÷ 0.0125 = 0.200 mol dm−3.
Earns marks n(NaOH) = 0.0250 × 0.100 = 2.50 × 10−3 mol; n(H2SO4) = ½ × 2.50 × 10−3 = 1.25 × 10−3 mol; c = 1.25 × 10−3 ÷ 0.0125 = 0.100 mol dm−3.
3. Picking the limiting reagent by mass (Reactivity 2.1)
The limiting reagent is decided by moles, scaled by the coefficients in the equation, never by which mass is smaller.
Example. 4.0 g of H2 reacts with 16.0 g of O2: 2H2 + O2 → 2H2O.
Loses marks H2 is limiting because 4.0 g is less than 16.0 g.
Earns marks n(H2) = 4.0 ÷ 2.02 = 1.98 mol; n(O2) = 16.0 ÷ 32.00 = 0.500 mol. 1.98 mol H2 would need 0.99 mol O2, so O2 is limiting. Mass of water = 2 × 0.500 × 18.02 = 18.0 g.
4. Mixing cm3 and dm3, or using the wrong molar volume (Structure 1.4, 1.5)
Concentrations are in mol dm−3, so divide cm3 by 1000 before multiplying. For gases at STP, use the molar volume in the data booklet (22.7 dm3 mol−1 at 273 K and 100 kPa), not 22.4 or 24 remembered from another course. In pV = nRT with p in Pa, V must be in m3.
Loses marks n = 25.0 × 0.100 = 2.50 mol.
Earns marks n = (25.0 ÷ 1000) dm3 × 0.100 mol dm−3 = 2.50 × 10−3 mol. (A titration giving 2.5 mol of anything should make you stop and check.)
Where do energetics answers go wrong?
5. Calorimetry: wrong mass, wrong unit, wrong sign (Reactivity 1.1)
Three slips in one calculation. The mass in q = mcΔT is the mass of the whole solution being heated. q comes out in joules but ΔH is quoted in kJ mol−1. And a temperature rise means the reaction released heat, so ΔH is negative.
Example. 50.0 cm3 of 1.00 mol dm−3 HCl is mixed with 50.0 cm3 of 1.00 mol dm−3 NaOH; the temperature rises by 6.8 K. Assume the solution has density 1.00 g cm−3 and c = 4.18 J g−1 K−1.
Loses marks q = 50.0 × 4.18 × 6.8 = 1421 J; ΔH = 1421 ÷ 0.0500 = +28 420 kJ mol−1.
Earns marks q = 100.0 × 4.18 × 6.8 = 2842 J = 2.842 kJ; n(H2O) = 0.0500 mol; ΔH = −2.842 ÷ 0.0500 = −57 kJ mol−1 (2 s.f., limited by ΔT).
6. Subtracting bond enthalpies the wrong way round (Reactivity 1.2)
Breaking bonds takes in energy; forming bonds releases it. So ΔH = Σ(bonds broken) − Σ(bonds formed). The trap is that the Hess's law route from enthalpies of formation goes the other way: products minus reactants. Write the formula at the top of your working each time so you never have to remember which is which under pressure.
Example. H2(g) + Cl2(g) → 2HCl(g), with H–H 436, Cl–Cl 242, H–Cl 431 kJ mol−1.
Loses marks ΔH = 2(431) − (436 + 242) = +184 kJ mol−1.
Earns marks ΔH = (436 + 242) − 2(431) = 678 − 862 = −184 kJ mol−1. Forming two strong H–Cl bonds releases more energy than breaking H–H and Cl–Cl takes in.
What do students get wrong about equilibrium and acids?
7. Thinking anything that shifts equilibrium changes K (Reactivity 2.3)
Le Châtelier's principle predicts the shift in position. The equilibrium constant K changes only with temperature. For N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1:
| Change | Position | K |
|---|---|---|
| Add N2 | Shifts right | Unchanged |
| Increase pressure | Shifts right (4 mol of gas → 2 mol) | Unchanged |
| Add a catalyst | No change; equilibrium is reached faster | Unchanged |
| Increase temperature | Shifts left (endothermic direction) | Decreases |
Loses marks Increasing the pressure increases K, so more ammonia forms.
Earns marks Increasing the pressure shifts the position to the right, the side with fewer moles of gas, so the yield of ammonia increases. K is unchanged because the temperature is unchanged.
8. Using "strong" to mean "concentrated" (Reactivity 3.1)
Strength is how fully an acid dissociates in water; concentration is how much of it is dissolved per dm3. 0.0010 mol dm−3 HCl is a dilute strong acid (pH 3.00). 10 mol dm−3 ethanoic acid is a concentrated weak acid.
Loses marks Ethanoic acid is weaker than hydrochloric acid because it is more dilute.
Earns marks Ethanoic acid is a weak acid because only a small fraction of its molecules dissociate into H+ and CH3COO− ions; HCl dissociates completely. At equal concentrations, ethanoic acid has a lower [H+] and a higher pH.
9. Taking −log of a weak acid's concentration (Reactivity 3.1, HL)
pH = −log10[H+], and for a weak acid [H+] is far smaller than the acid concentration. Use Ka.
Example. 0.100 mol dm−3 ethanoic acid, Ka = 1.74 × 10−5 (pKa = 4.76).
Loses marks pH = −log(0.100) = 1.00.
Earns marks [H+]2 ≈ Ka × 0.100 = 1.74 × 10−6; [H+] = 1.32 × 10−3 mol dm−3; pH = 2.88. (Assumes so little dissociates that [CH3COOH] stays ≈ 0.100.)
Which redox and atomic-structure slips are common?
10. Confusing the oxidising agent with what is oxidised (Reactivity 3.2)
The oxidising agent takes electrons, so it is itself reduced. Track oxidation states: oxidation is an increase, reduction a decrease. Write oxidation states with the sign first (+7) and ionic charges with the sign last (2+).
Example. MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O. Mn: +7 → +2 (reduced). Fe: +2 → +3 (oxidised).
Loses marks MnO4− is oxidised because it is the oxidising agent.
Earns marks MnO4− is the oxidising agent: manganese is reduced from +7 to +2 as it gains electrons from Fe2+, which is oxidised to Fe3+.
11. Writing Cr, Cu and transition-metal ions "by the rule" (Structure 1.3)
Two exceptions to learn: Cr is [Ar]3d54s1 and Cu is [Ar]3d104s1. And when transition metals form ions, the 4s electrons are lost before the 3d.
Loses marks Cu: [Ar]3d94s2. Fe2+: [Ar]3d44s2.
Earns marks Cu: [Ar]3d104s1. Fe2+: [Ar]3d6 (Fe is [Ar]3d64s2; both 4s electrons go first).
12. Multiplying E⦵ when you multiply a half-equation (Reactivity 3.2, HL)
Standard electrode potentials are read straight from the data booklet and never scaled by the number of electrons. Balancing electrons changes the equation, not the voltage.
Loses marks Zn/Cu cell: Ecell = 2 × 0.34 + 0.76 = +1.44 V because two electrons are transferred.
Earns marks E⦵cell = E⦵(cathode) − E⦵(anode) = +0.34 − (−0.76) = +1.10 V.
Which bonding mistakes keep coming back?
13. Breaking covalent bonds when a substance boils (Structure 2.2)
Melting and boiling a molecular substance overcome the forces between molecules; the covalent bonds inside them stay intact. Hydrogen bonding needs H bonded directly to N, O or F, so HCl has dipole–dipole forces and London dispersion forces, not hydrogen bonds.
Loses marks Water has a high boiling point because a lot of energy is needed to break the O–H bonds.
Earns marks Water has a high boiling point for its size because hydrogen bonds between molecules must be overcome; these are stronger than the London dispersion forces between molecules of a similar-sized non-polar substance such as methane.
14. Assuming polar bonds make a polar molecule (Structure 2.2)
A molecule is polar only if its bond dipoles do not cancel. CO2 (linear), BF3 (trigonal planar) and CCl4 (tetrahedral) all have polar bonds but are non-polar. NH3 (trigonal pyramidal, with a lone pair) and H2O (bent) are polar. Always draw or state the shape before deciding.
Which organic chemistry mistakes lose marks?
15. Curly arrows that start from the wrong place (Reactivity 3.4)
A curly arrow shows the movement of an electron pair, so it must start at a lone pair or at a bond, and end where that pair goes: at an atom, to form a bond, or on an atom that leaves with the pair. Check which mechanisms your level requires, then draw each one the same way every time.
Loses marks OH− + CH3CH2Br: an arrow from the "−" sign or from the H of OH− towards the molecule; an arrow from Br towards C.
Earns marks An arrow from a lone pair on the O of OH− to the δ+ carbon bonded to Br; a second arrow from the C–Br bond to the Br, which leaves as Br−.
16. Naming isomers without finding the longest chain (Structure 3.2)
Find the longest continuous carbon chain first, wherever it bends, then number from the end that gives the functional group (or the first branch) the lowest number.
Loses marks 2-ethylbutane; butan-3-ol for CH3CH2CH(OH)CH3.
Earns marks 3-methylpentane (the ethyl group extends the main chain to five carbons); butan-2-ol (numbering from the other end gives the OH the lower locant).
Why do "explain" answers score low?
17. Describing what happens without the particles
In Chemistry, "explain" almost always means: say what the particles are doing and why that leads to the observation. An answer that only restates the observation, or says particles "move faster", stops one step short.
Loses marks The rate increases at higher temperature because the particles have more energy and move faster.
Earns marks At higher temperature, particles have a higher average kinetic energy, so a greater proportion of collisions have energy equal to or greater than the activation energy (more of the Maxwell–Boltzmann distribution lies above Ea). The frequency of successful collisions increases, so the rate increases.
The same structure works for boiling points, ionic conductivity and equilibrium shifts: name the particles, name the interaction or energy involved, then link it to the result. The Chemistry Paper 2 guide has a full worked "explain" answer, and the command terms guide covers how explain differs from describe and outline.
How do you make these fixes stick?
- Write the formula, then substitute. ΔH = broken − formed, n = cV, q = mcΔT: written at the top of the working, every time. It costs ten seconds and it is where method marks live.
- Sense-check the size. A titration concentration of 20 mol dm−3, a pH of −1 for a weak acid, a ΔH of +28 420 kJ mol−1 for neutralisation: each is a signal to look for a missing 1000 or a missing ratio.
- Practise under exam conditions. In multiple choice the slips above are the distractors; see how to approach Paper 1A. For a broader overview of the papers, read what the Chemistry exams look like now.
Questions students ask
How many significant figures should I give in IB Chemistry?
Give the final answer to the same number of significant figures as the least precise piece of data you used, and keep full precision on your calculator until the end. If the data are 25.0 cm3 and 0.10 mol dm−3, the answer gets 2 significant figures.
Do I need to memorise the IB Chemistry data booklet?
No, you get a clean copy in Paper 1 and Paper 2. You do need to know what is in it and where, so you don't spend minutes hunting for a bond enthalpy, and you need to know which equations and relationships are not in it so you learn those instead. Practise with the same edition you will use in the exam.
How do I get a 7 in IB Chemistry?
Secure the calculation routines (moles, titrations, energetics, equilibrium) until they are automatic, write every explanation in terms of particles, and check each practice answer line by line against a mark scheme. Many of the marks between a 5 and a 7 go on slips like the ones on this page rather than on unfamiliar content.
Try one: Chemistry
Use these average bond enthalpies, in kJ mol−1: C–H 414, O=O 498, C=O 804, O–H 463.
What is the enthalpy change for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), in kJ mol−1?
- −1306
- −808
- +808
- −4
Show the answer and why
Answer: B
Bonds broken: 4 C–H + 2 O=O = 4(414) + 2(498) = 2652 kJ. Bonds formed: 2 C=O + 4 O–H = 2(804) + 4(463) = 3460 kJ. ΔH = broken − formed = 2652 − 3460 = −808 kJ mol−1. C reverses the subtraction. A forgets the 2 in front of O2, and D forgets the 2 C=O bonds in CO2. The standard enthalpy of combustion of methane (about −890 kJ mol−1) is more negative mainly because it forms liquid water, and also because average bond enthalpies are only approximations for any particular molecule.