Structure 1 – Models of the Particulate Nature of Matter: Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

Related: Structure 1 revision notes and the IB DP Chemistry syllabus guide.

Section A

1. State the number of protons, neutrons and electrons in an atom of chlorine-37 (atomic number 17). [2]

2. Write the full electron configuration of a phosphorus atom (atomic number 15). [2]

3. Distinguish between a physical change and a chemical change, giving one example of each. [2]

Section B

4. A hydrocarbon has an empirical formula of CH2 and a molar mass of 84.16 g mol-1.

(a) Calculate the empirical formula mass. [1] (b) Determine the molecular formula of the hydrocarbon. [2]

5. A sample of magnesium oxide contains 2.41 g of magnesium (Ar = 24.31) combined with oxygen.

(a) Calculate the amount, in moles, of magnesium present. [1] (b) Given that magnesium oxide has the formula MgO, calculate the mass of oxygen combined with this magnesium. [2] (c) Hence calculate the total mass of magnesium oxide formed. [1]

6. A fixed mass of an ideal gas occupies 2.00 dm3 at 300 K and 100 kPa.

(a) State two assumptions of the ideal gas model. [2] (b) Calculate the volume the gas would occupy at 450 K and 100 kPa, assuming ideal behaviour. [2]

Section C

7. Isotopes of the same element have identical chemical properties but different physical properties.

(a) Explain, in terms of sub-atomic particles, why this is the case. [2] (b) Chlorine exists as two isotopes, 35Cl (75.77% abundance) and 37Cl (24.23% abundance). Calculate the relative atomic mass of chlorine to two decimal places. [2] (c) Explain why the relative atomic mass of chlorine on the periodic table is not a whole number. [2]

Worked answers

1. Protons = 17, neutrons = 37 - 17 = 20, electrons = 17 (a neutral atom has equal protons and electrons). [1] for correct proton/electron count, [1] for correct neutron count.

2. 1s2 2s2 2p6 3s2 3p3. [1] for correct sublevels present, [1] for correct electron count in each (totalling 15).

3. A physical change does not alter chemical composition, e.g. melting ice to water. [1] A chemical change alters chemical composition via bond-breaking and bond-forming, e.g. burning magnesium in oxygen to form magnesium oxide. [1]

4. (a) CH2 mass = 12.01 + 2(1.01) = 14.03 g mol-1. [1] (b) 84.16 / 14.03 ~= 6, so molecular formula = C6H12. [2] (1 mark for correct ratio, 1 for correct formula)

5. (a) n = m/Ar = 2.41/24.31 = 0.0991 mol (3 s.f.). [1] (b) MgO has a 1:1 mole ratio, so moles of O = 0.0991 mol; mass of O = 0.0991 x 16.00 = 1.59 g (3 s.f.). [2] (1 mark for correct mole ratio reasoning, 1 for correct mass) (c) Total mass = 2.41 + 1.59 = 4.00 g. [1]

6. (a) Any two of: negligible particle volume compared with the container; no intermolecular forces between particles; perfectly elastic collisions between particles and with the container walls. [2] (b) At constant pressure, V1/T1 = V2/T2, so V2 = V1 x (T2/T1) = 2.00 x (450/300) = 3.00 dm3. [2] (1 mark for correct relationship/method, 1 for correct answer)

7. (a) Isotopes have the same number of protons and electrons, so their electron configuration – and therefore chemical bonding behaviour – is identical; they differ only in neutron number, which affects mass but not charge or electron arrangement. [2] (b) RAM = (35 x 0.7577) + (37 x 0.2423) = 26.5195 + 8.9651 = 35.48 (2 d.p.). [2] (1 mark for correct weighted-average method, 1 for correct numerical answer) (c) Because it is a weighted average of the masses of all naturally occurring isotopes of the element, weighted by relative abundance, not the mass of a single atom or isotope. [2]

Command term reminder

“Calculate” (questions 4a, 5, 6b and 7b) and “Determine” (question 4b) require a numerical answer or formula with full working shown – an unsupported final number rarely earns full marks even if correct, because the mark scheme awards marks for method as well as outcome. “Explain” (questions 7a and 7c) needs a reasoned chain of statements linking cause to effect, not just a one-line assertion – for example, question 7(a) requires both the shared electron configuration point and the consequence for chemical behaviour to earn both marks. “State” (question 1, 6a) wants a concise fact with no justification expected, so do not spend exam time explaining a “state” answer. “Distinguish between” (question 3) explicitly requires two contrasting definitions in the same answer, not two separate answers written as if the question asked for one thing at a time – examiners look for the contrast to be made explicit, usually by naming both terms in the same sentence or by using a connective such as “whereas” or “in contrast.”

Why this scope

These questions cover all five Structure 1 sub-topics – particulate matter, the nuclear atom, electron configurations, the mole and ideal gases. The guide gives teaching hours for Structure 1 as a whole (17 SL, 21 HL), not for each sub-topic, so the balance of questions here is Marlbridge’s own choice. Question 5 in particular starts with the same mass-to-moles step as the revision notes worked example (which runs mass to moles to particles), then continues through a mole ratio back to a mass, so it tests transfer of the method rather than memorisation of one specific chain.

Official syllabus

International Baccalaureate Organization, Chemistry guide, Diploma Programme, first assessment 2025 (published February 2023, updated May 2023) – copy consulted via a school-hosted mirror, not ibo.org. The same source is already cited by the Structure 1 revision notes and the full syllabus guide.