S4 Chemistry Paper: Atomic Structure, Chemical Bonding, Moles And Molar Mass, Solution Concentration, Empirical Formula

S4 Chemistry Paper: Atomic Structure, Chemical Bonding, Moles and Molar Mass, Solution Concentration, Empirical Formula

2026-11-13

S4 Chemistry – atomic structure, bonding and mole calculations

S4 Chemistry paper: 6 questions, 24 parts, 100 marks, 60 minutes on atomic structure, electron arrangement, bonding, moles, concentration and empirical formula.

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Paper details　S4 Chemistry practice paper with 6 questions and 24 parts on atomic structure, electron arrangement, bonding, moles, solution concentration and empirical formula, with marked answers.

Section A: Atomic Structure and Electron Arrangement (16 marks)

Answer the following questions on atomic structure.

#	Question	Answer and marking notes
1	An atom has 16 protons and 18 neutrons. Find its mass number and number of electrons, and write its electron arrangement.	Answer: mass number = 16 + 18 = 34; electrons = protons = 16; arrangement 2, 8, 6. (Marking: mass number 1, electrons 1, arrangement 2)
2	The ion X2- has 18 electrons and mass number 32. Find the proton and neutron numbers of X.	Answer: protons = 18 – 2 = 16; neutrons = 32 – 16 = 16. (Marking: protons 2, neutrons 2)
3	An element is in Period 3, Group VII. Write its symbol, electron arrangement, and the formula of the compound it forms with sodium.	Answer: Cl, arrangement 2, 8, 7, and the compound NaCl (one negative charge on chloride and one positive charge on sodium). (Marking: symbol 1, arrangement 2, formula 1)
4	An element has two isotopes: mass number 35 at 75%% abundance and mass number 37 at 25%%. Find its relative atomic mass.	Answer: 0.75 x 35 + 0.25 x 37 = 26.25 + 9.25 = 35.5. (Marking: each product 1.5, total 1)

Section B: Chemical Bonding and Compounds (16 marks)

Answer the following questions on chemical bonding.

#	Question	Answer and marking notes
5	Classify the bonding in sodium chloride, water, magnesium oxide and methane, and state the basis of your classification.	Answer: sodium chloride and magnesium oxide are ionic (electron transfer between metal and non-metal); water and methane are covalent (shared electron pairs between non-metal atoms). (Marking: 0.5 each, basis 2)
6	Describe the dot-and-cross diagram of a water molecule and explain why water is a polar molecule.	Answer: oxygen shares one pair of electrons with each hydrogen; oxygen is more electronegative so the pairs are pulled towards it, giving two polar bonds and a molecular dipole. (Marking: shared pairs 2, polarity 2)
7	Explain why solid sodium chloride does not conduct electricity at room temperature but molten sodium chloride does.	Answer: in the solid the ions are held in a lattice and cannot move; when molten the ions are free to move and carry charge, so it conducts. (Marking: lattice fixed 2, mobile ions 2)
8	The atomic numbers of potassium, calcium and bromine are 19, 20 and 35. Deduce the charge on each ion and write the formulae of potassium bromide and calcium bromide.	Answer: K loses one electron to form K+, Ca loses two to form Ca2+, Br gains one to form Br-. The formulae are KBr and CaBr2. (Marking: three charges 1 each, two formulae 0.5 each)

Section C: Moles and Molar Mass (18 marks)

Relative atomic masses: H = 1, C = 12, O = 16, Mg = 24, Cl = 35.5, K = 39; molar volume at room conditions is 24 dm3.

#	Question	Answer and marking notes
9	Find the molar mass of carbon dioxide and the number of moles in 8.8 g of it.	Answer: molar mass = 12 + 2 x 16 = 44 g/mol; moles = 8.8 / 44 = 0.2 mol. (Marking: molar mass 2, moles 2)
10	Find the number of moles and the mass of 3.01 x 10^23 water molecules.	Answer: moles = 3.01 x 10^23 / 6.02 x 10^23 = 0.5 mol; mass = 0.5 x 18 = 9.0 g. (Marking: moles 2, mass 2)
11	2.4 g of magnesium reacts completely with excess dilute hydrochloric acid. Find the moles and the volume of hydrogen formed at room conditions.	Answer: moles of Mg = 2.4 / 24 = 0.09999999999999999, so 0.09999999999999999 mol of hydrogen; volume = 0.09999999999999999 x 24 = 2.4 dm3. (Marking: Mg moles 2, hydrogen moles 1, volume 2)
12	12.25 g of potassium chlorate is heated until it decomposes completely (2KClO3 -> 2KCl + 3O2). Find the moles and mass of oxygen formed.	Answer: molar mass of KClO3 = 39 + 35.5 + 48 = 122.5 g/mol, moles = 12.25 / 122.5 = 0.1; oxygen = 0.1 x 3 / 2 = 0.15000000000000002 mol; mass = 0.15000000000000002 x 32 = 4.800000000000001 g. (Marking: moles 2, ratio 2, mass 1)

Section D: Solution Concentration and Dilution (18 marks)

Answer the following questions on concentration.

#	Question	Answer and marking notes
13	5.85 g of sodium chloride is dissolved in water to make 250 cm3 of solution. Find its concentration in mol/dm3.	Answer: moles = 5.85 / 58.5 = 0.09999999999999999; volume = 0.25 dm3; concentration = 0.09999999999999999 / 0.25 = 0.39999999999999997 mol/dm3. (Marking: moles 2, concentration 2)
14	From the 0.39999999999999997 mol/dm3 solution above, prepare 100 cm3 of 0.04 mol/dm3 solution. What volume of the original solution is needed?	Answer: moles needed = 0.04 x 0.1 = 0.004; volume = 0.004 / 0.39999999999999997 = 0.01 dm3 = 10.0 cm3. (Marking: moles 2, volume 2)
15	25.0 cm3 of 0.2 mol/dm3 sodium hydroxide is neutralised by 0.1 mol/dm3 sulphuric acid (2NaOH + H2SO4 -> Na2SO4 + 2H2O). Find the volume of acid needed.	Answer: moles of NaOH = 0.025 x 0.2 = 0.005000000000000001; acid = 0.005000000000000001 / 2 = 0.0025000000000000005; volume = 0.0025000000000000005 / 0.1 = 0.025 dm3 = 25.0 cm3. (Marking: NaOH moles 2, ratio 2, volume 1)
16	A salt has a solubility of 36 g per 100 g of water at 20 C. Find the percentage by mass of the saturated solution (one decimal place).	Answer: total mass = 36 + 100 = 136 g; percentage by mass = 36 / 136 x 100% = 26.5%. (Marking: total mass 2, percentage 2)

Section E: Empirical Formula and Percentage Composition (16 marks)

Answer the following questions on empirical formulae.

#	Question	Answer and marking notes
17	A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Find its empirical formula.	Answer: mole ratios C = 40 / 12 = 3.33, H = 6.7 / 1 = 6.7, O = 53.3 / 16 = 3.33, which simplifies to 1 : 2 : 1, so the empirical formula is CH2O. (Marking: three mole ratios 1 each, formula 1)
18	A compound has the empirical formula CH2O and a molar mass of 180 g/mol. Find its molecular formula.	Answer: empirical formula mass = 12 + 2 + 16 = 30; multiplier = 180 / 30 = 6; molecular formula C6H12O6. (Marking: formula mass 1, multiplier 2, formula 1)
19	Find the percentage by mass of iron in iron(III) oxide, Fe2O3 (Fe = 56, O = 16).	Answer: formula mass = 2 x 56 + 3 x 16 = 160; percentage of iron = 112 / 160 x 100% = 70.0%. (Marking: formula mass 2, percentage 2)
20	8.0 g of copper(II) oxide is reduced completely to give 6.4 g of copper. Find the relative atomic mass of copper (O = 16).	Answer: the moles of copper oxide equal the moles of copper, so 8.0 / (x + 16) = 6.4 / x; solving gives 8x = 6.4x + 102.4 and x = 64.00000000000001. (Marking: equation 2, solution 2)

Section F: Mixed Calculations (16 marks)

Answer the following questions and show your working.

#	Question	Answer and marking notes
21	12 g of carbon burns completely in excess oxygen. Find the mass of carbon dioxide formed.	Answer: moles of carbon = 12 / 12 = 1, so 1 mol of carbon dioxide; mass = 1 x 44 = 44 g. (Marking: moles 2, mass 2)
22	4.6 g of ethanol (C2H5OH, molar mass 46 g/mol) burns completely. Find the moles of carbon dioxide formed.	Answer: moles of ethanol = 4.6 / 46 = 0.09999999999999999; one mole gives two moles of carbon dioxide, so 0.19999999999999998 mol. (Marking: ethanol moles 2, ratio 2)
23	2.7 g of aluminium burns completely in oxygen (4Al + 3O2 -> 2Al2O3). Find the mass of aluminium oxide formed (Al = 27, O = 16).	Answer: moles of Al = 2.7 / 27 = 0.1; oxide = 0.1 / 2 = 0.05; mass = 0.05 x 102 = 5.1 g. (Marking: Al moles 1, ratio 2, mass 1)
24	5.0 g of calcium carbonate decomposes completely (CaCO3 -> CaO + CO2). Find the volume of carbon dioxide at room conditions (molar volume = 24 dm3).	Answer: moles of calcium carbonate = 5.0 / 100 = 0.05, so 0.05 mol of carbon dioxide; volume = 0.05 x 24 = 1.2 dm3. (Marking: moles 2, volume 2)

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