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S4 Chemistry Practice Paper: Atomic Structure, Bonding and Acid-Base Neutralisation (12 questions, 100 marks, 75 minutes)
S4 Chemistry: Atomic Structure, Bonding and Acid-Base Neutralisation
S4 chemistry practice paper: 12 questions on isotopes, ionic and covalent bonding, moles, titration, pH and electrolysis, with full answers and mark points.
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Paper details Level: S4 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Atomic Structure and Bonding (32 marks)
Answer all questions in Section A.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (Isotopes) Copper has two natural isotopes: copper-63 with a relative abundance of 69.2% and copper-65 with 30.8%. (a) Calculate the relative atomic mass of copper to 3 significant figures. (b) Explain why the value is not a whole number. (c) State the number of protons, neutrons and electrons in a copper-63 atom. | Answer: (a) Ar(Cu) = 63 x 0.692 + 65 x 0.308 = 43.596 + 20.020 = 63.6. (b) The relative atomic mass is the weighted average of the masses of the isotopes, and neither isotope has 100% abundance, so the value is not a whole number. (c) Copper-63: 29 protons, 34 neutrons, 29 electrons. Mark points: weighted average (1), answer (1), explanation (1), three particle numbers (3), significant figures and units (1), clarity (1). |
| 2 | (Atomic structure) The proton number of magnesium is 12 and that of aluminium is 13. (a) Write the electron configuration of a magnesium atom and of an aluminium atom. (b) Write the electron configuration of Mg2+ and of Al3+ and explain why the two ions have the same number of electrons. (c) State the number of protons, neutrons and electrons in a magnesium-24 atom and explain how it relates to magnesium-25. | Answer: (a) Mg: 2, 8, 2. Al: 2, 8, 3. (b) Mg2+: 2, 8. Al3+: 2, 8. A magnesium atom loses 2 electrons to leave 10 electrons; an aluminium atom loses 3 electrons to leave 10 electrons. Both then have the same electron configuration as neon, which has 10 electrons. (c) Magnesium-24 has 12 protons, 12 neutrons and 12 electrons. Magnesium-25 has the same proton number but a different neutron number, so the two are isotopes of magnesium. Mark points: two configurations (2), two ion configurations (2), explanation (1), particle numbers (1), isotope definition (2). |
| 3 | (Chemical bonding) Classify each substance as an ionic compound or a simple covalent molecule: sodium chloride, magnesium oxide, water, carbon dioxide, ammonium chloride. (a) Give the classification for all five. (b) State the two types of bond found in ammonium chloride. (c) Explain why the melting point of magnesium oxide is much higher than that of water. | Answer: (a) Ionic: sodium chloride, magnesium oxide, ammonium chloride. Covalent molecules: water, carbon dioxide. (b) Ammonium chloride contains ionic bonding between the ammonium ion and the chloride ion, and covalent bonding between nitrogen and hydrogen inside the ammonium ion. (c) Magnesium oxide is an ionic compound in which magnesium ions and oxide ions are held in a giant lattice by strong electrostatic attraction. Melting it requires breaking many strong ionic bonds at once, whereas water molecules are held together only by weak intermolecular forces. Mark points: classification (2), two bond types (2), lattice idea (2), energy comparison (1), clarity (1). |
| 4 | (Mole) 4.0 g of sodium hydroxide (relative formula mass 40) is dissolved in water to make 250.0 cm3 of solution. (a) Calculate the amount in moles of sodium hydroxide. (b) Calculate the concentration in mol dm-3. (c) Calculate the number of sodium hydroxide particles present in this amount (Avogadro constant 6.02 x 10^23 mol-1). | Answer: (a) n = 4.0 / 40 = 0.10 mol. (b) c = 0.10 / (250.0 / 1000) = 0.40 mol dm-3. (c) Number of particles = 0.10 x 6.02 x 10^23 = 6.02 x 10^22. Mark points: moles (2), concentration with unit conversion (3), particle number (2), units and significant figures (1). |
Section B: Mole Calculations and Acid-Base Titration (36 marks)
Answer all questions in Section B and show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (Neutralisation) 25.0 cm3 of 0.20 M sodium hydroxide solution is titrated with 0.10 M hydrochloric acid until the end point is reached. (a) Calculate the amount of sodium hydroxide in moles. (b) Calculate the volume of acid required in cm3. (c) Calculate the amount of water formed in moles. (d) Explain why the jet of the burette is kept just below the liquid surface and why the acid is added dropwise near the end point. | Answer: (a) n(NaOH) = (25.0 / 1000) x 0.20 = 0.005 mol. (b) The acid and the alkali react in a 1:1 mole ratio, so n(HCl) = 0.005 mol and the volume is 0.005 / 0.10 x 1000 = 50.0 cm3. (c) 0.005 mol of water is formed. (d) Keeping the jet just below the surface prevents splashing and stops drops clinging to the wall of the flask. Adding dropwise near the end point with swirling avoids overshooting by a whole drop, which would make the reading too large. Mark points: (a) 1, (b) 3, (c) 2, (d) 3. |
| 6 | (Titration data) 25.0 cm3 of hydrochloric acid of unknown concentration is titrated with 0.100 M sodium hydroxide solution. The three readings are 24.20, 24.25 and 24.25 cm3. (a) State which volume should be used and why. (b) Calculate the concentration of the acid in mol dm-3 to 3 significant figures. (c) Write the equation for this titration. | Answer: (a) Use 24.25 cm3, the mean of the second and third readings; the first reading differs from the others by more than 0.10 cm3 and is a rough titration, so it is discarded. (b) n(NaOH) = 0.100 x 25.0 / 1000 = 0.00250 mol, so n(HCl) = 0.00250 mol and c = 0.00250 / (24.25 / 1000) = 0.103 mol dm-3. (c) HCl + NaOH gives NaCl + H2O. Mark points: choice with reason (2), moles (2), concentration (3), equation (1), significant figures (1). |
| 7 | (Acid, alkali and pH) A solution has a hydrogen ion concentration of 1.0 x 10^-3 M. (a) Calculate its pH. (b) State the pH after the solution is diluted ten times. (c) State the pH of 0.010 M hydrochloric acid, which is fully ionised. (d) Explain why sodium hydroxide solution is an alkali and state how its pH relates to the hydrogen ion concentration. | Answer: (a) pH = -log(1.0 x 10^-3) = 3.0. (b) After tenfold dilution the hydrogen ion concentration is 1.0 x 10^-4 M, so pH = 4.0. (c) 0.010 M hydrochloric acid is fully ionised, so [H+] = 1.0 x 10^-2 M and pH = 2.0. (d) Sodium hydroxide is an alkali because it ionises completely in water to give hydroxide ions, which remove hydrogen ions, leaving a hydrogen ion concentration below 1.0 x 10^-7 M. The pH of an alkaline solution is greater than 7 and is related to the hydrogen ion concentration by pH = -log[H+]. Mark points: three pH values (6), dilution reasoning (1), alkali explanation (2). |
| 8 | (Electrolysis) A current of 2.0 A is passed through copper(II) sulphate solution for 965 seconds. (a) Calculate the quantity of electricity in coulombs. (b) Calculate the amount in moles and the mass of copper deposited at the cathode (relative atomic mass of copper 63.5; charge on 1 mol of electrons 96500 C). (c) Calculate the amount of oxygen produced at the anode in moles and its volume in cm3 at room conditions (molar volume 24.0 dm3 mol-1). | Answer: (a) Q = I x t = 2.0 x 965 = 1930 C. (b) n(e-) = 1930 / 96500 = 0.02 mol. Each copper ion needs 2 electrons, so n(Cu) = 0.010 mol and the mass is 0.010 x 63.5 = 0.635 g. (c) At the anode 2H2O gives O2 + 4H+ + 4e-, so n(O2) = 0.02 / 4 = 0.005 mol and the volume is 0.005 x 24000 = 120 cm3. Mark points: charge (2), electrons (1), copper moles and mass (3), oxygen moles (1), volume (2). |
Section C: Electrolysis, Empirical Formula and Applications (32 marks)
Answer all questions in Section C and give units with every numerical answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | (Empirical formula) 3.18 g of copper is heated completely in oxygen to give 3.98 g of an oxide. (a) Calculate the mass of oxygen that reacted. (b) Calculate the mole ratio of copper to oxygen. (c) Write the empirical formula and the chemical formula of the oxide and explain why the experimental ratio may not be exactly a whole number. | Answer: (a) Mass of oxygen = 3.98 – 3.18 = 0.80 g. (b) n(Cu) = 3.18 / 63.5 = 0.0501 mol and n(O) = 0.80 / 16.0 = 0.0500 mol, so the ratio is about 1:1. (c) The empirical formula is CuO and the chemical formula is also CuO. The ratio may not be exactly whole because some copper may be unreacted, some oxide may flake off, and there are weighing errors. Mark points: oxygen mass (1), two mole values (2), ratio and formula (3), error explanation (2). |
| 10 | (Molar volume of gas) 0.050 mol of hydrogen is collected at room conditions, where the molar volume is 24.0 dm3 mol-1. (a) Calculate the volume occupied, in dm3 and in cm3. (b) If the hydrogen was made by reacting magnesium with excess dilute hydrochloric acid (Mg + 2HCl gives MgCl2 + H2), calculate the mass of magnesium needed (relative atomic mass of magnesium 24.3). (c) Explain why the volume of gas collected is usually slightly less than the theoretical value. | Answer: (a) V = 0.050 x 24.0 = 1.2 dm3 = 1200 cm3. (b) One mole of magnesium gives one mole of hydrogen, so n(Mg) = 0.050 mol and the mass is 0.050 x 24.3 = 1.22 g. (c) The collected volume is usually slightly low because some gas remains in the apparatus, the syringe or gas jar has dead space, the gas may not have reached room temperature, or the apparatus leaks. Mark points: both units (3), mole ratio (1), mass (2), explanation (2). |
| 11 | (Percentage yield) The theoretical yield of a reaction is 8.0 g but only 6.4 g is obtained. (a) Calculate the percentage yield. (b) If 6.4 g is to be obtained with a yield of 80.0%, what theoretical yield is required? (c) Give two common reasons why the percentage yield is below 100%. | Answer: (a) Percentage yield = 6.4 / 8.0 x 100 = 80.0%. (b) Theoretical yield = 6.4 / 0.80 = 8.0 g. (c) Common reasons: the reaction is incomplete, a side reaction occurs, product is lost during transfer and filtration, or the product contains impurities and is not fully dry. Mark points: yield (2), reverse calculation (3), two reasons (2), clarity (1). |
| 12 | (Everyday applications) An antacid tablet contains calcium carbonate and neutralises excess stomach acid, which is hydrochloric acid. (a) Write the balanced equation for the reaction between calcium carbonate and hydrochloric acid. (b) Explain why lime (calcium oxide) can be used to treat acidic soil. (c) Explain why concentrated sulphuric acid must be added slowly to water, and never water to the acid. | Answer: (a) CaCO3 + 2HCl gives CaCl2 + H2O + CO2. (b) Calcium oxide reacts with water to form calcium hydroxide, which is alkaline and neutralises hydrogen ions in the soil, raising the pH into the range that suits plant growth and precipitating some harmful metal ions. (c) Concentrated sulphuric acid releases much heat when it dissolves. If water is poured into the acid it floats on top and may boil locally, spitting corrosive acid out of the vessel. Adding the acid slowly to a large volume of water lets the water absorb the heat so the temperature rises gently. Mark points: balanced equation (2), soil treatment (3), dilution safety (3). |
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