S4 Chemistry Paper: Mole Calculations, Solution Concentration And Gas Volume (100 Marks)

S4 Chemistry Paper: Mole Calculations, Solution Concentration and Gas Volume (100 marks)

2026-11-12

Chemistry – Moles and Concentration

S4 Chemistry: 6 questions, 24 parts, 100 marks, 75 minutes. Moles and mass, concentration and dilution, gas molar volume, limiting reagents, percentage yield and titration.

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Question 1 Moles and Mass (16 marks)

Answer each part independently and show your working.

#	Question	Answer and marking notes
1	Convert 8.8 g of carbon dioxide (Mr = 44.0) completely. Find the number of moles and the number of molecules (Avogadro constant = 6.02 x 10^23 per mol).	Moles = 8.8 / 44.0 = 0.2 mol; molecules = 0.2 x 6.02 x 10^23, about 1.20 x 10^23. (4 marks)
2	What is the mass of 0.25 mol of calcium carbonate (CaCO3, Mr = 100.0)? What amount of oxygen atoms does it contain?	Mass = 0.25 x 100.0 = 25.0 g; each formula unit has 3 oxygen atoms, so oxygen atoms = 0.25 x 3 = 0.75 mol. (4 marks)
3	Dissolve 5.0 g of sodium hydroxide (Mr = 40.0) in water and make up to 250.0 cm3 of solution. Find the molarity.	Moles = 5.0 / 40.0 = 0.125 mol; volume = 0.250 dm3; molarity = 0.125 / 0.250 = 0.5 mol/dm3. (4 marks)
4	Under the same conditions, compare 4.0 g of helium (Mr = 4.0) with 4.0 g of methane (Mr = 16.0). Which sample contains more particles, and by what factor?	Helium = 1.0 mol, methane = 0.25 mol; helium has more particles, by a factor of 4.0 . (4 marks)

Question 2 Concentration and Dilution (20 marks)

Answer each part independently; the amount of solute is unchanged on dilution.

#	Question	Answer and marking notes
5	To prepare 250.0 cm3 of 2.0 mol/dm3 hydrochloric acid (HCl, Mr = 36.5), how many moles and how many grams of solute are needed?	Moles = 2.0 x 0.250 = 0.5 mol; mass = 0.5 x 36.5 = 18.25 g, so weigh about 18.3 g. (5 marks)
6	Take 25.0 cm3 of that acid and dilute with water to a total volume of 500.0 cm3. Find the new molarity.	Solute = 0.0250 x 2.0 = 0.05 mol; new molarity = 0.05 / 0.500 = 0.1 mol/dm3. (5 marks)
7	Using 2.0 mol/dm3 acid, how many cm3 of the stock solution are needed to prepare 250.0 cm3 of 0.15 mol/dm3 acid?	Required solute = 0.15 x 0.250 = 0.0375 mol; stock volume = 0.0375 / 2.0 = 0.01875 dm3 = 18.75 cm3, then make up to 250.0 cm3. (5 marks)
8	Mix 100.0 cm3 of 0.50 mol/dm3 sodium chloride solution with 200.0 cm3 of 0.20 mol/dm3 sodium chloride solution. Find the molarity of the mixture.	Amount = 0.100 x 0.50 + 0.200 x 0.20 = 0.050 + 0.040 = 0.09 mol; total volume = 0.300 dm3; molarity = 0.09 / 0.300 = 0.3 mol/dm3. (5 marks)

Question 3 Molar Volume of Gases (18 marks)

Take the molar volume of a gas as 24.0 dm3/mol at room temperature and pressure.

#	Question	Answer and marking notes
9	At room temperature and pressure, find the amount and mass of 6.0 dm3 of oxygen (Mr = 32.0).	Moles = 6.0 / 24.0 = 0.25 mol; mass = 0.25 x 32.0 = 8.0 g. (5 marks)
10	What volume does 1.6 g of methane (Mr = 16.0) occupy at room temperature and pressure? What volume of oxygen is needed for complete combustion?	Methane = 0.1 mol, volume = 0.1 x 24.0 = 2.4 dm3; CH4 + 2O2 gives CO2 + 2H2O, so oxygen = 0.2 mol = 4.8 dm3. (5 marks)
11	What volume does 4.4 g of carbon dioxide (Mr = 44.0) occupy at room temperature and pressure, and how many molecules does it contain?	Moles = 0.1 mol; volume = 0.1 x 24.0 = 2.4 dm3; molecules = 0.1 x 6.02 x 10^23 = 6.02 x 10^22. (4 marks)
12	Under the same conditions, which has the greater mass: 2.0 dm3 of sulphur dioxide (Mr = 64.0) or 2.0 dm3 of oxygen (Mr = 32.0)? What is the mass ratio?	Equal volumes at the same temperature and pressure contain equal amounts, 0.0833 mol each; sulphur dioxide = 5.33 g and oxygen = 2.67 g, so sulphur dioxide is heavier with a mass ratio of 2.0 : 1. (4 marks)

Question 4 Limiting Reagent and Percentage Yield (16 marks)

The equation for the synthesis of ammonia is N2 + 3H2 gives 2NH3. An experiment uses 2.8 g of nitrogen and 0.90 g of hydrogen.

#	Question	Answer and marking notes
13	Find the amount of each reactant and identify the limiting reagent.	Nitrogen = 0.1 mol; hydrogen = 0.45 mol; 0.1 mol of nitrogen needs only 0.3 mol of hydrogen, so hydrogen is in excess and nitrogen is the limiting reagent. (4 marks)
14	After the reaction, how many moles and how many grams of hydrogen remain?	Remaining = 0.45 – 0.3 = 0.15 mol; mass = 0.15 x 2.0 = 0.3 g. (4 marks)
15	What amount and mass of ammonia (Mr = 17.0) should be formed in theory? If only 2.72 g is obtained, what is the percentage yield?	Ammonia = 0.2 mol = 3.4 g; percentage yield = 2.72 / 3.4 x 100% = 80.0%. (4 marks)
16	The reaction is exothermic. If the temperature is raised, how does the equilibrium yield of ammonia change? Explain in terms of the shift of equilibrium.	The yield decreases; raising the temperature shifts the equilibrium in the endothermic direction, that is towards the reactants, so the equilibrium amount of ammonia falls. (4 marks)

Question 5 Empirical and Molecular Formulae (16 marks)

An organic compound contains only carbon, hydrogen and oxygen: 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180.0.

#	Question	Answer and marking notes
17	Find the empirical formula and the molecular formula of the compound.	Mole ratio C : H : O = 3.33 : 6.7 : 3.33 = 1 : 2 : 1, so the empirical formula is CH2O (formula mass 30.0); n = 180.0 / 30.0 = 6.0, so the molecular formula is C6H12O6. (5 marks)
18	How many moles are there in 9.0 g of the compound (C6H12O6, Mr = 180.0)?	Moles = 9.0 / 180.0 = 0.05 mol. (3 marks)
19	When 0.05 mol of the compound burns completely (C6H12O6 + 6O2 gives 6CO2 + 6H2O), what mass of carbon dioxide is formed and what volume of oxygen at room temperature and pressure is used?	Carbon dioxide = 0.3 mol, mass = 13.2 g; oxygen = 0.3 mol, volume = 0.3 x 24.0 = 7.2 dm3. (5 marks)
20	Use the molecular formula C6H12O6 to check the percentage by mass of carbon, and state the relationship between the empirical and molecular formulae.	Carbon = 6 x 12.0 = 72.0, and 72.0 / 180.0 x 100% = 40.0%, matching the given data; the molecular formula is a whole-number multiple of the empirical formula (6.0 times here). (3 marks)

Question 6 Titration Data and Error Analysis (14 marks)

A student dissolves 5.30 g of anhydrous sodium carbonate (Na2CO3, Mr = 106.0) and makes up to 250.0 cm3, then titrates it with 0.20 mol/dm3 hydrochloric acid: Na2CO3 + 2HCl gives 2NaCl + CO2 + H2O.

#	Question	Answer and marking notes
21	Find the molarity of the sodium carbonate solution.	Moles = 5.30 / 106.0 = 0.050 mol; molarity = 0.050 / 0.250 = 0.2 mol/dm3. (4 marks)
22	Titrating 25.0 cm3 of this sodium carbonate solution, what volume of acid is required in theory?	Carbonate = 0.0250 x 0.2 = 0.005 mol; the ratio is 1 : 2, so acid needed = 0.01 mol; volume = 0.01 / 0.20 = 0.050 dm3 = 50.0 cm3. (4 marks)
23	If the student stops at 45.0 cm3 and records it as the end point, what is the percentage error?	Percentage error = (45.0 – 50.0) / 50.0 x 100% = -10.0%, that is 10% too little acid added. (3 marks)
24	Suggest two ways to reduce titration error, and explain which type of error cancels out when the volume used is calculated.	Use a white tile to judge the colour change, add dropwise with swirling, and repeat the titration for a mean value; a constant zero error of the burette is a systematic error that shifts both readings equally, so it cancels in the difference and does not affect the volume used. (3 marks)

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