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GCSE · Chemistry

C3 Quantitative Chemistry

AQA AQA 8462 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Concentration and units.

(a)Which unit is used to measure the concentration of a solution?(1)

A) g
B) mol
C) mol/dm3
D) dm3/mol

(b)A conical flask contains 500 cm3 of solution. State this volume in dm3.(1)

(Total for Question 1 is 2 marks)

A student heats magnesium ribbon in an open crucible until it burns completely to form a white powder, magnesium oxide.

(a)State the law of conservation of mass.(1)

(b)The total mass of magnesium oxide formed was 1.00 g. If 0.60 g of magnesium was used, calculate the mass of oxygen that combined with the magnesium.(2)

(c)Explain, in terms of particles, why the mass of the crucible and its contents increases during the reaction even though the crucible is open to the air.(2)

(Total for Question 2 is 5 marks)

Calcium carbonate, CaCO3, thermally decomposes when heated strongly. (Ar: Ca = 40, C = 12, O = 16)

(a)Calculate the relative formula mass (Mr) of calcium carbonate, CaCO3.(2)

(b)Write the balanced symbol equation for the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide.(2)

(c)Calculate the Mr of calcium oxide, CaO.(1)

(Total for Question 3 is 5 marks)

Sodium chloride, NaCl, has Mr = 58.5 (Na = 23, Cl = 35.5).

(a)Calculate the number of moles in 11.7 g of sodium chloride.(2)

(b)Calculate the mass of 0.25 mol of sodium chloride.(2)

(Total for Question 4 is 4 marks)

A student, Priya, prepares copper sulfate crystals in the lab. Her calculated theoretical yield was 12.5 g but she only obtained 9.8 g of dry crystals.

(a)Calculate the percentage yield of Priya's preparation.(2)

(b)Give two reasons why the percentage yield of a practical preparation is often less than 100%.(2)

(Total for Question 5 is 4 marks)

Ammonium nitrate, NH4NO3, is used as a fertiliser. Calculate the percentage by mass of nitrogen in ammonium nitrate. (Ar: N = 14, H = 1, O = 16)

(Total for Question 6 is 3 marks)

Magnesium reacts with hydrochloric acid: Mg + 2HCl -> MgCl2 + H2 (Ar: Mg = 24, Cl = 35.5; Mr of MgCl2 = 95)

(a)Calculate the number of moles of magnesium in 3.6 g of magnesium.(2)

(b)Assuming the hydrochloric acid is in excess, calculate the maximum mass of magnesium chloride, MgCl2, that could be produced from 3.6 g of magnesium.(3)

(Total for Question 7 is 5 marks)

A solution of sodium hydroxide contains 8.0 g of NaOH (Mr = 40) dissolved in 250 cm3 of solution.

(a)Calculate the number of moles of NaOH dissolved.(2)

(b)Calculate the concentration of the solution in mol/dm3.(2)

(c)Calculate the concentration of the solution in g/dm3.(2)

(Total for Question 8 is 6 marks)

Ethanol can be manufactured industrially by reacting ethene with steam: C2H4 + H2O -> C2H5OH. This reaction produces ethanol as the only product. (Mr: C2H4 = 28, H2O = 18, C2H5OH = 46)

(a)Calculate the atom economy of this reaction for producing ethanol.(2)

(b)Explain why a reaction with a high atom economy is generally considered more sustainable than one with a low atom economy.(2)

(Total for Question 9 is 4 marks)

Sodium reacts with chlorine gas: 2Na + Cl2 -> 2NaCl. A student, Kwame, reacts 4.6 g of sodium (Ar = 23) with 3.55 g of chlorine gas (Mr = 71).

(a)Calculate the number of moles of sodium and the number of moles of chlorine gas used.(2)

(b)Show that chlorine is the limiting reactant.(3)

(c)Calculate the mass of sodium chloride, NaCl (Mr = 58.5), formed in this reaction.(2)

(Total for Question 10 is 7 marks)

Required practical: Aisha carries out a titration to find the concentration of a hydrochloric acid solution using a standard solution of sodium hydroxide of known concentration. NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)

(a)Name the piece of apparatus used to add the hydrochloric acid to the conical flask so that the volume added can be measured accurately.(1)

(b)Name the piece of apparatus used to measure a fixed, accurate volume of the sodium hydroxide solution into the conical flask.(1)

(c)Name a suitable indicator for this titration and describe the colour change seen at the end point as the acid is added from the burette.(2)

(d)Describe how Aisha should carry out the titration to obtain accurate, concordant results for the volume of hydrochloric acid needed to neutralise the sodium hydroxide.(6)

(e)In one titration, 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide exactly neutralised 22.50 cm3 of the hydrochloric acid. Calculate the concentration of the hydrochloric acid in mol/dm3.(4)

(f)Calculate the concentration of the hydrochloric acid in g/dm3. (Ar: H = 1, Cl = 35.5)(2)

(Total for Question 11 is 16 marks)

A hydrocarbon is analysed and found to contain 85.7% carbon and 14.3% hydrogen by mass. (Ar: C = 12, H = 1)

(a)Calculate the empirical formula of the hydrocarbon.(3)

(b)The relative molecular mass of the hydrocarbon is 42. Determine its molecular formula.(2)

(Total for Question 12 is 5 marks)

At room temperature and pressure (RTP), one mole of any gas occupies 24 dm3 (24 000 cm3). Calcium carbonate reacts with excess hydrochloric acid: CaCO3 + 2HCl -> CaCl2 + H2O + CO2 (Mr of CaCO3 = 100)

(a)Calculate the number of moles of CO2 produced when 5.0 g of calcium carbonate reacts completely.(2)

(b)Calculate the volume of carbon dioxide gas produced at RTP.(2)

(Total for Question 13 is 4 marks)

The Avogadro constant is 6.02 x 10²³ per mole.

(a)Calculate the number of molecules in 0.50 mol of carbon dioxide.(2)

(b)Calculate the number of moles in 1.204 x 10²⁴ atoms of iron.(2)

(Total for Question 14 is 4 marks)

Zinc reacts with dilute sulfuric acid: Zn + H2SO4 -> ZnSO4 + H2. A student reacts 6.5 g of zinc (Ar = 65) with 100 cm3 of 1.20 mol/dm3 sulfuric acid. (Mr of ZnSO4 = 161; molar gas volume at RTP = 24 dm3/mol)

(a)Show that zinc is the limiting reactant.(3)

(b)Calculate the maximum mass of zinc sulfate, ZnSO4, that could be produced.(2)

(c)Calculate the volume of hydrogen gas produced at RTP.(2)

(Total for Question 15 is 7 marks)

Mark scheme · C3 Quantitative Chemistry

Question 1

(a) B1 C
(a) Answer: C
(b) B1 0.5 dm3 (cao)
(b) Answer: 0.5 dm3

Question 2

(a) B1 No atoms are created or destroyed in a chemical reaction (oe), so the total mass of reactants equals the total mass of products in a closed system
(a) Answer: Mass is conserved: total mass of reactants = total mass of products.
(b) M1 mass of oxygen = mass of MgO - mass of Mg (1.00 - 0.60) shown
(b) A1 0.40 g (cao)
(b) Answer: 0.40 g
(c) B1 Oxygen (particles/molecules) from the air enters the crucible and reacts with the magnesium (oe)
(c) B1 The oxygen becomes chemically bonded into the solid magnesium oxide, so extra mass is added to the solid that was not part of the original weighed magnesium (oe)
(c) Answer: Oxygen from the air combines with the magnesium and stays as part of the solid product, adding to the mass of the solid even though no mass is truly created.

Question 3

(a) M1 40 + 12 + (3 x 16) shown
(a) A1 100 (cao)
(a) Answer: 100
(b) B1 correct formulae: CaCO3, CaO, CO2
(b) B1 correctly balanced: CaCO3 -> CaO + CO2
(b) Answer: CaCO3 -> CaO + CO2
(c) A1 56 (cao)
(c) Answer: 56

Question 4

(a) M1 moles = mass / Mr = 11.7 / 58.5 shown
(a) A1 0.2 mol (cao)
(a) Answer: 0.2 mol
(b) M1 mass = moles x Mr = 0.25 x 58.5 shown
(b) A1 14.625 g (accept 14.6 g)
(b) Answer: 14.625 g (accept 14.6 g)

Question 5

(a) M1 percentage yield = (actual yield / theoretical yield) x 100 = (9.8 / 12.5) x 100 shown
(a) A1 78.4% (awrt)
(a) Answer: 78.4%
(b) B1 Product is lost during practical steps such as filtration, transferring between containers or evaporation (oe), e.g. crystals left on filter paper
(b) B1 The reaction may not go to completion, or unwanted side/reverse reactions occur, producing other products (oe)
(b) Answer: Any two of: product lost during transfer/filtration; reaction does not go to completion; unwanted side reactions occur.

Question 6

M1 Mr of NH4NO3 = 14 + (4 x 1) + 14 + (3 x 16) = 80 shown
M1 total mass of N in formula = 14 + 14 = 28 shown
A1 35% (cao)
Answer: 35%

Question 7

(a) M1 moles = mass / Ar = 3.6 / 24 shown
(a) A1 0.15 mol (cao)
(a) Answer: 0.15 mol
(b) M1 uses 1:1 mole ratio Mg : MgCl2 from the equation, so moles MgCl2 = 0.15 mol (ft from a)
(b) M1 mass = moles x Mr = 0.15 x 95 shown
(b) A1 14.25 g (cao)
(b) Answer: 14.25 g

Question 8

(a) M1 moles = mass / Mr = 8.0 / 40 shown
(a) A1 0.2 mol (cao)
(a) Answer: 0.2 mol
(b) M1 converts 250 cm3 to 0.25 dm3 and uses concentration = moles / volume
(b) A1 0.8 mol/dm3 (ft from a)
(b) Answer: 0.8 mol/dm3
(c) M1 concentration (g/dm3) = concentration (mol/dm3) x Mr = 0.8 x 40 shown
(c) A1 32 g/dm3 (ft from b)
(c) Answer: 32 g/dm3

Question 9

(a) M1 recognises ethanol is the only product formed, so atom economy = (Mr of desired product / sum of Mr of all products) x 100 = (46/46) x 100
(a) A1 100% (cao)
(a) Answer: 100%
(b) B1 A high atom economy means less of the reactant atoms end up as waste products (oe)
(b) B1 This makes better use of limited raw materials/reduces waste that needs disposing of, lowering environmental impact and cost (oe)
(b) Answer: High atom economy wastes fewer raw material atoms and produces less waste, making the process more resource-efficient and less environmentally damaging.

Question 10

(a) M1 moles Na = 4.6 / 23 = 0.2 mol
(a) M1 moles Cl2 = 3.55 / 71 = 0.05 mol
(a) Answer: 0.2 mol Na; 0.05 mol Cl2
(b) M1 uses the 2:1 mole ratio (Na:Cl2) from the equation to find the moles of Na needed to react exactly with 0.05 mol Cl2: 0.05 x 2 = 0.1 mol Na (ft from a)
(b) M1 compares 0.1 mol Na required with 0.2 mol Na available, showing Na is in excess (ft from a)
(b) A1 cso: correctly concludes chlorine (Cl2) is the limiting reactant because there is more sodium present than is needed to react with all the chlorine
(b) Answer: Chlorine (Cl2) is the limiting reactant.
(c) M1 uses limiting reactant (Cl2) with 1:2 mole ratio to find moles NaCl = 0.05 x 2 = 0.1 mol (ft from a/b)
(c) A1 5.85 g (cao, ft)
(c) Answer: 5.85 g

Question 11

(a) B1 burette
(a) Answer: Burette
(b) B1 (volumetric/graduated) pipette (with a pipette filler)
(b) Answer: Pipette (with a pipette filler)
(c) B1 phenolphthalein (or methyl orange, with a consistent colour change)
(c) B1 changes from pink (in the alkali) to colourless at the end point (oe, must match indicator named)
(c) Answer: Phenolphthalein; pink to colourless at the end point.
(d) L3 (5-6): A detailed, logically sequenced description covering measuring the alkali by pipette into the conical flask, adding indicator, carrying out a rough titration first, then accurate repeat titrations adding acid dropwise (swirling the flask) near the expected end point, repeating until concordant titres are obtained (within 0.10 cm3 of each other) and calculating a mean from the concordant results only.
(d) L2 (3-4): A description of most key steps of the method with some logical order, but with gaps in detail (e.g. no mention of a rough titration, dropwise addition, or checking for concordant results).
(d) L1 (1-2): Basic relevant points made, such as naming apparatus or the indicator, but the method is incomplete or lacks a logical structure.
(d) L0 (0): No relevant content.
(d) Answer: See levels/indicative content.
(e) M1 moles NaOH = concentration x volume = 0.100 x (25.0/1000) = 0.00250 mol
(e) M1 uses the 1:1 mole ratio from the equation, so moles HCl = 0.00250 mol
(e) M1 converts 22.50 cm3 to 0.02250 dm3 and uses concentration = moles / volume
(e) A1 0.111 mol/dm3 (awrt 0.111)
(e) Answer: 0.111 mol/dm3
(f) M1 Mr of HCl = 1 + 35.5 = 36.5, uses concentration (g/dm3) = concentration (mol/dm3) x Mr (ft from e)
(f) A1 4.06 g/dm3 (awrt, ft from e)
(f) Answer: 4.06 g/dm3

Question 12

(a) M1 moles C = 85.7/12 = 7.14; moles H = 14.3/1 = 14.3 (both shown)
(a) M1 divides both by the smaller value (7.14) to give a ratio of approximately 1 : 2
(a) A1 CH2 (cao)
(a) Answer: CH2
(b) M1 Mr of empirical unit CH2 = 12 + 2 = 14; divides 42 by 14 to get 3 (ft from a)
(b) A1 C3H6 (cao, ft)
(b) Answer: C3H6

Question 13

(a) M1 moles CaCO3 = 5.0 / 100 shown, and 1:1 mole ratio CaCO3:CO2 used
(a) A1 0.05 mol (cao)
(a) Answer: 0.05 mol
(b) M1 volume = moles x 24 = 0.05 x 24 shown (ft from a)
(b) A1 1.2 dm3 (accept 1200 cm3)
(b) Answer: 1.2 dm3 (1200 cm3)

Question 14

(a) M1 number of molecules = moles x Avogadro constant = 0.50 x 6.02 x 10²³ shown
(a) A1 3.01 x 10²³ (awrt)
(a) Answer: 3.01 x 10²³ molecules
(b) M1 moles = number of atoms / Avogadro constant = 1.204x10²⁴ / 6.02x10²³ shown
(b) A1 2.0 mol (cao)
(b) Answer: 2.0 mol

Question 15

(a) M1 moles Zn = 6.5/65 = 0.1 mol
(a) M1 moles H2SO4 = 0.100 x 1.20 = 0.12 mol
(a) A1 cso: 1:1 ratio means only 0.1 mol H2SO4 is needed to react with all the zinc, but 0.12 mol is available (excess acid), so zinc is the limiting reactant
(a) Answer: Zinc is the limiting reactant.
(b) M1 uses limiting reactant (Zn) with 1:1 ratio, so moles ZnSO4 = 0.1 mol (ft from a)
(b) A1 16.1 g (cao, ft)
(b) Answer: 16.1 g
(c) M1 1:1 ratio Zn:H2, so moles H2 = 0.1 mol; volume = moles x 24 (ft from a)
(c) A1 2.4 dm3 (cao, ft)
(c) Answer: 2.4 dm3

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