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A link to a video of this paper can be found in the link: OCR 2022 A level Paper 3, H432/03, Unified chemistry
1
These questions are from different areas of chemistry.
(a)
Ammonia, NH3, and ammonium nitrate, NH4NO3, are compounds of nitrogen.
(i)
The boiling point of NH3 is –33°C.
The boiling point of NH4NO3 is 210°C.
Explain why there is a large difference in boiling points.
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(ii)
Two students discuss the oxidation numbers in ammonium nitrate, NH4NO3.
One student claims that the two nitrogen atoms have the same oxidation number. The
other student disagrees and claims that the nitrogen atoms have different oxidation
numbers.
Explain with reasons which student is correct.
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If you want to refresh your memory about oxidation numbers, you can check How to balance redox equations
(b)
Brass is an alloy of copper and zinc.
The mass spectrum of a sample of brass is shown below.
The peaks at m/z = 63 and m/z = 65 are from the 63Cu and 65Cu isotopes of copper.
The remaining four peaks are from isotopes of zinc
(i)
What are the percentage compositions of copper and zinc in the brass sample?
Cu = …………. % Zn = …………. % [1]
(ii)
Calculate the relative atomic mass of zinc in the sample of brass.
Give your answer to 2 decimal places.
relative atomic mass = …………………………………………………. [2]
(c)
The structure of an organic compound is shown below.
The protons are in four different environments, which are labelled 1–4.
(i)
Fill in the table to predict the splitting patterns in the proton NMR spectrum of the
organic compound.
(ii)
The table shows the chemical shifts for the peaks in the proton NMR spectrum at
proton environments 2 and 3.
Suggest why the peaks for proton environments 2 and 3 have the chemical shifts which
are shown in the table.
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(d)
Glycine, H2NCH2COOH, is an α-amino acid.
(i)
Glycine reacts with NaOH to form the salt H2NCH2COONa.
Glycine reacts with HCl to form the salt HOOCCH2NH3Cl.
The salts have different H–N–H bond angles.
State the different H–N–H bond angles and explain why they are different.
H2NCH2COONa H–N–H bond angle = ……………….°
HOOCCH2NH3Cl H–N–H bond angle = ……………….°
explanation …………………………………………………………………………………………………………
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(ii)
Glycine reacts with aqueous copper(II) ethanoate to form copper(II) glycinate,
Cu(H2NCH2COO)2, and ethanoic acid. Copper(II) glycinate is a complex which exists
as two square planar isomers.
Write an equation for this reaction and draw the structures of the two square planar
isomers of the complex Cu(H2NCH2COO)2.
equation
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structures
[3]
2
This question is about redox reactions.
(a)
Calcium hypochlorite’, Ca(ClO)2, is an ionic compound used in ‘bleaching powder’.
The ClO– ion in Ca(ClO)2 is the active ingredient that kills bacteria.
Calcium hypochlorite is prepared by reacting chlorine gas with calcium hydroxide.
2Cl2(g) + 2Ca(OH)2(s) → Ca(ClO)2(s) + CaCl2(s) + 2H2O(l) Equation 2.1
(i)
420 dm3 of chlorine, measured at RTP, is reacted with an excess of Ca(OH)2.
The solid products are dissolved in water to form 4.00 m3 of solution.
Calculate the concentration of Ca(ClO)2(aq) in this solution, in mol dm–3.
Give your answer to an appropriate number of significant figures and in standard form.
concentration = …………………………………… mol dm–3 [3]
(ii)
Calcium hypochlorite, Ca(ClO)2, is heated. The Ca(ClO)2 decomposes to form CaCl2
and Ca(ClO3). This is a disproportionation reaction.
Write an equation for this decomposition and explain, using oxidation numbers, why this
is a disproportionation reaction.
equation …………………………………………………………………………………………………………….
explanation …………………………………………………………………………………………………………
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Disproportionation or dismutation is a redox reaction where one compound is oxidised and reduced in the same reaction. In this example we have the reaction:
Ca(ClO)2 → CaCl2 + Ca(ClO3)2
The equation might be difficult to adjust, and we might be tempted to add an extra O. If you are confident in adjusting the equation, skip the red part below.
To adjust the equation, I use the ‘algebraic method’ that I explain at How to balance chemical equations – (jctnotes.com). Let’s fix the number of mol of the reactant (Ca(ClO)2) to 1 and call a the number of mol of CaCl2 + and b the number of mol of Ca(ClO3)2:
1Ca(ClO)2 → aCaCl2 + bCa(ClO3)2
Counting and comparing the number of mol of each atom in the reactant and products we have:
Ca: 1 = a + b
Cl: 2 = 2a + 2b
O: 2 = 6b
Using the balance for the O, we can see that b = 2/6 = 1/3
With b = 1/3, we can go the Ca (or the Cl if you prefer) and find that 1 = a + 1/3. Where we can find that a = 1 – 1/3 = 2/3.
With this numbers we have:
1Ca(ClO)2 → (2/3)CaCl2 + (1/3)Ca(ClO3)2
Or multiplying by 3 the whole equation:
3Ca(ClO)2 → 2CaCl2 + Ca(ClO3)2
The adjusted equation is:
3Ca(ClO)2 → 2CaCl2 + Ca(ClO3)2
To explain the disproportionation we need to refer to the oxidation numbers. Let’s find these numbers:
- Cl in Ca(ClO)2. The Cl is in the form of ClO–. The O is in oxidation state -2, as the ion ClO– has a charge -1, the Cl only compensate 1 of the negative charges of the O, making the oxidation number of the Cl +1.
- Cl in CaCl2. The Ca is in oxidation state +2, as the CaCl2 is neutral, both atoms of Cl compensate the +2 of the Ca, making the oxidation number of the Cl -1.
- Cl in Ca(ClO3)2. The Cl is in the form of ClO3–. Each O has an oxidation state -2 and the contribution of the 3 atoms of O is -6. The ion ClO3– has a net charge of -1, meaning that 5 of the negative charges of the 3 atoms of O has been compensated by the Cl i.e. the oxidation number of the Cl +5.
With this information, we can answer the question:
Equation: 3Ca(ClO)2 → 2CaCl2 + Ca(ClO3)2
Explanation The Cl with oxidation number +1 in the in Ca(ClO)2 has been oxidised to oxidation number +5 in the Ca(ClO3)2 and reduced to oxidation number -1 in the CaCl2.
(b)
A student analyses the redox reactions shown below. State symbols have been omitted.
CH3CHO + 2H+ + 2e– ⇌ C2H5OH Eө = –0.197V
Cr2O72– + 14H+ + 6e– ⇌ 2Cr3+ + 7H2O Eө = +1.33V
FeO42– + 8H+ + 3e– ⇌ Fe3+ + 4H2O Eө = +2.20V
The student concludes that different ions containing chromium can act as oxidising or
reducing agents.
Using the terms oxidising agent and reducing agent, and ideas about electrode potentials
and equilibrium, explain how the student is correct.
Include overall equations.
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(c)
A student bubbles hydrogen sulfide gas, H2S(g), through an acidified solution containing
manganate(VII) ions, MnO4–(aq).
A redox reaction takes place, forming aqueous manganese(II) ions, a yellow precipitate and
one other product.
Construct the equation for this reaction. State symbols are not required.
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3
Information about 1-bromobutane and butan-1-ol is shown in the table.
A student prepares a sample of 1-bromobutane by refluxing 9.25g of butan-1-ol with sodium
bromide and sulfuric acid.
After reflux, the reaction mixture is purified.
The student obtains 6.10 cm3 of pure 1-bromobutane.
(a)*
Draw a diagram to show how the student would have carried out the reflux and calculate the
percentage yield of 1-bromobutane that the student obtains.
Describe how the student could have obtained pure 1-bromobutane from the reaction
mixture obtained after reflux.
[6]
(b)
Butan-1-ol reacts with sodium bromide and sulfuric acid to form 1-bromobutane by
nucleophilic substitution.
The mechanism for this reaction takes place by two steps.
Step 1 The oxygen atom of the alcohol group accepts a proton to form a positivelycharged intermediate.
Step 2 Bromide ions react with the intermediate from Step 1 by nucleophilic substitution
to form 1-bromobutane.
Show both steps in this mechanism.
[4]
4
Two students plan to investigate Equilibrium 4.1, shown below.
CoCl42−(aq) + 6H2O(l) ⇌ [Co(H2O)6]2+(aq) + 4Cl−(aq) Equilibrium 4.1
Blue Pink
(a)
The students are supplied with the equilibrium mixture in Equilibrium 4.1 at room
temperature.
- One student heats 20 cm3 of the mixture to 50°C.
- The other student heats 20 cm3 of the mixture to 90°C.
The students use colorimetry to observe how the colour of the equilibrium mixture changes
over time.
- The colorimeter is set up so that the greater the absorbance, the greater the
concentration of [Co(H2O)6]2+. - The initial absorbance is set to zero.
- The absorbance is recorded every 30 seconds.
The students plot the graph below from the results of the experiment.
Use the graph and relevant chemical theory to answer the following. Include all reasoning:
- Explain the different initial rates at 50°C and 90°C.
- Predict the sign of ∆H for the forward reaction in Equilibrium 4.1.
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(b)
The students investigate how addition of aqueous silver nitrate, AgNO3(aq), affects the
equilibrium position in Equilibrium 4.1.
The graph shows the changes in the equilibrium concentrations of CoCl42–, Cl– and
[Co(H2O)6]2+ after addition of the AgNO3(aq).
The AgNO3(aq) is added at time = t1
(i)
Explain why the Cl– concentration drops sharply at time = t1.
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(ii)
Explain the changes in concentration of CoCl42–, Cl– and [Co(H2O)6]2+ after time = t1.
Refer to Equilibrium 4.1 in your answer.
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5
This question is about energy changes.
(a)*
A student plans to determine the enthalpy change of hydration of calcium ions.
The student finds the information below from data tables.
The student carries out an experiment to find the enthalpy change of solution of calcium
chloride.
Student’s method:
- Weigh a bottle containing calcium chloride and weigh a polystyrene cup.
- Add water from a measuring cylinder to the polystyrene cup and measure its
temperature. - Add the calcium chloride, stir the mixture, and measure the maximum temperature of
the final solution. - Weigh the empty bottle and weigh the polystyrene cup with the final solution.
Calculate the enthalpy change of solution of calcium chloride and determine the enthalpy
change of hydration of calcium ions.
Show your working, including an energy cycle linking the energy changes.
Assume that the density and specific heat capacity, c, of the solution are the same as for
water.
[6]
(b)
Internal combustion engines have historically used fuels obtained from crude oil as a source
of power.
The environmental effects of fossil fuel use can be reduced by blending petrol with biofuels
such as ethanol.
A fuel is being developed using a 1:1 molar ratio of octane and ethanol.
(i)
Write the equation for the complete combustion of this fuel.
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For an explanation about how to balance chemical equations and practice exercises, check How to balance chemical equations – (jctnotes.com)
(ii)
Calculate the energy released, in kJ, by the complete combustion of 8.00kg of this fuel.
∆cH(C8H18) = –5470 kJ mol–1; ∆cH(C2H5OH) = –1367 kJ mol–1.
energy released = …………………………………………….. kJ [3]
(6)
A student carries out an investigation on vitamin C, C6H8O6.
(a)
The structure of vitamin C is shown below. Vitamin C is an optical isomer.
What is the total number of optical isomers with the structure of vitamin C?
total number of optical isomers = ………………………………………………… [1]
(b)
Vitamin C is extremely soluble in water. This means that vitamin C is removed rapidly from
the body. ‘Vitamin C ester’ is available in tablet form as a less soluble source of vitamin C
which stays in the body for longer.
(i)
Suggest why vitamin C is extremely soluble in water.
……The vitamin C molecule has 4 hydroxyl (-OH) groups that form hydrogen bonding with water.…………………………………………………………………………………………………………………….
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(ii)
A ‘vitamin C ester’ tablet contains an ester with the molecular formula C22H38O7.
This ester can be prepared by reacting vitamin C with a long chain carboxylic acid,
CxHyCOOH, in the presence of an acid catalyst.
Vitamin C and the long chain carboxylic acid react in a 1:1 molar ratio.
Determine x and y in the formula of this carboxylic acid.
x = ………….. y = ………….. [2]
(c)
Vitamin C, C6H8O6, is a weak acid (Ka = 7.94 x 10–5 (mol dm–3)), which is often referred to
as ascorbic acid.
(i)
In aqueous solution, vitamin C donates a proton to water:
C6H8O6 + H2O ⇌ C6H7O6– + H3O+
Add curly arrows to the diagram to suggest the mechanism for this process.
(ii)
The student dissolves 0.150 mol of vitamin C in water and makes the solution up to
250 cm3 in a volumetric flask.
Calculate the pH of this solution of vitamin C.
Give your answer to 2 decimal places.
pH = ………………………………………………… [3]
We can use the equilibrium constant to find the pH.
(d)
The label on a carton of orange juice lists the mass of vitamin C, in mg, in a typical serving
of 150 cm3.
The student carries out an investigation to check the vitamin C content in the orange juice.
Vitamin C can be oxidised by iodine:
C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2I–(aq) + 2H+(aq)
The student dilutes 150 cm3 of the orange juice with water to 250.0 cm3 in a volumetric flask.
The student then titrates 25.0 cm3 volume of this solution with 9.60 × 10–4 mol dm–3 iodine
solution, I2(aq).
The mean titre of I2(aq) is 22.50 cm3.
Determine the mass, in mg, of vitamin C in a 150 cm3 serving of the orange juice.
mass of vitamin C in the 150cm3 serving of orange juice = …………………………………………… mg [4]
END OF QUESTION PAPER
External links:
- Paper: A Level Chemistry A H432/03 June 2022 (ocr.org.uk)
- Mark scheme: Mark Scheme H432/03 Unified chemistry June 2022 (ocr.org.uk)
- OCR data sheet for the exam: AS GCE (H032) A GCE (H432) Data Sheet for Chemistry A (ocr.org.uk)
- OCR reaction pathways OCR A Level Chemistry A – Topic Exploration Pack – Reaction pathways