An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V1) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 11 (Multiple Choice)
There are forty questions on this paper. Answer all questions on the separate Answer Sheet.
40 Question · 40 marks
Question 1 · multiple_choice
1 marks
A mixture of 10 cm\(^3\) of a gaseous hydrocarbon, \(C_xH_y\), and 70 cm\(^3\) of oxygen (an excess) was exploded in a sealed vessel. After cooling to room temperature, the total volume of gas remaining was 55 cm\(^3\). Upon shaking this remaining gas with an excess of aqueous sodium hydroxide, the volume decreased to 25 cm\(^3\). What is the molecular formula of the hydrocarbon?
A.\(C_3H_4\)
B.\(C_3H_6\)
C.\(C_3H_8\)
D.\(C_4H_{10}\)
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Worked solution
Let the hydrocarbon react according to the equation: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\). At room temperature, the water produced is in the liquid state and has negligible volume. Shaking the remaining gas with excess aqueous NaOH absorbs the CO\(_2\) gas. Therefore, the decrease in volume is equal to the volume of CO\(_2\) gas produced: \(\text{Volume of } CO_2 = 55 \text{ cm}^3 - 25 \text{ cm}^3 = 30 \text{ cm}^3\). The remaining 25 cm\(^3\) of gas is the unreacted excess oxygen. The volume of oxygen that reacted is: \(\text{Volume of } O_2 \text{ reacted} = 70 \text{ cm}^3 - 25 \text{ cm}^3 = 45 \text{ cm}^3\). Since 10 cm\(^3\) of the hydrocarbon reacted, we can use the volumes (which are proportional to moles by Avogadro's hypothesis) to determine the coefficients: \(x = \frac{30}{10} = 3\) and \(x + \frac{y}{4} = \frac{45}{10} = 4.5\). Substituting \(x = 3\) into the second equation gives: \(3 + \frac{y}{4} = 4.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\). Thus, the molecular formula is \(C_3H_6\).
Marking scheme
1 mark for identifying the correct option (B) based on stoichiometric ratio calculations of the reacting volumes.
Question 2 · multiple_choice
1 marks
A 1.25 g sample of limestone (impure \(CaCO_3\), \(M_r = 100.1\)) was reacted completely with 50.0 cm\(^3\) of 0.500 mol dm\(^{-3}\) hydrochloric acid (an excess). The resulting solution was made up to 250.0 cm\(^3\) with distilled water in a volumetric flask. A 25.0 cm\(^3\) aliquot of this solution required 16.0 cm\(^3\) of 0.100 mol dm\(^{-3}\) sodium hydroxide solution for complete neutralisation. What is the percentage by mass of \(CaCO_3\) in the limestone sample? (Assume the impurities are chemically inert).
A.18.0%
B.36.0%
C.72.1%
D.93.7%
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Worked solution
1. Find initial moles of HCl: \(n(\text{HCl}) = 0.0500 \text{ dm}^3 \times 0.500 \text{ mol dm}^{-3} = 0.0250 \text{ mol}\). 2. Find moles of excess HCl in 25.0 cm\(^3\) aliquot using the titration with NaOH: \(n(\text{NaOH}) = n(\text{HCl})_{\text{excess, aliquot}} = 0.0160 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.00160 \text{ mol}\). 3. Calculate total excess HCl in 250.0 cm\(^3\): \(n(\text{HCl})_{\text{excess, total}} = 0.00160 \text{ mol} \times 10 = 0.0160 \text{ mol}\). 4. Calculate moles of HCl that reacted with CaCO\(_3\): \(n(\text{HCl})_{\text{reacted}} = 0.0250 \text{ mol} - 0.0160 \text{ mol} = 0.0090 \text{ mol}\). 5. Calculate moles of CaCO\(_3\) using the 1:2 stoichiometry (\(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\)): \(n(CaCO_3) = \frac{0.0090}{2} = 0.0045 \text{ mol}\). 6. Find mass of CaCO\(_3\): \(\text{mass} = 0.0045 \text{ mol} \times 100.1 \text{ g mol}^{-1} = 0.45045 \text{ g}\). 7. Calculate percentage: \(\% = \frac{0.45045 \text{ g}}{1.25 \text{ g}} \times 100 \approx 36.0\%\).
Marking scheme
1 mark for the correct back titration calculation leading to 36.0%.
Question 3 · multiple_choice
1 marks
Which statement correctly describes the behavior of a Period 3 chloride when added to excess water at room temperature?
A.Aluminium chloride dissolves to form a solution with a pH of 7.
B.Magnesium chloride reacts to produce a white precipitate of magnesium hydroxide.
C.Silicon(IV) chloride reacts to form an acidic solution and a white precipitate.
D.Sodium chloride reacts with water to form a solution with a pH of 12.
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Worked solution
A is incorrect: aluminium chloride hydrolyses in water because of the high charge density of the Al\(^{3+}\) ion, forming an acidic solution (pH around 3). B is incorrect: magnesium chloride dissolves to form a weakly acidic solution (pH around 6.5) and does not precipitate. C is correct: silicon(IV) chloride reacts vigorously (hydrolyses) with water to form silicon dioxide (a white precipitate) and hydrochloric acid (strongly acidic solution, pH 1-2). D is incorrect: sodium chloride dissolves neutral (pH 7) without any violent reaction.
Marking scheme
1 mark for choosing the correct period 3 chloride reaction behavior.
Question 4 · multiple_choice
1 marks
Two of the Period 3 oxides, X and Y, are added separately to water. Oxide X reacts with water to form an acidic solution. Oxide Y is insoluble in water, but reacts with both hot aqueous sodium hydroxide and dilute hydrochloric acid. Which row correctly identifies oxide X and oxide Y?
A.Oxide X is \(Al_2O_3\); Oxide Y is \(SiO_2\)
B.Oxide X is \(P_4O_{10}\); Oxide Y is \(Al_2O_3\)
C.Oxide X is \(SO_3\); Oxide Y is \(SiO_2\)
D.Oxide X is \(SiO_2\); Oxide Y is \(Al_2O_3\)
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Worked solution
Oxide X reacts with water to form an acidic solution, which means it must be a covalent non-metal oxide such as \(P_4O_{10}\) or \(SO_3\). Oxide Y is insoluble in water but amphoteric (reacts with both acids and bases), which uniquely identifies it as aluminium oxide, \(Al_2O_3\). Insoluble \(SiO_2\) is acidic and only reacts with strong alkalis, not acids. Therefore, option B correctly matches both oxides.
Marking scheme
1 mark for identifying oxide X as an acidic oxide that reacts with water and oxide Y as an amphoteric oxide.
Question 5 · multiple_choice
1 marks
An organic compound W has the molecular formula \(C_5H_{10}O\). Compound W reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate, but does not react with Fehling's solution. Compound W reacts with HCN in the presence of NaCN to form an organic product that does NOT exhibit optical isomerism. What is the identity of compound W?
A.3-methylbutan-2-one
B.pentan-2-one
C.pentan-3-one
D.pentanal
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Worked solution
1. Reacting with 2,4-DNPH means W is a carbonyl (aldehyde or ketone). 2. Not reacting with Fehling's means W is a ketone, which rules out pentanal. 3. Pentan-3-one (\(CH_3CH_2COCH_2CH_3\)) reacts with HCN to yield 2-ethyl-2-hydroxybutanenitrile (or 3-ethyl-3-hydroxypentanenitrile). The central carbon of this product is bonded to: -OH, -CN, and two identical -CH\(_2\)CH\(_3\) groups. Because it does not have four different groups attached, it lacks a chiral center and does not show optical isomerism. On the other hand, the addition of HCN to pentan-2-one or 3-methylbutan-2-one produces a compound with a chiral carbon.
Marking scheme
1 mark for identifying the correct ketone that forms an achiral cyanohydrin.
Question 6 · multiple_choice
1 marks
A compound reacts with alkaline aqueous iodine to yield a yellow precipitate. When this compound is reduced with \(NaBH_4\), the resulting organic product has a chiral centre. What is the identity of the starting compound?
A.propanal
B.propanone
C.butanal
D.butanone
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Worked solution
To give a yellow precipitate with alkaline aqueous iodine, the compound must contain a \(CH_3CO-\) group (or be oxidized to one). Propanone and butanone are ketones containing this group, while propanal and butanal are aldehydes that do not (acetaldehyde does, but not higher aldehydes). Upon reduction with \(NaBH_4\), propanone forms propan-2-ol, \(CH_3CH(OH)CH_3\), which is achiral. Butanone forms butan-2-ol, \(CH_3CH(OH)CH_2CH_3\), which has a chiral carbon bonded to -H, -OH, -CH\(_3\), and -CH\(_2\)CH\(_3\). Therefore, the starting compound is butanone.
Marking scheme
1 mark for determining the ketone that gives a positive iodoform test and reduces to a chiral alcohol.
Question 7 · multiple_choice
1 marks
An ester E with the molecular formula \(C_5H_{10}O_2\) is heated under reflux with dilute hydrochloric acid. The reaction mixture is then distilled, and two organic products, P and Q, are separated. Product P reacts with sodium carbonate to release carbon dioxide gas. Product Q can be oxidized by acidified potassium dichromate(VI) to form a compound that does not react with Fehling's solution, but does react with 2,4-dinitrophenylhydrazine. What is the structural formula of ester E?
A.\(CH_3CH_2COOCH_2CH_3\)
B.\(CH_3COOCH_2CH_2CH_3\)
C.\(CH_3COOCH(CH_3)_2\)
D.\(HCOOCH_2CH_2CH_2CH_3\)
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Worked solution
1. Acid hydrolysis of ester E yields a carboxylic acid P (reacts with \(Na_2CO_3\) to give \(CO_2\)) and an alcohol Q. 2. Since the oxidation product of Q reacts with 2,4-DNPH but not with Fehling's solution, the oxidation product is a ketone. This means Q is a secondary alcohol. 3. The simplest secondary alcohol is propan-2-ol, \(CH_3CH(OH)CH_3\) (3 carbons). 4. For E to have 5 carbons, P must have 2 carbons (ethanoic acid, \(CH_3COOH\)). 5. The ester formed from ethanoic acid and propan-2-ol is 1-methylethyl ethanoate, \(CH_3COOCH(CH_3)_2\).
Marking scheme
1 mark for deduction of the secondary alcohol fragment and identifying the corresponding isopropyl ester structure.
Question 8 · multiple_choice
1 marks
An organic compound X has the structural formula \(CH_3CH_2COOCH_2CH_2OOCCH_3\). Compound X is heated under reflux with an excess of aqueous sodium hydroxide. Which set of products is obtained after the reaction is complete?
A.\(CH_3CH_2COOH\), \(HOCH_2CH_2OH\), and \(CH_3COOH\)
B.\(CH_3CH_2COONa\), \(NaOCH_2CH_2ONa\), and \(CH_3COONa\)
C.\(CH_3CH_2COONa\), \(HOCH_2CH_2OH\), and \(CH_3COONa\)
D.\(CH_3CH_2COOCH_2CH_2OH\) and \(CH_3COONa\)
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Worked solution
Compound X is a diester. Heating under reflux with excess aqueous sodium hydroxide results in alkaline hydrolysis (saponification). Alkaline hydrolysis of esters yields carboxylate salts instead of free carboxylic acids, so we obtain sodium propanoate (\(CH_3CH_2COONa\)) and sodium ethanoate (\(CH_3COONa\)). The alcohol component is a diol, ethane-1,2-diol (\(HOCH_2CH_2OH\)). Unlike phenols, aliphatic alcohols are not acidic enough to react with aqueous sodium hydroxide, so the alcohol remains as a neutral diol and is not converted to a sodium alkoxide. Therefore, the products are \(CH_3CH_2COONa\), \(HOCH_2CH_2OH\), and \(CH_3COONa\).
Marking scheme
1 mark for correctly distinguishing between the salt forms of carboxylic acids and the unreacted neutral alcohol groups during alkaline hydrolysis.
Question 9 · multiple_choice
1 marks
A sample of \(10\text{ cm}^3\) of a gaseous hydrocarbon, \(C_xH_y\), was reacted with excess oxygen (\(80\text{ cm}^3\)) to ensure complete combustion. After combustion and cooling to room temperature, the remaining gas volume was \(60\text{ cm}^3\). This gas mixture was then passed through an excess of aqueous sodium hydroxide, which reduced the gas volume to \(30\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.\(C_3H_4\)
B.\(C_3H_6\)
C.\(C_3H_8\)
D.\(C_4H_{10}\)
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Worked solution
1. The volume of carbon dioxide produced is equal to the volume of gas absorbed by the aqueous sodium hydroxide: \(60\text{ cm}^3 - 30\text{ cm}^3 = 30\text{ cm}^3\). 2. Since \(10\text{ cm}^3\) of hydrocarbon produced \(30\text{ cm}^3\) of \(CO_2\), \(x = 3\). 3. The remaining gas after passing through NaOH is unreacted oxygen, which is \(30\text{ cm}^3\). 4. The volume of oxygen that reacted is \(80\text{ cm}^3 - 30\text{ cm}^3 = 50\text{ cm}^3\). 5. From the combustion equation: \(C_3H_y + (3 + \frac{y}{4}) O_2 \rightarrow 3CO_2 + \frac{y}{2}H_2O\). 6. The ratio of hydrocarbon to oxygen reacted is \(10 : 50 = 1 : 5\). Therefore, \(3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8\). 7. The molecular formula of the hydrocarbon is \(C_3H_8\).
Marking scheme
1 mark for identifying the correct option C. - Award 1 mark for calculating the volume of CO2 (30 cm3) and reacted oxygen (50 cm3), leading to the formula C3H8.
Question 10 · multiple_choice
1 marks
An anhydrous chloride of a Period 3 element, \(X\), is a liquid at room temperature. When added to water, it reacts vigorously with the evolution of white fumes and the formation of a white precipitate alongside an acidic solution.
What is the identity of element \(X\)?
A.Aluminium
B.Silicon
C.Phosphorus
D.Sulfur
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Worked solution
1. Silicon tetrachloride, \(SiCl_4\), is a liquid at room temperature. 2. It reacts vigorously with water in a hydrolysis reaction: \(SiCl_4(\text{l}) + 2H_2O(\text{l}) \rightarrow SiO_2(\text{s}) + 4HCl(\text{aq})\). 3. The white precipitate is silicon dioxide, \(SiO_2\)(s). 4. The white fumes are hydrogen chloride gas, \(HCl\)(g), which dissolves to form a highly acidic solution. 5. Therefore, the element \(X\) is silicon.
Marking scheme
1 mark for identifying the correct option B. - Award 1 mark for recognizing that silicon tetrachloride is a liquid at room temperature that hydrolyses to produce a precipitate of SiO2 and white fumes of HCl.
Question 11 · multiple_choice
1 marks
An organic compound, \(W\), has the molecular formula \(C_4H_8O\). \(W\) reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate, but does not react with Fehling's solution. Warming \(W\) with alkaline aqueous iodine produces a pale yellow precipitate.
What is the structure of \(W\)?
A.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
B.\(\text{CH}_3\text{COCH}_2\text{CH}_3\)
C.\(\text{CH}_3\text{CH(CH}_3\text{)CHO}\)
D.\(\text{CH}_2=\text{CHCH(OH)CH}_3\)
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Worked solution
1. The reaction with 2,4-dinitrophenylhydrazine (2,4-DNPH) indicates the presence of a carbonyl group (aldehyde or ketone). 2. The negative result with Fehling's solution indicates that \(W\) is a ketone rather than an aldehyde. 3. The positive tri-iodomethane (iodoform) test with alkaline aqueous iodine indicates the presence of a methyl ketone group, \(\text{CH}_3\text{CO}-\). 4. The only 4-carbon ketone is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).
Marking scheme
1 mark for identifying the correct option B. - Award 1 mark for analyzing functional group tests to establish that W is butanone.
Question 12 · multiple_choice
1 marks
An ester, \(E\), has the molecular formula \(C_6H_{12}O_2\). When \(E\) is heated under reflux with dilute sulfuric acid, it undergoes hydrolysis to yield two organic products, \(X\) and \(Y\). When product \(X\) is completely oxidized using a hot, acidified solution of potassium dichromate(VI), it forms product \(Y\).
What is the IUPAC name of ester \(E\)?
A.ethyl butanoate
B.propyl propanoate
C.butyl ethanoate
D.methyl pentanoate
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Worked solution
1. Acid hydrolysis of ester \(E\) (\(C_6H_{12}O_2\)) produces an alcohol \(X\) and a carboxylic acid \(Y\). 2. Complete oxidation of alcohol \(X\) with acidified \(K_2Cr_2O_7\) yields product \(Y\) (a carboxylic acid). 3. This implies that \(X\) is a primary alcohol and \(Y\) is the corresponding carboxylic acid with the same carbon chain length and structure. 4. Since the total number of carbon atoms in ester \(E\) is 6, both \(X\) and \(Y\) must contain 3 carbon atoms each. 5. Therefore, \(X\) is propan-1-ol and \(Y\) is propanoic acid. 6. The parent ester is propyl propanoate.
Marking scheme
1 mark for identifying the correct option B. - Award 1 mark for deducing that the alcohol and carboxylic acid must have equal carbon chain lengths (3 carbons each) to allow direct oxidation, identifying propyl propanoate.
Question 13 · multiple_choice
1 marks
An organic compound, \(Z\), contains only carbon, hydrogen, and oxygen. Complete combustion of a \(1.20\text{ g}\) sample of \(Z\) in excess oxygen produced \(1.76\text{ g}\) of carbon dioxide and \(0.72\text{ g}\) of water.
What is the empirical formula of \(Z\)?
A.\(\text{CHO}\)
B.\(\text{CH}_2\text{O}\)
C.\(\text{C}_2\text{H}_4\text{O}\)
D.\(\text{CH}_4\text{O}_2\)
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Worked solution
1. Calculate mass of Carbon: \(1.76\text{ g CO}_2 \times \frac{12.0}{44.0} = 0.48\text{ g of C}\). 2. Calculate mass of Hydrogen: \(0.72\text{ g H}_2\text{O} \times \frac{2.0}{18.0} = 0.08\text{ g of H}\). 3. Calculate mass of Oxygen by subtraction: \(1.20\text{ g} - (0.48\text{ g} + 0.08\text{ g}) = 0.64\text{ g of O}\). 4. Convert masses to moles: - Moles of C = \(0.48 / 12.0 = 0.04\text{ mol}\) - Moles of H = \(0.08 / 1.0 = 0.08\text{ mol}\) - Moles of O = \(0.64 / 16.0 = 0.04\text{ mol}\) 5. Find simplest whole-number ratio: \(C : H : O = 0.04 : 0.08 : 0.04 = 1 : 2 : 1\). 6. The empirical formula is \(\text{CH}_2\text{O}\).
Marking scheme
1 mark for identifying the correct option B. - Award 1 mark for calculating correct molar ratios of C, H, and O, leading to the empirical formula CH2O.
Question 14 · multiple_choice
1 marks
Which statement correctly describes a reaction of a Period 3 element or one of its oxides?
A.Sodium burns in oxygen with a lilac flame to form sodium oxide as the sole product.
B.Magnesium reacts rapidly with cold water to produce a solution of magnesium hydroxide and hydrogen gas.
C.Phosphorus(V) oxide reacts vigorously with water to form a strongly acidic solution with a pH of 1–2.
D.Sulfur dioxide reacts with water to form sulfuric(VI) acid, \(\text{H}_2\text{SO}_4\).
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Worked solution
1. Statement A is incorrect: Sodium burns in oxygen with a bright yellow-orange flame (not lilac, which is characteristic of potassium) to form a mixture of sodium oxide and sodium peroxide. 2. Statement B is incorrect: Magnesium reacts extremely slowly with cold water to produce magnesium hydroxide. It reacts rapidly with steam to produce magnesium oxide. 3. Statement C is correct: Phosphorus(V) oxide, \(P_4O_{10}\), reacts vigorously with water to form phosphoric(V) acid, \(H_3PO_4\), which is a strong acid that produces a highly acidic solution (pH 1–2). 4. Statement D is incorrect: Sulfur dioxide reacts with water to form sulfurous (sulfuric(IV)) acid, \(H_2SO_3\), which is a weak acid, rather than sulfuric(VI) acid, \(H_2SO_4\).
Marking scheme
1 mark for identifying the correct option C. - Award 1 mark for recognizing that phosphorus(V) oxide hydrolyses to produce a strongly acidic solution of pH 1-2.
Question 15 · multiple_choice
1 marks
Propanone reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of a sodium cyanide, \(\text{NaCN}\), catalyst to form 2-hydroxy-2-methylpropanenitrile.
Which statement about the mechanism of this reaction is correct?
A.The reaction is classified as electrophilic addition.
B.The intermediate is a carbocation formed after homolytic fission of the \(\text{C}=\text{O}\) bond.
C.The \(\text{CN}^-\) catalyst is regenerated in the final step of the mechanism by reaction of the intermediate with \(\text{HCN}\).
D.The first step involves the nucleophilic attack of a hydrogen ion, \(\text{H}^+\), on the carbonyl oxygen atom.
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Worked solution
1. The reaction is a nucleophilic addition. 2. In the first step, the nucleophile \(\text{CN}^-\)(aq) attacks the electron-deficient carbonyl carbon, breaking the \(C=O\) \(\pi\)-bond heterolytically to form a negatively charged intermediate, \((CH_3)_2C(CN)O^-\). 3. In the final step, this intermediate reacts with a molecule of \(\text{HCN}\) (or a proton), taking a hydrogen ion to form the hydroxyl group and regenerating the \(\text{CN}^-\) catalyst. 4. Therefore, option C is correct.
Marking scheme
1 mark for identifying the correct option C. - Award 1 mark for identifying that the catalyst CN- is regenerated in the final step of the nucleophilic addition mechanism.
Question 16 · multiple_choice
1 marks
A student attempts to prepare ethyl ethanoate by reacting \(0.100\text{ mol}\) of ethanoic acid with \(0.150\text{ mol}\) of ethanol in the presence of a concentrated sulfuric acid catalyst. After isolation and purification, \(5.28\text{ g}\) of ethyl ethanoate is obtained.
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Worked solution
1. Identify the limiting reactant: Ethanoic acid and ethanol react in a \(1:1\) molar ratio. Since \(0.100\text{ mol}\) of ethanoic acid is mixed with \(0.150\text{ mol}\) of ethanol, ethanoic acid is the limiting reactant. 2. Determine the theoretical yield of ethyl ethanoate: \(0.100\text{ mol}\). 3. Calculate the molar mass of ethyl ethanoate (\(C_4H_8O_2\)): \(4 \times 12.0 + 8 \times 1.0 + 2 \times 16.0 = 88.0\text{ g mol}^{-1}\). 4. Calculate the theoretical mass of ethyl ethanoate: \(0.100\text{ mol} \times 88.0\text{ g mol}^{-1} = 8.80\text{ g}\). 5. Calculate the percentage yield: \(\frac{5.28\text{ g}}{8.80\text{ g}} \times 100\% = 60.0\%\).
Marking scheme
1 mark for identifying the correct option B. - Award 1 mark for correctly determining the limiting reagent (ethanoic acid), calculating the theoretical yield (8.80 g), and establishing the yield as 60.0%.
Question 17 · multiple_choice
1 marks
A \(0.250\text{ g}\) sample of impure sodium hydrogencarbonate, \(\text{NaHCO}_3\), was dissolved in water and titrated with \(0.100\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\). \(\text{NaHCO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\) Exactly \(25.0\text{ cm}^3\) of the hydrochloric acid was required for complete reaction with the hydrogencarbonate in the sample. What is the percentage purity by mass of the sodium hydrogencarbonate in the sample? [\(M_{\text{r}}\text{ of NaHCO}_3 = 84.0\)]
A.10.5%
B.75.0%
C.84.0%
D.91.3%
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Worked solution
1. Calculate the moles of \(\text{HCl}\) used in the titration: \(n(\text{HCl}) = C \times V = 0.100\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.00250\text{ mol}\). 2. Use the stoichiometric ratio from the balanced equation (1:1): \(n(\text{NaHCO}_3) = n(\text{HCl}) = 0.00250\text{ mol}\). 3. Calculate the mass of pure \(\text{NaHCO}_3\) present: \(m(\text{NaHCO}_3) = n \times M_{\text{r}} = 0.00250\text{ mol} \times 84.0\text{ g mol}^{-1} = 0.210\text{ g}\). 4. Calculate the percentage purity by mass: \(\text{Percentage purity} = \frac{0.210\text{ g}}{0.250\text{ g}} \times 100\\% = 84.0\\%\).
Marking scheme
1 mark for the correct answer C.
Question 18 · multiple_choice
1 marks
A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon was exploded with \(70\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature and pressure, the remaining gas volume was \(50\text{ cm}^3\). When this remaining gas was passed through excess aqueous sodium hydroxide, the volume of gas decreased to \(20\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_6\)
C.\(\text{C}_3\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
1. The decrease in volume when passed through aqueous NaOH is due to the absorption of \(\text{CO}_2\) gas. Volume of \(\text{CO}_2\) produced = \(50\text{ cm}^3 - 20\text{ cm}^3 = 30\text{ cm}^3\). 2. Since \(10\text{ cm}^3\) of the hydrocarbon produced \(30\text{ cm}^3\) of \(\text{CO}_2\), there must be 3 carbon atoms per molecule of the hydrocarbon (ratio of 1 : 3). 3. The gas remaining after passing through NaOH is the unreacted excess oxygen. Volume of unreacted \(\text{O}_2 = 20\text{ cm}^3\). 4. Therefore, the volume of \(\text{O}_2\) that reacted is: \(70\text{ cm}^3 - 20\text{ cm}^3 = 50\text{ cm}^3\). 5. The ratio of hydrocarbon reacted to oxygen reacted is: \(10\text{ cm}^3 : 50\text{ cm}^3 = 1 : 5\). 6. The equation for the complete combustion of a hydrocarbon \(\text{C}_3\text{H}_y\) is: \(\text{C}_3\text{H}_y + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\). Comparing the moles of oxygen: \(3 + \frac{y}{4} = 5 \Rightarrow \frac{y}{4} = 2 \Rightarrow y = 8\). Thus, the hydrocarbon is \(\text{C}_3\text{H}_8\).
Marking scheme
1 mark for the correct answer D.
Question 19 · multiple_choice
1 marks
Two elements from Period 3, \(P\) and \(Q\), react separately with excess chlorine gas to form chlorides. - The chloride of \(P\) dissolves in water to form a solution with a pH of approximately 6.5. - The chloride of \(Q\) reacts violently with water to form a strongly acidic solution with a pH of approximately 2, accompanied by the evolution of dense white fumes. Which elements are \(P\) and \(Q\)?
A.\(P\) is sodium; \(Q\) is magnesium
B.\(P\) is magnesium; \(Q\) is silicon
C.\(P\) is aluminium; \(Q\) is phosphorus
D.\(P\) is silicon; \(Q\) is sulfur
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Worked solution
- Magnesium (\(P\)) reacts with chlorine to form magnesium chloride, \(\text{MgCl}_2\). When \(\text{MgCl}_2\) dissolves in water, the magnesium ions undergo slight hydrolysis, giving a weakly acidic solution with pH \(\approx 6.5\). - Silicon (\(Q\)) reacts with chlorine to form silicon tetrachloride, \(\text{SiCl}_4\). \(\text{SiCl}_4\) undergoes rapid and violent hydrolysis in water to produce silicon dioxide and hydrogen chloride gas (white fumes): \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g})\). The dissolved \(\text{HCl}\) forms a strongly acidic solution with pH \(\approx 2\).
Marking scheme
1 mark for the correct answer B.
Question 20 · multiple_choice
1 marks
Which element in Period 3 forms an oxide that is completely insoluble in water, does not react with dilute or concentrated hydrochloric acid, but reacts with hot concentrated sodium hydroxide?
A.aluminium
B.silicon
C.phosphorus
D.sulfur
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Worked solution
- Silicon forms silicon dioxide, \(\text{SiO}_2\), which is a giant covalent macromolecule. It is insoluble in water and does not react with hydrochloric acid because it is acidic. - However, being an acidic oxide, it reacts with strong bases like hot concentrated sodium hydroxide to form soluble silicates. - Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is amphoteric and reacts with both acids and bases. - Phosphorus and sulfur form simple molecular acidic oxides (such as \(\text{P}_4\text{O}_{10}\), \(\text{SO}_2\), \(\text{SO}_3\)) which react readily with water.
Marking scheme
1 mark for the correct answer B.
Question 21 · multiple_choice
1 marks
An organic compound \(W\) has the molecular formula \(C_4H_8O\). - Compound \(W\) reacts with 2,4-dinitrophenylhydrazine (2,4-DNPH) reagent to form an orange-red precipitate. - When heated with Fehling's solution, compound \(W\) produces no observable change. - When reacted with alkaline aqueous iodine, compound \(W\) forms a yellow precipitate. What is the IUPAC name of compound \(W\)?
A.butanal
B.butanone
C.cyclobutanol
D.2-methylpropanal
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Worked solution
1. The positive test with 2,4-DNPH indicates that \(W\) contains a carbonyl group (it is either an aldehyde or a ketone). 2. The negative test with Fehling's solution indicates that \(W\) is a ketone rather than an aldehyde. 3. The positive test with alkaline aqueous iodine (triiodomethane/iodoform reaction) indicates the presence of a methyl ketone group, \(\text{CH}_3\text{C}=\text{O}\) at carbon-2. 4. The only 4-carbon ketone is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\), which contains the \(\text{CH}_3\text{C}=\text{O}\) group and satisfies all the observations.
Marking scheme
1 mark for the correct answer B.
Question 22 · multiple_choice
1 marks
When a carbonyl compound is reduced by sodium tetrahydridoborate(III), \(\text{NaBH}_4\), a chiral alcohol can be formed. Which compound yields a chiral alcohol upon reduction with \(\text{NaBH}_4\)?
A.propanal
B.butanal
C.propanone
D.butanone
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Worked solution
Let us look at the reduction products of each compound: - Propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) is reduced to propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)), which has no chiral carbon. - Butanal (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)) is reduced to butan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)), which has no chiral carbon. - Propanone (\(\text{CH}_3\text{COCH}_3\)) is reduced to propan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_3\)), which is symmetrical and has no chiral carbon. - Butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) is reduced to butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)). The carbon at position 2 is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_3\). Therefore, it is a chiral center.
Marking scheme
1 mark for the correct answer D.
Question 23 · multiple_choice
1 marks
An ester with the molecular formula \(C_5H_{10}O_2\) is heated under reflux with dilute sulfuric acid to yield two organic products, \(A\) and \(B\). When product \(B\) is heated with acidified potassium dichromate(VI) solution, the orange solution turns green and a ketone is formed. Which structural formula represents the starting ester?
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Worked solution
1. Acid-catalyzed hydrolysis of an ester yields a carboxylic acid (\(A\)) and an alcohol (\(B\)). 2. The oxidation of alcohol \(B\) to a ketone indicates that \(B\) must be a secondary alcohol. 3. Let us evaluate the alcohol produced by each ester: - A: propyl ethanoate yields propan-1-ol (a primary alcohol, oxidizes to propanal/propanoic acid, not a ketone). - B: ethyl propanoate yields ethanol (a primary alcohol, oxidizes to ethanal/ethanoic acid, not a ketone). - C: 1-methylethyl ethanoate yields propan-2-ol (a secondary alcohol, oxidizes to propanone, which is a ketone). - D: methyl butanoate yields methanol (a primary alcohol, oxidizes to methanal, not a ketone). Therefore, the correct ester is \(\text{CH}_3\text{COOCH(CH}_3)_2\).
Marking scheme
1 mark for the correct answer C.
Question 24 · multiple_choice
1 marks
A student attempts to synthesize ethyl ethanoate by reacting \(15.0\text{ g}\) of ethanoic acid with \(9.2\text{ g}\) of ethanol in the presence of a concentrated sulfuric acid catalyst. \(\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\) After purification, the student obtains \(8.8\text{ g}\) of pure ethyl ethanoate. What is the percentage yield of ethyl ethanoate in this reaction? [\(M_{\text{r}}\text{ values: } \text{CH}_3\text{COOH} = 60.0;\ \text{C}_2\text{H}_5\text{OH} = 46.0;\ \text{CH}_3\text{COOC}_2\text{H}_5 = 88.0\)]
A.40.0%
B.50.0%
C.58.7%
D.80.0%
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Worked solution
1. Calculate the initial moles of both reactants to identify the limiting reactant: Moles of \(\text{CH}_3\text{COOH} = \frac{15.0\text{ g}}{60.0\text{ g mol}^{-1}} = 0.250\text{ mol}\); Moles of \(\text{C}_2\text{H}_5\text{OH} = \frac{9.2\text{ g}}{46.0\text{ g mol}^{-1}} = 0.200\text{ mol}\). 2. Since the stoichiometric ratio is 1:1, ethanol (0.200 mol) is the limiting reactant. The theoretical maximum moles of ethyl ethanoate that can be formed is 0.200 mol. 3. Calculate the theoretical mass of ethyl ethanoate: Theoretical mass = \(0.200\text{ mol} \times 88.0\text{ g mol}^{-1} = 17.6\text{ g}\). 4. Calculate the percentage yield: \(\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100\\% = \frac{8.8\text{ g}}{17.6\text{ g}} \times 100\\% = 50.0\\%\).
Marking scheme
1 mark for the correct answer B.
Question 25 · multiple_choice
1 marks
A \(10\text{ cm}^3\) sample of a gaseous hydrocarbon, \(C_xH_y\), is exploded with excess oxygen. After cooling to room temperature, the decrease in volume of the gas mixture is \(30\text{ cm}^3\). When the remaining gas is passed through aqueous sodium hydroxide, there is a further decrease in volume of \(30\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
A.\(C_3H_6\)
B.\(C_3H_8\)
C.\(C_4H_8\)
D.\(C_4H_{10}\)
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Worked solution
The general equation for the complete combustion of a hydrocarbon is: \(C_xH_y(g) + (x + \frac{y}{4})O_2(g) \rightarrow xCO_2(g) + \frac{y}{2}H_2O(l)\). Let the volume of the hydrocarbon be \(V = 10\text{ cm}^3\). The further decrease in volume of \(30\text{ cm}^3\) after passing through aqueous \(NaOH\) is due to the absorption of \(CO_2\). Therefore, the volume of \(CO_2\) produced is \(30\text{ cm}^3\). Since \(V(CO_2) = x \times V(C_xH_y)\), we have \(30 = x \times 10\), giving \(x = 3\). The initial contraction in volume after explosion and cooling (since water condenses to liquid and has negligible volume) is given by: \(\Delta V = V(\text{hydrocarbon}) + V(O_2\text{ reacted}) - V(CO_2\text{ produced})\). Given \(\Delta V = 30\text{ cm}^3\): \(30 = 10 + 10(3 + \frac{y}{4}) - 30\). Simplifying this: \(30 = 10 + 30 + 2.5y - 30 \Rightarrow 30 = 10 + 2.5y \Rightarrow 20 = 2.5y \Rightarrow y = 8\). The molecular formula of the hydrocarbon is therefore \(C_3H_8\).
Marking scheme
1 mark: Correctly identify that \(x = 3\) from the volume of \(CO_2\) absorbed by \(NaOH\), and use the contraction in volume equation to determine that \(y = 8\), yielding \(C_3H_8\).
Question 26 · multiple_choice
1 marks
A \(1.25\text{ g}\) sample of impure limestone (mainly \(CaCO_3\)) was reacted with \(50.0\text{ cm}^3\) of \(0.500\text{ mol dm}^{-3}\) hydrochloric acid (an excess). After the reaction was complete, the unreacted acid required \(30.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) sodium hydroxide for complete neutralisation. What is the percentage by mass of calcium carbonate, \(CaCO_3\), in the sample? [Assume the impurities do not react with acid. \(M_r\) of \(CaCO_3 = 100.1\)]
A.\(38.0\%\)
B.\(61.0\%\)
C.\(76.1\%\)
D.\(95.1\%\)
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Worked solution
First, calculate the initial moles of \(HCl\) added: \(n(HCl)_{\text{initial}} = \frac{50.0}{1000} \times 0.500 = 0.0250\text{ mol}\). Next, calculate the moles of \(NaOH\) used to neutralise the excess acid: \(n(NaOH) = \frac{30.0}{1000} \times 0.200 = 0.0060\text{ mol}\). Because \(HCl\) and \(NaOH\) react in a 1:1 ratio, the excess moles of \(HCl = 0.0060\text{ mol}\). The moles of \(HCl\) that reacted with \(CaCO_3\) are: \(n(HCl)_{\text{reacted}} = 0.0250 - 0.0060 = 0.0190\text{ mol}\). The reaction equation is: \(CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O\). The moles of \(CaCO_3\) in the sample are: \(n(CaCO_3) = \frac{0.0190}{2} = 0.0095\text{ mol}\). Find the mass of \(CaCO_3\): \(m(CaCO_3) = 0.0095 \times 100.1 = 0.95095\text{ g}\). Calculate the percentage by mass of \(CaCO_3\): \(\text{Percentage} = \frac{0.95095}{1.25} \times 100\% = 76.08\% \approx 76.1\%\).
Marking scheme
1 mark: Calculate the moles of acid reacting with calcium carbonate (0.0190 mol), use the 1:2 stoichiometric ratio to find moles of calcium carbonate (0.0095 mol), and calculate the correct percentage by mass (76.1%).
Question 27 · multiple_choice
1 marks
Which sequence shows the chlorides of Period 3 elements in order of decreasing pH of the resulting mixture when each chloride is added to water?
A.\(NaCl > MgCl_2 > AlCl_3 > SiCl_4\)
B.\(SiCl_4 > AlCl_3 > MgCl_2 > NaCl\)
C.\(NaCl > AlCl_3 > MgCl_2 > SiCl_4\)
D.\(MgCl_2 > NaCl > AlCl_3 > SiCl_4\)
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Worked solution
\(NaCl\) dissolves in water to form a neutral solution (pH \(\approx 7\)). \(MgCl_2\) undergoes slight hydrolysis due to the moderate charge density of the magnesium ion, forming a weakly acidic solution (pH \(\approx 6.5\)). \(AlCl_3\) undergoes significant hydrolysis because the highly polarizing aluminium ion decomposes water molecules to release \(H^+\), forming an acidic solution (pH \(\approx 3\)). \(SiCl_4\) undergoes rapid, complete hydrolysis with water to yield \(HCl(aq)\) and \(SiO_2\), forming a strongly acidic mixture (pH \(\approx 1-2\)). Therefore, the order of decreasing pH is: \(NaCl > MgCl_2 > AlCl_3 > SiCl_4\).
Marking scheme
1 mark: Correctly identify the pH trends and hydrolysis behavior of Period 3 chlorides to deduce the sequence of decreasing pH.
Question 28 · multiple_choice
1 marks
An element, \(X\), is in Period 3 of the Periodic Table. Its oxide, \(Y\), is insoluble in water but reacts with both hot aqueous sodium hydroxide and hot dilute hydrochloric acid. What is the identity of element \(X\)?
A.Magnesium
B.Aluminium
C.Silicon
D.Phosphorus
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Worked solution
Oxide \(Y\) reacts with both a strong acid (hydrochloric acid) and a strong base (sodium hydroxide). This shows that oxide \(Y\) is amphoteric. Among the elements in Period 3, only aluminium forms an amphoteric oxide, aluminium oxide (\(Al_2O_3\)), which is also insoluble in water. Therefore, element \(X\) is aluminium.
Marking scheme
1 mark: Deduce that the oxide is amphoteric because it reacts with both acids and bases, and identify aluminium as the Period 3 element with an insoluble, amphoteric oxide.
Question 29 · multiple_choice
1 marks
Which carbonyl compound reacts with hydrogen cyanide, \(HCN\), in the presence of sodium cyanide, \(NaCN\), to produce a product containing a chiral carbon atom, where the starting carbonyl compound itself is achiral?
A.Propanone
B.Butanone
C.Pentan-3-one
D.Methanal
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Worked solution
Nucleophilic addition of \(HCN\) to a carbonyl compound, \(R^1-CO-R^2\), yields a hydroxynitrile, \(R^1-C(OH)(CN)-R^2\). For the product to contain a chiral carbon atom, the four groups attached to the central carbon (\(-R^1\), \(-R^2\), \(-OH\), and \(-CN\)) must all be different. If we use butanone (\(CH_3COCH_2CH_3\)), where \(R^1 = -CH_3\) and \(R^2 = -CH_2CH_3\), the product is 2-hydroxy-2-methylbutanenitrile, in which the central carbon is bonded to four different groups (\(-CH_3\), \(-CH_2CH_3\), \(-OH\), and \(-CN\)). Hence, the product is chiral, whereas the starting material (butanone) is achiral. Propanone, pentan-3-one, and methanal do not produce chiral products because their respective products contain at least two identical groups on the central carbon.
Marking scheme
1 mark: Correctly identify that butanone reacts with HCN to form a chiral hydroxynitrile product, while the starting material is achiral.
Question 30 · multiple_choice
1 marks
Compound \(W\) has the molecular formula \(C_4H_8O\). It reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate, but does not give a silver mirror when warmed with Tollens' reagent. When \(W\) is treated with alkaline aqueous iodine, a yellow precipitate is formed. What is the IUPAC name of \(W\)?
A.Butanal
B.Butanone
C.Butan-1-ol
D.Methylpropanal
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Worked solution
The reaction of \(W\) with 2,4-dinitrophenylhydrazine to form an orange precipitate indicates that \(W\) contains a carbonyl (\(C=O\)) group (either an aldehyde or a ketone). The lack of reaction with Tollens' reagent indicates that \(W\) is a ketone and not an aldehyde. The reaction with alkaline aqueous iodine to form a yellow precipitate (the iodoform test) indicates the presence of a methyl ketone (\(CH_3CO-\)) group. For the molecular formula \(C_4H_8O\), the only ketone is butanone (\(CH_3COCH_2CH_3\)), which possesses the methyl ketone structure and reacts positively in the iodoform reaction.
Marking scheme
1 mark: Deduce that compound \(W\) is a ketone from the negative Tollens' test and positive 2,4-DNPH test, and identify it as butanone from the positive iodoform test.
Question 31 · multiple_choice
1 marks
An ester with the molecular formula \(C_5H_{10}O_2\) is heated under reflux with aqueous sodium hydroxide. The mixture is then distilled, and the distillate is found to contain an alcohol, \(Z\). When \(Z\) is warmed with acidified potassium dichromate(VI), the solution turns green and a carboxylic acid is formed. This carboxylic acid does not undergo any further oxidation. Which ester could this be?
A.Methyl butanoate
B.1-methylethyl ethanoate
C.Propyl ethanoate
D.1,1-dimethylethyl methanoate
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Worked solution
The ester has the formula \(C_5H_{10}O_2\). Alkaline hydrolysis produces an alcohol, \(Z\). Since \(Z\) can be oxidized to a carboxylic acid, it must be a primary alcohol. Additionally, the resulting carboxylic acid cannot be further oxidized, which rules out methanol (hydrolysis product of methyl butanoate) because methanol oxidizes to methanoic acid, which is readily oxidized further to carbon dioxide and water. Propyl ethanoate (\(C_5H_{10}O_2\)) undergoes hydrolysis to yield propan-1-ol (primary alcohol) and sodium ethanoate. Propan-1-ol oxidizes to propanoic acid, which is stable against further oxidation. 1-methylethyl ethanoate yields propan-2-ol (secondary, oxidizes to a ketone), and 1,1-dimethylethyl methanoate yields 2-methylpropan-2-ol (tertiary, does not oxidize).
Marking scheme
1 mark: Correctly identify that the alcohol must be a primary alcohol other than methanol to prevent further oxidation, and choose propyl ethanoate as the matching ester.
Question 32 · multiple_choice
1 marks
An organic compound, \(P\), has the molecular formula \(C_4H_8O_3\). \(P\) reacts with sodium carbonate to release carbon dioxide gas. When heated with a few drops of concentrated sulfuric acid, \(P\) undergoes intramolecular esterification to form a stable cyclic ester, \(Q\) (\(C_4H_6O_2\)), containing a five-membered ring. What is the IUPAC name of \(P\)?
A.2-hydroxybutanoic acid
B.3-hydroxybutanoic acid
C.4-hydroxybutanoic acid
D.2-hydroxy-2-methylpropanoic acid
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Worked solution
The reaction of \(P\) (\(C_4H_8O_3\)) with sodium carbonate to produce carbon dioxide shows that it contains a carboxylic acid group (\(-COOH\)). The intramolecular esterification to form a cyclic ester (lactone) indicates that \(P\) also contains a hydroxyl group (\(-OH\)). For the resulting cyclic ester to contain a stable five-membered ring, the ring must consist of four carbon atoms and one oxygen atom. This structural feature requires the hydroxyl group to be located on carbon-4 of the four-carbon chain (the \(\gamma\)-position). Thus, \(P\) is 4-hydroxybutanoic acid (\(HO-CH_2-CH_2-CH_2-COOH\)). Heating 4-hydroxybutanoic acid in acidic conditions leads to the formation of a five-membered lactone ring by losing a water molecule.
Marking scheme
1 mark: Identify that \(P\) has both carboxylic acid and hydroxyl groups, and deduce from the five-membered ring size of the cyclic ester that the hydroxyl group must be on carbon-4, identifying \(P\) as 4-hydroxybutanoic acid.
Question 33 · multiple_choice
1 marks
When \( 10\text{ cm}^3 \) of a gaseous hydrocarbon, \( \text{C}_x\text{H}_y \), is exploded with excess oxygen, there is a contraction in volume of \( 25\text{ cm}^3 \). After shaking the remaining gas with excess aqueous sodium hydroxide, there is a further contraction of \( 30\text{ cm}^3 \).
All gas volumes are measured at room temperature and pressure.
What is the molecular formula of the hydrocarbon?
A.\( \text{C}_3\text{H}_4 \)
B.\( \text{C}_3\text{H}_6 \)
C.\( \text{C}_3\text{H}_8 \)
D.\( \text{C}_4\text{H}_{10} \)
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Worked solution
1. The reaction for the complete combustion of a hydrocarbon \( \text{C}_x\text{H}_y \) is: \( \text{C}_x\text{H}_y(g) + \left(x + \frac{y}{4}\right)\text{O}_2(g) \rightarrow x\text{CO}_2(g) + \frac{y}{2}\text{H}_2\text{O}(l) \)
2. The contraction in volume is due to the loss of gaseous reactants and the fact that water is a liquid at r.t.p. The contraction per mole of \( \text{C}_x\text{H}_y \) is given by: \( \text{Contraction} = 1 + \left(x + \frac{y}{4}\right) - x = 1 + \frac{y}{4} \) moles of gas.
For \( 10\text{ cm}^3 \) of the hydrocarbon, the contraction is: \( 10 \left(1 + \frac{y}{4}\right) = 25\text{ cm}^3 \) \( 1 + \frac{y}{4} = 2.5 \implies y = 6 \).
3. Shaking with aqueous NaOH removes \( \text{CO}_2 \). The volume of \( \text{CO}_2 \) produced is \( 30\text{ cm}^3 \). Since \( 10\text{ cm}^3 \) of \( \text{C}_x\text{H}_y \) produces \( 10x\text{ cm}^3 \) of \( \text{CO}_2 \): \( 10x = 30 \implies x = 3 \).
Therefore, the molecular formula of the hydrocarbon is \( \text{C}_3\text{H}_6 \).
Marking scheme
1 mark for selecting B as the correct molecular formula based on stoichiometric calculations of volume changes.
Question 34 · multiple_choice
1 marks
A \( 25.0\text{ cm}^3 \) sample of a solution of iron(II) oxalate, \( \text{FeC}_2\text{O}_4 \), of concentration \( c \text{ mol dm}^{-3} \), requires \( 24.0\text{ cm}^3 \) of \( 0.0200\text{ mol dm}^{-3} \) acidified potassium manganate(VII), \( \text{KMnO}_4 \), for complete oxidation.
During this reaction, both \( \text{Fe}^{2+} \) and \( \text{C}_2\text{O}_4^{2-} \) ions are oxidized:
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Worked solution
1. Determine the number of moles of electrons released per mole of \( \text{FeC}_2\text{O}_4 \): - \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \) (1 mole of \( \text{e}^- \)) - \( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^- \) (2 moles of \( \text{e}^- \)) Total moles of electrons released per mole of \( \text{FeC}_2\text{O}_4 = 1 + 2 = 3 \).
2. Determine the number of moles of electrons accepted per mole of \( \text{MnO}_4^- \): - \( \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) (5 moles of \( \text{e}^- \)).
3. Set up the stoichiometric mole ratio: \( 5\text{ mol of FeC}_2\text{O}_4 \equiv 3\text{ mol of MnO}_4^- \).
6. Calculate the concentration \( c \): \( c = \frac{8.00 ' \times 10^{-4}\text{ mol}}{0.0250\text{ dm}^3} = 0.0320\text{ mol dm}^{-3} \).
Marking scheme
1 mark for B. Awarded for calculating the correct mole ratio of 5:3 for oxalate to manganate and obtaining the correct concentration.
Question 35 · multiple_choice
1 marks
A \( 3.66\text{ g} \) sample of a mixture of magnesium carbonate, \( \text{MgCO}_3 \), and barium carbonate, \( \text{BaCO}_3 \), is heated strongly until no further change occurs. After cooling, it is found that \( 1.32\text{ g} \) of carbon dioxide, \( \text{CO}_2 \), has been evolved.
What is the mass of \( \text{MgCO}_3 \) in the original mixture?
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Worked solution
1. Write down the molar masses of the substances: - \( M_r(\text{MgCO}_3) = 24.3 + 12.0 + (3 \times 16.0) = 84.3\text{ g mol}^{-1} \) - \( M_r(\text{BaCO}_3) = 137.3 + 12.0 + (3 \times 16.0) = 197.3\text{ g mol}^{-1} \) - \( M_r(\text{CO}_2) = 12.0 + (2 \times 16.0) = 44.0\text{ g mol}^{-1} \)
2. Let the mass of \( \text{MgCO}_3 \) in the mixture be \( m\text{ g} \). Then the mass of \( \text{BaCO}_3 \) is \( (3.66 - m)\text{ g} \).
3. Since both Group 2 carbonates decompose thermally to yield 1 mole of \( \text{CO}_2 \) per mole of carbonate: \( \text{Total moles of CO}_2 = \frac{m}{84.3} + \frac{3.66 - m}{197.3} \)
4. Calculate the actual moles of \( \text{CO}_2 \) evolved: \( n(\text{CO}_2) = \frac{1.32\text{ g}}{44.0\text{ g mol}^{-1}} = 0.0300\text{ mol} \).
5. Set up the equation: \( \frac{m}{84.3} + \frac{3.66 - m}{197.3} = 0.0300 \) \( 0.01186 m + 0.01855 - 0.00507 m = 0.0300 \) \( 0.00679 m = 0.01145 \) \( m = 1.686\text{ g} \approx 1.69\text{ g} \).
Marking scheme
1 mark for B. Calculated the relative formula masses of the carbonates, set up the simultaneous equation/algebraic expression, solved for the mass of magnesium carbonate correctly.
Question 36 · multiple_choice
1 marks
A \( 0.243\text{ g} \) sample of an unknown metal, \( \text{M} \), reacts completely with excess dilute hydrochloric acid to produce \( 240\text{ cm}^3 \) of hydrogen gas, measured at room temperature and pressure (r.t.p.).
Which metal is \( \text{M} \)?
[At r.t.p., 1.00 mol of gas occupies \( 24.0\text{ dm}^3 \)]
A.calcium
B.lithium
C.magnesium
D.sodium
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Worked solution
1. Calculate the number of moles of hydrogen gas produced: \( n(\text{H}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0100\text{ mol} \).
2. Test the stoichiometric ratios for different group metals: - For a Group 1 metal (e.g. Li, Na): \( 2\text{M} + 2\text{H}^+ \rightarrow 2\text{M}^+ + \text{H}_2 \) \( n(\text{M}) = 2 \times n(\text{H}_2) = 0.0200\text{ mol} \). \( A_r(\text{M}) = \frac{0.243\text{ g}}{0.0200\text{ mol}} = 12.15\text{ g mol}^{-1} \) (does not match Li (6.9) or Na (23.0)).
- For a Group 2 metal (e.g. Mg, Ca): \( \text{M} + 2\text{H}^+ \rightarrow \text{M}^{2+} + \text{H}_2 \) \( n(\text{M}) = n(\text{H}_2) = 0.0100\text{ mol} \). \( A_r(\text{M}) = \frac{0.243\text{ g}}{0.0100\text{ mol}} = 24.3\text{ g mol}^{-1} \). This matches Magnesium (\( \text{Mg} \), \( A_r = 24.3 \)) perfectly.
Marking scheme
1 mark for C. Determined the moles of hydrogen gas and evaluated the stoichiometry to find the matching relative atomic mass of Magnesium.
Question 37 · multiple_choice
1 marks
Two of the elements in Period 3, \( \text{X} \) and \( \text{Y} \), react with chlorine to form chlorides \( \text{XCl}_n \) and \( \text{YCl}_m \).
- Chloride \( \text{XCl}_n \) dissolves in water to give a solution with a pH of approximately 6.5. - Chloride \( \text{YCl}_m \) reacts violently with water to give a strongly acidic solution with a pH of approximately 2, and produces white fumes.
Which row correctly identifies elements \( \text{X} \) and \( \text{Y} \)?
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Worked solution
- Magnesium chloride, \( \text{MgCl}_2 \), dissolves in water and undergoes slight hydrolysis to give a slightly acidic solution with pH of about 6.5. - Silicon tetrachloride, \( \text{SiCl}_4 \), reacts violently with water to undergo complete hydrolysis, producing solid silicon dioxide and fumes of hydrogen chloride gas, which dissolves to form a strongly acidic solution with a pH of approximately 2. - Sodium chloride, \( \text{NaCl} \), dissolves to give a neutral solution with a pH of 7. - Aluminium chloride, \( \text{AlCl}_3 \), undergoes substantial hydrolysis to give an acidic solution with a pH of about 3. Thus, \( \text{X} \) is Mg and \( \text{Y} \) is Si.
Marking scheme
1 mark for B. Correctly matching the physical/chemical behavior of the chlorides of Period 3 elements when mixed with water.
Question 38 · multiple_choice
1 marks
Which statement about the oxides of Period 3 elements is correct?
A.Sulfur dioxide, \( \text{SO}_2 \), reacts with aqueous sodium hydroxide to form a salt and water, but does not react with water.
B.Aluminium oxide, \( \text{Al}_2\text{O}_3 \), reacts with both dilute hydrochloric acid and hot aqueous sodium hydroxide.
C.Phosphorus(V) oxide, \( \text{P}_4\text{O}_{10} \), reacts with water to form a solution that turns red litmus paper blue.
D.Silicon dioxide, \( \text{SiO}_2 \), reacts readily with cold dilute sodium hydroxide to form sodium silicate.
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Worked solution
A is incorrect: Sulfur dioxide is an acidic oxide and reacts with water to form sulfurous acid, \( \text{H}_2\text{SO}_3 \). B is correct: Aluminium oxide is amphoteric and reacts with both acids (forming \( \text{Al}^{3+} \)) and bases (forming aluminate, \( \text{[Al(OH)}_4\text{]}^- \)). C is incorrect: Phosphorus(V) oxide reacts with water to form phosphoric(V) acid, \( \text{H}_3\text{PO}_4 \), which turns blue litmus paper red. D is incorrect: Silicon dioxide has a giant covalent structure and is very unreactive. It only reacts with concentrated, hot aqueous NaOH or molten NaOH, not cold dilute NaOH.
Marking scheme
1 mark for B. Correctly identifying that aluminium oxide is amphoteric and is able to react with both acids and bases.
Question 39 · multiple_choice
1 marks
An organic compound, \( \text{W} \), has the molecular formula \( \text{C}_4\text{H}_8\text{O} \).
- \( \text{W} \) reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate. - \( \text{W} \) does not produce a silver mirror when warmed with Tollens' reagent. - When \( \text{W} \) is reduced with \( \text{NaBH}_4 \), compound \( \text{Z} \) is formed.
Which statement about \( \text{Z} \) is correct?
A.\( \text{Z} \) reacts with acidified potassium dichromate(VI) to form a carboxylic acid.
B.\( \text{Z} \) contains a chiral center and exhibits optical isomerism.
C.\( \text{Z} \) reacts with concentrated sulfuric acid at \( 170^\circ\text{C} \) to form only a single alkene.
D.\( \text{Z} \) is a tertiary alcohol and is resistant to oxidation.
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Worked solution
1. Compound \( \text{W} \) (\( \text{C}_4\text{H}_8\text{O} \)) is a carbonyl compound because it reacts with 2,4-DNPH. 2. Since it does not react with Tollens' reagent, it must be a ketone. The only 4-carbon ketone is butanone, \( \text{CH}_3\text{COCH}_2\text{CH}_3 \). 3. Reduction of butanone with \( \text{NaBH}_4 \) yields butan-2-ol, \( \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \), which is compound \( \text{Z} \). 4. Butan-2-ol is a secondary alcohol (ruling out D) and is oxidized to butanone, not a carboxylic acid (ruling out A). 5. It has a chiral carbon at C2 (bonded to H, OH, methyl, and ethyl), so it exhibits optical isomerism (B is correct). 6. Dehydration of butan-2-ol yields a mixture of alkenes (but-1-ene, cis-but-2-ene, and trans-but-2-ene), not just a single alkene (ruling out C).
Marking scheme
1 mark for B. Deducing that W is butanone and Z is butan-2-ol, then correctly identifying that butan-2-ol contains a chiral center.
Question 40 · multiple_choice
1 marks
An ester with the molecular formula \( \text{C}_5\text{H}_{10}\text{O}_2 \) is heated under reflux with dilute sodium hydroxide solution. The products of this hydrolysis are sodium ethanoate and an alcohol, \( \text{V} \).
When warmed with alkaline aqueous iodine, alcohol \( \text{V} \) produces a yellow precipitate.
What is the structural formula of the original ester?
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Worked solution
1. Hydrolysis of the ester \( \text{C}_5\text{H}_{10}\text{O}_2 \) with \( \text{NaOH} \) yields sodium ethanoate, which means the acid portion has 2 carbon atoms (\( \text{CH}_3\text{COO}^- \)). 2. The remaining part of the ester is the alcohol group, which must have \( 5 - 2 = 3 \) carbon atoms. Therefore, alcohol \( \text{V} \) is a propanol isomer. 3. Alcohol \( \text{V} \) reacts with alkaline aqueous iodine to give a yellow precipitate of triiodomethane (\( \text{CHI}_3 \)). This indicates the presence of the \( \text{CH}_3\text{CH(OH)}- \) group. 4. Propan-2-ol, \( \text{CH}_3\text{CH(OH)CH}_3 \), contains this group and gives a positive test, whereas propan-1-ol, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \), does not. 5. Hence, alcohol \( \text{V} \) is propan-2-ol, and the original ester is isopropyl ethanoate, \( \text{CH}_3\text{COOCH(CH}_3)_2 \).
Marking scheme
1 mark for B. Identifying the products of hydrolysis, applying the condition of the iodoform test to find the structure of alcohol V, and deducing the correct ester structure.
Paper 21 (AS Level Structured)
Answer all questions in the spaces provided on the question paper.
6 Question · 60 marks
Question 1 · structured
10 marks
Anhydrous Group 2 nitrate, \(\text{X(NO}_3\text{)}_2\), decomposes on heating to form a solid oxide, nitrogen dioxide gas, and oxygen gas.
(a) Write a balanced equation for the thermal decomposition of \(\text{X(NO}_3\text{)}_2\), including state symbols. [2]
(b) A 4.23 g sample of anhydrous \(\text{X(NO}_3\text{)}_2\) was completely decomposed by heating. The total volume of gas collected at room temperature and pressure (r.t.p.) was \(1.20\text{ dm}^3\).
(i) Calculate the total number of moles of gas produced. [Assume 1 mole of gas occupies \(24.0\text{ dm}^3\) at r.t.p.] [1] (ii) Calculate the number of moles of \(\text{X(NO}_3\text{)}_2\) that decomposed. [2] (iii) Calculate the relative formula mass of \(\text{X(NO}_3\text{)}_2\). [1] (iv) Identify element X. [1]
(c) Explain the trend in the thermal stability of Group 2 nitrates down the group. [3]
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Worked solution
(a) The balanced equation including state symbols is: \(2\text{X(NO}_3\text{)}_2(s) \rightarrow 2\text{XO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)\)
(b)(ii) According to the stoichiometry of the reaction, 2 moles of \(\text{X(NO}_3\text{)}_2\) produce 5 moles of gas (\(4\text{NO}_2 + \text{O}_2\)). \(\text{Moles of } \text{X(NO}_3\text{)}_2 = 0.050 \times \frac{2}{5} = 0.020\text{ mol}\)
(b)(iv) \(M_r\) of \(\text{X(NO}_3\text{)}_2 = A_r(\text{X}) + 2 \times [14.0 + (3 \times 16.0)]\) \(211.5 = A_r(\text{X}) + 124.0\) \(A_r(\text{X}) = 87.5\) This corresponds to Strontium (Sr) which has an \(A_r\) of 87.6.
(c) Thermal stability of Group 2 nitrates increases down the group. This is because: - The ionic radius of the Group 2 cation (\(\text{M}^{2+}\)) increases down the group while keeping the same \(2+\) charge. - This leads to a decrease in charge density of the cation. - The polarizing power of the cation decreases, meaning it distorts the electron cloud of the nitrate anion less, making the covalent N-O bonds harder to break.
Marking scheme
(a) - 1 mark for correct species and balancing: \(2\text{X(NO}_3\text{)}_2 \rightarrow 2\text{XO} + 4\text{NO}_2 + \text{O}_2\) - 1 mark for correct state symbols: \((s) \rightarrow (s) + (g) + (g)\)
(b)(i) - 1 mark for \(0.050\text{ mol}\)
(b)(ii) - 1 mark for identifying the \(2:5\) mole ratio (or multiplying gas moles by \(0.4\)) - 1 mark for \(0.020\text{ mol}\)
(b)(iii) - 1 mark for \(211.5\) (allow error-carried-forward from (b)(ii))
(b)(iv) - 1 mark for Strontium / Sr (allow error-carried-forward from (b)(iii))
(c) - 1 mark for stating that thermal stability increases down the group. - 1 mark for mentioning increasing cationic radius / decreasing charge density of the cation. - 1 mark for explaining that this leads to less polarization/distortion of the nitrate anion.
Question 2 · structured
10 marks
Hydrated iron(II) ammonium sulfate, \((\text{NH}_4)_2\text{Fe(SO}_4)_2 \cdot x\text{H}_2\text{O}\), is used as a standard primary reagent in redox titrations.
(a) Write the ionic equation for the reaction between \(\text{Fe}^{2+}\) and \(\text{MnO}_4^-\right.\) in acidic conditions. [2]
(b) A student dissolved 8.24 g of \((\text{NH}_4)_2\text{Fe(SO}_4)_2 \cdot x\text{H}_2\text{O}\) in dilute sulfuric acid and made the solution up to \(250\text{ cm}^3\) in a volumetric flask. A \(25.0\text{ cm}^3\) portion of this solution required exactly \(21.00\text{ cm}^3\) of \(0.0200\text{ mol dm}^{-3} \text{ KMnO}_4\) for complete oxidation.
(i) Calculate the number of moles of \(\text{MnO}_4^-\right.\) ions in \(21.00\text{ cm}^3\). [1] (ii) Determine the number of moles of \(\text{Fe}^{2+}\) ions in the \(25.0\text{ cm}^3\) sample. [1] (iii) Calculate the number of moles of \(\text{Fe}^{2+}\) ions in the \(250\text{ cm}^3\) volumetric flask. [1] (iv) Calculate the relative formula mass (\(M_r\)) of the hydrated salt. [2] (v) Determine the value of \(x\). [Ar: N = 14.0, H = 1.0, Fe = 55.8, S = 32.1, O = 16.0] [2]
(c) State the color change at the end-point of this titration. [1]
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(b)(v) \(M_r\) of anhydrous \((\text{NH}_4)_2\text{Fe(SO}_4)_2 = [2 \times (14.0 + 4.0)] + 55.8 + [2 \times (32.1 + 64.0)] = 36.0 + 55.8 + 192.2 = 284.0\). Mass of water of crystallization per mole of salt = \(392.4 - 284.0 = 108.4\). \(x = \frac{108.4}{18.0} = 6.02\), which rounds to 6.
(c) The titration is self-indicating: the color changes from colorless to permanent pale pink.
Marking scheme
(a) - 1 mark for correct species: reactants and products. - 1 mark for correct balancing: \(1:5:8 \rightarrow 1:5:4\).
(b)(i) - 1 mark for \(4.20 \times 10^{-4}\text{ mol}\).
(b)(ii) - 1 mark for \(2.10 \times 10^{-3}\text{ mol}\) (allow ecf from b(i)).
(b)(iii) - 1 mark for \(2.10 \times 10^{-2}\text{ mol}\) (allow ecf from b(ii)).
(b)(iv) - 1 mark for formula \(M_r = \text{mass}/\text{moles}\). - 1 mark for \(392.4\) (allow ecf from b(iii)).
(b)(v) - 1 mark for calculating the molar mass of anhydrous salt as \(284.0\). - 1 mark for \(x = 6\) (must be an integer, allow ecf from b(iv)).
(c) - 1 mark for colorless to permanent pale pink.
Question 3 · structured
10 marks
This question is about Period 3 elements and their compounds.
(a) State and explain the trend in electronegativity across Period 3 from sodium to sulfur. [2]
(b) Silicon and phosphorus are two adjacent elements in Period 3.
(i) Describe the structures and bonding of silicon(IV) oxide and phosphorus(V) oxide. [2] (ii) Write an equation for the reaction of phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), with water, and state the pH of the resulting solution. [2]
(c) Aluminium oxide shows amphoteric behavior.
(i) Write an ionic equation to show the reaction of aluminium oxide with dilute hydrochloric acid. [1] (ii) Write an ionic equation to show the reaction of aluminium oxide with hot aqueous sodium hydroxide. [2]
(d) State the observation when a small piece of sodium is added to water. [1]
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Worked solution
(a) Electronegativity increases from sodium to sulfur. This is because the nuclear charge (number of protons) increases, while shielding remains relatively constant (as electrons are added to the same main energy level). Consequently, there is a stronger electrostatic attraction between the nucleus and bonding electrons.
(b)(i) Silicon(IV) oxide (\(\text{SiO}_2\)) has a giant covalent macromolecular structure with strong covalent bonds extending in three dimensions. Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) has a simple molecular structure with strong covalent bonds within molecules and weak intermolecular forces (London dispersion forces) between molecules.
(b)(ii) The reaction with water is: \(\text{P}_4\text{O}_{10}(s) + 6\text{H}_2\text{O}(l) \rightarrow 4\text{H}_3\text{PO}_4(aq)\) The resulting solution is strongly acidic with a pH of 1 or 2.
(c)(i) Reaction with acid (acting as a base): \(\text{Al}_2\text{O}_3(s) + 6\text{H}^+(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{H}_2\text{O}(l)\)
(c)(ii) Reaction with hot alkali (acting as an acid): \(\text{Al}_2\text{O}_3(s) + 2\text{OH}^-(aq) + 3\text{H}_2\text{O}(l) \rightarrow 2[\text{Al(OH)}_4]^-(aq)\)
(d) Sodium melts into a silvery sphere, floats and darts/moves rapidly across the water surface, fizzing/bubbling, and eventually disappears.
Marking scheme
(a) - 1 mark for stating electronegativity increases. - 1 mark for explaining: nuclear charge increases and shielding is constant, leading to stronger attraction to bonding electrons.
(b)(i) - 1 mark for silicon(IV) oxide: giant covalent. - 1 mark for phosphorus(V) oxide: simple molecular / covalent molecules.
(b)(ii) - 1 mark for correct equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\). - 1 mark for pH 1 or 2.
(c)(i) - 1 mark for balanced ionic equation: \(\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}\).
(c)(ii) - 1 mark for correct products: \([\text{Al(OH)}_4]^-\right.\) or \([\text{Al(H}_2\text{O)}_2\text{(OH)}_4]^-\right.\). - 1 mark for balancing: \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-\).
(d) - 1 mark for any one valid observation: melts into a ball / moves on surface / effervescence / disappears.
Question 4 · structured
10 marks
Compounds A and B are structural isomers with the molecular formula \(\text{C}_4\text{H}_8\text{O}\).
(a) Compound A reacts with Tollens' reagent to form a silver mirror.
(i) Deduce the functional group present in A. [1] (ii) Draw the displayed formula of the two structural isomers of A that give this positive test with Tollens' reagent. [2]
(b) Compound B does not react with Tollens' reagent, but it is reduced by \(\text{NaBH}_4\) to form a secondary alcohol, C.
(i) Identify B and draw its skeletal formula. [1] (ii) Name the type of reaction that occurs when B is converted to C. [1]
(c) Compound B reacts with hydrogen cyanide, HCN, in the presence of a NaCN catalyst.
(i) State the role of NaCN in this reaction. [1] (ii) Outline the mechanism for the nucleophilic addition reaction between B and HCN. Include curly arrows, lone pairs, and relevant dipoles. [4]
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Worked solution
(a)(i) The functional group is an aldehyde group (\(-\text{CHO}\)).
(a)(ii) The two structural isomers of \(\text{C}_4\text{H}_8\text{O}\) that are aldehydes are butanal and 2-methylpropanal. Their displayed formulas must show all bonds (including C-H and O-H bonds if any): - Butanal: \(\text{H}-\text{C}(\text{H})_2-\text{C}(\text{H})_2-\text{C}(\text{H})_2-\text{C}(=\text{O})-\text{H}\) - 2-methylpropanal: \(\text{H}-\text{C}(\text{H})_2-\text{C}(\text{H})\big(\text{C}(\text{H})_3\big)-\text{C}(=\text{O})-\text{H}\)
(b)(i) Compound B is butanone. Its skeletal formula shows a four-carbon zig-zag chain with a double-bonded oxygen atom on carbon-2.
(b)(ii) Reduction (or nucleophilic addition).
(c)(i) NaCN acts as a catalyst (providing the nucleophilic cyanide ion, \(\text{CN}^-\), to initiate the reaction).
(c)(ii) Mechanism steps: 1. The nucleophile \(\text{CN}^-\right.\) (showing a lone pair on the carbon atom) attacks the electron-deficient carbonyl carbon of butanone (which has a \(\delta+\right.\) charge, and the oxygen has a \(\delta-\right.\) charge). 2. A curly arrow is drawn from the lone pair on the carbon of \(^-\text{CN}\) to the carbonyl carbon. A second curly arrow starts from the carbonyl double bond and points to the oxygen atom. 3. The intermediate is a tetrahedral alkoxide ion: \(\text{CH}_3\text{CH}_2\text{C}(\text{O}^-)(\text{CN})\text{CH}_3\). The oxygen atom must carry a negative charge and a lone pair. 4. A curly arrow is drawn from the lone pair of the negative oxygen intermediate to the hydrogen of HCN. Another curly arrow shows the breaking of the H-C bond in HCN to reform the \(\text{CN}^-\right.\) catalyst. The final product is 2-hydroxy-2-methylbutanenitrile.
Marking scheme
(a)(i) - 1 mark for aldehyde / \(-\text{CHO}\).
(a)(ii) - 1 mark for correct displayed formula of butanal. - 1 mark for correct displayed formula of 2-methylpropanal.
(b)(i) - 1 mark for drawing the correct skeletal formula of butanone.
(b)(ii) - 1 mark for reduction / nucleophilic addition.
(c)(i) - 1 mark for stating that NaCN acts as a catalyst / source of \(\text{CN}^-\right.\) nucleophile.
(c)(ii) - 1 mark for showing correct dipole on carbonyl group (\(\text{C}^{\delta+}=\text{O}^{\delta-}\)) and curly arrow from lone pair of \(^-\text{CN}\) to the carbon. - 1 mark for curly arrow from \(\text{C}=\text{O}\) bond to the oxygen. - 1 mark for correct structure of the intermediate with a negative charge on the oxygen atom. - 1 mark for curly arrow from the lone pair of the intermediate's oxygen to the hydrogen of HCN (or \(\text{H}^+\)) and arrow showing bond cleavage of H-CN.
Question 5 · structured
10 marks
Esters are organic compounds widely used as flavorings and perfumes.
(a) Ethyl propanoate can be prepared from a carboxylic acid and an alcohol.
(i) Write an equation for this reaction. State the catalyst and conditions required. [3] (ii) State the name of an alternative organic reactant that could react with ethanol to produce ethyl propanoate more rapidly, without the need for a catalyst. [1]
(b) A sample of ethyl propanoate is heated under reflux with dilute aqueous sodium hydroxide.
(i) Write the balanced chemical equation for this alkaline hydrolysis. [2] (ii) Explain why alkaline hydrolysis of an ester is preferred to acid hydrolysis for obtaining a high yield of the alcohol product. [1]
(c) An unknown ester, Y, has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\). - The infrared (IR) spectrum of Y shows a strong absorption peak at \(1740\text{ cm}^{-1}\) and no broad absorption peak around \(2500\text{--}3000\text{ cm}^{-1}\) or \(3200\text{--}3600\text{ cm}^{-1}\). - On acid hydrolysis, Y yields methanol and a carboxylic acid, Z.
(i) Deduce the structures of Y and Z. Draw their skeletal formulas. [2] (ii) Identify the bond responsible for the absorption peak at \(1740\text{ cm}^{-1}\). [1]
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(b)(ii) Alkaline hydrolysis is a non-reversible reaction (it goes to completion) because the carboxylic acid is converted to a carboxylate salt which cannot react back with the alcohol. Acid hydrolysis is a reversible equilibrium reaction.
(c)(i) Since Y undergoes hydrolysis to form methanol (a 1-carbon alcohol) and a carboxylic acid, and the total number of carbon atoms in Y is 4, the carboxylic acid Z must contain 3 carbon atoms (propanoic acid). Therefore, the ester Y is methyl propanoate. - Skeletal formula of Y (methyl propanoate): a 3-carbon chain carbonyl connected to an oxygen bonded to a methyl group. - Skeletal formula of Z (propanoic acid): a 3-carbon carboxylic acid chain.
(c)(ii) The absorption peak at \(1740\text{ cm}^{-1}\) is due to the \(\text{C}=\text{O}\) double bond of the ester carbonyl group.
Marking scheme
(a)(i) - 1 mark for correct equation: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\). - 1 mark for catalyst: concentrated sulfuric acid / \(\text{H}_2\text{SO}_4\). - 1 mark for condition: heat / reflux.
(a)(ii) - 1 mark for propanoyl chloride.
(b)(i) - 1 mark for correct products: sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)) and ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)). - 1 mark for correct reactants and balancing.
(b)(ii) - 1 mark for stating that the reaction is non-reversible / goes to completion.
(c)(i) - 1 mark for drawing correct skeletal structure of Y (methyl propanoate). - 1 mark for drawing correct skeletal structure of Z (propanoic acid).
(c)(ii) - 1 mark for identifying the \(\text{C}=\text{O}\) bond (carbonyl).
Question 6 · structured
10 marks
A gaseous organic compound, W, is a hydrocarbon.
(a) W has a composition of 85.7% carbon and 14.3% hydrogen by mass.
(i) Calculate the empirical formula of W. [2] (ii) A mass of 0.210 g of W occupied a volume of \(153\text{ cm}^3\) at a temperature of \(373\text{ K}\) and a pressure of \(101\text{ kPa}\). Calculate the relative molecular mass (\(M_r\)) of W. [Gas constant, \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)] [3] (iii) Deduce the molecular formula of W. [1]
(b) When W is reacted with cold, dilute, alkaline \(\text{KMnO}_4\), a diol is formed.
(i) State the class of hydrocarbon to which W belongs. [1] (ii) Draw the structural formula of the diol formed when W reacts with cold, dilute, alkaline \(\text{KMnO}_4\). [1] (iii) State the color change observed in this reaction. [2]
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Worked solution
(a)(i) Moles of Carbon: \(\frac{85.7\text{ g}}{12.0\text{ g mol}^{-1}} = 7.14\text{ mol}\) Moles of Hydrogen: \(\frac{14.3\text{ g}}{1.0\text{ g mol}^{-1}} = 14.3\text{ mol}\) Ratio C : H = \(\frac{7.14}{7.14} : \frac{14.3}{7.14} = 1 : 2\) Empirical formula is \(\text{CH}_2\).
(a)(ii) Using the ideal gas equation, \(pV = nRT\): Convert units to SI units: - \(p = 101\text{ kPa} = 101 \times 10^3\text{ Pa}\) - \(V = 153\text{ cm}^3 = 153 \times 10^{-6}\text{ m}^3 = 1.53 \times 10^{-4}\text{ m}^3\) - \(T = 373\text{ K}\)
(a)(iii) Empirical formula mass of \(\text{CH}_2 = 12.0 + (2 \times 1.0) = 14.0\). Ratio \(\frac{M_r}{\text{empirical mass}} = \frac{42.1}{14.0} \approx 3\). Thus, the molecular formula is \((\text{CH}_2)_3 = \text{C}_3\text{H}_6\).
(b)(i) Alkene.
(b)(ii) The alkene is propene, \(\text{CH}_3\text{CH}=\text{CH}_2\). Reaction with cold, dilute, alkaline \(\text{KMnO}_4\) oxidizes the double bond to form propane-1,2-diol: \(\text{CH}_3\text{CH(OH)CH}_2\text{OH}\).
(b)(iii) The purple solution of potassium manganate(VII) turns to a colorless solution (or forms a brown precipitate of \(\text{MnO}_2\)).
Marking scheme
(a)(i) - 1 mark for dividing percentages by \(A_r\) values (C: 7.14, H: 14.3). - 1 mark for correct empirical formula: \(\text{CH}_2\).
(a)(ii) - 1 mark for converting pressure to Pa and volume to \(\text{m}^3\). - 1 mark for calculating moles: \(n = 4.99 \times 10^{-3}\text{ mol}\). - 1 mark for calculating \(M_r = 42.0\) (accept range 41.9 - 42.2).
(a)(iii) - 1 mark for deducing molecular formula \(\text{C}_3\text{H}_6\) based on \(M_r\) and empirical formula.
(b)(i) - 1 mark for alkene.
(b)(ii) - 1 mark for correct structural formula: \(\text{CH}_3\text{CH(OH)CH}_2\text{OH}\).
(b)(iii) - 1 mark for stating the initial color is purple. - 1 mark for stating the final color is colorless (or brown precipitate).
Paper 31 (Practical Skills)
Answer all questions. Show your working and use appropriate significant figures and units.
3 Question · 39.99 marks
Question 1 · practical
13.33 marks
A student is required to determine the water of crystallisation in hydrated sodium carbonate, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\), by volumetric analysis.
The student dissolves a sample of hydrated sodium carbonate in distilled water and makes it up to \(250.0\text{ cm}^3\) in a volumetric flask. A \(25.0\text{ cm}^3\) aliquot of this solution is pipetted into a conical flask and titrated against \(0.100\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\), using methyl orange indicator.
The experimental results obtained by the student are shown below: - Mass of weighing bottle + hydrated sodium carbonate = \(11.45\text{ g}\) - Mass of empty weighing bottle = \(6.30\text{ g}\)
(a) Identify the concordant titres and calculate the mean titre of hydrochloric acid to be used in the calculations. Show your working.
(b) Calculate: (i) the number of moles of \(\text{HCl}\) in the mean titre. (ii) the number of moles of \(\text{Na}_2\text{CO}_3\) present in \(25.0\text{ cm}^3\) of the solution. (iii) the number of moles of \(\text{Na}_2\text{CO}_3\) present in the \(250.0\text{ cm}^3\) volumetric flask. (iv) the mass of anhydrous \(\text{Na}_2\text{CO}_3\) present in the sample. (v) the mass of water of crystallisation and hence the value of \(x\) in \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\). [Assume \(M_r\): \(\text{Na} = 23.0\), \(\text{C} = 12.0\), \(\text{O} = 16.0\), \(\text{H} = 1.0\)]
(c) Calculate the percentage uncertainty in the \(25.0\text{ cm}^3\) volume measured by the pipette, given that the tolerance of the pipette is \(\pm 0.06\text{ cm}^3\).
(d) Suggest one modification to the experimental procedure to improve the accuracy of the mass measurement, explaining your choice.
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Worked solution
(a) Titrations 2 and 3 are concordant because they are within \(0.10\text{ cm}^3\) of each other. Mean titre = \(\frac{36.00 + 36.00}{2} = 36.00\text{ cm}^3\).
(b) (i) Moles of \(\text{HCl} = 0.100\text{ mol dm}^{-3} \times \frac{36.00}{1000}\text{ dm}^3 = 0.00360\text{ mol}\). (ii) Reaction equation: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\) Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = \frac{0.00360}{2} = 0.00180\text{ mol}\). (iii) Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3 = 0.00180 \times 10 = 0.0180\text{ mol}\). (iv) Mass of anhydrous \(\text{Na}_2\text{CO}_3 = 0.0180\text{ mol} \times 106.0\text{ g mol}^{-1} = 1.908\text{ g}\). (v) Total mass of sample \(= 11.45 - 6.30 = 5.15\text{ g}\). Mass of water of crystallisation \(= 5.15 - 1.908 = 3.242\text{ g}\). Moles of water \(= \frac{3.242}{18.0} = 0.1801\text{ mol}\). Ratio \(x = \frac{0.1801}{0.0180} = 10.0\). Therefore, \(x = 10\).
(d) To improve accuracy of the mass measurement, weigh by difference: weigh the bottle with the solid, empty the solid into the beaker, and then reweigh the empty bottle. This accounts for any residual solid sticking to the weighing bottle.
Marking scheme
- **Part (a):** [1 mark] Selects titrations 2 and 3 as concordant and calculates mean titre as \(36.00\text{ cm}^3\) (must show 2 decimal places). - **Part (b):** - (i) [1 mark] Moles of \(\text{HCl} = 0.00360\text{ mol}\). - (ii) [1 mark] Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = 0.00180\text{ mol}\). - (iii) [1 mark] Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3 = 0.0180\text{ mol}\). - (iv) [1 mark] Mass of anhydrous \(\text{Na}_2\text{CO}_3 = 1.91\text{ g}\) (or \(1.908\text{ g}\)). - (v) [2 marks] Mass of water \(= 3.242\text{ g}\) and moles of water \(= 0.180\text{ mol}\) (1 mark), and integer ratio value of \(x = 10\) (1 mark). - **Part (c):** [2 marks] Percentage uncertainty calculation: \(\frac{0.06}{25.0} \times 100 = 0.24\%\) (1 mark for correct working, 1 mark for correct value to 2 s.f.). - **Part (d):** [2.33 marks] Suggests weighing by difference (1.33 marks) and explains that this accounts for solid residue remaining in the weighing bottle (1 mark).
Question 2 · practical
13.33 marks
An investigation is carried out to determine the enthalpy change of reaction, \(\Delta H_{\text{r}}\), for the neutralisation of magnesium oxide (a Period 3 oxide) with hydrochloric acid. \[\text{MgO(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)}\]
A student added \(1.00\text{ g}\) of magnesium oxide, \(\text{MgO}\), to \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) hydrochloric acid (an excess) in a polystyrene cup. The temperature of the acid was recorded every minute for 3 minutes. At the 4th minute, the \(\text{MgO}\) was added and the mixture stirred. Temperature readings were continued from 5 minutes to 10 minutes.
(a) Use the data to estimate the theoretical maximum temperature rise, \(\Delta T\), at 4 minutes by extrapolation. Show how you obtained this value.
(b) Calculate: (i) the heat energy, \(q\), released in the reaction in \(\text{kJ}\). [Assume density of the solution \(= 1.00\text{ g cm}^{-3}\), and specific heat capacity, \(c = 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1}\)] (ii) the number of moles of \(\text{MgO}\) used in the experiment. [Assume \(M_r\) of \(\text{MgO} = 40.3\)] (iii) the enthalpy change of the reaction, \(\Delta H_{\text{r}}\), in \(\text{kJ mol}^{-1}\), including the correct sign.
(c) Explain why the student used an excess of hydrochloric acid. Show by calculation that the acid was indeed in excess.
(d) A student suggested that using a copper cup instead of a polystyrene cup would improve the accuracy of the experiment. State and explain whether you agree or disagree with this suggestion.
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Worked solution
(a) The initial temperature is constant at \(21.5\ ^\circ\text{C}\). To extrapolate the cooling curve back to \(t = 4\text{ min}\): The cooling rate from \(5\text{ min}\) to \(10\text{ min}\) is constant at \(0.6\ ^\circ\text{C}\) per minute: - \(t = 5\text{ min}\), \(T = 35.8\ ^\circ\text{C}\) - \(t = 6\text{ min}\), \(T = 35.2\ ^\circ\text{C}\) Extrapolating back by 1 minute to \(t = 4\text{ min}\): \(T_{\text{max}} = 35.8 + 0.6 = 36.4\ ^\circ\text{C}\). Therefore, \(\Delta T = 36.4 - 21.5 = 14.9\ ^\circ\text{C}\).
(b) (i) \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 14.9\ ^\circ\text{C} = 3114.1\text{ J} = 3.11\text{ kJ}\). (ii) Moles of \(\text{MgO} = \frac{1.00}{40.3} = 0.0248\text{ mol}\). (iii) \[\Delta H_{\text{r}} = -\frac{3.1141\text{ kJ}}{0.0248\text{ mol}} = -125.6\text{ kJ mol}^{-1} \approx -126\text{ kJ mol}^{-1}\] (or \(-125\text{ kJ mol}^{-1}\) depending on rounding of moles).
(c) An excess of hydrochloric acid ensures that all the magnesium oxide reacts completely. Moles of \(\text{HCl}\) available \(= 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). From the stoichiometry, \(1\text{ mol}\) of \(\text{MgO}\) reacts with \(2\text{ mol}\) of \(\text{HCl}\). Thus, \(0.0248\text{ mol}\) of \(\text{MgO}\) requires \(2 \times 0.0248 = 0.0496\text{ mol}\) of \(\text{HCl}\). Since \(0.0500\text{ mol} > 0.0496\text{ mol}\), \(\text{HCl}\) is in excess.
(d) Disagree. Copper is a good conductor of heat and will lose heat much faster to the surroundings than a polystyrene cup (which is a good thermal insulator). This would increase heat loss, leading to a smaller recorded temperature rise and a less accurate, less exothermic value of \(\Delta H_{\text{r}}\).
Marking scheme
- **Part (a):** [2 marks] Extrapolates cooling curve back to 4 minutes to get \(36.4\ ^\circ\text{C}\) (\(\pm 0.1\ ^\circ\text{C}\)) (1 mark) and calculates \(\Delta T = 14.9\ ^\circ\text{C}\) (\(\pm 0.1\ ^\circ\text{C}\)) (1 mark). - **Part (b):** - (i) [1 mark] Correctly calculates \(q = 3.11\text{ kJ}\) (allow ecf from \(\Delta T\)). - (ii) [1 mark] Calculates \(\text{mol of MgO} = 0.0248\text{ mol}\). - (iii) [2 marks] Calculates \(\Delta H_{\text{r}} = -125\text{ to }-126\text{ kJ mol}^{-1}\) (1 mark for correct magnitude, 1 mark for negative sign). - **Part (c):** [3.33 marks] States that excess acid ensures complete reaction of \(\text{MgO}\) (1 mark). Calculates moles of \(\text{HCl} = 0.0500\text{ mol}\) (1 mark). Compares stoichiometry to show \(\text{HCl}\) is in excess (1.33 marks). - **Part (d):** [4 marks] Disagree with the student (1 mark). Explains that copper is a good thermal conductor (1 mark), which increases heat loss to surroundings (1 mark), resulting in a lower measured \(\Delta T\) and less exothermic calculated enthalpy change (1 mark).
Question 3 · practical
13.33 marks
You are provided with three unknown organic liquids, **X**, **Y**, and **Z**. Each liquid contains only one functional group. The possible functional groups are: - alcohol (\(-\text{OH}\)) - aldehyde (\(-\text{CHO}\)) - ketone (\(-\text{CO}-\)) - ester (\(-\text{COOR}\))
The student performs a series of tests to identify the functional groups.
**Test 1:** Add 5 drops of 2,4-dinitrophenylhydrazine (2,4-DNPH) reagent to 1 cm³ of each organic liquid. - **X**: Yellow-orange precipitate formed. - **Y**: No change / stays clear. - **Z**: Yellow-orange precipitate formed.
**Test 2:** Add 1 cm³ of Tollens' reagent to 1 cm³ of each organic liquid and warm gently in a water bath. - **X**: No change. - **Y**: No change. - **Z**: A silver mirror is formed on the walls of the test tube.
**Test 3:** Add 1 cm³ of dilute sulfuric acid and a few drops of potassium dichromate(VI) solution to 1 cm³ of liquid **Y** and warm. - **Y**: The orange solution remains orange.
(a) Complete the table below to deduce the functional group present in each organic liquid, based on the observations. | Liquid | Deduced functional group | Reason / Evidence | | :--- | :--- | :--- | | **X** | | | | **Y** | | | | **Z** | | |
(b) Write a chemical equation for the reaction of \(\text{Z}\) with Tollens' reagent. Use \(\text{RCHO}\) to represent liquid \(\text{Z}\) and \([\text{O}]\) to represent the oxidising agent.
(c) Describe one chemical test (reagents and observations) that could be used to confirm that \(\text{Y}\) is an ester rather than an alcohol, since both gave negative results in Test 1 and Test 2, and Test 3 ruled out a primary/secondary alcohol.
(d) A student suggested that liquid \(\text{X}\) could be propanone. Describe a further chemical test that would confirm the presence of a \(\text{CH}_3\text{CO}-\) group in \(\text{X}\), stating the reagents used and the positive observation.
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Worked solution
(a) - **X**: Ketone. Reason: Positive 2,4-DNPH test shows it is a carbonyl (aldehyde or ketone), negative Tollens' test confirms it is a ketone. - **Y**: Ester. Reason: Negative 2,4-DNPH and Tollens' show it is not a carbonyl. Negative acidified dichromate test shows it is not a primary/secondary alcohol, and its formula represents an ester. - **Z**: Aldehyde. Reason: Positive 2,4-DNPH shows it is a carbonyl, positive Tollens' test (silver mirror) confirms it is an aldehyde.
(c) To confirm **Y** is an ester, heat the sample with aqueous sodium hydroxide (hydrolysis), then acidify. The characteristic sweet, fruity smell of the ester disappears as it is hydrolyzed into a carboxylic acid and alcohol, or test for the carboxylic acid product by adding sodium carbonate to observe effervescence.
(d) Triiodomethane (iodoform) test: - Reagents: Iodine (\(\text{I}_2\)) and aqueous sodium hydroxide (\(\text{NaOH}\)). - Observation: A yellow precipitate of triiodomethane (\(\text{CHI}_3\)) forms.
Marking scheme
- **Part (a):** [6 marks] - 1 mark for each correct functional group (3 marks total). - 1 mark for each correct justification based on experimental evidence (3 marks total). - **Part (b):** [2 marks] Write balanced equation \(\text{RCHO} + [\text{O}] \rightarrow \text{RCOOH}\) (or correct ionic equivalent). - **Part (c):** [2 marks] Explains hydrolysis with \(\text{NaOH(aq)}\) followed by acidification (1 mark) and the change in smell or identification of product carboxylic acid via effervescence with carbonate (1 mark). - **Part (d):** [2.33 marks] Reagents: Iodine and aqueous sodium hydroxide (1.33 marks). Observation: Yellow precipitate (1 mark).
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