Analytical Chemistry Answer

Structured Questions - Answers 1

(a) Pb2+ ions should not be present because Pb2+ ions could react with sodium sulphate to form lead(II) sulphate which is insoluble in water. [1] Therefore, a white precipitate could be observed if it is present. [1] Pb2+(aq) + SO42(aq) PbSO4(s) [1]

(b) (i) The student could not identify the cation by the results of test (2). [1] Both Zn2+ ions and Al3+ ions could react with a little sodium hydroxide solution to give white precipitates [1] and both precipitates redissolve in excess sodium hydroxide solution. [1] Therefore, it is impossible to identify whether the cation is Zn2+ ion or Al3+ ion.

(ii) In test (2), the student should add a little and then excess ammonia solution to X. [1] If Zn2+ ions are present, a white precipitate will form first and then redissolve to give a clear solution. [1] If Al3+ ions are present, a white precipitate will form and remain unchanged after adding excess ammonia solution. [1]

2

(a) (i) Zinc ions are present in solution B. [1] (ii) Zn2+(aq) + 2OH(aq) Zn(OH)2(s) [1]

(b) Potassium ions are present in solution A. [1] Potassium ions give lilac flame when heated in a Bunsen flame. [1] Calcium ions are present in solution C. [1] Calcium ions give brick-red flame when heated in a Bunsen flame. [1]

3

(a) (i) Copper(II) hydroxide [1] (ii) Pale blue precipitate [1]

(b) (i) Copper(II) sulphate solution [1] (ii) Cu(OH)2(s) + H2SO4(aq) CuSO4(aq) + 2H2O(l) [1]

(c) (i) Metal C is iron. [1] Solution D is iron(II) sulphate solution. [1] Solid E is copper. [1]

(ii) Displacement reaction [1] (d) Iron(II) hydroxide [1]

4 The four unknown solutions can be identified by the following procedure: 1. Add a little ammonia solution to each of the solutions. [1] Only the solution containing

calcium ions shows no observable change. [1] All the other three solutions will form white precipitate. [1]

2. Add excess ammonia solution to the remaining three solutions. [1] Only the precipitate in zinc solution will redissolve to give a clear colourless solution. [1] The precipitates in the other two solutions remain unchanged. [1]

3. Add excess sodium hydroxide solution to the remaining two solutions. [1] Only the precipitate in aluminium solution will redissolve to give a clear colourless solution. [1] Therefore, the remaining solution is magnesium solution. [1]

5

(a) (i) Dissolve ammonium chloride salt in water. [1] Divide ammonium chloride solution into two equal portions. [1] Identify ammonium ions: Warm the first portion with sodium hydroxide solution. [1] If a gas with a characteristic pungent smell is detected and the gas turns moist red litmus paper blue, ammonium ions should be present in the solution. [1] Identify chloride ions: Add acidified silver nitrate solution to the second portion. [1] If white precipitate is observed, chloride ions should be present in the solution. [1]

(ii) Identify ammonium ions: NH4+(aq) + OH(aq) NH3(g) + H2O(l) [1] Identify chloride ions: Ag+(aq) + Cl(aq) AgCl(s) [1]

(b) (i) Divide sodium hypochlorite solution into two equal portions. [1] Identify sodium ions: Dip the wire into the first portion and heat the wire in a Bunsen flame. [1] If brilliant golden yellow flame is observed, sodium ions should be present in the solution. [1] Identify hypochlorite ions: Add several drops of solution from the second portion to a piece of litmus paper. [1] If the paper turns white, hypochlorite ions should be present in the solution. [1]

(ii) Identify hypochlorite ions: OCl(aq) + dye Cl(aq) + (dye + O) [1]

6

(a) (i) A green precipitate (FeCO3) would be observed. [1] (ii) K2CO3(aq) + Fe(NO3)2(aq) 2KNO3(aq) + FeCO3(s) [1]

(b) His statement is incorrect [1] because potassium carbonate salt will not decompose on heating. [1]

(c) His statement is incorrect [1] because both potassium hydroxide and sodium carbonate are soluble in water. [1]

7 The three unknown solutions can be identified by the following procedure: 1. Add barium nitrate solution to each of the solutions. [1] Only sodium sulphate solution

forms a precipitate with barium nitrate solution. [1] Na2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NaNO3(aq) [1]

2. Add a little sodium hydroxide solution to each of the two remaining unknown solutions. [1] Both will form a white precipitate. [1] MgCl2(aq) + 2NaOH(aq) Mg(OH)2(s) + 2NaCl(aq) [1] AlCl3(aq) + 3NaOH(aq) Al(OH)3(s) + 3NaCl(aq) [1]

3. Add excess sodium hydroxide solution to each of them. [1] Only the precipitate present in aluminium chloride solution redissolves to become colourless solution. [1] Al(OH)3(s) + OH(aq) [Al(OH)4](aq) [1] The solution with the precipitate is magnesium chloride solution. [1]

8

(a) Solution X contains NH4+ ions. [1] NH4+ ions react with warm sodium hydroxide solution to give ammonia gas which has a characteristic pungent smell. [1] NH4+(aq) + OH(aq) NH3(g) + H2O(l) [1]

(b) (i) Gas Y is ammonia. [1] (ii) It turns moist red litmus paper blue. [1]

(c) (i) NH3(g) + HCl(g) NH4Cl(s) [1] (ii) A white fume is observed. [1]

9

(a) (i) Chlorine gas [1] (ii) 2H+(aq) + Cl(aq) + OCl(aq) Cl2(g) + H2O(l) [1] (iii) Test chlorine with moist blue litmus paper. [1] It turns moist blue litmus paper red

and then white. [1] (b) (i) Carbon dioxide [1]

(ii) CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) [1] (iii) Test carbon dioxide with limewater. [1] It turns limewater milky. [1]

(c) (i) Sulphur dioxide [1] (ii) SO32(aq) + 2H+(aq) SO2(g) + H2O(l) [1] (iii) Test sulphur dioxide with bromine water. [1] It decolorizes bromine water. [1]

(d) (i) Hydrogen [1] (ii) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) [1] (iii) Test hydrogen with a burning splint. [1] It burns with a pop sound. [1]

10

(a) (i) Bromide ions are present in solution B. [1]

(ii) Ag+(aq) + Br(aq) AgBr(s) [1] (b) (i) Sulphite ions are present in solution A. [1]

(ii) SO32(aq) + 2H+(aq) SO2(g) + H2O(l) [1] (iii) It turns bromine water from brown to colourless. [1]/

It turns acidified potassium dichromate solution from orange to green. [1]/ It decolorizes acidified potassium permanganate solution. [1]

(c) (i) Hypochlorite ions are present in solution C. [1] (ii) 2H+(aq) + Cl(aq) + OCl(aq) Cl2(g) + H2O(l) [1] (iii) It turns moist blue litmus paper red, then bleaches it. [1]

11

(a) NH3 should not be the gas in the jar. [1] When NH3 gas dissolves in water, it will give an alkaline solution which turns moist red litmus paper blue quickly. [1] NH3(g) + H2O(l) NH4+(aq) + OH(aq) [1]

(b) HCl should be the gas in the jar. [1] When HCl is placed near ammonia gas, a white fume will form. [1] HCl(g) + NH3(g) NH4Cl(s) [1] On the other hand, SO2 will not form a white fume when placed near ammonia gas. [1]

12

(a) methylpropanal

[1]

methylpropan-1-ol

[1]

methylpropan-2-ol

[1]

methylpropene

[1]

(b) The four unknown compounds can be identified by the following procedure: 1. Add bromine in organic solvent to each of the compounds. [1] Only methylpropene

decolorizes the orange bromine. [1] 2. Add 2,4-dinitrophenylhydrazine solution to each of the three remaining unknown

compounds. [1] Only methylpropanal forms an orange-yellow precipitate. [1] 3. Add acidified potassium dichromate solution to each of the two remaining

unknown compounds. [1] Only methylpropan-1-ol changes acidified potassium dichromate solution from orange to green. [1]

4. methylpropan-2-ol, the remaining compound, has no reaction with acidified potassium dichromate solution. [1]

13

(a) Acidified potassium dichromate solution changes from orange to green. [1] (b) (i) Ethanoic acid [1]

(ii) It is tested by reacting with sodium carbonate or sodium hydrogencarbonate solution to give carbon dioxide which turns limewater milky. [2]

(iii) 2CH3COOH + Na2CO3 2CH3COONa+ + H2O + CO2 [1] (c) (i) Intermediate A is ethanal. [1]

(ii) It is tested by reacting with Tollens reagent to give a silver mirror. [2]

14 (a) 3-methylpent-1-ene:

[1]

Pentan-3-one:

[1]

Propan-2-ol:

[1]

Pentanal:

[1]

(b) The four unknown organic compounds can be identified by the following procedure: 1. Add bromine in organic solvent to each of the compounds. [1] Only

3-methylpent-1-ene decolorizes the orange bromine. [1]

2. Add Tollens reagent to each of the three remaining unknown compounds. [1] Only pentanal gives a silver mirror. [1]

3. Add acidified potassium dichromate solution to each of the two remaining unknown compounds. [1] Only propan-2-ol turns the solution from orange to green. [1]

4. Pentan-3-one, the remaining compound, can be identified by reaction with 2,4-dinitrophenylhydrazine solution [1] to give an orange-yellow precipitate. [1]

15

(a) C18H34O2 [1] (b) Carbon-carbon double bond (C=C) [1]

Carboxyl group (COOH) [1] (c) Test for C=C bond:

(i) Test it with bromine water. [1] (ii) It decolorizes bromine water. [1] (iii) CH3(CH2)7CH=CH(CH2)7COOH + Br2

C

2

H3(CH2)7CHBrCHBr(CH2)7COOH [1](accept other appropriate answers)

Test for COOH group: (i) Test it with sodium carbonate. [1] (ii) It gives carbon dioxide which turns limewater milky. [1] (iii) 2CH3(CH2)7CH=CH(CH2)7COOH + Na2CO3

CH3(CH2)7CH=CH(CH2)7COONa+ + H2O + CO2 [1](accept other appropriate answers)

16

(a) Using Tollens reagent. [1] Propanal reacts with Tollens reagent to give a silver mirror. [1] On the other hand, propanone has no reaction with Tollens reagent. [1]

(b) CH3CH2CHO + 2[Ag(NH3)2]+ + 3OH CH3CH2COO + 2Ag + 4NH3 + 2H2O [1]

17 (a) (i) A silver mirror is observed. [1]

(ii) CH3CH2CHO + 2[Ag(NH3)2]+ + 3OH C

H3CH2COO + 2Ag + 4NH3 + 2H2O [1](b) (i) Acidified potassium dichromate solution changes from orange to green. [1]

(ii) 3SO32(aq) + Cr2O72(aq) + 8H+(aq) 3SO42(aq) + 2Cr3+(aq) + 4H2O(l) [1](c) (i) A white fume is observed. [1]

(ii) NH3(g) + HCl(g) NH4Cl(s) [1] (d) (i) Calcium hydroxide solution turns milky. [1]

(ii) Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l) [1]

18

(a) (i) Propene should be present in bottle A. [1] Propene has reaction with acidified potassium permanganate solution but has no reaction with acidified potassium dichromate solution. [1]

(ii) CH2=CHCH3 + H2O + [O] CH2OHCHOHCH3 (can be further oxidized) [1]

(b) 2-methylpropanol should be present in bottle B. [1] It is a tertiary alcohol which cannot be oxidized by either acidified potassium dichromate solution or acidified potassium permanganate solution. [1]

(c) (i) The compound present in bottle C may be propan-1-ol [1] or propan-2-ol. [1] Each of them could be oxidized by either acidified potassium dichromate solution or acidified potassium permanganate solution. [1]

(ii) For propan-1-ol, CH3CH2CH2OH + [O] CH3CH2COOH [1] For propan-2-ol, CH3CHOHCH3 + [O] CH3COCH3 [1]

19

The ions can be separated according to the following procedure: 1. Add a little ammonia solution to the mixture. All three ions will form a precipitate. [1] 2. When excess ammonia solution is added, only zinc ions will redissolve to give a

colourless solution. [1] 3. Filter the solution to obtain zinc complex in the filtrate. Aluminium hydroxide and

iron(II) hydroxide precipitates are obtained as residues. [1] 4. Acidify the filtrate with dilute sulphuric acid to obtain zinc ions. [1] 5. Add excess dilute sodium hydroxide solution to the residue. Only aluminium hydroxide

precipitate will redissolve to form a colourless solution. [1] 6. Filter the mixture again to obtain aluminium complex in the filtrate. [1] Iron(II)

hydroxide precipitate is obtained as the residue. [1] 7. Acidify the filtrate with dilute sulphuric acid to obtain aluminium ions. [1] 8. Add dilute sulphuric acid to iron(II) hydroxide precipitate to give a solution containing

iron(II) ions. [1]

20 The separation scheme:

[12M]

The ions can be separated according to the following procedure: 1. Add excess hydrochloric acid to the mixture. Only Ag+ ions will form a precipitate. [1] 2. Filter the solution to obtain Fe3+ ions, Al3+ ions, Ba2+ ions and K+ ions in the filtrate. [1]

The insoluble silver chloride is obtained as residue. 3. Add excess sulphuric acid to the filtrate. Only Ba2+ ions will form a precipitate. [1] 4. Filter the solution to obtain Fe3+ ions, Al3+ ions and K+ ions in the filtrate. [1] The

insoluble barium sulphate is obtained as residue. 5. Add excess ammonia solution to the filtrate. [1] Fe3+ ions and Al3+ ions will form a

precipitate. [1] 6. Filter the solution to obtain K+ ions in the filtrate. [1] Iron(III) hydroxide and aluminium

hydroxide precipitates are obtained as residues. 7. Add excess sodium hydroxide solution to the residues. Only aluminium hydroxide will

redissolve to form a colourless solution. [1] 8. Filter the mixture to obtain aluminium complex in the filtrate. [1] Iron(III) hydroxide

precipitate is obtained as residue.

21 The ions can be separated according to the following procedure: 1. Add excess hydrochloric acid to the mixture. Only silver ions will form a precipitate. [1]

Ag+(aq) + Cl(aq) AgCl(s) [1] 2. Filter the solution to obtain ammonium ions, copper(II) ions and calcium ions in the

filtrate. [1] The insoluble silver chloride is obtained as residue. 3. Add excess sodium hydroxide solution to the filtrate. Copper(II) ions and calcium ions

will form a precipitate. [1] Cu2+(aq) + 2OH(aq) Cu(OH)2(s) [1] Ca2+(aq) + 2OH(aq) Ca(OH)2(s) [1]

4. Filter the solution to obtain ammonium ions in the filtrate. [1] Copper(II) hydroxide and calcium hydroxide precipitates are obtained as residues.

5. Add excess ammonia solution to the residue. Only copper(II) hydroxide precipitate will redissolve to form a deep blue solution. [1] Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH(aq) [1]

6. Filter the mixture again to obtain copper(II) complex in the filtrate. [1] Calcium hydroxide is obtained as residue.

22

(a) Both the cation and anion are colourless. [1] The compound may contain Group I ions, Group II ions, aluminium ions or zinc ions. [1]

(b) The compound contains zinc ions. [1] (c) The compound contains carbonate ions. [1] (d) Zinc carbonate [1] (e) (i) Y dissolved and a gas which turns limewater milky was evolved. [2]

(ii) ZnCO3(s) + H2SO4(aq) ZnSO4(aq) + CO2(g) + H2O(l) [1]

23 (a) Chlorine gas turns moist blue litmus paper red, then bleaches it. [1] (b) Cl2(aq) + H2O(l) HCl(aq) + HOCl(aq) [1]

OCl(aq) + dye Cl(aq) + (dye + O) [1] (c) Chlorine gas is toxic. [1]

Safety precautions: Performing the experiment in fume cupboard. [1] Avoid breathing in chlorine gas. [1] Wear gloves. [1]

24

(a) Crystallization by slow evaporation of a solution [1] (b) As the solvent in a solution evaporates at room temperature, the remaining solution

becomes more and more concentrated. [1] As a result, the solution becomes saturated. [1] Further evaporation causes crystallization to occur. [1]

(c) The impurity present in only a small amount will remain dissolved in the solution. [1] (d) If hot distilled water or a large amount of water is used for washing, the crystal will

dissolve and the yield will decrease. [1]

25 (a) (i) ZnCO3(s) + H2SO4(aq) ZnSO4(aq) + H2O(l) + CO2(g) [1]

(ii) Test carbon dioxide with limewater [1] and it turns limewater milky. [1] (b) The reaction is complete when no more gas bubbles are evolved. [1] (c) Filter the mixture to obtain zinc sulphate solution [1] and unreacted zinc carbonate is

obtained as residue. [1] (d) (i) As the solvent in a solution evaporates at room temperature, the remaining solution

becomes more and more concentrated. [1] Then, the solution becomes saturated. [1] Further evaporation causes crystallization to occur. [1]

(ii) 1. Filter the mixture to obtain zinc sulphate crystals as residue. [1] 2. Wash the crystals two or three times with a little cold distilled water. [1] 3. Take out the crystals and dry them by gently pressing them between filter

papers. [1]

26 1. Dissolve the sample in the water. [1] 2. Filter the solution to obtain magnesium sulphate in the filtrate. [1] Insoluble lead(II)

sulphate is obtained as residue. [1] 3. Heat the filtrate until it becomes saturated. [1] 4. Allow the saturated solution to cool down at room temperature. [1] 5. Filter the mixture to obtain magnesium sulphate crystals as residue. [1] 6. Wash the crystals two or three times with a little cold distilled water. [1] 7. Take out the crystals and dry them by gently pressing them between filter papers. [1]

27 (a) The crystals obtained by slow evaporation are larger, more regular in shape and more

pure than those obtained by cooling a hot concentrated solution. [3] (b) Slow evaporation can give a more pure crystal. [1] It is because solute particles need

sufficient time to arrange themselves regularly inside the crystals. [1]

(c) Separate the crystals from the saturated solution by filtration. [1] After filtration, wash the crystals two or three times with a little cold distilled water in order to remove any soluble impurities on the surface of the crystals. [1] Then, take out the crystals and dry them by gently pressing them between filter papers. [1]

28

(a) Fractional distillation [1] (b) The two miscible liquids are separated into different fractions based on their different

boiling points. [1] It involves repeated vaporization and condensation of the mixture. [1](c) Apparatus X: fractionating column [1]

Apparatus Y: Liebig condenser [1] (d) The beads provide a large surface area for repeated vaporization and condensation of the

mixture. [1] (e) Liquid A [1] (f) It is used in the refining of petroleum. [1]

29 The mixture can be separated by the following procedure: 1. Add equal volumes of water and heptane to the mixture. [1] 2. Shake the mixture. [1] Sodium chloride salt will dissolve in water while iodine solid will

dissolve in heptane. [1] 3. Filter the mixture to obtain sodium chloride and iodine in the filtrate. [1] The insoluble

sand is obtained as residue. [1] 4. Transfer the filtrate to a separating funnel. [1] Since water and heptane are immiscible,

they remain as two separate layers. [1] Heptane forms the top layer while water forms the bottom layer. [1]

5. Run the water layer out from the separating funnel. [1] Boil the water layer to obtain a saturated solution of sodium chloride. [1]

6. Allow the solution to cool down at room temperature in order to obtain sodium chloride crystals. [1]

7. Separate sodium chloride crystals from the saturated solution by filtration. [1] 8. Wash the crystals two or three times with a little cold distilled water in order to remove

any soluble impurities on the surface of the crystals. [1] 9. Take out the crystals and dry them by filter paper. [1] 10. Run the heptane layer into a pear-shaped flask and distil off heptane in order to collect

the iodine solid. [1]

30 (a) Liquid-liquid extraction/Solvent extraction [1]

(b) The experimental procedure is shown below: 1. Add 20 cm3 aqueous solution of iodine and 20 cm3 heptane to a separating funnel.

[1] 2. Stopper the separating funnel and shake it gently. [1] 3. Invert the funnel and slowly open the tap. [1] 4. Close the tap and shake the funnel vigorously for a few seconds. [1] 5. Repeat steps 3 to 4 for three times. [1] 6. Remove the stopper of the flask. Leave the funnel to stand until two separate layers

appear. [1] 7. Place a beaker below the separating funnel. Open the tap to allow the lower layer to

drain into the beaker. [1] 8. Close the tap when the upper layer just reaches the tap. Remove the upper layer by

pouring it out from the top of the funnel to another beaker. [1] (c) The upper layer contains most of iodine [1] because iodine is more soluble in heptane.

[1] Heptane forms the upper layer because it is less dense than water. [1]

31 (a) Paper chromatography [1]

Thin-layer chromatography [1] Column chromatography [1]

(b) (i) Paper chromatography [1] and thin-layer chromatography [1] (ii) Column chromatography [1]

(c) (i) Paper chromatography [1]

(ii) Rf = cm5.0cm2.4 = 0.48 [1]

32 (a) A: solvent front [1]

B: baseline [1]

(b) Rf of spot C = cm6.3cm5.2 = 0.83 [1]

Rf of spot D = cm6.3cm3.8 = 0.60 [1]

Rf of spot E = cm6.3cm2.6 = 0.41 [1]

(c) They are different compounds [1] because they have different Rf values. [1]

33 (a) The set-up is applied to determine the boiling point of propan-2-ol. [1] (b) The thermometer bulb should be immersed in the liquid. [1] (c) A: anti-bumping granule [1]

B: Liebig condenser [1] (d) Propan-2-ol is flammable. [1] It is dangerous to heat it in a naked flame. [1] (e) If propan-2-ol is pure, its boiling point should be quite sharp. [1]

34 (a) P4O10 [1] (b) Place a little of the dry powdered phosphorus pentoxide solid in a thin-walled glass

melting point tube. [1] Attach the thermometer to the tube. [1] Then, heat the paraffin oil gently. [1] Record the temperature (t1) at which the solid starts to melt and the temperature (t2) at which the solid melts completely. [1] As a result, the melting point of phosphorus pentoxide is between t1 and t2. [1]

(c) If the white phosphorus pentoxide solid is pure, it has a sharp melting point, i.e. t2 t1 is smaller than 0.5C. [1] If the white phosphorus pentoxide solid is impure, it melts gradually over a wide temperature range. [1]

35 (a) Determining their boiling points. [1] (b) They should have a sharp boiling point if they are pure. [1] (c) The mixture of methanol and ethanol can be separated by fractional distillation [1]

because the difference in their boiling points is between 10C and 25C. [1] (d) Toxic [1]

36 (a) Structure of X

[1]

Structure of Y

[1]

(b) X: butanal [1] Y: methylpropanal [1]

(c) Butanal has a straight chain which is long and thin [1] while methylpropanal has a branched chain which is more rounded. [1] Therefore, butanal has a larger surface area than methylpropanal. [1] The van der Waals forces between butanal molecules are stronger. [1] Thus, butanal has a higher boiling point than methylpropanal.

(d) Butanal and methylpropanal can be separated by fractional distillation [1] because they are two miscible liquids which have very close boiling points. [1]

37 (a) Add excess potassium hydroxide solution to the mixture to precipitate all magnesium

ions. [1] Filter the mixture to obtain potassium chloride in the filtrate and insoluble magnesium hydroxide is obtained as residue. [1] Add hydrochloric acid to the filtrate to neutralize any remaining potassium hydroxide. [1] Allow the solution to evaporate at room temperature until it becomes saturated. Potassium chloride crystals will come out as water is further evaporated. [1] Filter the mixture to obtain potassium chloride crystals. [1]

(b) Separate the mixture by fractional distillation [1] as there is a small difference in boiling points between hexane and heptane. [1]

38 (a) W is butanoic acid. [1]

Z is ethyl ethanoate. [1] (b) They are functional group isomers. [1] (c) Butanoic acid contains the carboxyl group which can participate in hydrogen bond

formation. [1] The butanoic acid molecules are held together by strong hydrogen bonds [1] while the ethyl ethanoate molecules are held together by weak van der Waals forces only. [1] Therefore, butanoic acid has a higher boiling point than ethyl ethanoate.

(d) Simple distillation [1] can be used because they have a large difference in boiling points. [1]

39 (a) Paper chromatography [1] (b) The experimental procedure is shown below:

1. Dissolve the coating of each brand of chocolate beans into equal amounts of water. [1]

2. Draw a baseline with a pencil across the bottom (about 1 cm from the edge) of a

strip of filter paper. [1] 3. Place a drop of the banned pigment on the baseline. [1] 4. Place a drop of each brand of chocolate beans coating solution on the baseline. [1] 5. Allow those spots to dry. [1] 6. Pour a suitable solvent into a beaker to a depth of not more than 0.5 cm. [1] 7. Suspend the paper in the solvent, with the spots above the liquid level. [1] 8. Wrap the beaker with a piece of polyethene film. [1] 9. Remove and dry the paper until the solvent is near the top of the paper. [1]

(c) (i) Brand C contains the banned pigment [1] because one spot from brand C has the same Rf value as the banned pigment. [1]

(ii) Rf = cm5.0cm2.8 = 0.56 [1]

40 (a) Paper chromatography [1] (b) It is based on different solubilities and adsorption characteristics of different pigments.

[1] Therefore, different pigments move up the paper at different speeds. [1] (c) Pigment B will have a larger Rf value. [1]

Since pigment B is more soluble in the running solvent, it is carried away by the solvent to a greater extent. [1]

41 (a) It can be speeded up by using a Buchner funnel connected to a suction pump. [1] (b) An electronic oven [1] (c) (i) It keeps the content inside out of contact with the moisture from the air. [1]

(ii) Silica gel/anhydrous calcium chloride [1] (d) The precipitate must be cooled to room temperature before weighing. [1]

This avoids any unsteadiness of reading during weighing. [1] (e) An analytical balance [1]

42 The mass of copper in 500.0 cm3 of the sample solution can be determined by the following procedure: 1. Weigh the filter paper using an electronic balance. [1] 2. Add excess sodium hydroxide solution to the 500.0 cm3 of sample solution. [1] 3. Filter the precipitate (Cu(OH)2) formed by suction filtration. [1] 4. Dry the filter paper with the precipitate in an electronic oven. [1] 5. Put the filter paper with the precipitate in a desiccator for cooling. [1]

6. Weigh the filter paper with the precipitate using the electronic balance. Find the mass of the precipitate by weighing by difference. [1]

7. Calculate the percentage by mass of Cu in Cu(OH)2. [1] 8. Determine the mass of Cu in the sample solution by multiplying the mass of the

precipitate by the percentage by mass of Cu in Cu(OH)2. [1]

43 (a) MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq) [1] (b) Number of moles of AgCl formed = 1molg35.5)(108

g0.840 = 0.00585 mol [1]

From the equation, mole ratio of MgCl2 : AgCl = 1 : 2

Number of moles of MgCl2 = 2mol0.00585 = 0.00293 mol [1]

Mass of MgCl2 in the sample = 0.00293 mol (24.3 + 35.5 2) g mol1 = 0.279 g [1] Percentage purity of magnesium chloride in the original sample

= 100%g0.450g0.279 = 62.0% [1]

44 (a) (i) An excess of oxalic acid is added to a sample of water. [1] The precipitate formed

between calcium ions and oxalate ions is filtered, dried and weighed. [1] From the mass of precipitate obtained, the concentration of calcium ions in the water sample can be calculated. [1]

(ii) Ca2+(aq) + C2O42(aq) + H2O(l) CaC2O4 H2O(s) [1] (b) Water can be classified as hard water if its calcium content is over 20 mg dm3. [1] (c) (i) Mass of CaC2O4 H2O obtained = 0.0548 g

Mass of Ca2+ in the water sample

= 0.0548 g 16.0)2(1.0416.0212.040.1

40.1

= 0.0150 g = 15.0 mg [1] Concentration of Ca2+ in the water sample

= 3dm0.5mg15.0 = 30.0 mg dm3 [1]

(ii) The water sample is hard water because the concentration of Ca2+ in it is over 20 mg dm3. [1]

45

(a) An analytical balance [1] (b) The pan of the analytical balance should be enclosed. [1] (c) The mass of precipitate formed = 0.3500 g 0.3400 g = 0.0100 g [1] (d) Ca2+(aq) + C2O42(aq) + H2O(l) CaC2O4 H2O(s) [1] (e) Number of moles of CaC2O4 H2O formed

= mol16.021.0416.0212.040.1

0.0100

= 6.84 105 mol [1] Since 1 mole of CaC2O4 H2O contains 1 mole of calcium ions, concentration of calcium ions in the natural water sample

= 35

cm25.0mol106.84

= 2.74 106 mol cm3 [1] (f) Concentration of calcium ions in the natural water sample is mg dm3

= 2.74 106 40.1 1000 1000 mg dm3 = 110 mg dm3 [1] the natural water is hard water as 110 mg dm3 > 20 mg dm3. [1]

46 (a) (i) HPO42(aq) + NH4+(aq) + Mg2+(aq) + OH(aq) + 5H2O(l)

MgNH4PO4 6H2O(s) [1](ii) A white precipitate can be observed. [1] (iii) Magnesium ammonium phosphate hexahydrate [1]

(b) Percentage by mass of P in the fertilizer sample: Mass of MgNH4PO4 6H2O obtained = 0.97 g Mass of P in MgNH4PO4 6H2O

= 0.97 g 616.0)2(1.0416.031.041.014.024.3

31.0

= 0.123 g

% by mass of P in the fertilizer = 100%3.01

0.123 = 4.09% [1] Percentage by mass of P2O5 in the fertilizer sample:

Number of moles of P in MgNH4PO4 6H2O = 1molg31.0g0.123

= 0.00397 mol

each mole of P2O5 contains 2 moles of P.

number of moles of P2O5 = 2mol0.00397 = 0.00199 mol [1]

Mass of P2O5 = 0.00199 mol (31.0 2 + 16.0 5) g mol1 = 0.283 g

% by mass of P2O5 = 100%g3.01g0.283 = 9.40% [1]

47 (a) Pipette [1] (b) (i) Methyl orange [1]

(ii) The reaction mixture changed from yellow to red. [1]

(c) Number of moles of HCl = 0.1 M 3dm100020.00 = 0.002 mol [1]

2HCl + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g) From the equation, mole ratio of HCl : Na2CO3 = 2 : 1 Number of moles of Na2CO3 in 25.0 cm3 of solution

= 2

0.002 mol = 0.001 mol [1]

Number of moles of Na2CO3 in 250.0 cm3 of solution

= 0.001 mol 25.0250.0 = 0.01 mol [1]

Mass of Na2CO3 in 250.0 cm3 of solution = 0.01 mol (23.0 2 + 12.0 + 16.0 3) g mol1 = 1.06 g [1] Mass of impurities = 1.20 1.06 g = 0.14 g percentage by mass of impurities = 100%

1.200.14 = 11.7% [1]

48 (a) HnA(aq) + nNaOH(aq) NanA(aq) + nH2O(l) [1] (b) Number of moles of HnA in 250.0 cm3 of solution

= mol90.01.80 = 0.0200 mol [1]

Number of moles of HnA in 25.0 cm3 of solution

= 0.0200 mol250.025.0 = 0.00200 mol [1]

Number of moles of NaOH used = 0.14 mol dm3 3dm100028.60 = 0.0040 mol

12

0.002000.0040

acidtheofmolesofNumberNaOHofmolesofNumber [1]

n = 2

Therefore, the basicity of the acid is 2. [1]

49 (a) 5H2C2O4(aq) + 6H+(aq) + 2MnO4(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l) [1] (b) The reaction mixture changed from colourless to purple. [1]

(c) Number of moles of MnO4 = 0.034 M 3dm100023.00 = 7.82 104 mol [1]

From the equation, mole ratio of H2C2O4 : MnO4 = 5 : 2 number of moles of H2C2O4 in 25.0 cm3 of solution

= 7.82 104 mol 25 = 1.96 103 mol [1]

number of moles of H2C2O4 in 250.0 cm3 of solution

= 1.96 103 mol 25.0250.0 = 1.96 102 mol [1]

mass of H2C2O4 in 250.0 cm3 of solution = 1.96 102 mol (1.0 2 + 12.0 2 + 16.0 4) g mol1 = 1.76 g [1]

50 (a) Burette [1] (b) Phenolphthalein [1] (c) From pink to colourless [1]

(d) Number of moles of NaOH used = 1molg1.0)16.0(23.0g2.24

= 0.0560 mol [1] Molarity of the 250.0 cm3 NaOH solution

= 3dm250.01000mol0.0560 = 0.224 M [1]

Number of moles of NaOH in 25.0 cm3 of solution

= 3dm100025.0M0.224 = 0.00560 mol [1]

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) From the equation, mole ratio of NaOH : HCl = 1 : 1 Number of moles of HCl in 28.00 cm3 of solution = 0.00560 mol [1]

molarity of HCl solution = 3dm28.001000mol0.00560 = 0.200 M [1]

51

(a) 2MnO4(aq) + 5SO32(aq) + 6H+(aq) 2Mn2+(aq) + 5SO42(aq) + 3H2O(l) [1] (b) The reaction mixture changed from colourless to purple. [1] (c) Number of moles of MnO4 in 15.25 cm3 of 0.01 M KMnO4(aq)

= 3dm100015.25M0.01 = 1.53 104 mol [1]

From the equation, mole ratio of MnO4 : SO32 = 2 : 5 [1] number of moles of SO32 in 25.0 cm3 of solution

= 1.53 104 mol 25

= 3.83 104 mol [1] Molarity of sulphite ions in the solution

= 3.83 104 mol 3dm25.01000

= 0.0153 M [1] (d) each mole of Na2SO3 contains 2 moles of Na+ ions.

number of moles of Na+ ions in 25.0 cm3 of solution = 3.83 104 mol 2 = 7.66 104 mol [1] Molarity of Na+ ions in the solution

= 7.66 104 mol 3dm25.01000

= 0.0306 M [1]

52 (a) CaCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Ca(NO3)2(aq) [1] (b) Number of moles of AgNO3 in 25.60 cm3 of 0.150 M AgNO3

= 0.150 M 3dm100025.60

= 3.84 103 mol From the equation, mole ratio of AgNO3 : CaCl2 = 2 : 1 [1] number of moles of CaCl2 in 25.0 cm3 of solution

= mol2

103.84 3 = 1.92 103 mol [1] number of moles of CaCl2 in 250.0 cm3 of solution

= 1.92 103 25.0250.0 mol

= 0.0192 mol [1] the mass of CaCl2 present in 250.0 cm3 of solution

= 0.0192 mol (40.1 + 35.5 2) g mol1 = 2.13 g [1]

53 (a) Potassium chromate [1] (b) A reddish-brown precipitate can be observed. [1] (c) 2Ag+(aq) + CrO42(aq) Ag2CrO4(s) [1] (d) Ag+(aq) + Cl(aq) AgCl(s)

Number of moles of AgNO3 in 20.60 cm3 of 0.105 M AgNO3 solution

= 0.105 mol dm3 3dm100020.60

= 2.16 103 mol [1] From the equation, mole ratio of Ag+ : Cl = 1 : 1 number of moles of Cl ions in 25.0 cm3 of unknown water sample = 2.16 103 mol [1] Molarity of chloride ions in the unknown water sample

= 2.16 103 mol 3dm25.01000

= 0.0864 M [1]

54 (a) Precipitation titration [1] (b) Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2NaOH(aq) [1] (c) Measure the conductivity of the solution. [1] (d) During the titration, record the conductivity and the corresponding volume of sodium

sulphate solution added. [1] Plot a graph of conductivity against the volume of sodium sulphate solution added. [1] The intersection of two straight lines corresponds to the end point. [1]

(e) Number of moles of Na2SO4 = 0.02 M 3dm100030.00 = 6.0 104 mol [1]

From the equation, mole ratio of Ba(OH)2 : Na2SO4 = 1 : 1 Number of moles of Ba(OH)2 = 6.0 104 mol [1]

Molarity of Ba(OH)2 = 34 dm25.01000mol106.0 = 0.024 M [1]

55 (a) A standard solution is a solution of accurately known molarity. [1]

(b) Phenolphthalein [1] (c) From pink to colourless [1]

(d) Number of moles of HCl = 0.050 100027.90 mol = 1.40 103 mol [1]

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) From the equation, mole ratio of NaOH : HCl = 1 : 1. number of moles of NaOH = 1.40 103 mol [1]

The molarity of NaOH = 1.40 103 25.01000 M = 0.0560 M [1]

56

(a) A reddish-brown precipitate can be observed. [1] (b) If the pH is too high, the silver ions may be precipitated by the hydroxide ions. [1] If the

pH is too low, the chromate ions will change to hydrogen chromate ions or dichromate ions. [1] This will affect the accuracy of the results. [1]

(c) Ag+(aq) + Cl(aq) AgCl(s) Number of moles of Ag+ in 14.50 cm3 of 0.85 M AgNO3 solution

= 0.85 M 3dm 100014.50

= 0.0123 mol [1] From the equation, mole ratio of Ag+ : Cl = 1 : 1 number of moles of Cl in 25.0 cm3 of solution = 0.0123 mol [1] Concentration of Cl in the sample solution

= 0.0123 mol 3dm25.01000

= 0.492 M [1]

57 (a) (i) Oxidizing agent:

potassium permanganate/potassium dichromate/potassium iodate [1] (ii) Reducing agent:

iron(II) sulphate/oxalic acid/potassium oxalate/sodium thiosulphate [1] (b) (i) 6Fe2+(aq) + Cr2O72(aq) + 14H+(aq) 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) [1]

(ii) Number of moles of Cr2O72 = 0.02 M 3dm100032.50 = 6.50 104 mol [1]

From the equation, mole ratio of Cr2O72 : Fe2+ = 1 : 6 Number of moles of Fe2+ in 25.0 cm3 of the solution = 6.50 104 6 mol = 3.90 103 mol [1]

molarity of Fe2+ ions in the iron tablet solution

= 3.90 103 mol 3dm25.01000 = 0.156 M [1]

(iii) Number of moles of Fe2+ in 250.0 cm3 of the solution

= 3.90 103 mol 25.0250.0 = 0.0390 mol [1]

Mass of Fe in the tablet = 0.0390 mol 55.8 g mol1 = 2.18 g

percentage by mass of Fe in the tablet = 100%2.902.18 = 75.2% [1]

58 (a) Pipette [1] (b) OCl(aq) + Cl(aq) + 2H+(aq) Cl2(g) + H2O(l) [1] (c) (i) The colourless solution changes to brown. [1]

(ii) Cl2(aq) + 2I(aq) I2(aq) + 2Cl(aq) [1] (d) (i) Starch solution [1]

(ii) Starch reacts with iodine irreversibly to form a dark blue complex. [1] If starch solution is added to the reaction mixture too early, some of the iodine will react with starch. [1] As a result, the volume of sodium thiosulphate used for titration with iodine will be smaller. [1]

(e) (i) I2(aq) + 2S2O32(aq) 2I(aq) + S4O62(aq) [1] (ii) Number of moles of Na2S2O3 reacted with I2

= 0.05 M 3dm100020.12

= 1.006 103 mol [1] From the equation, mole ratio of I2 : S2O32 = 1 : 2 number of moles of I2 liberated from the reaction between Cl2 and I

= mol2

101.006 3 = 5.03 104 mol [1] From the equation, mole ratio of Cl2 : I2 = 1 : 1 number of moles of Cl2 formed in 25.0 cm3 diluted bleach solution = 5.03 104 mol [1] From the equation, mole ratio of OCl : Cl2 = 1 : 1 number of moles of OCl in 25.0 cm3 diluted bleach solution = 5.03 104 mol [1] Molarity of OCl in 25.0 cm3 diluted bleach solution

= 5.03 104 mol 3dm25.01000 = 0.0200 M

molarity of OCl in the original bleach = 0.0200 M 33

cm25.0cm250.0

= 0.200 M [1]

59 (a) Step 2: SO2(aq) + 2OH(aq) SO32(aq) + H2O(l) [1]

Step 3: SO32(aq) + 2H+(aq) SO2(aq) + H2O(l) [1] (b) (i) Starch solution [1]

(ii) The reaction mixture changes from colourless to dark blue. [1] (c) (i) SO2(aq) + 2H2O(l) + I2(aq) SO42(aq) + 4H+(aq) + 2I(aq) [1]

(ii) Number of moles of I2 = 0.006 M 3dm10006.85 = 4.11 105 mol [1]

From the equation, mole ratio of I2 : SO2 = 1 : 1 Number of moles of SO2 = 4.11 105 mol [1] Mass of SO2 = 4.11 105 1 mol (32.1 + 16.0 2) g mol = 2.63 mg concentration of SO2 in the wine

= 2.63 mg 3dm25.01000 = 105.2 mg dm3 [1]

60 (a) Permanganate index shows the mass of potassium permanganate required to oxidize all

the reducing agents in one dm3 of water sample. [1] (b) 2MnO4(aq) + 5C2O42(aq) + 16H+(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l) [1] (c) The reaction mixture shows a permanent pink colour. [1] (d) Number of moles of MnO4 that reacted with the remaining C2O42

= 0.00200 M 3dm10006.05 = 1.21 105 mol

From the equation, mole ratio of MnO4 : C2O42 = 2 : 5 number of moles of the remaining C2O42

= 1.21 105 mol 25

= 3.03 105 mol [1] Number of moles of C2O42 that reacted with excess MnO4

= 0.00500 M 3dm100025.0 3.03 105 mol

= 9.47 105 mol [1] From the equation, mole ratio of MnO4 : C2O42 = 2 : 5 number of moles of excess MnO4 being removed by C2O42

= 9.47 105 mol 52

= 3.79 105 mol [1] Number of moles of MnO4 that reacted with the reducing agents in the water sample

= 0.00200 M 3dm100025.0 3.79 105 mol

= 1.21 105 mol [1] Mass of KMnO4 used to react with the reducing agents in the water sample = 1.21 105 mol (39.1 + 54.9 + 16.0 4) g mol1 = 1.91 103 g [1]

Permanganate index = 1.91 103 g 50.01000 dm3 [1]

= 0.0382 g dm3 the permanganate index of the water sample is 0.0382 g dm3. [1]

(e) The permanganate index of the water sample is 38.2 mg dm3 (0.0382 1000). [1] It is higher than the national standard (4 mg dm3). Therefore, the water sample should not be used as consumable raw water. [1]

61 (a) The experimental procedure is shown below:

1. Crush the weighed iron tablets using the pestle and mortar. [1] 2. Add dilute sulphuric acid to dissolve the iron tablets, forming Fe2+ solution. [1] 3. Transfer the Fe2+ solution into a volumetric flask and make up to 250.0 cm3

solution. [1] 4. Transfer 25.0 cm3 of the Fe2+ solution into a conical flask. [1] 5. Titrate the Fe2+ solution with a standard solution of potassium permanganate. [1]

(b) (i) MnO4(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) [1] (ii) No indicator is required in the titration. [1]

The reaction mixture is yellow in colour just before the end point. [1] When an additional drop of potassium permanganate solution is added, the colour of the mixture changes from yellow to light purple. [1] This colour change can indicate the end point of the titration. [1]

(iii) Number of moles of MnO4 = 0.005 M 3dm100012.60 = 6.30 105 mol [1]

From the equation, mole ratio of MnO4 : Fe2+ = 1 : 5 Number of moles of Fe2+ = 6.30 105 mol 5 = 3.15 104 mol [1] molarity of iron(II) ions in the iron tablet solution

= 3.15 104 mol 3dm25.01000

= 0.0126 M [1]

62 (a) The higher colour intensity of the riboflavin solution, the larger amount of light is

absorbed, i.e. higher absorbance. [1] (b)

Correct drawing [2] Correct labeling [2]

(c) From the calibration curve, the concentration of the riboflavin solution is equal to 3.7 105 M. [1]

63 (a)

Concentration of riboflavin ( 105 M)

Abs

orba

nce

Correct drawing [2] Correct labelling [2]

(b) From the calibration curve, the concentration of copper(II) sulphate solution is equal to 0.400 mg dm . [1] 3

64 (a)

Correct drawing [2]

Abs

orba

nce

Concentration of CuSO4(aq) (mg dm3)

Concentration of Fe2+(aq) (mg dm3)

Abs

orba

nce

Correct labelling [2] (b) From the calibration curve, the concentration of the iron(II) solution is equal to 1.48 mg

dm3. [1]

65 (a)

Correct drawing [2] Correct labelling [2] From the (b) calibration curve, the concetration of sunset yellow in the sample is 1.78 mg

(c) f sunset yellow for a 23.0 kg child

of soft drink

dm3. [1] The maximum acceptable daily intake o= 1.0 mg kg1 23.0 kg = 23.0 mg [1] The amount of sunset yellow in a can

= 1.78 mg 1000

= 0.668 mg [1]

Since the amount of sunset yellow in a can of soft drink is much lower than the maximum accepta

375

ble daily intake, it is safe for a 23.0 kg child to drink a can of soft drink per day. [1]

66 (a)

Let the mass of compound A be 100 g. Thus, the mass of carbon in the compound = 60.0 g

Abs

orba

nce

Concentration of sunset yellow solutions (mg dm3)

the mass of hydrogen in the compound = 13.3 g the mass of oxygen in the com

Oxygen

pound = 26.7 g

Carbon Hydrogen Mass (g) 60.0 13.3 26.7

Number of moles (mol) 50.12 0.60 3.13

0.1 3.13 67.1

0.16 7.26

Mole ratio 99.267.15 96.7

67.13.13 1

6.167.1

7Simplest mole ratio 3 8 1

[1]

(b) e (C3H8O)n. .0

n (12.0 3 + 1.0 8 + 16.0) = 60.0 [1]

(c) (i)

the empirical formula of compound A is C3H8O. [1] Let the molecular formula of the compound bRelative molecular mass of (C3H8O)n = 60

n = 1 the molecular formula of compound A is C3H8O. [1]

Since compound A has an oxygen atom and can be oxidized by potassium dichromate solution, it is likely to be an aldehyde or alcohol. The saturated hydrocarbon with 3 carbon atoms, there will be 8 hydrogen atoms. Degree of

unsaturation = 02

. Therefore, com88 pound A is an alcohol. [1] Compound A can be propan-1-ol or propan-2-ol. [1]

or [1]

(d) und 3300 cm1 can be observed because of the absorption of OH bond in alcohols. [2]

(ii) They are position isomers. [1] A strong and broad absorption band at aro

67 (a)

e solution or Tollens reagent. [2] It should be a primary or

9 correspond to CH3CH2CH(OH)+, CH3CH(OH)+ and CH3CH2+

Compound X should not be an aldehyde or a ketone as it does not react with 2,4-nitrophenylhydrazinsecondary alcohol. [1] In the mass spectrum, the peak at m/e = 74 corresponds to the molecular ion. The peaks at m/e = 59, 45 and 2ions respectively. [3] Therefore, the possible structure of compound X is CH3CH(OH)CH2CH3. [1]

(b) (i) (ii) nd X by

mass spectroscopy [1] as they produce the same fragmentation pattern. [1]

Compound X exhibits enantiomerism [1] as it contains a chiral carbon. [1] It is impossible to distinguish between the pair of enantiomers of compou

68 (a)

tes all ethene has n ol. [1] It shows that the reaction is completed.

(b) (ii) 1 would be observed

1 would be observed because of the presence of the C=O bond in ethanoic acid. [2]

If ethene is present, the absorption peak between 16101680 cm1 can be observed. [1] The absence of the absorption peak between 16101680 cm1 indicabee converted to ethan(i) Ethanoic acid [1]

A strong and broad absorption peak between 25003300 cmbecause of the presence of the OH bond in ethanoic acid. [2] A strong absorption peak between 16801750 cm

69 (a)

road peak at 3350 cm1 corresponds to the OH bond. [1] (b)

ound 1720 cm1 corresponds to the C=O bond. [1] (c)

at around 3000 cm1 and 1710 cm1 correspond to the OH bond and C=O bond. [2]

Ethanol [1] A strong and bEthanal [1] A strong peak at arEthanoic acid [1] The peaks

70 (a)

aporized to a gaseous form. [1]

ions [1] and then the molecular ions undergo fragmentation to give fragment

e accelerated by an electric field towards a magnetic field. [1]

ns are deflected by the magnetic field. [1]

(b) correspond to CH3CH35Cl37Cl+,

Vaporization [1] The sample is vIonization [1] The gaseous sample is bombarded by a beam of fast-moving electrons to form molecularions. [1] Acceleration [1] Positive ions arDeflection [1] The positive ioDetection [1] The detector produces electrical signals proportional to the number of ions detected. [1] The peaks at m/e = 100, 98, 65 and 63

CH3CH35Cl35Cl+, CH3CH37Cl+ and CH3CH35Cl+ ions respectively. [4]

71 (a)

the mass of oxygen in the com

Carbon Hydrogen Oxygen

Let the mass of compound Y be 100 g. Thus, the mass of carbon in the compound = 72.0 g the mass of hydrogen in the compound = 12.0 g

pound = 16.0 g

Mass (g) 72.0 12.0 16.0

Number of moles (mol) 60.120.72 12

0.10.12 1

0.160.16

Mole ratio 616 12

112 1

11

Simplest mole ratio 6 12 1 [1]

(b) e (C6H12O)n. 00

n (12.0 6 + 1.0 12 + 16.0) = 100 [1]

(c) [1] because only aldehydes could react with Tollens reagent to

) The peak at m/e = 29 represents HCO+ ion. [1]

the empirical formula of compound Y is C6H12O. [1] Let the molecular formula of the compound bRelative molecular mass of (C6H12O)n = 1

n = 1 the molecular formula of compound Y is C6H12O. [1] Compound Y is hexanal give a silver mirror. [1]

(d

72

(c) 237ClCH237Cl+, CH235ClCH237Cl+ and CH235ClCH235Cl+ respectively. [3]

(a) 35Cl and 37Cl [2] (b) The relative abundance of 35Cl and 37Cl are 75% and 25% respectively. [2]

The peaks at m/e = 102, 100 and 98 represent the molecular ions CH

73 (a) und 1690 cm1 is due to the absorption of C=O bonds in

peak at around 2920 cm1 is due to the absorption of OH bond in

A strong absorption peak at arocarboxyl and ester groups. [2] A strong absorption carboxyl group. [1]

(b) as IR spectrum can only give the information about the functional groups in aspirin. [1] It is impossible to deduce the structure of aspirin by interpreting IR spectrum only [1]

74

(b) (i) lecular ions,

d 93 correspond to the fragment ions, CH281Br+ and

(ii) It is because the relative abundances of 79Br and 81Br are very close. [1]

(a) Vaporization, ionization, acceleration, deflection and detection [5]

The peaks at m/e = 176, 174 and 172 correspond to the moCH281Br81Br+, CH281Br79Br+ and CH279Br79Br+ respectively. [3] The peaks at m/e = 95 anCH279Br+ respectively. [2]

75 (a)

the mass of chlorine in the compound = 54.7 g

Carbon Hydrogen Chlorine

Let the mass of compound Z be 100 g. Thus, the mass of carbon in the compound = 37.5 g the mass of hydrogen in the compound = 7.80 g

Mass (g) 37.5 7.80 54.7

Number of moles (mol) 13.30.125.37 80.7

0.180.7 54.1

5.357.54

Mole ratio 03.254.113.3 06.5

54.18.7 1

5.154.1 4


(b) (C2H5Cl)n. 64

n (12.0 2 + 1.0 5 + 35.5) = 64 [1]

(c)

Cl+ respectively. [2] (d)

(ii) served in the IR spectrum [1] because of the presence of OH bond in ethanol. [1]

the empirical formula of compound Z is C2H5Cl. [1] Let the molecular formula of compound Z beRelative molecular mass of (C2H5Cl)n =

n = 1 the molecular formula of compound Z is C2H5Cl. [1] The peaks at m/e = 66 and 64 correspond to the molecular ions, CH3CH237Cl+ and CH3CH235Cl+ respectively [2] while the peaks at m/e = 51 and 49 correspond to thefragment ions, CH237Cl+ and CH235

(i) Compound W is ethanol. [1] A strong and broad absorption peak at around 3300 cm1 is ob

76

(a)


Oxygen

Let the mass of compound X be 100 g. Thus, the mass of carbon in the compound = 66.7 g the mass of hydrogen in the compound = 11.1 g

pound = 22.2 g



0.11.11 39.1

0.162.22

Mole ratio 43.156.5 9

99.739.1

1.11 13.139.1


[1]

(b) lecular ion of

(c)

hen it is tested with Tollens reagent. [1] Thus,

(d) = 57, 43 and 29 correspond to CH3CH2CO+, CH3CO+ and CH3CH2+ respectively. [3]

the empirical formula of compound X is C4H8O. [1] From the mass spectrum, the peak at m/e = 72 corresponds to the mocompound X. [1] Therefore, the molecular mass of compound X is 72. [1] Since the molecular mass of compound X is 72, its molecular formula is C4H8O. [1] Compound X should contain C=O group as it gives orange-yellow precipitate when it reacts with 2,4-dinitrophenylhydrazine solution. [1] Compound X should not be an aldehyde as no observable changes wcompound X should be butanone. [1] The peaks at m/e

77

(b) 3

+, CH3CH2CO+ and CH3CH2+ ions respectively. [3]

+, CH3CH(CH3)CO+ and CH3CO+ ions

respectively. [3] Therefore, B is methylbutanone. [1]

(a) They are chain isomers. [1] By interpreting the mass spectrum A, the prominent peaks at m/e = 86, 57 and 29 correspond to CH3CH2COCH2CHTherefore, A is pentan-3-one. [1] By interpreting the mass spectrum B, the prominent peaks at m/e = 86, 71 and 43 correspond to CH3CH(CH3)COCH3

78 (a)

Let the mass of compound Y be 100 g. Thus, the mass of carbon in the compound = 60.0 g

the mass of hydrogen in the compound = 13.3 g the mass of oxygen in the compound = 26.7 g

Oxygen Carbon Hydrogen

Mass (g) 60.0 13.3 26.7

Number of moles (mol) 5 0.120.60 3.13

0.1 3.13 67.1

0.167.26

Mole ratio 99.267.15 96.7

67.13.13 1

6.167.1

7

Simplest mole ratio 3 8 1

[1]

(b) (C3H8O)n. .0.

n (12.0 3 + 1.0 8 + 16.0) = 60.0 [1]

(c) X

compound Y e, compound X should be propan-2-ol. [1]

) Compound Y is propanone. [1]

the empirical formula of compound Y is C3H8O. [1] Let the molecular formula of the compound beRelative molecular mass of (C3H8O)n = 60

n = 1 the molecular formula of compound Y is C3H8O. [1] A strong and broad absorption peak at around 3300 cm1 indicates that compoundcontains OH bond. [1] Thus, compound X should be propan-1-ol or propan-2-ol. [1] Since compound Y does not react with Tollens reagent, it shows that should be a ketone. [1] Therefor

(d

79 (a)

the mass of oxygen in the compound = 10.8 g


Let the mass of compound A be 100 g. Thus, the mass of carbon in the compound = 81.1 g the mass of hydrogen in the compound = 8.1 g

Mass (g) 81.1 8.1 10.8


0.11.8 675.0

0.168.10

Mole ratio 106.076.6 75

1267.0

1.8 167.0675.0

5 5Simplest mole ratio 10 12 1

[1]

(b) ecular ion of the empirical formula of compound Z is C10H12O. [1] From the mass spectrum, the peak at m/e = 148 corresponds to the mol

compound Z. [1] Therefore, the molecular mass of compound Z is 148. [1] Since the molecular mass of compound Z is 148, its molecular formula is C(c) 10H12O. [1] For saturated hydrocarbon with 10 carbon atoms, there will be 22 hydrogen atoms.

Degree of unsaturation = 52

1222 . Therefore, other and the benzene ring (degree of unsaturation = 4), there is one unsaturated bond. [1] Referring to the IR spectrum, the presence of a strong absorption round 1720

and 43 correspond to C6H5CH2CH2+, C6H5CH2+ and

Therefore, the possible structure of compound Z is C6H5CH2CH2COCH3. [1]

peak at acm1 indicates that compound Z contains a C=O group. [1] Referring to the mass spectrum, the peak at m/e = 148 corresponds to the molecular ion. The peaks at m/e = 105, 91 CH3CO+ ions respectively. [3]

80 (a)



Let the mass of compound A be 100 g. Thus, the mass of carbon in the compound = 69.8 g the mass of hydrogen in the compound = 11.6 g

Mass (g) 69.8 11.6 18.6


0.16.11 16.1

0.166.18

Mole ratio 51.18.5 6

101.16.11 1

1.116.1 6


[1]

(b) lecular ion of

(c) sorption peak at around 1720 cm1. [1] It

t m/e = 71 and 43 correspond to [2]

Therefore, compound A is pentan-2-one. [1]

the empirical formula of compound A is C5H10O. [1] From the mass spectrum, the peak at m/e = 86 corresponds to the mocompound A. [1] Therefore, the molecular mass of compound A is 86. [1] From the IR spectrum, there is a strong abindicates the presence of the C=O bond. [1] From the mass spectrum, the peak at m/e = 86 corresponds to the molecular ion CH CH CH COCH [1] while the peaks a3 2 2 3+

CH3CH2CH2CO+ and CH3CO+ respectively.

81

(a)


Oxygen

Let the mass of compound X be 100 g. Thus, the mass of carbon in the compound = 48.7 g the mass of hydrogen in the compound = 8.1 g

pound = 43.2 g


1.40.127.48 1.8

0.11.8 Number of moles (mol) 7 .2

0.162.43

Mole ratio 5.17.21.4 3

7.21.8 1

7.27.2


(b) lecular ion of

(c)

the empirical formula of compound X is C3H6O2. [1] From the mass spectrum, the peak at m/e = 74 corresponds to the mocompound X. [1] Therefore, the molecular mass of compound X is 74. [1] Referring to the IR spectrum, the presence of a strong absorption peak at around 1750 cm1 indicates that compound X contains a C=O group. [1] The absence of a broad absorption peak in the region of 25003500 cm1 indicates that compound X does not contain a OH group. [1] Referring to the mass spectrum, the peak at m/e = 74 corresponds to the molecular ion. Therefore, the molecular mass of compound X is 74. The molecular formula of compound X is C3H6O2. [1] For saturated hydrocarbon with 3

carbon atoms, there will be 8 hydrogen atoms. Degree of unsaturation = 12

. This indicates that compound X should be an ester. [1] The peaks at m/e

68

= 59, 43 and 15

Therefore, the possible structure of compound X is CH3COOCH3. [1] should correspond to CH3COO+, CH3CO+ and CH3+ respectively. [3]

82 (a)



Let the mass of compound Y be 100 g. Thus, the mass of carbon in the compound = 54.5 g the mass of hydrogen in the compound = 9.1 g

Mass (g) 54.5 9.1 36.4


0.11.9 3.2

0.164.36

Mole ratio 96.3.25.4 1 96.

3.21.9 3 1

3.23.2


(b) lecular ion of und Y is 88. [1]

(c) e (C2H4O)n. 88

n (12.0 2 + 1.0 4 + 16.0) = 88 [1]

(d)

ion of 25003500 cm1 indicates that compound Y does not

the empirical formula of compound Y is C2H4O. [1] From the mass spectrum, the peak at m/e = 88 corresponds to the mocompound Y. [1] Therefore, the molecular mass of compoLet the molecular formula of compound Y bRelative molecular mass of (C2H4O)n =

n = 2 the molecular formula of compound Y is C4H8O2. [1] Referring to the IR spectrum, the presence of a strong absorption peak at around 1740 cm1 indicates that compound Y contains a C=O group. [1] The absence of a broad absorption peak in the regcontain a OH group. [1] For saturated hydrocarbon with 4 carbon atoms, there will be 10 hydrocarbon atoms.

Degree of unsaturation = 12

810 [1] There is only one unsaturated bond and two oxygen atoms in compound Y. It is likely to be an ester. [1] Referring to the mass spectrum, the peak at m/e = 88 corresponds to the molecular ion. The peaks at m/e = 73, 45 and 43 correspond to CH3COOCH2+, CH3CH2O+ and

Therefore, the possible structure of compound Y is CH3COOCH2CH3. [1] CH3CO+ respectively. [3]

83 (a) ]

+ 2H2O(l) 2HI(aq) + H2SO4(aq) [1] (b)

colourles

(c)

Step 1: SO2(aq) + 2OH(aq) SO32(aq) + H2O(l) [1Step 2: SO32(aq) + 2H+(aq) SO2(aq) + H2O(l) [1] Step 3: SO2(aq) + I2(aq)(i) Starch solution [1] (ii) The reaction mixture changes from s to dark blue. [1]

Number of moles of I2 = 0.00180 M 3dm100010.5 = 1.89 105 mol [1]

1 + 16.0 2) g mol1 = 1.21 mg tion of SO2

From the equation, mole ratio of SO2 : I2 = 1 : 1Number of moles of SO2 = 1.89 105 mol [1] Mass of SO2 = 1.89 105 mol (32. concentra in the wine

= 1.21 mg 3dm25.0

= 48.4 mg dm [1]

Since the concentration of SO

1000 3

(d) mple (48.4 mg dm3) is much lower than 450 mg dm3, it does not exceed the limit. [1]

2 in the wine sa

84

(b) ] while the sample from drug 2 gives much

he same Rf value as pure compound A. [1] ) Gas chromatography-mass spectrometry [1]

(a) Thin-layer chromatography [1] Drugs 1, 2 and 3 should be fake drugs. [1] The samples from drugs 1 and 3 give much smaller Rf value than pure compound A [1larger Rf value than pure compound A. [1] Only drug 4 is the genuine drug [1] as it has t

(c

85

(a) The Rf value of the spot from active ingredient A =

cm8.2cm2.3 0.28 [1]

The Rf value of the spot from active ingredient B = cm8.2cm3.6 0.44 [1]

The R value of the spot from active ingredient C = f cm8.2cm5.4 0.55 [1]

(c) ngredients of Renshen (A, B, D and C). [1]

) Gas chromatography-mass spectrometry [1]

(b) Product 1 contains both active ingredients A and B. [1] Product 1 is better to buy as it contains four active iE) while product 2 only contains two (B and

(d

86 hanol to ethanoic acid by atmospheric oxygen. [1]

(c) t be taken into calculations. erage volume of I2 solution used

=

(a) It is used to prevent the oxidation of et(b) It is toxic and is pungent in odour. [1]

The data from the first trial should noAv

3cm3

10.4010.3010.35 = 10.35 cm3 [1]

(d) = 0.00200 M Number of moles of I2 3dm1000 = 2.010.35 7 105 mol [1]

(aq) [1]

.1 + 16.0 2) g mol1 = 1.33 mg 2 in the w

SO2(aq) + I2(aq) + 2H2O(l) 2HI(aq) + H2SO4From the equation, mole ratio of SO2 : I2 = 1 : 1Number of moles of SO2 = 2.07 105 mol [1] Mass of SO2 = 2.07 105 mol (32 concentration of SO ine

= 1.33 mg 3dm25.0

= 53.2 mg dm [1] 1000 3

(e) mg dm3) is much lower than the limit (450 mg dm3), it does not exceed the limit. [1] Since the concentration of SO2 in the wine sample (53.2

87 lychlorinated dibenzodioxins [1]/Polychlorinated dibenzofurans [1]

(b) (a) Po

[1] or [1] They are by-pr(c) oducts of combustion of chlorine-containing material and many industrial


processes. [1] (d) They are carcinogenic. [1] (e

88 (a) (i)

(ii) sorption band is linearly proportional

act with an alkaline solution of a

arbon monoxide can be obtained from the absorbance of the ]

(b) (ii) Infrared spectroscopy [1] and colorimetry [1]

Infrared spectroscopy [1] and colorimetry [1] For infrared spectroscopy, carbon monoxide has a characteristic absorption band at around 2170 cm1. [1] The intensity of the abto the concentration of carbon monoxide. [1] For colorimetry, carbon monoxide is allowed to resilver salt, forming a silver-coloured solution. [1] The concentration of ccoloured solution. [1

(i) Wood furniture [1]

89

g material. [2] troscopy [2]

(d) he following:

on ghing

[2]

(a) CH2O [1] (b) Formaldehyde is used an adhesive or coatin(c) Colorimetry and infrared spec

Any TWO of tCarcinogenic Skin irritatiCou

90 (a) (i)

contains

(ii)

ticides. [1] Therefore, it is not organic. [1]


Thin-layer chromatography can be used to identify some artificial food dyes by checking the Rf value. [1] If the spots from the artificial food dyes have the same Rf value as those from the fruit juice, it indicates that the fruit juiceartificial food dyes. [1] Therefore, it is not 100% natural and free of additives. [1] Mass spectrometry can be used to identify pesticides. The mass spectrum and fragmentation pattern of some pesticides are known [1]. If the mass spectrum shows the peaks of the molecular ions or fragment ions of the pesticides, itindicates that the fruit juice contains pes

(b

91

(b)

e of the colour change is directly related to the alcohol content in

(c) (ii) Identifying fingerprints by iodine sublimation. [1]

(a) A breathalyser [1] In the breathalyser, a small tube is filled with orange potassium dichromate crystals. [1] If the breathed air containing alcohol vapour enters the tube, some of the crystals turn green. [1] The degrethe breathed air. [1] (i) Collecting fingerprints inside the car. [1]

92

(b) hanges from orange to green, it . [1]

(c) 3CH3CH2OH(aq) + 2Cr2O72(aq) + 16H+(aq

(d) number of Cr decreases from +6 to +3 while

(e) 3300 cm1 should be observed [1] because of the presence of OH bond in ethanol. [1]

(a) Potassium dichromate crystals [1] If the colour of the potassium dichromate crystals cindicates that the drivers breath contains ethanol

) 3CH3COOH(aq) + 4Cr3+(aq) + 11H2O(l) [1]

It is a redox reaction. [1] The oxidationethanol is oxidized to ethanoic acid. [1] A strong and broad absorption peak between 2500

93

liva [1] (b)

(ii) m them have the same Rf value, it indicates that


(a) Urine, [1] blood [1], sweat [1] and sa(i) Thin-layer chromatography [1]

Both illegal drugs and the samples obtained from the bodies develop spots on chromatogram. [1] If the spots frothe person took illegal drugs. [1]

(c

94

(a) an object, a layer bject and exist as an invisible fingerprint. [1]

(c) limation. [1] The iodine vapour then

t, making it brown. [1] ) A 1% starch solution [1]

There is oil (grease) on our fingers. [1] When we press our fingers onof oil will stick on the o

(b) Iodine sublimation [1] Placing the suspected material in a closed container with iodine crystals. [1] The crystals are then heated gently, causing them to undergo subdissolves in the fingerprin

(d

95 (a)

minary test for urine, blood or tissue samples from a person. [1]

rmine what specific proteins are present in cancer cells but not in

y abnormalities of the inside of our bodies. [1] (b)

(iii) Mass spectrometry [1]

Thin-layer chromatography [1] It is used to give a preliMass spectrometry [1] It is used to detenormal cells. [1] Magnetic resonance imaging [1] It is used to identif(i) Insulin [1] (ii) Chromatography [1]

96

] scopy [1]


(a) Colorimetry [1] (b) Gas chromatography-mass spectrometry [1(c) Colorimetry [1]/Infrared spectro(d) Thin-layer chromatography [1] (e) Gas chromatography-mass spectrometry [1] (f

Analytical Chemistry Answer

Documents

Transcript of Analytical Chemistry Answer