Topic 16 Analytical Chemistry Part A Unit-based exercise Unit 53 Qualitative analysis — detecting the presence of inorganic chemical species Fill in the blanks Test Observation Deduction a) dilute nitric acid; aqueous solution of silver nitrate — — b) — orange; green sulphite c) i) — ii) — i) reddish brown ii) orange; green bromine; sulphur dioxide d) — white precipitate e) — — chlorine; hypochlorite f) — A white precipitate — g) — — sulphate h) — — ammonium i) — white; dissolves; colourless — j) — white — True or false 2 F 3 F 4 T 5 F 6 T 7 T 8 F 9 T 0 T F
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�
Topic 16 Analytical Chemistry
Part A Unit-basedexercise
Unit 53 Qualitative analysis — detecting the presence of inorganic chemical species
Fillintheblanks
�Test Observation Deduction
a)dilute nitric acid; aqueous solution of silver nitrate
— —
b) — orange; green sulphite
c) i) — ii) —
i) reddish brown ii) orange; green
bromine; sulphur dioxide
d) — white precipitate
e) — — chlorine; hypochlorite
f) — A white precipitate —
g) — — sulphate
h) — — ammonium
i) — white; dissolves; colourless —
j) — white —
Trueorfalse
2 F
3 F
4 T
5 F
6 T
7 T
8 F
9 T
�0 T
�� F
2
Multiplechoicequestions
�2 A
�3 C
�4 A
�5 C
�6 A
�7 D
�8 A
�9 A
20 B
2� C
22 D
23 C
24 C
25 B
26 D
27 C
28 D
29 B
30 D
3� B
32 C
33 C
34 B
3
35 A
36 B
37 C
38 D
39 C
40 D
4� B
42 A
43 C
44 D
45 B
46 C
47 A
48 D
49 D
50 A
5� D
52 A
53 A
54 D
55 D
56 B
4
Unit 54 Tests for functional groups, separation and purification of compounds
Fillintheblanks
�Test Observation Deduction
a) — — C=C; C≡C
b) — orange; green aldehyde
c) 2,4-dinitrophenylhydrazine — ketone
d) Tollens’ reagent — aldehyde
e) — — tertiary
f) — —
CHI3;
O
C CH3 ;
OH
C
H
CH3
g) silver nitrate — silver bromide; bromine
h) — Effervescence —
i) — — ester; alcohol
Trueorfalse
2 F
3 T
4 F
5 T
6 T
7 T
8 F
9 T
5
Multiplechoicequestions
�0 B
�� C
�2 B
�3 D
�4 D
�5 B
�6 A
�7 D
�8 A
�9 C
20 B
2� A
22 B
23 D
24 C
25 D
26 B
27 D
28 C
29 C
30 A
3� C
32 C
6
33 C
34 D
35 A
36 B
37 B
38 A
39 C
40 A
4� C
42 A
43 B
44 D
45 D
Unit 55 Quantitative methods of analysis
Fillintheblanks
� a) instrumental error
b) operative error
c) error of method
2 a) silver nitrate; chromate
b) silver chromate
3 a) starch solution
b) colourless; dark blue
4 a) iodine; iodine; sodium thiosulphate
b) starch solution; dark blue; colourless
7
Trueorfalse
5 T
6 T
7 T
8 F
9 F
�0 F
Multiplechoicequestions
�� D
�2 B
�3 B
�4 C
�5 C
�6 C
�7 D
�8 B
�9 C
20 A
2� C
22 A
23 A
24 B
25 A
26 B
27 D
8
28 C
29 B
30 A
3� A
32 B
33 D
34 C
35 D
36 A
37 C
38 D
39 C
40 D
4� D
42 C
43 C
Unit 56 Instrumental analytical methods
Fillintheblanks
� a) Tollens’ reagent / acidified aqueous solution of potassium dichromate
b) iodine; sodium hydroxide
c) carbonyl / C=O
d) i) CH3CH2+
ii) CH3CO+
2 a) mixing with aqueous solution of sodium hydrogencarbonate / warming with an alcohol in the presence
of concentrated sulphuric acid
9
b) i) O–H
ii) (�) C–O
(2) C=O
c) i) COOH+
ii) CH3CO+
Trueorfalse
3 F
4 F
5 T
6 T
7 T
8 F
9 T
�0 T
�� F
�2 F
Multiplechoicequestions
�3 C
�4 A
�5 D
�6 C
�7 D
�8 C
�9 D
20 C
2� C
22 B
�0
23 A
24 D
25 B
26 D
27 C
28 A
29 B
30 B
3� A
32 B
33 A
34 C
35 D
36 B
37 B
38 D
39 D
40 A
4� A
42 C
Unit 57 Contribution of analytical chemistry to our society
Fillintheblanks
�Phase
Paper chromatography
Thin layer chromatography
Gas-liquid chromatography
Mobile — — carrier gas
Stationary water in the paper fibres fine layer of alumina or silica gel —
��
Trueorfalse
2 T
3 F
4 F
5 T
6 T
Multiplechoicequestions
7 D
8 D
9 B
�0 A
�� B
�2 D
�3 C
�4 C
�5 D
�6 A
�7 A
�8 A
�9 B
20 C
2� D
�2
Part B Topic-based exercise
Multiplechoicequestions
� A
2 C
3 B
4 B
5 C
6 C
7 C
8 A
9 C
�0 A
�� D
�2 C
�3 A
�4 B
�5 D
�6 D
�7 C
�8 C
�9 B
20 D
2� C
�3
22 A
23 C
24 C
25 D
26 B
27 B
28 A
29 D
30 D
3� B
32 A
33 D
34 C
35 B
36 B
37 A
38 D
39 D
40 B
4� D
42 B
43 B
44 C
45 A
�4
46 C
47 A
48 B
49 B
50 D
5� C
Shortquestions
52 a) i) Dip a clean nichrome wire into concentrated hydrochloric acid. (�)
Dip the nichrome wire into the solid. (�)
Heat the wire in a Bunsen flame. (�)
ii) Green (�)
b) Potassium flame colour would be masked by the strong sodium flame colour. (�)
53 a) White precipitate (�)
insoluble in dilute hydrochloric acid (�)
b) The precipitate would dissolve in dilute hydrochloric acid. (�)
54 a)Silver chloride Silver bromide Silver iodide
Dilute aqueous ammonia
soluble (0.5) insoluble (0.5) insoluble (0.5)
Concentrated aqueous ammonia
soluble (0.5) soluble (0.5) insoluble (0.5)
b) i) White fumes (�)
ii) NH3(g) + HCl(g) NH4Cl(s) (�)
55 a) Any two of the following:
• Ethoxyethane is a good solvent for most carbon compounds. (�)
• Ethoxyethane is immiscible with water. (�)
• Ethoxyethane is volatile. It can easily be removed by distillation. (�)
• Ethoxyethane is chemically unreactive. (�)
b) Avoid naked flame / carry out the evaporation in a fume cupboard. (�)
�5
56Tests Observation
a) Shake a few drops of compound X with aqueous bromine.
The yellow-brown aqueous bromine becomes colourless quickly. (�)
b) Add anhydrous zinc chloride in concentrated hydrochloric acid to compound X and warm.
Cloudiness appears. (�)
c) Add iodine in an aqueous solution of sodium hydroxide to compound X and warm the mixture.
A bright yellow precipitate forms. (�)
57Tests Observation with X Observation with Y
Aqueous solution of sodiumcarbonate
effervescence occurs (0.5) no observable change (0.5)
2,4-dinitrophenylhydrazine no observable change (0.5) yellow to red precipitate forms (0.5)
Aqueous solution of potassiumdichromate + dilute sulphuricacid
no observable change (0.5)the dichromate solution changes from orange to green (0.5)
58 a)
(� mark for correct set-up; � mark for correct labels; award 0 mark if the set-up is not workable) (2)
b)
(� mark for correct set-up; � mark for correct labels; award 0 mark if the set-up is not workable) (2)
�6
59 a) When the aqueous solution of silver nitrate is added to a solution of chloride ions, silver chloride precipitate forms. (�)
When all the chloride ions are precipitated, the first excess silver nitrate solution gives a reddish brown silver chromate precipitate with the chromate indicator. This signals the end point of the titration. (�)
2Ag+(aq) + CrO42–(aq) Ag2CrO4(s) (�)
reddish brown
b) If the solution is too acidic, then part of the indicator will be present as hydrogen chromate ions and more silver ions will be required to form the silver chromate precipitate. (�)
If the solution is too basic, silver oxide may be precipitated. (�)
Structuredquestions
60 a)Test Procedure Observation Deduction
� — — sodium (�)
2 — — oxygen (�)
3 — — chloride (�)
b) Sodium chlorate / NaClO3 (Sodium hypochlorite / NaOCl may be accepted.) (�)
6� a)Test Procedure Observations Deduction
� — — transition metal (0.5)
2 — — ammonia; (0.5) ammonium ions (0.5)
3
— — —
a) — — copper(II) ions (0.5)
b) — — barium sulphate; (0.5) sulphate ions (0.5)
b) Cations – ammonium ion and copper(II) ion
Anion – sulphate ion
62 We need to look at the properties of the cations:
• Ag+(aq) ions form an insoluble chloride. (�)
• PbCl2(s) is insoluble in cold water, but soluble in hot water. (�)
• Al3+(aq) and Zn2+(aq) ions form insoluble hydroxides with dilute aqueous solution of sodium hydroxide and the hydroxides are soluble in excess alkali. (�)
(�)
�7
• Fe2+(aq) ions form an insoluble hydroxide with dilute aqueous solution of sodium hydroxide. (�)
• Al3+(aq) and Fe2+(aq) ions form insoluble hydroxides with dilute aqueous ammonia. (�)
• Zn2+(aq) ions form an insoluble hydroxide with dilute aqueous ammonia and the hydroxide is soluble in excess alkali. (�)
precipitate of
Al(OH)3(s)solution containing
[Zn(NH3)4]2+(aq)
Step 3 — add excess NaOH(aq)
Step 4 — add HCl(aq)
followed by
excess NH3(aq)
Step 1 — add HCl(aq)
precipitate of
AgCl(s)
precipitate of
Fe(OH)2(s)PbCl2(aq)
Step 2 — add hot water
solution containing Ag+(aq), Al3+(aq), Fe2+(aq),
Pb2+(aq), Zn2+(aq)
precipitate of
AgCl(s), PbCl2(s)
solution containing
Al3+(aq), Fe2+(aq), Zn2+(aq)
solution containing
[Al(OH4)]–(aq), [Zn(OH4)]
2–(aq)
(� mark for each step with correct resulting species stated; appropriate steps in a different order is also acceptable) (4)
63 A and E reacted to give a white precipitate (Zn(OH)2(s)) that was soluble in excess A. Thus, A was NH3(aq) and E was ZnCl2(aq). (�)
The white precipitate dissolved due to the formation of complex ions.
A and C reacted to give a white precipitate (Ba(OH)2(s)). Thus, C was BaCl2(aq). (�)
Ba2+(aq) + 2OH–(aq) Ba(OH)2(s) (�)
B gave a white precipitate that was soluble in NH3(aq) with C (BaCl2(aq)). The precipitate was AgCl(s). Thus, B was AgNO3(aq). (�)
Ag+(aq) + Cl–(aq) AgCl(s) (�)
�8
B and F reacted to give a yellow precipitate (AgI(s)). Thus, F was KI(aq). (�)
Ag+(aq) + I–(aq) AgI(s) (�)
C and D reacted to give a white precipitate (BaSO4(s)). Thus, D was Na2SO4(aq). (�)
Ba2+(aq) + SO42–(aq) BaSO4(s) (�)
64 a) Any one of the following:
• Treat with 2 ,4-dinitrophenylhydrazine. (�)
Only compound P gives a yellow to red precipitate. (�)
• Warm with the Tollens’ reagent. (�)
Only compound P gives a silver deposit on the wall of the reaction vessel. (�)
• Warm with acidified aqueous solution of potassium dichromate. (�)
Only compound P turns the dichromate solution from orange to green. (�)
• Add aqueous solution of sodium hydrogencarbonate. (�)
Effervescence occurs for compound Q only. (�)
b) Any one of the following:
• Warm with acidified aqueous solution of potassium dichromate. (�)
Only compound R turns the dichromate solution from orange to green. (�)
• Mix with phosphorus pentachloride. (�)
Only compound R gives steamy fumes of hydrogen chloride. (�)
• Treat with anhydrous zinc chloride in concentrated hydrochloric acid. (�)
Cloudiness appears in 5 minutes for compound R only. (�)
Any other appropriate reaction is also acceptable.
c) Any one of the following:
• Warm with acidified aqueous solution of potassium dichromate. (�)
Only compound T turns the dichromate solution from orange to green. (�)
• Treat with anhydrous zinc chloride in concentrated hydrochloric acid. (�)
Cloudiness appears in � minute for compound U only. (�)
d) Warm with dilute aqueous solution of silver nitrate in ethanol. (�)
Compound V gives a white precipitate. (0.5)
Compound W gives a yellow precipitate. (0.5)
�9
65 a) Mobile phase – the developing solvent (�)
Stationary phase – water in the paper fibres (�)
b)
(� mark for correct set-up; � mark for correct labels; award 0 mark if the set-up is not workable) (2)
c) Each amino acid distributes itself between the mobile phase and the stationary phase. (�)
Amino acids that are more soluble in the mobile phase (or developing solvent) travel up more quickly. Thus, the amino acids separate. (�)
d) Rf = 3.� cm5.4 cm
= 0.57
e)
66 a) i) Any one of the following:
• The distance a component travels relative to the solvent. (�)
• Rf = distance travelled by the componentdistance travelled by the solvent
(�)
ii) Retention time is the time between injection and detection of a component / the time a component held in a column. (�)
b) Mobile phase – carrier gas (�)
Stationary phase – liquid coated onto the inside of the column (�)
20
c) i) X (�)
It is the first to emerge. / It has the shortest retention time. (�)
ii) Total area of the peaks = �2
x �0 x 4 + �2
x 30 x 4 + �2
x 20 x 8
= �60 (�)
Percentage abundance of component Y = 60�60
x �00%
= 37.5% (�)
67 a) To enable reactants to be heated for a long time. (�)
To prevent any loss of the reaction mixture. (�)
b)
(� mark for correct drawing of separating funnel with tap; � mark for showing the � -bromobutane layer below water; award 0 mark if the funnel is not workable) (2)
c) To dry the � -bromobutane. (�)
d) i)
(� mark for correct set-up; � mark for correct labels; � mark for correct direction of water flow in condenser; award 0 mark if the set-up is not workable) (3)
2�
ii) �00 °C – �04 °C (�)
e) i) AgBr (�)
ii) C4H9Br + H2O C4H9OH + H+ + Br– (�)
iii) As a common solvent for �-bromobutane and silver ions (�)
iv) Any one of the following:
• The reaction is slow at room temperature. (�)
• Heat speeds up the reaction. (�)
• The reactant(s) are flammable. (�)
f) CH3CH2CH2CH2+ / CH2Br+ / CH2CH2CH2Br+ (�)
68 a)
(�)
b) Wear protective gloves. / Carry out in a fume cupboard. (�)
c) The reaction is exothermic / vigorous. (�)
d)
(� mark for Buchner funnel and Buchner flask; � mark for suction applied; � mark for correct labels; award 0 mark if the set-up is not workable) (3)
e) Wash the solid with water. (�)
Add an aqueous solution of silver nitrate to the water obtained. (�)
No creamy precipitate forms. (�)
f) Dissolve the crude sample in minimum amount of hot methanol. (�)
Filter the mixture while hot. (�)
Allow the filtrate to cool and collect the crystals by filtration. (�)
Wash the crystals and dry. (�)
22
69 a) Bromine (�)
b) Aqueous solution of sodium hydroxide (�)
c) –OH group / hydroxyl group (�)
d) Heat with concentrated sulphuric acid / aluminium oxide. (�)
e) i)
C Br or
H
C
H
HH
C
H
H
H C H
H
C
H
HBr
C
H
H
H
(�)
ii)
C OH or
H
C
H
HH
C
H
H
H C H
H
C
H
HOH
C
H
H
H
(�)
iii)
C H
H
C
H
H
C
H
H
(�)
70 a) i) CHI3 (�)
ii) Compound A contains a CH3C
O
group. (�)
iii)
(CH3)2CHC CH3
O
(�)
methylbutanone (�)
b) i) A yellow / red precipitate forms. (�)
ii)N
H
N
NO2
NO2
(CH3)2CH
CH3
C
(�)
c) i)
C4H9 — C — H + 2[Ag(NH3)2]+ + 3OH–
O
C4H9 — C — O– + 2Ag + 4NH3 + 2H2O
O
(�)
ii) Compound A (ketone) cannot be oxidized. (�)
23
7� a) Test Proceduce Observation Deduction
�Ignite �0 drops of X in a crucible.
X burns with a sooty flame.
X has a high carbon content. (�)
2To 2 cm3 of X, add 8 drops of aqueous bromine.
T h e y e l l o w - b r o w n aqueous bromine becomes colourless quickly.
X is unsaturated. (�)
3
To � cm3 of X, add 8 drops of acidified aqueous solution of potassium dichromate and warm.
The dichromate solution changes from orange to green.
X may be an alcohol or aldehyde. (�)
4To 4 d r o p s o f X , add 2 cm 3 o f 2 ,4 -dinitrophenylhydrazine.
An orange precipitate forms.
X may be an aldehyde or a ketone. (�)
5
Mix 4 drops of X with 2 cm3 of Tollens’ reagent in a test tube. Warm in a water bath at 60 °C.
A silver mirror forms on the wall of the test tube.
X is an aldehyde. (�)
b) i) Compound X contains identical groups / atoms on the same side of a double bond. (�)
ii)
H
CHO
HC C
(�)
72 a) i) Suppose we have �00 g of compound X, so there are 58.8 g of carbon, 9.80 g of hydrogen and 3�.4 g of oxygen.
Carbon Hydrogen Oxygen
Mass of element in the compound
58.8 g 9.80 g 3�.4 g
Number of moles of atoms that combine
58.8 g�2.0 g mol–� = 4.90 mol
9.80 g�.0 g mol–� = 9.80 mol
3�.4 g�6.0 g mol–� = �.96 mol (�)
Simplest ratio of atoms
4.90 mol�.96 mol
= 2.509.80 mol�.96 mol
= 5.00�.96 mol�.96 mol
= �.00
Simplest whole number ratio of atoms
2.50 x 2 = 5.00 5.00 x 2 = �0.0 �.00 x 2 = 2.00 (�)
∴ the empirical formula of X is C5H�0O2.
24
ii) Let (C5H�0O2)n be the molecular formula of X.
Relative molecular mass of X = n(5 x �2.0 + �0 x �.0 + 2 x �6.0) = �02n
i.e. �02n = �02.0 n = � (�)
∴ the molecular formula of X is C5H�0O2.
b) Y is propan-2-ol. (�)
Y must contain a CH3
OH
C
H
group (�)
as it gives triiodomethane when warmed with iodine in an aqueous solution of sodium hydroxide.
Z is sodium ethanoate. (�)
Z must contain two (i.e. 5 – 3) carbon atoms. (�)
X is �-methylethyl ethanoate / CH3COOCH(CH3)2 (�)
c) i) Iodine (�)
ii) Hydrolyze with aqueous solution of sodium hydroxide. (0.5)
Acidify with nitric acid. (0.5)
Add aqueous solution of silver nitrate. (0.5)
A yellow precipitate forms. (0.5)
73 a) As compound X has a linear structure, compound Z should also have a linear structure.
As compound Z gives cloudiness with the Lucas reagent in 5 minutes, it is probably a secondary alcohol. (�)
As compound Z has no chiral carbon, the –OH group should be attached to the central carbon atom. (�)
C
H
C
H
HH
C
H
H
H C H
H
C
H
HH
C
H
H
OH
H
C
(�)
25
b) Compound Y undergoes hydrogenation to give compound Z which has two more hydrogen atoms, thus compound Y must contain a C=C bond. (�)
Compound X undergoes reduction to give compound Y which is a secondary alcohol, thus compound X is probably a ketone with (�)
the C=O group attached to the central carbon atom. (�)
As compound X exists as a mixture of geometrical isomers, the C=C bond should not be at the end of the carbon chain. (�)
Structure of X:
C
H
C
H
C
H
H
H C H or
H
C
H
HH
C
H
H
O
C C C
H
H
C
H
H
H C H
H
C
H
HH
C
H
H
O
C
(�)
Structure of Y:
C
H
C
H
C
H
H
H C H
H
C
H
HH
C
H
H
OH
H
C C C
H
H
C
H
H
H C H
H
C
H
HH
C
H
H
OH
H
Cor
chiral carbon chiral carbon (�)
74 a) i) (NH4)2C2O4(aq) + 2H+(aq) 2NH4+(aq) + H2C2O4(aq) (�)
ii) Urea decomposes to form hydroxide ions. (�)
The hydroxide ions neutralize the oxalic acid to liberate oxalate ions. (�)
b)
(� mark for Buchner flask and sintered glass filtering crucible; � mark for suction applied; � mark for correct labels; award 0 mark if the set-up is not workable) (3)
26
c) Number of moles of CaC2O4•H2O = 0.462 g�46.� g mol–�
= 3.�6 x � 0–3 mol (�) = number of moles of Ca in mineral
Mass of Ca in mineral = 3.�6 x �0–3 mol x 40.� g mol–�
= 0.�27 g
Percentage by mass of Ca in mineral = 0.�27 g0.599 g
x �00%
= 2�.2% (�)
∴ the percentage by mass of calcium in the mineral is 2�.2%.
d) Any two of the following:
• As no ionic substance is completely insoluble in water, a little calcium oxalate will remain dissolved in the solution. (�)
• A little of the precipitate will be lost as the precipitate is washed. (�)
• Little splashes, inefficient rinsing out of the beaker, and inefficient filtering can cause some loss of the precipitate. (�)
Any one of the following:• Mix with aqueous solution of sodium hydrogencarbonate. (�)• Warm with ethanol in the presence of concentrated sulphuric acid. (�)
ii) The CH3CH2CH2CO+ ion is responsible for the peak at m/e = 7�. (�)
The CH3CO+ ion is responsible for the peak at m/e = 43. (�)
36
88 a)
C CH3
CH3
OH
H3C
(�)
methylpropan-2-ol (�)
b)
C
H
C
CH3
HH
C
H
H
H OH
(�)
methylpropan-�-ol (�)
c) Any one of the following:
• Iodine in an aqueous solution of sodium hydroxide (i.e. iodoform test) (�)
Only butan-2-ol gives a yellow precipitate when warmed with the reagent. (�)
• A solution of anhydrous zinc chloride in concentrated hydrochloric acid (i.e. Lucas reagent) (�)
Cloudiness appears in 5 minutes when butan-2-ol is treated with the Lucas reagent. (0.5)
Cloudiness appears in hours when butan-�-ol is treated with the Lucas reagent. (0.5)
d) i) The absorption at 3 230 – 3 670 cm–� (�)
ii) Refer to the fingerprint region. (�)
Match with spectra of known compounds in database. (�)
89 a) i) Any one of the following:
• Warm each chemical with acidified aqueous solution of potassium dichromate. (�)
Only CH3CH2CHO turns the dichromate solution from orange to green. (�)
• Warm each chemical with acidified aqueous solution of potassium permanganate. (�)
Only CH3CH2CHO turns the permanganate solution from purple to colourless. (�)
• Warm each chemical with the Tollens’ reagent. (�)
Only CH3CH2CHO gives a silver mirror on the wall of the reaction vessel. (�)
• Treat each chemical with iodine in an aqueous solution of sodium hydroxide. (�)
Only CH3COCH3 gives a yellow precipitate of iodoform. (�)
ii) Mix each chemical with aqueous solution of sodium hydrogencarbonate. (�)
Effervescence occurs for CH3CH2COOH only. (�)
iii) Treat each chemical with iodine in an aqueous solution of sodium hydroxide. (�)
Only CH3CH2CH2COCH3 gives a yellow precipitate of iodoform. (�)
37
b) i) Only D has a broad absorption at 2 500 – 3 300 cm–�. (�)
ii) Confirm the identity of a compound by comparing the fingerprint region with those of known compounds. (�)
Look for an exact match. (�)
c) i) Any one of the following:
• Major peak due to CH3CH2CO+ (�)
m/e = 57 (�)
• Major peak due to CH3CH2+ (�)
m/e = 29 (�)
ii) • [CH3CH2CO CH2CH3]+· CH3CH2CO+ + ·CH2CH3 (�)
• [CH3CH2CO CH2CH3]+· CH3CH2
+ + ·OCCH2CH3 (�)
90 a)
C
H
C
H
HH
C
H
H
H OH propan-1-ol/ (�)
C
H
C
H
HOH
C
H
H
H OH propan-2-ol/ (�)
b) The dichromate solution changed from orange to green. (�)
c) Theoretical yield of C3H6O = 0.300 x 58.0 g mol–�
= �7.4 g (�)
Precentage yield of C3H6O = 6.09 g�7.4 g
x �00%
= 35.0% (�)
d) i) The presence of C=O bond (�)
ii) The presence of O–H bond in an acid (�)
ii) Propan-�-ol (�)
It was oxidized to propanoic acid besides the carbonyl compound C3H6O. (�)
9� a)
H
OH
H
CC C
HOO
OHHOC + CO2
malic acid
* (1)
H
OH
H
CH C
HO
OHC
lactic acid
38
b) i) The peak at m/e = �34. (�)
ii) The m/e value of the peak marked X is 89. The difference between � 34 and 89 is 45, (�)
so it is probable that a COOH radical is lost from the molecular ion, producing the peak at m/e = 89.
Thus,
H
OH
CC C
H
H +
O
HO is responsible for the peak at m/e = 89. (�)
c) i)
C
H
C
H
O
OH
H
C C
H
C H
H
H
H
H
O
(�)
ii) The wine is more fruity due to the esters. / The wine is less sour as acids are used up. (�)
92 a) i) Place � cm3 of aqueous solution of silver nitrate in a very clean test tube. Add a little dilute aqueous solution of sodium hydroxide. (�)
Add dilute aqueous ammonia dropwise until the precipitate of silver oxide nearly dissolves. (�)
Add � – 2 drops of compound X. Place in a beaker of hot water. (�)
Silver metal deposits on the wall of the test tube. (�)
ii) Silver ions are reduced to metal silver. (�)
X is oxidized to CH3CH2CH2CH2COO–, a carboxylate ion. (�)
b) i) 2,4-dinitrophenylhydrazine (�)
A yellow to red precipitate forms. (�)
ii)N
H
N
NO2
NO2
C4H9
HC
(�)
c) Any one of the following:
• Mix with phosphorus pentachloride. (�)
Compound Y gives steamy fumes of hydrogen chloride. (�)
• Warm with ethanoic acid in the presence of concentrated sulphuric acid. (�)
Compound Y reacts to give a product with a pleasant smell. (�
d) i) The molecular ion peak is at m/e = 86. The difference between 86 and 57 is 2 9. (�)
It is probable that a CHO radical is lost from the molecular ion. (�)
39
ii) The base peak is at m/e = 57.
Losing a CHO radical from the molecular ion probably produces a very stable tertiary carbocation. (�)
Thus, this is probably the mass spectrum of compound Z. (�)
93 a) i) Suppose we have � 00 g of compound X, so there are 54.5 g of carbon, 9.�0 g of hydrogen and 36.4 g of oxygen.
Carbon Hydrogen Oxygen
Mass of element in the compound
54.5 g 9.�0 g 36.4 g
Number of moles of atoms that combine
54.5 g�2.0 g mol–� = 4.54 mol
9.�0 g�.0 g mol–� = 9.�0 mol
36.4 g�6.0 g mol–� = 2.28 mol (�)
Simplest ratio of atoms
4.54 mol2.28 mol
= �.999.�0 mol2.28 mol
= 3.992.28 mol2.28 mol
= �.00 (�)
∴ the empirical formula of X is C2H4O.
ii) From the mass spectrum, the molecular ion peak is at m/e = 88. Thus, the relative molecular mass of X is 88.0.
Let (C2H4O)n be the molecular formula of X.
Relative molecular mass of X = n(2 x �2.0 + 4 x �.0 + �6.0) = 44n
i.e. 88 = 44n n = 2 (�)
∴ the molecular formula of X is C4H8O2.
b) O–H bond (in an alcohol) is responsible for the absorption at 3 500 cm–�. (�)
C=O bond is responsible for the absorption at � 700 cm–�. (�)
c) X contains a ketone functional group (�)
but not an aldehyde functional group. (�)
d) CH3CO+ (�)
C3H7+ ions, which also give a peak at m/e = 43, are unlikely to be formed from a ketone with four carbon
atoms. (�)
40
e)
C CH3
H
H
C C
H
H
HO
O
4-hydroxybutanone
(�) (�)
or
C CH3
OH
H
C C
H
H
H
O
3-hydroxybutanone
(�) (�)
94 a) The base peak is the peak with the greatest relative intensity. / The base peak corresponds to the most stable species. (�)
b) The molecular ion peak is at m/e = �36. (0.5)
As there are two oxygen atoms in compound X, the carbon and hydrogen atoms must add up to �04 mass units. (0.5)
As compound X is a mono-substituted aromatic compound, it must contain a C6H5 group, adding up to 77 mass units. (0.5)
(�04 – 77) mass units = 27 mass units. Thus, compound X should also contain two carbon and three hydrogen atoms. (0.5)
The molecular formula of compound X is C8H8O2. (�)
c) CH2+
(�)
d) � 680 – � 750 cm–� due to C=O bond (�)
2 500 – 3 300 cm–� due to O–H bond (�)
e) CH2COOH
(�)
4�
95 The infrared absorption at about � 250 cm–� indicates the presence of a C–O group. (�)
The infrared absorption at about � 750 cm–� indicates the presence of a C=O group. (�)
The absence of a broad absorption band at 2 500 cm–� – 3 300 cm–� indicates that X is not an alcohol or a carboxylic acid. (�)
X is probably an ester. (�)
In the mass spectrum, the molecular ion peak is at m/e = 74.
The peak at m/e = 43 is probably due to the CH3CO+ ion. (�)
The difference between 74 and 43 is 3�. So, it is probable that a OCH3 fragment has been lost from the molecular ion, giving rise to an ion responsible for the peak at m/e = 43. (�)
Thus, compound X should have a –OCH3 group. (�)
A possible structure of compound X is CH3COOCH3. (�)
96 a) SO2(g) + H2O2(aq) SO42–(aq) + 2H+(aq) (�)
Ba2+(aq) + SO42–(aq) BaSO4(s) (�)
b) i) Number of moles of BaSO4(s) formed = 0.0607 g233.4 g mol–�
= 2.60 x �0–4 mol (�) = number of moles of SO4
2–(aq) ions in resulting solution
As � mole of SO2(g) reacts with H2O2(aq) to give � mole of SO42–(aq) ions,
i.e. number of moles of SO2(g) in air sample = 2.60 x �0–4 mol
Mass of SO2(g) in air sample = 2.60 x �0–4 mol x 64.� g mol–�
= �.67 x �0–2 g (�)
Concentration in μg m–3 of SO2(g) in air sample = �.67 x �04 μg200.0� 000
m3
= 83 500 μg m–3 (�)
ii) Volume of SO2(g) in air sample = 2.60 x �0–4 mol x 24.0 dm3 mol–�
= 6.24 x �0–3 dm3 (�)
Concentration in ppm of SO2(g) in air sample = 6.24 x �0–3 dm3
200.0 dm3 �06 ppm
= 3�.2 ppm (�)
97 a) Butane-�,4-dioic acid (�)
b) HOCH2CH2CH2CH2OH (�)
42
c) (CH2)4O CO (CH2)2
O
C
O (�)
d) Less waste / less landfill / easier to dispose of / less pollution from burning (�)
e) Any two of the following:
• These plastics are much more expensive. (�)
• These plastics may begin to degrade before the right time, causing inconvenience or even safety concerns. (�)
• These plastic articles look like other plastic articles. If the two types of articles are mixed up in the recycling process, articles made from the recycled plastic may become weaker. (�)
f) i) Base peak (�)
ii) C4H4O3+ (�)
iii) A COOH radical (�)
98 a) Mobile phase – the developing solvent (�)
Stationary phase – fine layer of alumina / silica gel coated onto a glass plate (�)
b) Separation depends on the adsorption of the components onto the stationary phase and the relative solubility of the components in the developing solvent. (�)
Components that adsorb more strongly onto the stationary phase travel up less quickly. Thus, the components separate. (�)
c) i) C4H9+ or CCOOH+ (�)
ii) Any one of the following:
Peak at m/e value of Ion producing it
59 (�) OCOCH3+ (�)
�2� (�) C6H4COOH+ (�)
�37 (�) C6H4(COOH)O+ (�)
�80 (�) C9H8O4+ (molecular ion) (�)
d) OH
NHCOCH3
Y
+
OH
NaOH(aq)
X
NH2
(1)
CH3COONa(1)
43
e) O–H bond (�) 3 230 – 3 670 cm–� (�)
99 CH3CH2CH2CHO
Warm with the Tollens’ reagent. (0.5)
A silver mirror forms on the wall of the reaction vessel. (0.5)
CH3CH2COCH3
Warm with iodine in an aqueous solution of sodium hydroxide. (0.5)
A bright yellow precipitate forms. (0.5)
CH3CH2CH2COOH
Any one of the following:
• Mix with an aqueous solution of sodium hydrogencarbonate. (0.5)
Effervescence occurs. (0.5)
• Heat with ethanol in the presence of concentrated sulphuric acid. (0.5)
A pleasant / fruity smell can be detected. (0.5)
CH3CH2CH2CH2Br
Warm with dilute aqueous solution of silver nitrate in ethanol. (0.5)
A creamy precipitate forms. (0.5)
(CH3)3COH
Treat with anhydrous zinc chloride in concentrated hydrochloric acid. (0.5)
Cloudiness appears quickly. (0.5)
CH3CH2CH2CH2OH
Heat with acidified aqueous solution of potassium dichromate. (0.5)
The dichromate solution changes from orange to green. (0.5)
(3 marks for organization and presentation)
44
�00 Iron(III) ions react with thiocyanate ions to give deep red [Fe(SCN)]2+(aq) ions. (�)
This species absorbs most strongly in the green region of the visible spectrum. Thus, choose a green filter in the colorimeter. (�)
Next prepare a series of standard solutions of [Fe(SCN)]2+(aq) ions. Record the absorbance of each solution. Prepare a calibration graph by plotting absorbance against concentration. (�)
Dissolve a known mass of the foil in concentrated hydrochloric acid to convert the iron to iron(II) ions. (�)
Add aqueous hydrogen peroxide solution / concentrated nitric acid to oxidize the iron(II) ions to iron(III) ions. (�)
Add excess aqueous solution of potassium thiocyanate to the iron(III) ions to give deep red [Fe(SCN)]2+(aq) ions. Measure the absorbance of the solution. Read off the concentration of [Fe(SCN)]2+(aq) ions in the solution from the calibration curve. (�)
Calculate the mass of iron in the aluminium foil and thus the percentage of iron in the foil.
(3 marks for organization and presentation)
�0� Prepare standard aqueous solution of potassium iodate by dissolving a known mass of KIO3(s) in distilled water and dilute the solution to a known volume. (�)
Add excess aqueous solution of potassium iodide and dilute sulphuric acid to a known volume of the standard IO3
–(aq) to obtain standard iodine solution. (�)
Titrate the standard iodine solution against the aqueous solution of sodium thiosulphate. (�)
When the solution becomes pale yellow, (�)
add starch solution and continue the titration. (�)
The end point is marked by the disappearance of the blue colour. (�)