GCE
Chemistry B
H433/01: Fundamentals of chemistry
Advanced GCE
Mark Scheme for Autumn 2021
Oxford Cambridge and RSA Examinations
H433/01 Mark Scheme October 2021
1. Annotations
Annotation Meaning
Correct response
Incorrect response
Omission mark
Benefit of doubt given
Contradiction
Rounding error
Error in number of significant figures
Error carried forward
Level 1
Level 2
Level 3
Benefit of doubt not given
Noted but no credit given
Ignore
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H433/01 Mark Scheme October 2021
2. Abbreviations, annotations and conventions used in the detailed Mark Scheme (to include abbreviations and subject-specific
conventions).
Annotation Meaning
/ alternative and acceptable answers for the same marking point
Separates marking points
DO NOT ALLOW Answers which are not worthy of credit
IGNORE Statements which are irrelevant
ALLOW Answers that can be accepted
() Words which are not essential to gain credit
__ Underlined words must be present in answer to score a mark
ECF Error carried forward
AW Alternative wording
ORA Or reverse argument
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H433/01 Mark Scheme October 2021
Section A
Q Key Mark AO
1 D 1 1.1
2 B 1 1.2
3 A 1 1.2
4 B 1 1.2
5 A 1 1.2
6 D 1 1.2
7 A 1 1.2
8 B 1 1.1
9 A 1 2.2
10 C 1 1.1
11 B 1 1.1
12 D 1 1.1
13 C 1 1.1
14 C 1 2.8
15 D 1 2.7
16 A 1 1.2
17 A 1 2.4
18 D 1 2.5
19 D 1 2.1
20 C 1 2.8
21 A 1 2.6
22 B 1 1.2
23 C 1 2.2
24 C 1 2.3
25 C 1 2.8
26 B 1 1.2
27 A 1 1.2
28 C 1 1.1
29 B 1 1.1
30 D 1 1.1
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H433/01 Mark Scheme October 2021
Question Answer Mark AO Guidance
31 a 1 1.1 IGNORE brackets
31 b i (Warm with) sodium hydroxide AND 1 1.2
gas/ammonia/fumes turns indicator paper/red litmus paper
blue
31 b ii NH 4 + -3 2 2.6 x 2 Must have the sign
NO 3 –, +5
31 b iii NH 4 + + 2O 2 NO 3 – + H 2 O + 2H+ 1 2.6 ALLOW multiples
31 b iv (colourless gas turns to) a brown gas 1 1.2
31 c i (Ba ions are) Ba2+ and (sulfate is) SO 4 2– 1 1.2
OR both ions have same/2 charge
31 c ii CHECK ANSWER LINE 5 2.4 x 4
If answer = 62.2% award 5 marks
(Moles of Ba2+ = moles of SO 4 2- in 25 cm3= 0.0155 x 0.2) ECF throughout
= 3.1 x 10 -3
Moles of Ba2+ = moles of SO 4 2- stated or implied AND moles
of SO 4 2- in 0.250 dm3= 0.031 moles
Mass ammonium sulfate = 0.031 x 132.1 (= 4.0951)
% ammonium sulfate is 4.0951/6.58 x 100 = 62.2% (to any
sf) 3.1
62.2 % (3sf) ALLOW any calculated answer to 3 sf for last MPt.
31 c iii CHECK ANSWER LINE 3 2.2 x 3 6.02 x 1018 scores 2
If answer = 2.4(38) x 1017 award 3 marks
[SO 4 2-] =√(1x 10-10) = (1 x 10-5) in 1 dm3
No of moles of in 40.5 cm3 = (0.0405 x 1 x 10-5)
= 4.05 x 10-7
No of ions = (4.05 x 10-7 x 6.02 x 1023) = 2.4(38) x 1017 ALLOW a calculated value x 6.02 x 1023 correctly
evaluated for last point.
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H433/01 Mark Scheme October 2021
Question Answer Mark AO Guidance
32 a H H H H 2 2.1 x 2
Cl H C C H C C H + HCl
H H H H ALLOW absence of right-hand arrow to match text-
Half-headed arrows as shown book page 207.
HCl and ethyl radical product
32 b i CH 3 CH 2 Cl + NH 3 CH 3 CH 2 NH 2 + HCl 1 1.1 ALLOW any unambiguous structures or molecular
formulae.
32 b ii Nucleophilic 2 1.1 x 2
Substitution
32 c i White AND 1 1.1
(pale) yellow
32 c ii C 2 H 5 I + H 2 O C 2 H 5 OH + HI 2 2.7 IGNORE state symbols ALLOW H+ + I- for HI
Ag+ + I– AgI 1.1 ALLOW other equations forming AgI such as:
AgNO 3 + HI AgI + HNO 3
32 c iii C-Cl is more polar than C-I 3 3.1 x 3 Mark independently
Bond strength more important and C-I bond ORA throughout
is weaker than C-Cl AW
(ppt forms) faster with iodoethane
32 d i 1,1,1-trichloroethane 1 1.1 IGNORE commas and dashes
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H433/01 Mark Scheme October 2021
32 d ii Refer to marking instructions on page 5 of mark scheme for 6 1.1 x 2 Indicative scientific points include:
guidance on marking this question. 2.5 x2
3.2 x 2 Strongest type of intermolecular bonds:
Level 3 (5-6 marks) • Fats have pd-pd imbs
Correctly identifies the imbs in all three compounds with • Substance A has pd-pd forces
some explanations • Hexane has only id-id forces.
AND Explains why fat is most soluble in A in terms of (Polar and non-polar must be correctly linked
strength of imbs broken and made. to imbs)
There is a well-developed line of reasoning which is clear and How intermolecular bonds arise:
logically structured. • Pd-pd form due to differences in atoms’
electronegativity in a covalent bond
Level 2 (3-4 marks) • Cδ+=Oδ- and Cδ+-Clδ-
Makes a reasonable attempt to describe the points in all • Causing electrostatic attractions
three sections or detailed description of 2 areas • Id-id forces due to uneven distributions of
There is a line of reasoning presented with some electrons creating temporary dipoles that
structure induce dipoles in adjacent molecules OR
Mention of instantaneous dipole-induced
Level 1 (1-2 marks) dipole forces at least once in full
Makes some relevant points but may contain errors. Ability to dissolve fat molecules:
There is an attempt at a logical structure with a line of • A substance will dissolve if the strength of
reasoning. The information is in the most part relevant and intermol bonds broken < intermol bonds
correct. made OR energy required to break
compared with energy released in making.
Level 0 (0 marks) • Fat’s pd-pd bonds stronger than id-id bonds
No response or nothing worthy of credit. with hexane so less soluble.
• Fat and A form pd-pd forces of comparable
strength so soluble.
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H433/01 Mark Scheme October 2021
Question Answer Mark AO Guidance
33 a Blue/green(white flashes) 2 1.2
(Heat energy) excites e- so they emit light when they fall AW 2.5
33 b i Ammonia acts as a base 3 1.1
NH 3 + H 2 O ⇌ NH 4 + + OH- 2.3
Pale blue ppt is Cu(OH) 2 AND Cu2+ + 2OH- Cu(OH) 2 2.3
OR Cu2+(aq) + 2OH-(aq) Cu(OH) 2 (s)
33 b ii [Cu(NH 3 ) 4 (H 2 O)2+]/[NH 3 ]4[Cu(H 2 O) 6 2+] 2 2.1 ALLOW as shown without multiple square backets.
IGNORE state symbols, charges on ions must be
Units: dm12 mol-4 present.
ALLOW mol-4 dm12 ALLOW units matching
incorrect expression of K c
33 b iii CHECK ANSWER ON ANSWER LINE 3 3 x 2.8
If K c = 1(.0) x 1013 award 3 marks
Moles Cu2+ at eqm (= 0.022 - 0.02) = 0.002.
Moles of ammonia at eqm (= 0.081 – (4 x 0.02))= 0.001
K c = (0.02/(0.0014 x 0.002) = 1.0 x 1013 ALLOW ecf from first two mpts and (b)(ii).
INGORE units
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H433/01 Mark Scheme October 2021
33 b iv Refer to marking instructions on page 5 of mark scheme for 6 1.2 x4 Indicative scientific points include:
guidance on marking this question. 3.3 Setting up colorimeter:
Level 3 (5-6 marks) 3.4 • Select a filter
There is a clear outline of the experiment which includes • Orange filter/ complementary colour/ filter
detailed descriptions of most points in each section and which with greatest absorbance
would produce valid results. • Zero with a cuvette of distilled water.
There is a well-developed line of reasoning which is clear and Calibration:
logically structured. • Make up set of solutions of the complex of
known concentration. AO 3.3
Level 2 (3-4 marks) • calc of mass to produce conc ≥ 0.02M
Most of the experimental details are described and there is • Use concentrations more and less
reference to all three sections or two in detail. concentrated than the unknown (details of
There is a line of reasoning presented with some structure dilution to get the concs). AO3.4
• Measure absorbance of the solutions.
Level 1 (1-2 marks) AO1.2
Some details are described, but one of the sections may be • Plot a graph of concentration against
omitted. absorbance. AO1.2
There is an attempt at a logical structure with a line of • Draw a line of best fit.
reasoning. The information is in the most part relevant and Measuring the concentration of the unknown
correct. sample: AO1.2
• Place cuvette of sample in the machine
Level 0 (0 marks) • Record absorbance
No response or nothing worthy of credit. • Use the graph to read concentration of that
absorbance
Calibration and measuring may be from a diagram.
ALLOW % transmittance for absorbance.
33 c i Bidentate 1 1.1 ALLOW polydentate .
c ii 6 3 1.1 x 3 Mark independently
octahedral
90o
33 d Conc H 2 N CH 2 CH 2 NH 2 = 4.5/60 = 0.075 mol dm–3 4 2.8 ALLOW alternative method involving ratios for
Amt H 2 N CH 2 CH 2 NH 2 in 20 cm3 = 1.5 x 10–3 mol MP2 and 3
Amt Cu2+ = 7.5 x 10-4 mol H 2 N CH 2 CH 2 NH 2 :Cu2+ 20x 0.075 :15 x 0.05
Ratio 2:1 so formula is [Cu(H 2 N CH 2 CH 2 NH 2 ) 2 ]2+ multiples or any correct ratio
1.5:0.75
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H433/01 Mark Scheme October 2021
Question Answer Mark AO Guidance
34 a i Condensation 1 1.1 ALLOW acylation
34 a ii (Secondary) amide/ peptide 1 1.1 Primary or tertiary are CON
34 b HCl 1 1.1
34 c products are: 4 3.1 x 4 NOT names
NH 3 +CH(CH 3 )COOH , NOT -NH 3 without a charge
(one mark if NH 2 not protonated)
C 2 H 5 OH
CH 3 CH 2 COOH
34 d ANY THREE FROM 3 1.2 x 3
Select a solvent in which the product is much more soluble at high
temperature
Dissolve solid in minimum volume of hot solvent
Filter when hot to remove insoluble impurities
then cool and collect purer solid by filtration
34 e i SOCl 2 1 2.5
e ii The product molecules have functional groups on both ends so reactions 1 3.2
can continue AW
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H433/01 Mark Scheme October 2021
Question Answer Mark AO Guidance
35 a i Rate is not proportional to concentration AW 1 3.1 ALLOW numerical examples
35 a ii Scales and labels on axes, must take up half the graph paper 3 2.6 x 3
Correct calculation of squares look for values on the table for this
marking point can be subsumed by
Plotting of points on a correct numerical scale and line of best fit correct numbers on x-axis
ALLOW MP1 and 3 as ECF if conc not
squared or rate squared.
35 a iii Rate = k x [N2O]2 1 2.6
35 a iv Evidence of use of graph to find gradient (lines on graph or working 3 2.2.x 3
consistent with the graph)
An answer that rounds to 4 x 104 ALLOW 2.5 x 10-5 as ECF if axes
reversed in (a)(ii). No other ECF
Units dm3mol-1 s-1
35 b i Zero order 2 3.2 x 2
Constant gradient so rate is constant AW
35 b ii The surface of the catalyst is completely covered so adding more gas 1 3.1
doesn’t increase the rate. AW
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