Mark schemes
Q1.
(a)     Mg    S    O4
24 + 32 + 16 (×4) or 64 / evidence of all Ar’s
gains 1 mark
          but (Mr) = 120
gains 2 marks
2
(b)     evidence that 24(g) magnesium would produce 120(g) mapesiurn sulphate
gains 1 mark
          or correct scaling by 1/6
          but 20(g) magnesium sulphate
gains 2 marks
[credit error carried forward from (a) with full marks in (b)]
2
[4]
Q2.
(a)     add excess copper carbonate (to dilute hydrochloric acid)
accept alternatives to excess, such as ‘until no more reacts’
1
filter (to remove excess copper carbonate)
reject heat until dry
1
heat filtrate to evaporate some water or heat to point of crystallisation
accept leave to evaporate or leave in evaporating basin
1
leave to cool (so crystals form)
until crystals form
1
must be in correct order to gain 4 marks
(b)     Mr CuCl2 = 134.5
correct answer scores 4 marks
1
moles copper chloride = (mass / Mr = 11 / 134.5) = 0.0817843866
1
Mr CuCO3= 123.5
1
Mass CuCO3 (=moles × M2= 0.08178 × 123.5) = 10.1(00)
1
accept 10.1 with no working shown for 4 marks
(c)     
or
11.0 × 0.791
1
8.70 (g)
1
accept 8.70(g) with no working shown for 2 marks
(d)     Total mass of reactants = 152.5
1
134.5
152.5
allow ecf from step 1
1
88.20 (%)
1
allow 88.20 with no working shown for 3 marks
(e)     atom economy using carbonate lower because an additional product is made or carbon dioxide is made as well
allow ecf
1
[14]
Q3.
(a)     produces H+ / hydrogen ions in aqueous solution
1
(but is) only partially / slightly ionised
1
(b)     indicator changes colour
1
from blue to yellow
allow from blue to green
1
(when) the acid and alkali are (exactly) neutralised
or
(when) no excess of either acid or alkali
1
(c)     pipette measures one fixed volume (accurately)
1
(but) burette measures variable volumes (accurately)
1
(d)      
1
(mean titre =) 12.13(3) (cm 3)
1
(moles NaOH = conc × vol) = 0.00255
1
(moles citric acid =  moles NaOH) = 0.00085
1
(conc acid = moles / vol) = 0.0701 (mol / dm 3)
allow ecf from steps 1, 2, 3 and / or 4
allow an answer of 0.0701 (mol / dm 3) without working for 1 mark only
1
[12]
Q4.
(a)     must be description of a titration no titration = no marks
          NaOH in burette
do not accept biuret etc
1
          add NaOH until (indicator) changes colour
if specific colour change mentioned, must be correct – colourless to pink / red or ‘goes pink / red’
do not accept ‘clear’ for colourless
1
          note (burette) volume used or final reading
accept ‘work out the volume’
1
          one other point: eg repeat
accept:
(white) tile or add dropwise / slowly or white background or swirling / mix or read meniscus at eye level or wash apparatus
1
(b)     0.054
for 2 marks
(0.1 × 13.5)/25 for 1 mark
(c)     don’t know – insufficient evidence to decide
owtte
any sensible answer
          or
          depends on whether acid level is considered safe or unsafe
          yes, safe – acid level low / weak acids / low compared with stomach acid
owtte
any sensible answer
2
          no, unsafe – acid level (too) high / other substances or bacteria may be
present / insufficient evidence to decide
owtte
any sensible answer
1
(d)     (methyl orange) would have changed colour (well) before the end-point / pH7 / neutral
owtte
1
          weak acid present
weak acid-strong base (titration)
allow methyl orange used for strong acid-weak base titration
1
[9]
Q5.
Divide by Ar:
Na = 22.8 / 23
B = 21.8 / 11
O = 55.4 / 16
if student has calculated moles upside down they can score mp 3 mp 4 and mp 5 as follows:
Na 23 / 22.8
B 11 / 21.8
O 16 / 55.4
1
Values
0.991
1.01
1.98
0.505
3.46
0.289
1
Divide by the smallest
1 : 2 : 3.5
Divide by the smallest (1)
3.5 : 1.75 : 1
1
Whole number ratio
2 : 4 : 7
Whole number ratio (1)
14 : 7 : 4
1
Empirical formula
Na2B4O7
Empirical formula (1)
Na14B7O4
if no working shown allow 4 marks for Na2B4O7
1
[5]
Q6.
(a)     (i)      fizz / effervescence / bubbles
allow calcium carbonate decreases in size or dissolves
1
because carbon dioxide produced / released
allow because gas produced / released
1
limewater turns cloudy / milky / white
1
because (a precipitate of or solid) calcium carbonate forms
allow because of carbon dioxide if not already credited
1
(ii)       
allow -OH
do not allow lower case ‘h’
1
(iii)     acid
must be in this order
ignore any name of an acid
1
ester(s)
1
(b)     white (precipitate)     no change
no change                    no change
all four correct 2 marks
any two correct 1 mark
2
(c)     (i)      lilac
allow purple
1
red
1
must be in this order
(ii)     colours are masked / changed by each flame colour
1
[12]
Q7.
Marks awarded for this answer will be determined by the Quality of Communication (QC) as well as the standard of the scientific response. Examiners should also apply a ‘best-fit’ approach to the marking.
0 marks
No relevant content
Level 1 (1 – 2 marks)
Any description of a method used and / or a result given
Level 2 (3 – 4 marks)
Description of workable methods used, with results to identify positive or negative ions
Level 3 (5 – 6 marks)
Description of methods used to identify both positive and negative ions, with relevant results
examples of the points made in the response
extra information
Test: add (platinum / nichrome) wire (for the flame test)
accept any method of introducing the solution into the flame, eg a splint soaked in the solution or sprayed from a bottle
Result: the sodium compounds result in a yellow / orange / gold flame or the potassium compound results in a lilac / purple / mauve flame
student could state that potassium carbonate gives a different colour to the three sodium compounds as long as it is clear that the flame test colour comes from Na+ or K+
Test: add dilute nitric acid to all four solutions
allow any acid
Result: sodium carbonate and potassium carbonate will effervesce or sodium chloride and sodium iodide will not effervesce
Test: add dilute nitric acid followed by silver nitrate
Result: sodium chloride and sodium iodide produce a precipitate or sodium chloride produces a white precipitate and sodium iodide produces a yellow precipitate
accept sodium carbonate and potassium carbonate do not produce a precipitate
[6]

Q8.
(a)     magnesium loses two electrons and chlorine gains one electron
accept magnesium loses electrons and chlorine gains electrons for 1 mark
ignore oxidation and reduction
2
one magnesium and two chlorines
accept MgCl2
1
noble gas structure
or
eight electrons in the outer shell
accept full outer shell (of electrons)
or
(electrostatic) attraction between ions
or
forms ionic bonds
do not accept covalent bonds
1
reference to incorrect particles or incorrect bonding or incorrect structure = max 3
(b)     (i)      because ions can move
ignore ions attracted
do not accept molecules / atoms moving
do not accept incorrect reference to electrons moving
1
(and ions move) to the electrodes
or
(and ions) carry charge
1
accept converse for solid
(ii)     magnesium (ions) attracted (to the electrode)
1
so magnesium ions gain electrons
accept magnesium ions are reduced
ignore oxidised
1
2 electrons
accept a correct half equation for 2nd and 3rd marking points
1
(iii)     hydrogen
allow H2
1
(iv)     magnesium is more reactive than hydrogen
accept converse
allow magnesium is high in the reactivity series or magnesium is very/too reactive.
do not accept magnesium ions are more reactive than hydrogen ions
1
(v)     2 Cl- → Cl2 + 2e-
must be completely correct
1
(c)     layers (of particles/atoms/ions)
1
(particles/atoms/ions/layers) can slide
1
any mention of intermolecular / weak bonds/forces = max 1
[14]

Answer/Indicative content	Marks	Guidance	
Q9		

Mole ratio C : H is 0.1 : 0.4 (1)

Empirical formula is CH4 / H4C(1)	2	LOOK AT THE ANSWER FIRST IF CH4 / H4C AWARD 2 MARKS

allow moles of C = 0.1 and moles of hydrogen = 0.4
allow moles of C = 1.2 / 12 and moles of hydrogen = 0.4/1

allow C1H4

allow full marks despite any working out for correct empirical formula 

Examiner's Comments

Candidates found this empirical formula calculation very challenging and only the best candidates were able to get the answer CH4. Typically candidates did show their working out for this question and this often showed up their misconceptions. One misconception was that candidates used the atomic number rather than the relative atomic mass to calculate the mole ratio of carbon to hydrogen. Other candidates did not use the mole ratio at all and used a mass ratio instead getting an answer of C3H. A significant proportion of the candidates did not attempt this question.	
		Total	2		
Q10		A	1		
		Total	1		
Q11		FIRST LOOK AT THE ANSWER
IF ANSWER = Fe2O3 AWARD 3 MARKS

mole ratio (1)

simplest mole ratio (1)

empirical formula is Fe2O3 (1)	3	

If fraction is the wrong way around = 0 marks for the question

If divide by atomic number = 0 marks for the question

If just use ratio of masses = 0 for the question

allow ecf from mole ratio

allow ecf from simplest ratio

allow FeO1.5 = 2 marks for the question

?Examiner's Comments??

Candidates found this empirical formula calculation very challenging. Even candidates who calculated the mole ratio often rounded the ratio incorrectly to end up with FeO2 or stated the formula was FeO1.5. Common misconceptions included using the mass ratio rather than the mole ratio or using atomic number to calculate the mole ratio rather than the relative atomic mass. A significant proportion of candidates left this question blank.	
		Total	3		
Q12	i	Moles = 0.0025 / 2.5 × 10?3 (1)	1	?Examiner's Comments??

Many candidates did not understand the relationship between volume, molar volume and moles and as a result did not get the correct answer of 0.0025 moles. A typical incorrect answer was 1.44 moles. 	
	ii	Mass = 0.25 (g) / 2.5 × 10?1 (1)	1	allow ecf from number of moles, i.e. moles × 100

?Examiner's Comments??

Many candidates could not sure the relationship between molar mass, moles and mass in (ii) and as a result did not multiply the answer to (i) by 100. A significant proportion of candidates left (ii) blank.	
		Total	2		
Q13	i	idea that average only uses titrations 2, 3 and 4 / titration 1 is not used to calculate the average (1)

titrations 2, 3 and 4 have a consistent value / titration 1 is a rough estimate / titration 1 could be an anomalous value / titration 1 is a range-finder / titration 1 is an outlier (1)	2	allow calculation of the mean using the values from 2, 3 and 4

not just use titration 3
not titration 3 is the mid-value 

Examiner's Comments

Although many candidates appreciated that titre 1 was an anomaly or an outlier they did not always go on to state that the other three were used to calculate the mean. Only a small proportion of the candidates commented on the closeness of the last three titres. The best answers indicated why titre 1 was not used and then showed how the other three titres are used to get the mean titre. A common misconception was that 25.2 cm3 was used as a titre because it was the mid-range value of the three best titres.	
	ii	0.0015 or 1.5 × 10?3 (1)	1	ignore trailing zeroes 

Examiner's Comments

Many candidates could not calculate the amount in moles as 0.0015. The most common misconception was to forget to convert cm3 into dm3 and calculate the number of moles as 1.5.	
	iii	

moles of HNO3 = 0.0015 or 1.5 × 10?3 (1)

concentration of HNO3 = 0.0595 (1)	2	LOOK FOR THE ANSWER FIRST IF IT IS 0.0595 AWARD 2 marks

allow ecf from (ii)

allow one mark for 0.06 / 0.05952 or answers with more significant figures

allow ecf providing answer has three significant figures i.e. mole/volume 

Examiner's Comments

A significant proportion of the candidates did not attempt this question often because they could not do (ii). Only the very best answers showed the working out for the calculation. Centres should advise candidates that it is important to show the working out to allow examiner's to award error carried forward marks. Many candidates did not quote the final answer, 0.0595 mol/dm3, to three significant figures as required in the question and gave the answer as 0.06 mol/dm3.	
		Total	5		
Q14	i	any answer in the range 
21.5 − 22.5 (cm3) (1)	1		
	ii	20 (cm3) (1)	1		
	i		1	allow 0.030 × 0.3

allow 30 × 10−3 × 0.3

allow other substitution and rearrangement of moles = conc × volume e.g.
"0.009/0.30 = 0.03dm3 = 30cm3, and 
0.009 × 30 = 0.3"	
	ii	0.45 (mol / dm3) (2)

but if answer incorrect then

	2	allow ecf from (a)(ii) i.e.

		Total	5		
Q15	i		1	ALLOW any correct multiple

DO NOT ALLOW & or and instead of +	
	ii	Ions cannot move (1)	1	IGNORE electrons cannot move	
		Aluminium is above carbon in the reactivity series so cannot be obtained by reaction of oxide with carbon (1)

Copper is below carbon in the reactivity series (1)	2		
		Total	4		
Q16		iron or top reaction loses electrons which is oxidation (1)

oxygen or bottom reaction gains electrons which is reduction (1)	2	no mark for identifying which reaction is oxidation and which is reduction

allow water gains electrons which is reduction 

Examiner's Comments

Most candidates used the concept of Oil Rig and were able to identify that iron is oxidised and oxygen is reduced. Only a small proportion of candidates did not give a reason for their choice.	
		Total	2		
Q17	i	at the anode electrons are lost which is oxidation (1)

at the cathode electrons are gained which is reduction (1)	2	allow 1 mark if oxidation is described as electron loss and reduction as electron gain without identification of the electrodes or with incorrect identification of the electrodes 

Examiner's Comments

Good responses described the reaction at the anode as oxidation because electrons are lost and the reaction at the cathode as reduction because electrons are gained. One mark was awarded to candidates who described oxidation as electron loss and reduction as electron gain, without identification of the electrodes (or with incorrect identification of the electrodes).	
	ii	the anode loses mass because copper ions go into solution (1)

the cathode gains mass because the copper ions gain electrons and become copper (1)	2	if ion is missed out in both marking points then allow one mark

allow copper ions move from the anode to the cathode for 1 mark if no other mark awarded 

Examiner's Comments

The aspects of the specification was not well known in this part. To gain credit candidates were required to describe copper ions going into solution at the anode and then those copper ions gaining electrons and becoming copper at the cathode.	
		Total	4		
Q18		Level 3 (5–6 marks)
Manipulates the data to describe and explain how the position of equilibrium changes with pressure
AND
Manipulates the data to describe and explain how the position of equilibrium changes with temperature
Quality of written communication does not impede communication of the science at this level.

Level 2 (3–4 marks)
Manipulates the data to describe and explain how the position of equilibrium changes with pressure
OR
Manipulates the data to describe and explain how the position of equilibrium changes with temperature
Quality of written communication partly impedes communication of the science at this level.

Level 1 (1–2 marks)
Manipulates the data to describe how the position of equilibrium changes with pressure AND with temperature
Quality of written communication impedes communication of the science at this level.

Level 0 (0 marks)
Insufficient or irrelevant science. Answer not worthy of credit.	6	This question is targeted at grades up to A*.

Relevant points at levels 2 and 3 include:
•	increasing the pressure moves the equilibrium to the right because there are fewer molecules (or number of moles)on the rhs
•	increasing the temperature moves the equilibrium to the left because the forward reaction is exothermic or the backward reaction is endothermic.

Relevant points at level 1 include:
•	as the pressure increases the position of equilibrium moves to the right or vice versa
•	as pressure increases percentage of ammonia increases
•	as the temperature increases the position of equilibrium moves to the left or vice versa
•	as temperature increases the percentage of ammonia decreases

ignore references to rate 

Examiner's Comments

This was a very challenging six mark question that involved data analysis and le Chatelier's principle. It was targeted up to grade A*.

Some candidates only described the trends shown in the two tables and found it very difficult to explain the trends. Other candidates contradicted themselves in terms of percentage yield and the position of equilibrium, incorrectly stating that with a low percentage yield the position of equilibrium is on the right. A significant proportion of candidates referred to changing the temperature and pressure but did not specify the direction of the change.

Candidates were more likely to be able to explain the effect of increasing temperature on the position of equilibrium rather than the effect of increasing the pressure. The best answers for increasing temperature referring to the reaction being exothermic or the reverse reaction being endothermic. The best answers for increasing pressure referred to the reaction moving to the side with the least number of moles.

A significant proportion of the candidates answered a completely different question to the one set, since they tried to explain the conditions used in the reaction and referred to optimum conditions and rate of reaction.	
		Total	6		
Q19		copper is deposited at the cathode (1)
copper anode dissolves / copper ions produced at anode (1)	2	ALLOW higher level answers in terms of half equations
e.g. at cathode Cu2+ + 2e− → Cu (1)
e.g. at anode Cu → Cu2+ + 2e− / Cu − 2e− → Cu2+ (1)	
		Total	2		
Q20		Rate of forward reaction equals the rate of the backward reaction (1)

Concentration of reactants and products do not change (1)	2	ALLOW concentration of reactant and product do not change

DO NOT ALLOW concentration of reactant and products are the same	
		Total	2		
Q21		catalyst / vanadium(V) oxide / V2O5 (1)	1	allow vanadium pentoxide

if a named catalyst is given it must be correct including oxidation number except allow vanadium oxide catalyst

if formula and name given both must be correct

Examiner's Comments

This question was about the Contact Process.

The need for a catalyst was well known but many candidates gave the name of the wrong catalyst for example iron or nickel. A small proportion of candidates referred to a high concentration as a condition.	
		any three from:
catalyst increases rate of reaction (1)

catalyst does not change position of equilibrium (1)

increasing temperature - increases rate of reaction / temperature used to have a high rate of reaction (1)

but increasing temperature position of equilibrium to left / temperature used to not shift the equilibrium to the left (1)

at low pressure position of equilibrium is already on right (1)

so expensive high pressures are not needed / at low pressure rate is low so reaction is easier to control (1)	3	allow ora where appropriate

allow catalyst does not change (percentage) yield

allow increasing temperature decreases (percentage) yield / increasing temperature favours backward reaction (1)

reference to it is a compromise temperature is not sufficient

allow good product (percentage) yield at low pressure

not use low pressure to shift equilibrium to the right

Examiner's Comments

This question was about the Contact Process.

There were many different marking points that could be accessed by the candidates. In terms of temperature the idea that the reaction would be fast and was low enough to keep the equilibrium on the right was sufficient – many candidates only mentioned the reference to rate. In terms of the pressure many candidates thought that the low pressure was to force the position of equilibrium to the right. Only the best answers appreciated that even at low pressure the position of equilibrium was on the right. Many candidates appreciated that a higher pressure would be more expensive. Most candidates appreciated that the catalyst increases the rate of reaction but the candidates did not often mention that the position of equilibrium was unaffected.	
		Total	4		
Q22	i	addition reaction (1)	1	allow bromination

Examiner's Comments

Were challenging and only a small proportion of the candidates described the addition reaction and named the compound type correctly.	
	ii	a dibromocompound (1)	1	allow saturated / halogenocompound

Examiner's Comments

Were challenging and only a small proportion of the candidates described the addition reaction and named the compound type correctly.	
		Total	2		
Q23		Level 3 (5–6 marks)Includes an explanation of both a reversible reaction and an equilibrium and correctly describes the effect of pressure and temperature on the position of equilibrium.
Quality of written communication does not impede communication of the science at this level.

Level 2 (3–4 marks)
Includes an explanation of either a reversible reaction or an equilibrium and correctly describes the effect of pressure or temperature on the position of equilibrium
OR
includes an explanation of both a reversible reaction and an equilibrium
OR
correctly describes the effect of pressure and temperature on the position of equilibrium
Quality of written communication partly impedes communication of the science at this level.

Level 1 (1–2 marks)
Explains what is meant by either a reversible reaction or an equilibrium
OR
includes one correct description of the effect of either pressure or temperature on the position of equilibrium.
Quality of written communication impedes communication of the science at this level.

Level 0 (0 marks)
Insufficient or irrelevant science. Answer not worthy of credit.	6	This question is targeted at grades up to grade C.

Relevant points include:
•	idea that reversible reactions can proceed in both directions / go forwards and backwards
•	idea that, at equilibrium, rate of forward reaction equals the rate of the backward reaction
•	idea that, at equilibrium, concentrations of reactants and products do not change
•	as the pressure increases the position of equilibrium moves to the right or concentration of products increases or vice versa
•	as the temperature increases the position of equilibrium moves to the left or the concentration of reactants increases or vice versa.
•	increase temperature less ammonia
•	increase pressure more ammonia

Examiner's Comments

The Haber process appeared to be fairly well known, but the principles behind it were not. Descriptions of reversible reactions were poor at best, and there were very few even partly correct descriptions of equilibrium. The better candidates were able to state the effect of changes in temperature and pressure in terms of the amount of ammonia.	
		Total	6