Amount of Substance, Acids and Redox UNIT 2 Flashcards
Give the equation of calculating percentage composition by mass [1]
Percentage by mass: (Mass of element/Mass of the compund) x 100
Suggest two modifications to the method that would reduce the percentage uncertainty
in the mass of the residue.[2]
Use balance that weighs to 3/more decimal
places ✓
Use a larger mass (of hydrated strontium
chloride) ✓
Examiner’s Comments
Most candidates identified either using a larger mass
or a more accurate balance, not many stated both.
The most common incorrect answers involved
heating for longer or taking less measurements.
Zinc reacts with hydrochloric acid, HCl(aq), as shown in the following equation.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
A student investigates the rate of this reaction.
The student plans to react 50.0 cm3 of 0.100 mol dm–3 HCl with 0.200 g of zinc (an excess).
Calculate the volume, in cm3, of hydrogen that should be produced at RTP. [3]
FIRST CHECK THE ANSWER ON ANSWER
LINE
If answer = 60 cm3 award 3 marks
————————————————–
Volume = 2.5(0) × 10–3 × 24.0 × 1000
= 60(.0) cm3 ✓
Examiner’s Comments
This was a well answered question, with the majority
of candidates obtaining all 3 marks
This question is about compounds of magnesium and phosphorus.
A student plans to prepare magnesium phosphate using the redox reaction of magnesium with
phosphoric acid, H3PO4.
3Mg(s) + 2H3PO4(aq) → Mg3(PO4)2(s) + 3H2(g)
i. In terms of the number of electrons transferred, explain whether magnesium is being
oxidised or reduced. [1]
Oxidised
AND
(Mg) transfers/loses/donates 2 electrons ✓
2 essential
Examiner’s Comments
Despite the question clearly asking for a response in
terms of the number of electrons transferred, most
candidates answered in terms of oxidation number
changes. Candidates are recommended to read the
question and to answer in terms of its requirements.
Underlining ‘number of electrons’ may have helped
candidates to answer the question that had been set.
This question is about compounds of magnesium and phosphorus.
A student plans to prepare magnesium phosphate using the redox reaction of magnesium with
phosphoric acid, H3PO4.
3Mg(s) + 2H3PO4(aq) → Mg3(PO4)2(s) + 3H2(g)
The student plans to add magnesium to 50.0 cm3 of 1.24 mol dm−3 H3PO4.
Calculate the mass of magnesium that the student should add to react exactly with the
phosphoric acid.
Give your answer to three significant figures. [3]
FIRST CHECK ANSWER ON THE ANSWER
LINE
IF answer = 2.26 (3 SF) award 3 marks
———————————–
✓
mass of Mg = 0.0930 × 24.3 = 2.26 (g) ✓
Examiner’s Comments
Most candidates are competent at answering
questions based on the mole. Almost all candidates
were able to calculate the amount of H3PO4 as
0.062 mol. Candidates then needed to use the 2:3
mole stoichiometric ratio to show that 0.093 mol of
Mg reacts, which has a mass of 2.26 g to the
required 3 significant figures. The commonest errors
were use of the inverse 3:2 ratio to obtain 1.00 g Mg,
or to omit the ratio to obtain 1.51 g Mg, as shown in
the exemplar. Candidates are advised to show clear
working so that credit can be awarded for such
responses by applying error carried forward.
How could the student obtain a sample of magnesium phosphate after reacting
magnesium with phosphoric acid? [2]
Separation of solid
Filter to obtain solid/precipitate ✓
Requires realisation that solid is filtered
off.
Solid may be stated within in ‘removal of
water’
Removal of water
Dry (solid)
OR Evaporate (water/solution/liquid) ✓
Examiner’s Comments
Candidates often struggle with questions based on
practical work. There were many random responses
to this question, with relatively few candidates
identifying that solid magnesium phosphate could be
obtained by filtration, followed by drying.
i. Phosphine, PH3, is a gas formed by heating phosphorous acid, H3PO3, in the absence of air.
4H3PO3(s) → PH3(g) + 3H3PO4(s)
i. 3.20 × 10−2 mol of H3PO3 is completely decomposed by this reaction.
Calculate the volume of phosphine gas formed, in cm3
, at 100 kPa pressure and 200 °C. [4]
ii.When exposed to air, phosphine spontaneously ignites, forming P4O10 and water.
Construct an equation for this reaction.[1]
i. FIRST CHECK ANSWER ON THE ANSWER
LINE
IF answer = 315 (cm3) award 4 marks
———————————————————
Amount of PH3
Unit conversions
p conversion → Pa = 100 × 103 (Pa)
AND
T conversion → K = 473 (K) ✓
Evidence of use of rearranged gas equation
OR V = 3.15 × 10−4 ✓
Calculator: = 3.1460176 × 10−4
V conversion of m3 → cm3 ✓
V =3.15 × 10−4 × 106 = 315 cm3 ✓
Calculator from unrounded cm3: 314.60176 cm3
Requires 3 OR MORE SF, correctly rounded
Examiner’s Comments
Almost all candidates realised that the calculation
required the ideal gas equation. Most candidates
correctly rearranged the equation and used the data
from the question to obtain a value for the volume of
phosphine. The most common errors were with
conversion of units into Pa and m3. It is
recommended that candidates learn how to carry out
these conversions. In their calculations, many
candidates used the amount of phosphoric acid, 3.20
× 10−3 mol, rather than 8.00 × 10−3 mol of phosphine,
obtaining a volume of 1258 cm3. Error carried
forward ensured that 3 of the available 4 marks could
be credited, provided that the working was clear. The
exemplar shows such a response.
ii. 4PH3 + 8O2 → P4O10 + 6H2O ✓
Examiner’s Comments
Most candidates were able to write a correctly
balanced equation for this reaction.
1.00 tonne of ammonia is reacted with carbon dioxide to prepare the fertiliser urea, NH2CONH2.
2NH3(g) + CO2(g) → NH2CONH2(s) + H2O(1)
1.35 tonnes of urea are formed.
Calculate the percentage yield of urea.
Show all your working [3]
FIRST CHECK THE ANSWER ON THE
ANSWER LINE
IF answer = 76.5 (%) award 3 marks
n(NH3) = (1 × 106) / 17 = 5.88 × 104 (58824)
(mol)
AND
Theoretical yield:
n(NH2CONH2) = 5.88 × 104 / 2 = 2.94 × 104
(29412) (mol) (1)
Actual yield:
n(NH2CONH2) = 1.35 × 106 / 60 = 2.25 × 104
(22500) (mol) (1)
% yield = (2.94 × 104 / 2.25 × 104) × 100% =
76.5(%) (1)
How many molecules are present in 1.3535 g of PBr3? [3]
FIRST CHECK THE ANSWER ON THE
ANSWER LINE
If answer = 3.01 × 1021 award 3 marks
Mr(PBr3) = 270.7 (g mol−1) (1)
n(PBr3) = 1.3535 / 270.7 = 5.000 × 10−3 mol (1)
number of molecules = 5.000 × 10−3 × 6.02 ×
1023 = 3.01 × 1021 molecules (1)
An alkene D is a liquid at room temperature and pressure but can easily be vaporised.
When vaporised, 0.1881 g of D produces 82.5 cm3 of gas at 101 kPa and 373 K.
Determine the molar mass and molecular formula of alkene D.
Show all your working. [5]
FIRST check the molar mass on answer line
MUST be derived from pV = nRT,
Award 4 marks for calculation for:
- answer = 70
- OR answer that rounds to 69.9 OR
70.0
———————————————————
Rearranging ideal gas equation to make
n subject
n= pV/RT
Substituting all values including conversion to
Pa and m3
n= (101x 10(3))X(82.5x10(-6))/8.314x373
Calculation of molar mass, M
Molecular formula : C5H10
Examiner’s Comments
Most candidates realised the need to use the ideal
gas equation. The equation was usually rearranged
correctly, with substituted values for p, V, R and T
being added. Pressure and volume were not always
converted correctly into Pa and m3, creating
problems for subsequent parts. Many candidates
attempted to convert from cm3 to m3 by multiplying
by 10−3 rather than 10−6.
Candidates usually obtained a value for n, although
those who had struggled with unit conversion
obtained values that differed by powers of 10.
Finally, candidates needed to derive the molar mass
using their value of n and the mass of the alkene.
Some candidates over-rounded their value of n,
introducing an error in calculating the molar mass.
Surprisingly, an appreciable number of candidates
wrote their value of n on the answer line rather than
the molar mass indicated by the answer prompt. This suggested that some candidates do not understand
the term molar mass.
Candidates who had obtained a molar mass of 70.0
usually determined that the alkene had the formula
C5H10.
Answer: 70.0 g mol−1
Barium combines with oxygen, chlorine and nitrogen to form ionic compounds.
Barium oxide, BaO, has a giant ionic lattice structure.
i. State what is meant by the term ionic bond.
[1]
ii. Draw a ‘dot-and-cross’ diagram to show the bonding in barium oxide.
Show outer electrons only. [2]
iii. Calculate the number of barium ions in 1.50 g of barium oxide.
Give your answer in standard form and to three significant figures. [2]
i. Electrostatic attraction between positive and
negative ions ✔
Examiner’s Comments
The specification describes ionic bonding as an
electrostatic attraction and a small proportion of
answers were missing this key phrase.
ii.Ba shown with either 0 or 8 electrons
AND
O shown with 8 electrons with 6 dots and
2 crosses (or vice versa)
✔
Correct charges on both ions ✔
Examiner’s Comments
Covalent bonding diagrams were not common and
this question was well answered by the vast majority
of candidates.
iii.FIRST CHECK THE ANSWER ON THE
ANSWER LINE
IF answer = 5.89 × 1021 award 2 marks for
calculation
Moles of barium oxide
n(BaO) =1.50/153.3 OR 9.78 × 10-3 ✔
Number of barium ions
(9.78 × 10-3 × 6.02 × 1023) = 5.89 × 1021
✔
3 SF AND standard form required
Examiner’s Comments
Use of the relative mass of barium to calculate moles
of barium oxide was a common error but these
candidates were usually able to pick up one mark for
correctly multiplying their moles by the Avogadro
constant. Some candidates correctly calculated
moles but then divided by two thus losing the final
mark.
i. Barium chloride, BaCl2, is soluble in water.
i. Compare the electrical conductivities of solid and aqueous barium chloride.
Explain your answer in terms of the particles involved.[2]
ii. Describe the use of aqueous barium chloride in qualitative analysis. [2]
iii. Hydrated barium chloride can be crystallised from solution.
Hydrated barium chloride has the formula BaCl2xH2O and a molar mass of 244.3 g mol −1.
Determine the value of x in the formula of BaCl2xH2O.
Show your working[2]
i.Barium chloride does not conduct electricity
when solid
AND
because it has ions which are fixed (in position /
in lattice) ✔
Barium chloride conducts when in aqueous
solution
AND
because it has mobile ions ✔
Examiner’s Comments
Many precise answers gained full marks by
describing the fixed position of ions in a lattice and
the mobility of ions in aqueous solution. Delocalised
or free electrons were occasionally mentioned.
Vague answers often used the terms ‘free’ instead of
mobile, ‘charge carrier’ instead of ion and
‘carry a charge’ instead of conduct electricity.
ii. Test for sulfate / SO4
2- ✔
White precipitate forms (when barium
chloride solution is mixed with a solution
containing sulfate ions) ✔
Examiner’s Comments
There was some confusion with the displacement
reactions of halogens, the test for halide ions and the
use of silver nitrate but the majority of students could
recall the use of aqueous barium chloride to test for
sulfate ions. Occasionally candidates described the
use of dilute hydrochloric acid to remove carbonate
ions from solution before their creditworthy
description of the sulfate test.
iii. FIRST CHECK THE ANSWER ON THE
ANSWER LINE
IF answer = 2 award 2 marks
M(BaCl2) = ((137.3 + (35.5 × 2))
= 208.3 (g mol-1)
244.3 - 208.3 = 36
AND
36/18 = 2
Examiner’s Comments
Very well answered, the majority of candidates
scored full marks for this simple calculation.
By 2020, the EU has regulated that a car must emit less CO2 per kilometre than in 2015. A typical
car will need to emit 5.6 × 105 g less CO2 in 2020 compared with 2015.
Calculate how much less petrol would be consumed by a typical car in 2020 to meet this
regulation.
Give your answer in litres of petrol (1 litre of petrol has a mass of 700 g).
Assume that petrol is liquid octane and that complete combustion takes place, as in the equation
below.
C8H18 (l) + 12.5 O2 (g) → 8CO2 (g) + 9H2O (l) [4]
IF answer = 259 (litres), award 4 marks
………………………………………………….
….
(n(CO2) decrease = 5.6 × 105/44.0) =
12727.27273 (mol) ✓
(n(C8H16) decrease = 12727 ÷ 8) = 1590.909091
(mol) ✓
(mass of C8H18 decrease) = 1591 × 114 =
181363.6364 (g) ✓
(C8H18 decrease) = 181363.6364 ÷ 700 g = 259
(litres) ✓
Examiner’s Comments
In general candidates coped well with this
unstructured calculation. The majority chose to
convert the mass of CO2 into moles and use the
balanced equation to determine the mass of octane,
before obtaining the reduction in petrol consumption.
However, alternative approaches were also seen
and awarded full credit where due. Error carried
forward marks were awarded, and most candidates scored three or four marks. Weaker candidates often
divided the mass of CO2 by 700 and failed to achieve
a meaningful answer. Candidates should be
encouraged to start multistep calculations by
considering amounts in moles, rather than just
experimenting with the data provided in the question.
Answer: 259 litres
Suggest one modification that the student could make to their method to reduce the
percentage error in the mass of water removed[1]
Use balance weighing to 3/more decimal places
OR
Use a larger mass/amount □ ✓
Examiner’s Comments
Correct answers suggested using a larger mass of
the salt or a more accurate balance with more
decimal places. Many responses instead discussed
repeating the experiment and taking an average, or
using a lid.
One coin has a mass of 5.00 g and contains 84.0% of copper, by mass.
Calculate the number of copper atoms in one coin.
Give your answer in standard form and to three significant figures. [2]
FIRST CHECK ANSWER ON THE ANSWER
LINE
If answer = 3.97 × 1022 (from 63.62) award 2
marks
If answer = 3.98 × 1022 (from 63.5) award 2
marks
_____________________________
Using 63.62: correct Ar of Cu from 21(b)(i)
See bottom of answer zone
Cu atoms = 0.0660 × 6.02 × 1023 = 3.97 × 23
1022 ✓
Must be calculated in standard form AND to 3
SF
OR_____________________________
Using 63.5: Ar of Cu from periodic table
Cu atoms = 0.0661 × 6.02 × 1023 = 3.98 × 1022
✓
Must be calculated in standard form AND to 3
SF
Examiner’s Comments
This part was generally well answered with most
candidates processing the data correctly. Candidates
sometimes failed to consider 84% or rounded
incorrectly in places.
Answer = 3.97 × 1022 atoms
This question is about several salts.
A hydrated salt, compound A, is analysed and has the following percentage composition by
mass:
Cr, 19.51%; Cl, 39.96%; H, 4.51%; O, 36.02%.
Calculate the formula of compound A, showing clearly the water of crystallisation.
Show your working.
[3]
Initial ratios
OR
Cr, 0.375; Cl,1.126; H,4.51; O, 2.25 ✓
Whole number ratios
Cr, 1; Cl, 3; H, 12; O, 6 ✓
Formula with water of crystallisation
CrCl36H2O ✓
Examiner’s Comments
Many candidates were able to calculate the empirical
formula of the hydrated salt. While the majority went
on to shown the formula as CrCl36H2O to score all
three marks, a significant minority failed to convert
12 H and 6 O into 6H2O
A student reacts 35.0 cm3 of 3.00 × 10−2 mol dm−3 H2SO4(aq) with an excess of Al.
An equation for this reaction is shown.
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
Calculate the mass, in g, of Al 2(SO4)3 formed in solution.
Give your answer to three significant figures.
Show your working. [4]
First check the answer line. If answer = 0.120
award 4 marks.
M1 Mol of H2SO4 = 3.00 x 10–2 x 35.0/1000= 1.05 x
10–3 mol ✓
M2 Mol of Al2(SO4)3 = 1.05x10(-3)/3= 3.5(0) x 10–4
mol ✓
M3 = 342.3 ✓
M4 Mass Al2(SO4)3 = 3.5(0) x 10–4 x 342.3
and
= 0.120 g ✓
Answer must be 3 sf
Examiner’s Comments
This open style calculation would have usually
proved difficult for the typical AS candidate but this
year a significant majority of candidates were able to
secure all four marks.
When magnesium nitrate, Mg(NO3)2, is heated, it decomposes as shown.
2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)
A student heats 2.966 g of Mg(NO3)2, which decomposes as above.
Calculate the total volume of gas formed, in cm3, at room temperature and pressure, RTP. [3]
First check the answer line.
If answer = 1200 cm3 award 3 marks.
Mol of Mg(NO3)2 = = 2(.00) x 10–2 OR
0.02(00) mol✓
Mol of gas = 2(.00) x 10–2 x 5/2 = 5(.00) x 10–2
OR 0.05(00) mol ✓
Vol of Gas = 0.05 x 24 000 = 1200 cm3 ✓
Examiner’s Comments
This seemingly difficult calculation was answered
successfully by all but a relatively small handful of
candidates.
Calculate the amount, in mol, of nitrogen atoms in 5.117 × 1020 nitrogen molecules.
Give your answer in standard form. [2]
First check the answer line.
If answer = 1.7(0) × 10–3 award 2 marks.
…………………………………………………
..
M1 (Dividing by 6.02 x 1023)
Number of N2 molecules = = 8.5. x
10–4
OR 0.85 x 10–3 OR 0.085 x 10–2 OR 0.0085 x
10–1 OR 0.00085 ✓
M2 (Correct conversion of molecules to atoms +
standard form)
M1 x 2 and in standard form ✓
From 0.0085, answer = 2 x 0.00085 = 0.00170
= 1.7(0) x 10–3
Examiner’s Comments
This proved to be one of the more difficult questions
on the paper.
N2O3 reacts with water to form an acid as the only product. This reaction is not a redox reaction.
The empirical formula of the acid formed is the same as the molecular formula.
i. State what is meant by the term molecular formula.
[1]
ii. Suggest the empirical formula of the acid formed.[1]
i.(Actual) number of atoms of each element
present in a molecule ✓
Examiner’s Comments
Many candidates were successful in describing the
term ‘molecular formula’ but weaker candidates gave
answers which confused terms such as atoms and
molecules. By far the most common erroneous
response was ‘The number of atoms in a molecule’.
ii.HNO2 ✓
Examiner’s Comments
Weaker candidates convinced themselves that the
acid formed when water is added to nitrogen dioxide
was HNO3. Better candidates were able to work out
the product would have the formula H2N2O4 but
failed to convert this to its simplest form.
Nitrogen forms several different oxides.
N2O is a useful anaesthetic and NO has been linked to the depletion of ozone in the stratosphere.
N2O is supplied as a compressed gas in steel cylinders for use as an anaesthetic.
The cylinders are stored at 20.0 °C.
Calculate the gas pressure, in Pa, in a 2.32 dm3 steel cylinder containing 187 g of N2O gas.
Give your answer in standard form to three significant figures [4]
FIRST CHECK ANSWER ON THE ANSWER L
IF answer = 4.46 × 106 (Pa) award 4 marks
If there is an alternative answer, check to see if
there is any ECF credit possible
Amount of N2O
1
ALLOW ECF from incorrect amount of N2O
e.g. use of incorrect Mr for N2O could still score 3
marks
Unit conversion
Volume conversion to m3 = 2.32 × 10−3 (m3) ✓
1
Ideal gas equation / temperature conversion
AND
Use of T = 293 K ✓
Final answer
p = 4.46 × 106 (Pa) ✓
Must be calculated in standard form AND to 3
SF
Examiner’s Comments
This was a new addition to the OCR specification as
part of the curriculum changes. The vast majority of
candidates made a good attempt at this calculation
which required both the rearrangement of a formula
and the conversion of units of temperature and
volume. The conversions and calculation did not
prove that difficult for many candidates however
answers were often not given to three significant
figures or quoted in standard form resulting in the
loss of one mark. Candidates clearly need to develop
their mathematical skills in order to access the 20%
of marks available for quantitative work.
Answer = 4.46 × 106(Pa)
Europium, atomic number 63, reacts with oxygen at room temperature.
4Eu + 3O2 → 2Eu2O3
Calculate the volume of oxygen, in cm3, required to fully react with 9.12 g of europium at room
temperature and pressure. [2]
Check the answer line.
If answer = 1080 cm3 award 2 marks
Amount of Eu = 9.12 / 152.0 = 0.06(00) mol ✔
Amount of O2 = 0.0600 × 3 / 4 = 0.045(0) mol
and
Volume of O2 = 0.0450 × 24000 = 1080 cm3 ✔
Examiner’s Comments
This potentially difficult calculation was well
addressed by candidates and many scored both
marks available.
i. A compound of thulium, atomic number 69, has the following composition by mass:
O 30.7% S 15.4% Tm 53.9%
i. State what is meant by the term empirical formula.
[1]
ii. Determine the empirical formula of the compound.
Show your working. [2]
i.The simplest whole number ratio of atoms (of
each element) present in a compound ✔
ii.Check the answer line.
If answer = O12S3Tm2 award 2 marks
O = 30.7 / 16.0 S 15.4 / 32.1 Tm = 53.9 / 168.9
OR
1.9(2) mol 0.480 mol 0.319 mol ✔
O12S3Tm2 ✔
Examiner’s Comments
This question perhaps demonstrated the extent to
which candidates rely upon rote application of a
‘mathematical’ method without fully understanding
what they are actually attempting to do.
Nearly all candidates were able to convert a ratio by
mass to a ratio by moles of atoms, by dividing the
mass ratios by the relevant relative atomic masses.
These candidates were further able to obtain a unit
value for one atom by the mathematical operation of
dividing all values by the smallest number.
This gave a formula of TmS1.5O6 and many
candidates were convinced that increasing the value
of S atoms from 1.5 to 2 (the nearest whole number)
would meet the requirements that an empirical
formula has to have whole number values of atoms.
Only the stronger candidates were able to realise
that the initial ratio calculated needed to be doubled
to obtain integer values which kept the same ratio of
atoms.
Hydrated strontium chloride, SrCl2·6H2O, has a molar mass of 266.6 g mol−1.
A student heats 5.332 g of SrCl2·6H2O.
The SrCl2·6H2O loses some of its water of crystallisation forming 3.892 g of a solid product.
Use the information above to determine the formula of the solid product.
Show your working. [3]
FIRST CHECK THE ANSWER ON THE
ANSWER LINE
IF answer = SrCl22H2O award 3 marks
M1 Correctly calculates
Mol of SrCl26H2O = (5.332 / 266.6) = 0.02 mol
✔
M2 Correctly calculates
Mol of water given off [(5.332 − 3.892) / 18] =
0.08 mol ✔
M3 Correctly calculates
0.08 / 0.02 = 4 mol of water lost from one mol of
SrCl26H2O Therefore
Answer = SrCl22H2O ✔
Examiner’s Comments
Many of the more able candidates were able to give
the correct formula here and did so with very clear
working, which revealed that they understood the
path that lay behind their calculations. Less able
candidates converted the mass of the hydrate and
the mass of water lost into the respective mol of
substance (0.02 and 0.08). This is perhaps not
surprising as these steps are common to the more
familiar problem of working out the number of waters
of crystallisation in a hydrated salt that is then fully
dehydrated by the action of heat. However the
degree of difficulty caused many to become unclear
as to what to do with these numbers and hence
SrCl2*G4H2O was a common incorrect answer.
This question is about compounds used in fertilisers.
A compound used as a fertiliser has the following composition by mass:
C, 20.00%; H, 6.67%; N, 46.67%; O, 26.66%.
Calculate the empirical formula of this compound.[2]
FIRST CHECK THE ANSWER ON THE
ANSWER LINE
IF answer = CH4N2O award 2 marks
Examiner’s Comments
Calculating empirical formulae is a skill which most
candidates are familiar with and consequently the
vast majority of candidates were awarded both
marks.
. An aqueous solution of aluminium chloride can be prepared by the redox reaction between
aluminium metal and dilute hydrochloric acid.
A student reacts 0.0800 mol of aluminium completely with dilute hydrochloric acid to form an
aqueous solution of aluminium chloride.
The equation for this reaction is shown below.
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
Calculate the volume of hydrogen gas formed, in dm3, at room temperature and pressure. [2]
FIRST CHECK THE ANSWER ON THE
ANSWER LINE
IF answer = 2.88 dm3 award 2 marks
Mol of H2 = 0.12 ✔
Volume of H2 = 0.12 × 24.0 = 2.88 dm3 ✔
Examiner’s Comments
Weaker candidates forgot to consider the
stoichiometric ratio between Al and H2 but were still
able to gain credit for the correct use of the molar
gas volume, leading to an answer of 1.92 cm3, rather
than the expected 2.88 cm3.
1-Bromobutane (Mr, 136.9) can be made from a reaction of butan-1-ol, C4H9OH, as shown in the
equation below.
C4H9OH + KBr + H2SO4 → C4H9Br + KHSO4 + H2O
i. Calculate the atom economy for the formation of 1-bromobutane in this reaction. [1]
ii. Suggest a reactant, other than a different acid, that could be used to improve the atom
economy of making 1-bromobutane by the same method.
iii. A student prepares a sample of 1-bromobutane.
5.92 g of butan-1-ol are reacted with an excess of sulfuric acid and potassium bromide.
After purification, 9.72 g of 1-bromobutane are collected.
Calculate the percentage yield.
Give your answer to three significant figures. [3]
i. (136.x100)/291.1= 47%
Examiner’s Comments
This was a very well answered question and most
candidates were able to calculate to the atom
economy for the reaction.
ii.NaBr OR LiBr ✔
Examiner’s Comments
This novel question required candidates to suggest a
way of increasing the atom economy by using an
alternative reactant. The most able correctly
identified that either sodium or lithium bromide would
be an appropriate replacement for potassium
bromide. The most common response was HBr
which was not credited as the question specified a
chemical other than an acid should be suggested.
iii.Look at answer if 88.8% AWARD 3 marks if
88.75% AWARD 2 marks (not 3 sig. fig.)
Moles of butan-1-ol = 0.08(00) ✔
Moles of 1-bromobutane = 0.071(0) ✔
% yield = 88.8% ✔
Examiner’s Comments
This was a very well answered question and the
majority of responses were clearly laid out.
Consequently most of the candidates scored two or
three marks. Some candidates gave their final
answer to more than three significant figures,
despite the prompt in the question. Other candidates
decided to over-round the actual yield of 1-
bromobutane to one significant figure which led to a
yield of 87.5%.
Butane, C4H10, is a highly flammable gas, used as a fuel for camping stoves. Butane reacts with
oxygen as in the equation below:
C4H10(g) + 6.5O2(g) → 4CO2(g) + 5H2O(l)
i. The use of portable heaters in enclosed spaces can result in potential dangers if
incomplete combustion takes place.
Explain the potential danger of incomplete combustion.[1]
ii. A portable heater is lit to heat a room.
The heater burns 600 g of butane and consumes 1.50 m3 of O2, measured at room
temperature and pressure.
Determine whether this portable heater is safe to use.
Show all your working. [3]
i. CO is toxic
ii.Calculation:
n(butane) = 600/58.0 = 10.34 (mol)
AND n(O2) required = 6.5 × 10.34 = 67.2 (mol)
(1)
n(O2) consumed = 1.50 × 103 / 24.0 = 62.5 (mol)
OR
volume O2 required for complete combustion =
67.2 × 24.0/1000 = 1.61 m3 (1)
Conclusion:
incomplete combustion / stove not safe to use
AND
62.5 < 67.2 OR 1.61 > 1.50 (1)
