A Level Chemistry: Amount of Substances Exam Questions
Q1.
A student does an investigation to determine the relative formula mass, Mr, of a solid
unknown diprotic acid, H2A
H2A + 2 NaOH → Na2A + 2 H2O
•
•
250 cm3 of aqueous solution are prepared using 1300 mg of H2A
A pipette is used to add 25.0 cm3 of 0.112 mol dm–3 aqueous sodium hydroxide to a
conical flask.
This aqueous sodium hydroxide is titrated with the acid solution.
•
The titration results are shown in the table.
Rough
1
2
3
Final volume / cm3
27.35
26.75
38.90
35.70
Initial volume / cm3
0.00
0.35
12.15
9.20
Titre / cm3
27.35
26.40
26.75
26.50
(a)
Use the results to calculate the Mr of H2A
Mr of H2A _______________________
(5)
(b)
The uncertainty in using the pipette in this experiment is ±0.06 cm3
Calculate the percentage uncertainty in using the pipette.
% uncertainty _________________________
(1)
Page 1 of 16
(c)
Before adding the solution from the burette in the rough titration, there was an air
bubble below the tap.
At the end of this titration the air bubble was not there.
Explain why this air bubble increases the final burette reading of the rough titration.
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(1)
(d)
During the titration the student washed the inside of the conical flask with some
distilled water.
Suggest why this washing does not give an incorrect result.
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(1)
Q2.This question is about the reactions of magnesium and its compounds.
(a) Magnesium is used in one of the stages in the extraction of titanium. Give an equation for the
reaction between titanium(IV) chloride and magnesium. State the role of magnesium in this
reaction.
Equation
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Role of magnesium ___________________________________________________
(2)
(b)
A mixture of magnesium oxide and magnesium hydroxide has a mass of 3200 mg
This mixture is reacted with carbon dioxide to form magnesium carbonate and
water. The mass of water produced is 210 mg
Mg(OH)2 + CO2 → MgCO3 + H2O
MgO + CO2 → MgCO3
Calculate the percentage by mass of magnesium oxide in this mixture.
% of magnesium oxide ____________________________
(4)
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Q3. A sample of strontium ore is known to contain strontium oxide, strontium carbonate and some
inert impurities. To determine the mass of strontium carbonate present, a student weighed a sample
of the solid ore and then heated it in a crucible for 5 minutes. The sample was allowed to cool and
then reweighed. This heating, cooling and reweighing was carried out three times.
The results are set out in the table.
(a)
Mass of crucible / g
9.85
Mass of crucible and ore sample / g
16.11
Mass of crucible and sample after first
heating / g
14.66
Mass of crucible and sample after second
heating / g
14.58
Mass of crucible and sample after third
heating / g
14.58
When strontium carbonate is heated it decomposes according to the following
equation.
SrCO3 ⟶ SrO + CO2
Give a reason why the mass of the solid sample changed during the experiment.
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(1)
(b)
Use the data in the table to calculate the mass of strontium carbonate in the original
ore sample. Give your answer to an appropriate precision.
Mass of strontium carbonate = ___ g
(5)
(c)
Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.
Percentage error = ____________ %
(1)
Page 3 of 16
(d)
The mass of inert impurities in the sample was 347 mg.
Deduce the mass of SrO in the sample and justify any assumption made in
calculating your answer.
(If you have been unable to answer part (b), assume the mass of strontium
carbonate was 4.85 g. This is not the correct answer.)
Mass of SrO = _______________
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(2)
(e)
Strontium metal can be extracted by heating strontium oxide with aluminium metal.
In this reaction, strontium vapour and solid aluminium oxide are formed.
Write an equation for the reaction and state the role of the aluminium in the process.
Explain why strontium forms a vapour but aluminium oxide is formed as a solid.
Equation
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Role of aluminium ___________________________________________________
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Explanation ________________________________________________________
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(5)
(Total 14 marks)
Page 4 of 16
Q4.This question is about reactions of calcium compounds.
(a)
A pure solid is thought to be calcium hydroxide. The solid can be identified from its
relative formula mass.
The relative formula mass can be determined experimentally by reacting a
measured mass of the pure solid with an excess of hydrochloric acid. The equation
for this reaction is
Ca(OH)2 + 2HCl
CaCl2 + 2H2O
The unreacted acid can then be determined by titration with a standard sodium
hydroxide solution.
You are provided with 50.0 cm3 of 0.200 mol dm−3 hydrochloric acid.
Outline, giving brief practical details, how you would conduct an experiment to
calculate accurately the relative formula mass of the solid using this method.
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(8)
(b)
A 3.56 g sample of calcium chloride was dissolved in water and reacted with an
excess of sulfuric acid to form a precipitate of calcium sulfate.
The percentage yield of calcium sulfate was 83.4%.
Calculate the mass of calcium sulfate formed.
Give your answer to an appropriate number of significant figures.
Mass of calcium sulfate formed = ____________ g
(3)
Page 5 of 16
Q5.
Phosphoric(V) acid (H3PO4) is an important chemical. It can be made by two methods.
The first method is a two-step process.
(a)
In the first step of the first method, phosphorus is burned in air at 500 ºC to produce
gaseous phosphorus(V) oxide.
P4(s)
+
5O2(g)
→
P4O10(g)
220 g of phosphorus were reacted with an excess of air.
Calculate the volume, in m3, of gaseous phosphorus(V) oxide produced at a
pressure of 101 kPa and a temperature of 500 ºC.
The gas constant R = 8.31 J K–1 mol–1
Give your answer to 3 significant figures.
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(4)
(b)
In the second step of the first method, phosphorus(V) oxide reacts with water to
form phosphoric(V) acid.
P4O10(s)
+
6H2O(l)
→
4H3PO4(aq)
Calculate the mass of phosphorus(V) oxide required to produce 3.00 m3 of
5.00 mol dm–3 phosphoric(V) acid solution.
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(3)
Page 6 of 16
(c)
In the second method to produce phosphoric(V) acid, 3.50 kg of Ca3(PO4)2 are
added to an excess of aqueous sulfuric acid.
Ca3(PO4)2(s)
+
3H2SO4(aq)
→
2H3PO4(aq)
+
3CaSO4(s)
1.09 kg of phosphoric(V) acid are produced.
Calculate the percentage yield of phosphoric(V) acid.
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(4)
(d)
Explain whether the first method or the second method of production of phosphoric
acid has the higher atom economy.
You are not required to do a calculation.
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(1)
(Total 12 marks)
Page 7 of 16
Q6.
(a)
Calcium phosphate reacts with aqueous nitric acid to produce phosphoric acid and
calcium nitrate as shown in the equation.
Ca3(PO4)2
(i)
+
6HNO3
2H3PO4
+
3Ca(NO3)2
A 7.26 g sample of calcium phosphate reacted completely when added to an
excess of aqueous nitric acid to form 38.0 cm3 of solution.
Calculate the concentration, in mol dm–3, of phosphoric acid in this solution.
Give your answer to 3 significant figures.
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(5)
(ii)
Calculate the percentage atom economy for the formation of calcium nitrate in
this reaction.
Give your answer to 1 decimal place.
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(2)
(b)
Write an equation to show the reaction between calcium hydroxide and phosphoric
acid to produce calcium phosphate and water.
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(1)
Page 8 of 16
(c)
Calcium dihydrogenphosphate can be represented by the formula Ca(H2PO4)x
where x is an integer.
A 9.76 g sample of calcium dihydrogenphosphate contains 0.17 g of hydrogen, 2.59
g of phosphorus and 5.33 g of oxygen.
Calculate the empirical formula and hence the value of x.
Show your working.
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(4)
(Total 12 marks)
Page 9 of 16
Mark schemes
Q1.
(a)
Average titre = 26.45 cm3
M1 = average of concordant titres
1
n(NaOH) = (25 × 0.112 / 1000) = 2.80 × 10–3 mol
M2 – this value only
1
n(acid in titre) = 2.80 × 10–3 / 2 = 1.40 × 10–3 mol
M3 = M2/2
1
n(acid in 250 cm3) = 1.40 × 10–3 × 250/26.45 = 0.0132 mol
M4 = M3 × 250/M1
1
Mr = mass / moles = 1.300/0.0132 = 98.2-98.5
M5 = (1.300/M4) = answer
Mr must be given to at least 1dp
1
Alternatives:
98.6 – scores 4
92.9 – scores 4
87.8 – scores 3
49.3 – scores 3
49.1 – scores 4
(b)
% uncertainty = 0.06/25.0 × 100 = 0.24 %
1
(c)
Some solution/acid replaces air bubble /
Solution/acid fills below the tap /
Air bubble takes up volume that would be filled by solution/acid
Score for the idea that:
Acid/solution replaces air/bubble/fills jet space
Allow acid/solution fills the bubble/gap
‘The final reading is higher than the volume added’ is not enough.
1
(d)
Does not react (with the alkali) / does not change the number of moles (of
alkali)
Allow water is a product / water is not a reagent
1
[8]
Q2.
(a)
Equation: 2 Mg + TiCl4 → Ti + 2 MgCl2
Allow multiples / ignore ss
Page 10 of 16
1
Role: Reducing agent
Allow electron donor (not electron pair donor)
1
(b)
M1: moles of water in 210 mg
= mass / mr = 0.210 / 18
= 0.0117 mol ONLY
Equal to moles of magnesium hydroxide produced in stage one
M2: mass of Mg(OH)2 = 0.0117 x 58.3 = 0.680 g
M3: mass of MgO
=
3.2 – 0.68
= 2.52 g
M1 = moles of water
M2 = mass of Mg(OH)2 = M1 x 58.3
M3 = subtraction = 3.2 – M2
M4 = answer to M3 x 100/3.2
Accept correct alternative methods such as
M1 = moles of water
M2 = mass of Mg(OH)2 = M1 x 58.3
M3 = M2 x 100/3.2
M4 = 100 – M3
M4: % of MgO = 2.52/3.2 x 100 = 78.7%
M4: Allow 78.7 – 78.8 or 79 %
4
[6]
Q3.
(a)
CO2 gas escapes or is lost
1
(b)
Mass CO2 = 16.11 − 14.58 = 1.53 g
1
Mr CO2 = 44.0
Mol CO2 = 1.53 / 44.0 = 3.48 × 10−2
1
Mol SrCO3 = 3.48 × 10−2
1
Mass SrCO3 = mol × Mr = 3.48 × 10−2 × 147.6
Mass SrCO3 = 5.13 (g)
1 mark for the answer and 1 for 3 sf precision
Allow 5.14 g (as a result of rounding)
2
(c)
Percentage error =
Page 11 of 16
= 0.160 (%)
1
(d)
Original Mass SrO = 6.26 − 0.347 − 5.13
= 0.783 g (or 783 mg)
OR 6.26 − 0.347 − 4.85 = 1.063 g
Allow 0.773 g or 773 mg (from rounding error in part (b)
1
Justification: All SrCO3 reacted because heated to constant mass.
1
(e)
3SrO + 2Al ⟶ Al2O3 + 3Sr
1
Al acts as a reducing agent
1
Sr is collected as a vapour because
1
Al2O3 is an ionic lattice and so has strong ionic attractions
1
Than Sr which is a metallic structure with (relatively) weaker bonding
1
[14]
Q4.
(a)
Stage 1: appreciation that the acid must be in excess and calculation of amount of
solid that permits this
Statement that there must be an excess of acid
1
Moles of acid = 50.0 × 0.200 / 1000 = 1.00 × 10–2 mol
1
2 mol of acid react with 1 mol of calcium hydroxide therefore moles of solid
weighed out must be less than half the moles of acid = 0.5 × 1.00 × 10–2 =
5.00 × 10–3 mol
1
Mass of solid must be –3 × 74.1 =
1
Stage 2: Experimental method
Measure out 50 cm3 of acid using a pipette and add the weighed amount of
solid in a conical flask
1
Titrate against 0.100 (or 0.200) mol dm–3 NaOH added from a burette and
record the volume (v) when an added indicator changes colour
1
Stage 3: How to calculate Mr from the experimental data
Page 12 of 16
Moles of calcium hydroxide = 5.00 × 10–3 – (v/2 × conc NaOH) / 1000 = z mol
1
Mr = mass of solid / z
1
Extended response
Maximum of 7 marks for answers which do not show a
sustained line of reasoning which is coherent, relevant,
substantiated and logically structured.
(b)
Moles of calcium chloride = 3.56 / 111.1 = 3.204 × 10–2
1
Moles of calcium sulfate = 3.204 × 10–2 × 83.4 / 100 = 2.672 × 10–2
1
Mass of calcium sulfate = 2.672 × 10–2 × 136.2 = 3.6398 = 3.64 (g)
Answer must be to 3 significant figures
1
[11]
Q5.
(a)
Correct conversion of temperature and pressure (773 and 101 × 103)
Correct answer with or without working scores 4 marks
1
No moles P = (220 / 4 × 31.0) = 1.77
Max 2 (M1 and M3) if 31.0 used
(=0.451 m3 or if 220/31 rounded to 2 sf ie 7.1 then 0.452)
1
V = nRT/P (correct rearrangement or insert of values V =
1.77 × 8.31 × 773/101 × 103 =0.1128 m3)
Max 2 (M1 and M3) if 284 (P4O10) used then 0.0493
1
V = 0.113 (m3)
Must be 3 sig figs
1
(b)
No moles H3PO4 = 3 × 103 (dm3) × 5 = 15,000 (mols)
Correct answer with or without working scores 3 marks
If M1 incorrect then can only score M2
1
No moles phosphorus(V) oxide =
( = 3,750 mols)
If M2 incorrect can only score M1
1
1.1 × 106 or1.07 × 106 or 1.065 × 106 (g)
or 1,100 or 1,070 or 1065 kg
or 1.1 or 1.07 or 1.065 tonne
= (3.75 × 103 × 284.0)
Min 2 sig fig
1
Page 13 of 16
(c)
No moles Ca3(PO4)2 (= 3.50kg =)
= 11.28
Correct answer with or without working scores 4 marks
If M1 incorrect can only score M2 and M3
1
Theoretical No. moles H3PO4 = 11.28 × 2 = 22.56
If M2 incorrect can only score M1 and M3
1
Theoretical mass H3PO4 = 22.56 × 98(.0) = 2211
If M3 incorrect can only score M1and M2
1
or Actual No. moles H3PO4 produced =
49 – 49(.312) (%)
= 11.12
1
(d)
Method 1 / (a) & (b) because only one product / no other
products formed / atom economy = 100% (even though two
steps)
Allow calculations
Do not allow if P2O5 is formed
Allow converse explanation
1
[12]
Q6.
(a)
(i)
M1 - Mr calcium phosphate = 310(.3)
If Mr wrong, lose M1 and M5.
1
M2 - Moles calcium phosphate =
(= 0.0234)
0.0234 moles can score M1 and M2.
If Mr incorrect, can score M2 for
.
Allow M2 and / or M3 to 2 significant figures here but will
lose M5 if answer not 1.23.
1
M3 - Moles phosphoric acid = 2 × 0.0234 = 0.0468
Allow student’s M2 × 2. If not multiplied by 2 then lose M3
and M5.
1
M4 - Vol phosphoric acid = 0.038(0) dm3
If not 0.038(0) dm3 then lose M4 and M5.
1
Page 14 of 16
Conc phosphoric acid =
M5 = 1.23 (mol dm−3)
This answer only – unless arithmetic or transcription error
that has been penalised by 1 mark.
Allow no units but incorrect units loses M5.
1
× 100
(ii)
× 100
OR
1 mark for both Mr correctly placed.
= 71.5%
2
(b)
3Ca(OH)2 + 2H3PO4
Allow multiples.
Ca3(PO4)2 + 6H2O
1
(c)
If x = 2 with no working, allow M4 only.
Ca = 1.67 g (M1).
1
Mark for dividing by correct Ar in Ca and P (M2).
If M1 incorrect can only score M2.
1
Correct ratio (M3).
1
CaH4P2O8
OR Ca(H2PO4)2 OR x = 2
Value of x or correct formula (M4).
1
Alternative
Ca
H2PO4
Ca = 1.67 g (M1).
Mark for dividing by correct Ar / Mr in Ca and H2PO4 (M2).
If M1 incorrect can only score M2.
Correct ratio (M3).
CaH4P2O8
OR
Ca(H2PO4)2
OR
x=2
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Value of x or correct formula (M4).
[12]
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