Edexcel A-Level Chemistry AS Paper 1, June 2018: Question 5
15 marks · Hard difficulty · Practical Techniques and Data Analysis
Analyse the preparation of a hydrated metal chloride from its carbonate, identify the metal using stoichiometry and flame testing, and calculate the percentage yield.
Practise this questionQuestion
Question text
5 A student made crystals of a metal chloride, JCl2.6H2O, by reacting the
metal carbonate, JCO3, with hydrochloric acid, HCl(aq). The product was purified.
Procedure
Step 1 150 cm3 of hydrochloric acid, concentration 0.80 mol dm−3, was transferred to a
400 cm3 conical flask. The flask was warmed gently using a Bunsen burner.
A spatula measure (about 1.0g) of metal carbonate was added to the acid.
Step 2 When the reaction in Step 1 was finished, more metal carbonate was added
until the metal carbonate was in excess.
Step 3 The resulting mixture was filtered into an evaporating basin.
Step 4 The evaporating basin was heated using a Bunsen burner to concentrate the
solution. The concentrated solution was allowed to cool and crystallise.
Step 5 Once crystal formation was complete, the resulting mixture was filtered for a
second time.
Step 6 The resulting white crystals were rinsed with a small volume of ice-cold water.
The equation for the reaction between the metal carbonate and hydrochloric acid is
JCO3(s) + 2HCl(aq) → JCl2(aq) + H2O(l) + CO2(g)
(a) (i) Describe two observations that the student might make which show that the
reaction in Step 1 has finished.
(2)
(ii) State the purpose of the filtration in Step 3.
(1)
(iii) Explain the use of a small volume of ice-cold water in Step 6.
(2)
… 12
… *P51459A01224*
(b) The student obtained a mass of 14.26g of hydrated crystals.
Assuming that the percentage yield is 100%, use the information in the procedure
to give a possible identity of J.
(5)
(c) The student was surprised by the white colour of the crystals of JCl2.6H2O in Step 6.
This did not agree with the possible identity for J from the calculation in (b).
The student decided to perform a flame test on the crystals.
(i) Explain why the student was surprised and decided to carry out a flame test.
(2)
(ii) The flame test colour was crimson red. Identify J.
(1)
A barium
B calcium
C lithium
D strontium
(iii) Calculate the actual percentage yield of the reaction, which produced 14.26g
of crystals.
Give your answer to two significant figures.
(2)
Mark scheme
Show the mark scheme
Question
Acceptable Answer Additional Guidance Mark
Number
5(a)(i) A description making reference to the following points: (2)
fizzing / effervescence stops (1) Allow stops frothing / no more bubbles
(all) metal carbonate / solid disappears (1) Allow metal carbonate / solid
“dissolved”
OR just ‘a clear solution forms’ for M2
Ignore colourless
Question
Acceptable Answer Additional Guidance Mark
Number
5(a)(ii) (1)
remove excess / unreacted metal carbonate Allow to remove excess / unreacted
solid
Allow “removes insoluble solid”
Ignore just “to remove impurities”
Question
Acceptable Answer Additional Guidance Mark
Number
5(a)(iii) An explanation that makes reference to the following (2)
points:
so as little product dissolves as possible (1) Allow product might dissolve in large
volumes / warm water
to remove any soluble impurities (1) Ignore rinse / wash / clean the
crystals
Ignore hydration of crystals
Question
Acceptable Answer Additional Guidance Mark
Number
5(b) Example of calculation: (5)
M1 calculate moles of acid (1) 150/1000 x 0.800 = 0.12(0) (mol)
M2 finds moles of JCl2 / 6H2O (1) 0.12 / 2 = 0.06(00) (mol)
Either
M3 finds M of JCl (1) M = 14.26 / 0.0600 = 237.7 (g mol-1)
r 2 r
M4 finds A of J (1) A = 237.7 - (71 + 108) = 58.7 (g mol-1)
r r
Or J is Ni
Allow TE for M5 on the Ar calculated
M3 finds mass of water and finds mass of
JCl2 by subtraction (1)
M4 finds mass and Ar of J (1) Mass of water = 0.06 x 6 x 18 = 6.48 (g)
Mass of JCl2 = 14.26 – 6.48 = 7.78 (g)
Mass of J = 7.78 – (0.06 x 71) = 3.52 (g)
A of J = 3.52 = 58.66667 / 58.7 (g mol-1)
r
0.06
Or
Mr of JCl2 = 7.78 = 129.6667 / 129.7
M5 identifies J (1) 0.06
A = 129.7 - 71 = 58.7 (g mol-1)
r
J is Ni
Allow TE for M5 on the Ar calculated
Ignore SF except 1SF
Question
Acceptable Answer Additional Guidance Mark
Number
5(c)(i) An explanation which makes the following points: (2)
M1
transition metals form coloured compounds / are not Allow any stated colour as long as the
normally white presence of a transition metal (in the
compound) is stated
or
crystals are white suggesting (compound of) an s-block Do not award compound of a group 1
element / group 2 element (1) element
M2
flame test to identify cation / metal ion (1)
Question
Answer Mark
Number
5(c)(ii) The only correct answer is D (1)
A is not correct because barium gives a green flame colour
B is not correct because calcium gives an orange-red flame colour
C is not correct because lithium is not in Group 2
Question
Acceptable Answer Additional Guidance Mark
Number
5(c)(iii) Example of Calculation (2)
Method 1 Correct answer with no working scores (2)
calculate molar mass of SrCl2.6H2O (1) 266.6
EITHER
calculates the percentage yield (1) 237.7 / 266.6 x 100 = 89.16%
= 89% (to 2 S.F.)
or
calculates maximum mass of SrCl2.6H2O and hence Maximum mass = 0.0600 x 266.6
percentage yield (1) = 15.996 (g)
Percentage yield
or = 14.26 x 100 = 89.147 %
15.996
= 89% (to 2 S.F.)
finds moles of SrCl2.6H2O and hence percentage yield 14.26 = 0.0534883 / 0.0535 (mol)
(1) 266.6
Moles of SrCO3 / SrCl2 (calculated in 5(b))
= 0.06 (mol)
Percentage yield
= 0.0534883 x 100 = 89.147 %
0.0600
= 89% (to 2 S.F.)
Allow TE on an incorrect choice of metal
only
(Total for Question 5 = 15 marks)
How to answer it
Preparation, Identification, and Yield of a Hydrated Salt
This 15-mark practical and quantitative analysis question assesses your core laboratory and theoretical skills:
- Inorganic synthesis observations: Identifying the end of an acid-carbonate reaction and understanding purification steps (filtration, washing with ice-cold solvent).
- Stoichiometric deduction: Working backwards from practical data to determine the formula mass of an unknown metal chloride hydrated salt ( JCl₂·6H₂O ) and identifying the metal J.
- Transition metal vs. Group 2 characteristics: Relating compound colour to electron configuration and using flame test colours for cation identification.
- Quantitative yield analysis: Calculating the percentage yield of a hydrated crystal to two significant figures.
Observations Showing Reaction Completion in Step 1
Recognising when an acid has reacted with a limited amount of carbonate
✅ Mark Scheme Answers
- Point 1: Fizzing / effervescence stops (or no more bubbles / frothing stops). [1 mark]
- Point 2: All the metal carbonate / solid dissolves / disappears (or a clear solution forms). [1 mark]
🧠 Exam Technique & Guidance
Remember that in Step 1, only about 1.0 g of metal carbonate was added to excess acid. Therefore, the solid should completely react and dissolve!
- Be precise: write "effervescence stops" rather than just "gas produced".
- Do not confuse Step 1 with Step 2 (where carbonate is in excess). In Step 1, no solid remains!
❌ Common Errors & Examiner Traps
- Stating solid remains: Stating that "solid sits at the bottom" refers to Step 2, not Step 1.
- Vague colour changes: Mentioning "turns colourless" is ignored unless linked to a clear solution forming without solid.
Purification Techniques in Salt Preparation
Filtration and Washing of Hydrated Crystals
✅ Acceptable Answers
Part (a)(ii) [1 mark]:
- To remove excess / unreacted metal carbonate (allow: remove excess / unreacted insoluble solid).
Part (a)(iii) [2 marks]:
- Ice-cold / small volume: So that as little product dissolves as possible / to minimise loss of crystals by dissolving. [1 mark]
- Water: To wash away / remove soluble impurities (e.g. unreacted acid). [1 mark]
💡 Key Knowledge
- Step 3 filtration: Hot gravity filtration separates the insoluble excess reagent ( JCO₃ ) from the aqueous product solution ( JCl₂ ).
- Step 6 washing: Hydrated salts are water-soluble. Ice-cold water lowers product solubility ( low T = low solubility ), while the small volume prevents significant dissolving.
❌ Common Errors
- (a)(ii): Simply writing "to remove impurities" is too vague and scores 0 marks. You must specify that it removes the unreacted / excess solid carbonate.
- (a)(iii): Forgetting the word soluble when explaining why water is used, or saying it "hydrates the crystals" (which is incorrect and ignored).
Deducing the Identity of Metal J
Step-by-Step Stoichiometric Calculation
📐 Calculation Breakdown
Step 1: Calculate moles of HCl used
n(HCl) = volume × concentration = (150 / 1000) dm³ × 0.80 mol dm⁻³ = 0.120 mol
Step 2: Calculate theoretical moles of JCl₂·6H₂O formed
From the balanced equation: 2 HCl ≡ 1 JCl₂ ≡ 1 JCl₂·6H₂O (ratio 2 : 1)
n(JCl₂·6H₂O) = 0.120 / 2 = 0.0600 mol
Step 3: Calculate the apparent molar mass (Mᵣ) of JCl₂·6H₂O
Assuming 100% yield, mass = 14.26 g:
Mᵣ(JCl₂·6H₂O) = mass / moles = 14.26 g / 0.0600 mol = 237.67 g mol⁻¹
Step 4: Calculate the relative atomic mass (Aᵣ) of metal J
Subtract mass of 2 chlorines and 6 water molecules:
Mass of non-metal parts = (2 × 35.5) + (6 × 18.0) = 71.0 + 108.0 = 179.0
Aᵣ(J) = 237.67 - 179.0 = 58.67 ≈ 58.7
Step 5: Identify the metal
Looking at the Periodic Table for an element with Aᵣ ≈ 58.7:
J is Nickel (Ni) (Aᵣ of Ni = 58.7).
❌ Common Mistakes
- Forgetting to divide moles of HCl by 2 (missing the 1:2 stoichiometric ratio).
- Forgetting to include the 6 waters of crystallisation ( 6 × 18.0 = 108.0 ) when finding Aᵣ.
- Rounding intermediate values excessively, leading to incorrect identification (e.g. confusing Ni with Co).
🧠 Exam Tip: Transfer of Error (TE)
Even if you made an arithmetic error in calculating Aᵣ, you can still gain M5 if you correctly identify the element on the Periodic Table that closest matches your calculated value!
Inorganic Analysis: Colour and Flame Testing
✅ (c)(i) Why the Student Was Surprised [2 marks]
- Reason for surprise [1 mark]: Nickel is a transition metal, and transition metal compounds are usually coloured (or nickel compounds are green), whereas the crystals obtained were white (indicating an s-block / Group 2 compound).
- Reason for flame test [1 mark]: A flame test is used to identify the metal cation (specifically group 1 or group 2 metal ions).
✅ (c)(ii) Flame Test Result [1 mark]
Correct Answer: D (strontium)
- A (barium): Produces an apple-green flame.
- B (calcium): Produces a brick-red / orange-red flame.
- C (lithium): Produces a crimson flame, but forms Li₂CO₃ and LiCl (Group 1, +1 oxidation state), not JCO₃ or JCl₂ (+2 state).
- D (strontium): Group 2 metal forming SrCO₃ and gives a crimson-red flame.
Calculating the Actual Percentage Yield
Percentage Yield using Strontium Chloride Hexahydrate
📐 Calculation Steps (Target: 2 Significant Figures)
Step 1: Calculate the true Mᵣ of hydrated strontium chloride, SrCl₂·6H₂O
Mᵣ(SrCl₂·6H₂O) = 87.6 + (2 × 35.5) + (6 × 18.0) = 87.6 + 71.0 + 108.0 = 266.6 g mol⁻¹
Step 2: Calculate theoretical maximum mass
From 5(b), maximum theoretical moles of salt = 0.0600 mol.
Theoretical mass = 0.0600 mol × 266.6 g mol⁻¹ = 15.996 g
Step 3: Calculate percentage yield
% Yield = (Actual Mass / Theoretical Mass) × 100
% Yield = (14.26 g / 15.996 g) × 100 = 89.147%
Step 4: Round to 2 significant figures (as explicitly requested!)
89.147% → 89%
💡 Alternative Calculation: Moles Ratio
Actual moles = 14.26 / 266.6 = 0.05349 mol
% Yield = (0.05349 / 0.0600) × 100 = 89.15% → 89%
❌ Significant Figures Penalty
The question strictly demands: "Give your answer to two significant figures."
- Writing 89.1% or 89.15% loses M2 completely.
- Always double-check significant figure requirements before moving to the next question!
Topics
Physical Chemistry · Inorganic Chemistry · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.