OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 25
8 marks · Medium difficulty · Calculations
Determine the empirical formula of scandium oxide from mass data, suggest a practical modification to ensure complete reaction, and calculate the molar mass and identity of a gas using the ideal gas equation.
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Question text
25 This question is about the analysis of unknown compounds.
(a) Scandium (atomic number 21) reacts with oxygen to form an oxide of scandium.
A student carries out an experiment to determine the empirical formula of the scandium
oxide.
A diagram of the apparatus used by the student is shown below.
crucible
pipeclay triangle
The student’s method is outlined below.
• Weigh an empty crucible.
• Add scandium to the crucible and reweigh.
• Heat the crucible and contents for 10 minutes.
• Allow to cool and reweigh.
The student’s results are shown below.
Mass of crucible / g 12.165
Mass of crucible + scandium / g 12.435
Mass of crucible + scandium oxide / g 12.579
(i) Determine the empirical formula of the scandium oxide.
empirical formula = … [2]
(ii) The student was unsure that all of the scandium had reacted.
Suggest one modification that the student could make to their method to be confident
that all the scandium had reacted. Explain your reasoning.
… [1]
(b) A gas cylinder has a gas volume of 9.39 dm3.
The gas cylinder holds 1.69 kg of a gas at a pressure of 1.37 × 107 Pa at 20 °C.
Determine the molar mass and possible identity of the gas.
molar mass = … g mol−1
identity of gas = …
[5]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
25 (a) (i) Moles Sc OR moles O 2 AO2.8
0.27 –3 ×2
n(Sc) = = 6 × 10 (mol)
OR
0.144 –3
n(O) = = 9 × 10 (mol)
16.0
Empirical formula
Sc2O3 NO ECF
(a) (ii) Heat to constant mass 1 AO3.4 ALLOW response that implies heating to
constant mass,
e.g. Heat again until mass does not change
IGNORE ‘heat for longer’
No link to constant mass
(b) Rearranging ideal gas equation 5 ALLOW ECF throughout
pV AO1.2 pV
n = RT IF n = RT is omitted, ALLOW when values are
pV substituted into rearranged ideal gas equation.
Unit conversion AND substitution into n = RT :
• R = 8.314 OR 8.31 ALLOW ECF from incorrectly rearranged ideal
• V = 9.39 × 10–3 m3 RT
gas equation, e.g. n = pV → 0.0189361411
• T in K: 293 K
1.37 × 107 × 9.39 × 10–3 M → 89247 (Likely to be 3/5 max)
e.g. n = AO2.4
8.314 × 293
×3 ALLOW use of 8.31 for R, which gives:
Calculation of n
n = 52.80906994 (mol) n = 52.83448947
M = 31.98668175
Calculation of M
1.69 × 103 ALLOW 3 SF or more, e.g. 52.8
M = = 32.00207847
52.80906994
Using 52.8, M = 32.00757576
ALLOW 2 SF or more
ALLOW ECF for a ‘reasonable gas’ that matches
Gas AO3.2
calculated molar mass
O2 OR oxygen
How to answer it
Analysis of Unknown Compounds Study Guide
This question assesses core quantitative chemistry skills required for AS Level: calculating empirical formulas from mass data obtained via combustion/reaction in a crucible, understanding practical improvements for complete reactions, and applying the ideal gas equation ( pV = nRT ) with strict unit conversions to determine molar mass and identify an unknown gas.
Part (a)(i): Empirical Formula Determination
Determine the empirical formula of the scandium oxide.
✅ Correct Answer
Sc₂O₃
💡 Key Knowledge
- Mass of Sc = (Crucible + Sc) - (Empty crucible) = 12.435 - 12.165 = 0.27 g
- Mass of O = (Crucible + Scandium oxide) - (Crucible + Sc) = 12.579 - 12.435 = 0.144 g
📐 Step-by-Step Calculation
- Find moles of Sc: 0.27 / 45.0 = 6 × 10⁻³ mol
- Find moles of O: 0.144 / 16.0 = 9 × 10⁻³ mol
- Find simplest ratio: Divide by smallest ( 6 × 10⁻³ ): Sc = 1 , O = 1.5
- Scale up to whole numbers: Multiply by 2 to get Sc : O = 2 : 3
❌ Common Errors
- Using the mass of the crucible directly in molar mass calculations instead of finding the difference.
- Dividing the wrong way around when determining the ratio (e.g., O moles divided by Sc moles instead of using standard molar ratio steps).
Part (a)(ii): Practical Modifications
Suggest one modification to ensure complete reaction. Explain your reasoning.
✅ Correct Answer
Modification: Heat to constant mass.
Reasoning: Reheat the crucible and contents until two consecutive weighings give the same mass, proving all the scandium has fully reacted.
🧠 Exam Technique
Always use standard chemical terminology in practical descriptions. "Heat to constant mass" is a mandatory phrase expected by examiners when determining if a reaction has gone to completion in thermal decomposition or combustion setups.
Part (b): Ideal Gas Equation & Gas Identification
Determine the molar mass and possible identity of the gas.
📐 Step-by-Step Calculation
- Rearrange ideal gas equation: n = pV / RT
- Convert units:
• p = 1.37 × 10⁷ Pa
• V = 9.39 dm³ = 9.39 × 10⁻³ m³
• T = 20 °C + 273 = 293 K
• R = 8.314 (or 8.31) J mol⁻¹ K⁻¹ - Substitute values for moles (n):
n = (1.37 × 10⁷ × 9.39 × 10⁻³) / (8.314 × 293) = 52.81 mol - Calculate Molar Mass (M = mass / n):
Mass in grams = 1.69 kg = 1690 g
M = 1690 / 52.81 = 32.0 g mol⁻¹
✅ Correct Answer & Identity
Molar Mass: 32.0 g mol⁻¹ (Accept 32 or 32.0)
Identity: O₂ OR oxygen
❌ Common Calculation Traps
- Volume units: Forgetting to convert dm³ to m³ by multiplying by 10⁻³ when pressure is in Pascals ( Pa ).
- Temperature: Forgetting to add 273 to convert Celsius to Kelvin.
- Mass units: Forgetting to convert kg to g ( 1.69 kg = 1690 g ) when calculating molar mass.
Topics
Module 2: Foundations in chemistry · Practical Activity Groups · PAG 1: Moles determination · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.