OCR A-Level Chemistry AS Depth in chemistry (02), June 2023: Question 1

11 marks · Medium difficulty · Structured Questions

Calculate the relative atomic mass of titanium, determine electron configuration and ion subatomic particles, calculate mass of sodium needed for titanium extraction, and describe separation of titanium from sodium chloride.

Practise this question

Question

Question text

1 This question is about titanium (atomic number 22) and its compounds.

(a) Titanium exists as a mixture of five isotopes.

A chemist analyses a sample of titanium using mass spectrometry.

The results are shown in the table below.

Isotope Abundance (%)

46Ti 8.30

47Ti 7.40

48Ti 73.70

49Ti 5.40

50Ti 5.20

(i) Calculate the relative atomic mass of titanium in the sample.

Give your answer to 2 decimal places.

relative atomic mass = … [2]

(ii) Complete the electron configuration of a titanium atom.

1s2 … [1]

(iii) Complete the table to show the number of protons, neutrons and electrons in a

48Ti2+ ion.

Protons Neutrons Electrons

48Ti2+ ion

[1]

(b) An ore of titanium contains impure TiO2.

Titanium is manufactured from TiO2 in a two-stage process.

Stage 1 TiO2 + 2C + 2Cl2 TiCl4 + 2CO Reaction 1.1

Stage 2 TiCl4 + 4Na Ti + 4NaCl Reaction 1.2

(i) The common name for TiO2 is titanium dioxide.

What is the systematic name of TiO2?

… [1]

(ii) In Reaction 1.2, the percentage yield of titanium from TiCl4 is 72.0%.

Calculate the minimum mass, in kg, of sodium that is needed to produce 1.00 kg of

titanium.

Give your answer to 3 significant figures.

mass of sodium = … kg [4]

(iii) Reaction 1.2 produces a mixture of titanium and sodium chloride.

Suggest how titanium could be separated from this mixture at room temperature.

Explain your answer.

… [2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

1 (a) (i) FIRST CHECK ANSWER ON ANSWER LINE 2 ALLOW one mark for ECF from seen

If answer = 47.92 (to 2 DP) seen award 2 marks AO2.2 incorrect sum provided final answer

between 46 and 50 and to 2 DP

(46 x 8.3) + (47 x 7.4) + (48 x 73.7) + (49 x 5.4) + (50 x 5.2)

381.8 + 347.8 + 3537.6 + 246.6 + 260 AO1.1

4791.8 ✓

4791.8/100

= 47.92 ✓ 2DP required

(ii) (1s2)2s22p63s23p63d24s2 ✓ 1 AO1.1 ALLOW subscripts

Look carefully at (1s2) 2s22p63s23p6 ALLOW 4s before 3d i.e.

– there may be a mistake (1s2)2s22p63s23p64s23d2

ALLOW upper case D, etc and subscripts,

e.g … 3S 3P6

DO NOT ALLOW [Ar] as shorthand for

1s22s22p63s23p6

(iii) Protons Neutrons Electrons 1 AO2.1

48Ti2+ 22 26 20 ✓

ALL 3 numbers required for the mark

(b) (i) Titanium (IV) oxide ✓ 1 AO2.5 DO NOT ALLOW titanium dioxide

element

1 (b) (ii) FIRST CHECK ANSWER ON ANSWER LINE 4 AO2.2 ALLOW ECF throughout

If answer = 2.67 kg award 4 marks × 4 TAKE CARE: values shown may be

-------------------------------------------------- truncated calculator values.

1000

n(Ti) = 47.9 OR 20.8768… (mol) ✓ Steps can be calculated in any order

which will change the intermediate

n(Na) for 72% yield = 20.88 × 4 OR 83.5073… (mol) ✓ answers. Marks are for the processing of

the data.

n(Na) for 100% yield = 83.51 × 72 OR 115.98237… (mol) ✓ ALLOW 3SF up to calculated value

throughout

mass Na = 115.98 x 23.0 = 2667.659… (g)

= 2.67 (kg) ✓ IGNORE rounding errors past 3SF

3 SF AND kg required

Common Errors for 3 marks:

1.92 (missing yield )

1.38 (yield wrong way round)

0.673 (use of Mr 189.9 for TiCl4 instead

47.9 for Ti)

(iii) Add water AND filter ✓ 2 AO3.3 ALLOW dissolve in water

× 2

Ti does not dissolve OR NaCl does dissolve ✓ ALLOW Ti is insoluble OR NaCl is

soluble/aqueous

ALLOW Ti is the residue OR NaCl is the

filtrate

How to answer it

Titanium and its Compounds — AS Level Chemistry Study Guide

What this question tests

This multi-part question assesses foundational atomic structure and chemical calculations. Core skills tested include calculating relative atomic mass from isotopic abundances, writing sub-shell electron configurations, determining subatomic particle counts in ions, applying IUPAC nomenclature for transition metal oxides, executing complex multi-step stoichiometry involving percentage yield and mass conversions (kg to g), and designing practical separation techniques based on physical solubility properties.

Question 1 (a) (i)

Relative Atomic Mass Calculation

✅ Correct Answer

47.92

Marks: 2 marks (AO1.1 / AO2.2)

📐 Step-by-Step Calculation

Multiply mass by percentage abundance for each isotope:
(46 × 8.30) + (47 × 7.40) + (48 × 73.70) + (49 × 5.40) + (50 × 5.20) = 381.8 + 347.8 + 3537.6 + 246.6 + 260 = 4791.8
Divide by 100:
4791.8 / 100 = 47.918
Round to required precision:
47.92 (Strictly 2 decimal places)

❌ Common Errors

Failing to give the final answer to 2 decimal places as explicitly requested in the stem.
Dividing by numbers other than 100 when percentages are used.

Question 1 (a) (ii)

Electron Configuration of a Titanium Atom

✅ Correct Answer

1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²

(Note: 4s² 3d² is also accepted by the mark scheme)

Marks: 1 mark (AO1.1)

💡 Key Knowledge

Titanium has atomic number 22, meaning it has 22 electrons. Remember that the 4s sub-shell fills before the 3d sub-shell due to energy level ordering, but both conventions are commonly accepted at AS level.

❌ Common Errors

Using noble gas shorthand like [Ar] when the question explicitly starts the line with 1s² .
Miscounting electrons (e.g., leaving a sub-shell partially filled incorrectly or miscounting p-block electrons as 5 instead of 6).

Question 1 (a) (iii)

Subatomic Particles in an Ion

✅ Correct Answer

Protons	Neutrons	Electrons
22	26	20

Marks: 1 mark — all 3 numbers required (AO2.1)

🧠 Exam Technique

Protons: Equal to the atomic number (22).
Neutrons: Mass number minus atomic number (48 - 22 = 26).
Electrons: Atomic number minus charge for cations (22 - 2 = 20 electrons).

Question 1 (b) (i)

Systematic Nomenclature

✅ Correct Answer

Titanium(IV) oxide

Marks: 1 mark (AO2.5)

❌ Common Errors

Writing "titanium dioxide" instead of the requested systematic name containing Roman numerals.
Incorrect Roman numeral formatting or missing brackets.

Question 1 (b) (ii)

Multi-Step Stoichiometry and Percentage Yield Calculation

✅ Correct Answer

2.67 kg

Marks: 4 marks (AO2.2 × 4)

📐 Step-by-Step Calculation

Find moles of Ti needed:
Mass in grams = 1.00 kg × 1000 = 1000 g.
Moles of Ti = 1000 / 47.9 = 20.8768 mol.
Use reacting ratios from Equation 1.2:
Equation shows TiCl₄ + 4Na → Ti + 4NaCl (1 mole of Ti requires 4 moles of Na).
Moles of Na for 100% yield = 20.8768 × 4 = 83.5073 mol.
Account for percentage yield (72.0%):
Since yield is less than 100%, more reactant must be added.
Adjusted moles = 83.5073 × (100 / 72.0) = 115.982 mol of Na.
Convert moles of Na back to mass in kg:
Mass = Moles × Ar (Na = 23.0) = 115.982 × 23.0 = 2667.659 g = 2.67 kg (to 3 SF).

❌ Common Errors

Yield direction trap: Multiplying by 0.72 instead of dividing by 0.72 (applying the yield the wrong way round).
Unit conversion errors: Forgetting to convert kg to g at the start or end, leading to magnitude errors.
Using incorrect molar masses (e.g., using Mr of TiCl₄ instead of Ar for Ti).

Question 1 (b) (iii)

Separation Technique

✅ Correct Answer

Add water (to dissolve the sodium chloride mixture).
Filter (to separate solid titanium from the aqueous solution).

Marks: 2 marks (AO3.3 × 2)

💡 Key Knowledge & Explanation

Titanium is a metal and is insoluble in water, whereas sodium chloride (NaCl) is an ionic salt that is soluble in water. Adding water creates a solution of NaCl while leaving titanium as a solid residue that can be isolated via filtration.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.