OCR A-Level Chemistry AS Breadth in chemistry (01), November 2021: Question 21

8 marks · Medium difficulty · Structured Questions

Complete tables on electron configurations and isotope subatomic particles, calculate relative atomic mass from mass spectrometry data, draw a halothene molecule, and calculate the number of fluorine atoms in a given mass.

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Question

Question text

21 This question is about atomic structure.

(a) Complete the table to show the maximum number of electrons that can occupy each shell

and sub-shell. Some boxes may need to be left blank.

Total number Sub-shell

Shell

of electrons s p d

1st

2nd

3rd

[2]

(b) Selenium, Se, has the atomic number 34.

76Se and 82Se are two isotopes of selenium.

Complete the table to show the numbers of protons, neutrons and electrons in these two

isotopes.

Protons Neutrons Electrons

76Se

82Se

[1]

The table shows the results of a mass spectrum of a sample of sulfur, S.

Isotope Abundance (%)

32S 94.93

33S 0.78

34S 4.29

Calculate the relative atomic mass of the sample of sulfur.

Give your answer to 3 decimal places.

relative atomic mass = … [2]

(d) Halothane, C2HBrClF3, (Mr = 197.4) is used as a general anaesthetic in medicine.

(i) The systematic name for halothane is 2-bromo-2-chloro-1,1,1-trifluoroethane.

Draw the structure of a halothane molecule.

[1]

(ii) What is the number of fluorine atoms in 7.896 g of halothane, C2HBrClF3?

number of fluorine atoms = … [2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

21 (a) 2 AO1.1

Sub-shell ×2

Total number of

Shell

electrons s p d

ALLOW

2 (1)s2

1st 2 2 6

(2)s (2)p

2nd 8 2 6 2 6 10

(3)s (3)p (3)d

3rd 18 2 6 10

DO NOT ALLOW extra numbers

1st 2 rows correct → 1 mark

3rd row correct → 1 mark

(b) 1 AO1.2

Protons Neutrons Electrons

76Se 34 42 34

82Se 34 48 34

ALL 6 entries correct for mark

IF answer = 32.094 (to 3 DP) award 2 marks ×2

(32 × 94.93) + (33 × 0.78) + (34 × 4.29) For 1 mark: ALLOW ECF → to 2 DP if:

100 • %s used with wrong isotopes ONCE

OR 32.0936 OR

• transposed decimal places for ONE %

= 32.094 (to 3 DP)

element

(d) (i) 1 AO2.5 ALLOW any combination of skeletal OR

structural OR displayed formula as long

as unambiguous, e.g. CF3CHClBr

(ii) FIRST, CHECK ANSWER 2 AO2.2 Alternative approaches

IF answer = 7.224 × 1022, award 2 marks ×2 7.896

n(F atoms) = 197.4 × 3 = 0.12

--------------------------------------------------------------------

7.896 F atoms = 0.12 × 6.02 × 1023

n(C2HBrClF3) = OR 0.04(00) (mol) 22

197.4 = 7.224 × 10

F atoms = 3 × 0.0400 × 6.02 × 1023 3 mol F atoms

= 3 × 6.02 × 1023 = 1.806 × 1024

= 7.224 × 1022 F atoms = 1.806 × 1024 × 0.04

Minimum 3 SF required = 7.224 × 1022

Mass F in 7.896 g

= × 7.896 = 2.28 (g)

197.4

2.28 23

F atoms = × 6.02 × 10

= 7.224 × 1022

ALLOW ECF from incorrect n(C2HBrClF3)

ALLOW use of 6.022 × 1023

OR 6.023 × 1023

-------------------------------------------------------

Common error

2.408 × 1022 OR 2.41 × 1022 → 1 mark

No × 3

1.806 × 1024 → 1 mark No n(C HBrClF )

Total 8

How to answer it

Atomic Structure and Moles Revision Guide

OCR AS Level Chemistry • Question 21

What this question tests

This multi-part question tests foundational chemical knowledge from Module 1 and Module 2 of the OCR AS specification. It covers electron shells and sub-shells, subatomic particle counting in isotopes, calculating relative atomic mass from abundance data, interpreting organic nomenclature into structural/skeletal formulas, and advanced amount-of-substance calculations involving Avogadro's constant.

Part (a): Electron Shells and Sub-shells

Question Part (a) [2 Marks]

✅ Correct Answer Table

Shell	Total e⁻	s	p	d
1st	2	2	-	-
2nd	8	2	6	-
3rd	18	2	6	10

• 1st 2 rows correct = 1 mark
• 3rd row correct = 1 mark

💡 Key Knowledge

Maximum electrons per shell follow the formula 2n² .
Sub-shells hold fixed maximum capacities: s = 2 , p = 6 , d = 10 .
The 1st shell only has an s sub-shell. The 2nd shell has s and p . The 3rd shell has s , p , and d .

❌ Common Errors

Leaving unnecessary boxes filled incorrectly instead of blank or writing zeros where sub-shells don't exist (though blank or dash is accepted, ensure sub-shells that do not exist for a given shell are left empty).
Miscalculating the 3rd shell total as 8 due to confusion with octet rules in bonding.

Part (b): Isotope Subatomic Particles

Question Part (b) [1 Mark]

✅ Correct Answer

Isotope	Protons	Neutrons	Electrons
⁷⁶Se	34	42	34
⁸²Se	34	48	34

All 6 entries must be correct to award the 1 mark.

🧠 Exam Technique

Protons = Atomic number ( Z ), which is 34 for Selenium.
Electrons = Number of protons in a neutral atom (also 34).
Neutrons = Mass number ( A ) minus Atomic number ( Z ). For ⁷⁶Se : 76 - 34 = 42. For ⁸²Se : 82 - 34 = 48.

Part (c): Relative Atomic Mass Calculation

Question Part (c) [2 Marks]

✅ Correct Answer

Final Answer: 32.094

First check answer line. If correct to 3 decimal places, award full 2 marks automatically.

📐 Step-by-Step Calculation

Identify formula: Ar = Σ(isotopic mass × abundance) / 100
Substitute values:
(32 × 94.93) + (33 × 0.78) + (34 × 4.29) / 100
Calculate numerator: 3037.76 + 25.74 + 145.86 = 3209.36
Divide by 100: 32.0936
Apply rounding: Round to 3 decimal places to get 32.094 .

❌ Common Errors & Examiner Guidance

Forgetting to divide by 100 when abundances are given as percentages.
Rounding too early in intermediate steps leading to accuracy loss on the final decimal place.
ECF policy: 1 mark max can be awarded if percentages were used with the wrong isotopes once, or if decimal places were transposed on one percentage.

Part (d)(i): Organic Structure Drawing

Question Part (d)(i) [1 Mark]

✅ Correct Answer

Structural, displayed, or skeletal formula showing 2-bromo-2-chloro-1,1,1-trifluoroethane unambiguously.

Displayed layout description:
Carbon-1 on the right bonded to three fluorine atoms ( -CF₃ ).
Carbon-2 on the left bonded to one Hydrogen ( H ), one Bromine ( Br ), one Chlorine ( Cl ), and bonded to Carbon-1.

Accept any unambiguous combination of skeletal, structural, or displayed formula (e.g., CF₃CHClBr).

🧠 Exam Technique

Check valencies carefully: Carbon must form exactly 4 bonds, Hydrogen 1, Halogens (F, Cl, Br) 1.
Ensure carbon-carbon single bonds connect the two carbons clearly without ambiguous bond connections.

Part (d)(ii): Amount of Substance Calculation

Question Part (d)(ii) [2 Marks]

✅ Correct Answer

Final Answer: 7.224 × 10²² (or 7.22 × 10²² )

First check answer line. Minimum 3 significant figures required.

📐 Step-by-Step Calculation

Calculate Molar Mass (Mr) of Halothane ( C₂HBrClF₃ ):
(2 × 12.0) + 1.0 + 79.9 + 35.5 + (3 × 19.0) = 197.4 g mol⁻¹
Find moles of halothane molecules:
Moles = mass / Mr = 7.896 / 197.4 = 0.0400 mol
Account for reacting ratio / stoichiometry:
Each molecule of C₂HBrClF₃ contains 3 fluorine atoms.
Moles of F atoms = 0.0400 × 3 = 0.120 mol
Convert moles of atoms to number of atoms:
Number = moles × Avogadro constant (6.022 × 10²³)
= 0.120 × 6.022 × 10²³ = 7.2264 × 10²² (rounds to 7.224 × 10²² depending on intermediate rounding).

❌ Common Errors & Examiner Traps

Forgetting to multiply by 3: Failing to account for the 3 fluorine atoms per molecule is the most frequent place students drop the second mark (resulting in 2.408 × 10²² , which scores 1 mark).
Significant figures: Answers given to 1 or 2 SF will lose marks; ensure at least 3 SF are maintained.

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 4.1 Basic concepts and hydrocarbons

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.