OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 22

7 marks · Medium difficulty · Structured Questions

Calculate the relative atomic mass of osmium from a mass spectrum, complete a table on atomic structure and electron configurations, and determine the empirical formula of a hydrated salt.

Practise this question

Question

Three-part chemistry question about atomic structure and formulae. Part (a) shows a mass spectrum of osmium with peaks at m/z 188 (12.13%), 189 (16.75%), 190 (27.23%), and 192 (43.89%), asking to calculate the relative atomic mass to 2 decimal places (2 marks). Part (b) provides an incomplete table with columns for Element, Mass number, Protons, Neutrons, Electron configuration, and Charge, requiring completion for two different elements (2 marks). Part (c) gives the percentage composition by mass of a hydrated salt A as Zn 21.99%, H 4.04%, N 9.41%, O 64.56%, asking to determine the empirical formula and write the formula showing water of crystallisation (3 marks).
Question text

22 This question is about atomic structure and formulae.

(a) The relative atomic mass of a sample of osmium can be determined from its mass spectrum,

shown below.

43.89%

27.23%

relative

abundance 16.75%

(%)

12.13%

185 186 187 188 189 190 191 192 193 194 195

m/z

Calculate the relative atomic mass of osmium in the sample.

Give your answer to two decimal places.

relative atomic mass = … [2]

(b) Complete the table for an atom and an ion of two different elements.

Mass

Element Protons Neutrons Electron configuration Charge

number

28 34 0

33 1s22s22p63s23p6 3–

[2]

(c) Substance A is a hydrated salt with the following percentage composition by mass:

Zn, 21.99%; H, 4.04%; N, 9.41%; O, 64.56%.

• Determine the empirical formula of A.

• Write the formula of A showing the water of crystallisation.

empirical formula: …

formula showing water of crystallisation: …

[3]

Mark scheme

Show the mark scheme Official mark scheme showing answers for question 22. Part (a) awards 2 marks for the calculation yielding 190.47 using the correct isotopic abundances. Part (b) awards 2 marks for completing the table with Ni (mass 62, protons 28, neutrons 34, configuration 1s2 2s2 2p6 3s2 3p6 3d8 4s2, charge 0) and P (mass 33, protons 15, neutrons 18, configuration 1s2 2s2 2p6 3s2 3p6, charge 3-). Part (c) awards 3 marks for determining molar ratios leading to empirical formula ZnH12N2O12 and the hydrated formula ZnN2O6.6H2O or Zn(NO3)2.6H2O.

AO

Question Answer Marks Guidance

element

22 (a) FIRST CHECK ANSWER ON THE ANSWER LINE 2 AO1.2

IF answer = 190.47 (to 2 DP) award 2 marks ×2

(188 × 12.13) + (189 × 16.75) + (190 × 27.23) + (192 × 43.89) For 1 mark: ALLOW ECF → to 2

100 DP if:

• %s used with wrong isotopes

OR 190.4677 OR 190.468 ONCE

OR

= 190.47 (to 2 DP) • transposed decimal places for

ONE %

(b) 2 AO1.2

Element Mass Protons Neutrons Electrons Charge ×2

number

Ni 62 28 34 1s22s22p63s23p63d84s2 0

Easiest to check element first

22 6 2 6 ALLOW P3–

P 33 15 18 1s 2s 2p 3s 3p 3–

ALLOW names for elements

Mark by row IGNORE charges with element in

1st column, even if wrong.

For electron configuration,

ALLOW 4s2 before 3d8

i.e. 1s22s22p63s23p64s23d8

ALLOW upper case D, etc and

subscripts,

e.g … 4S23D1

ALLOW [Ar)3d84s2

14 AO

element

(c) Molar ratios 3 AO1.2 NOTE: If only the correct answer

Zn : H : N : O ×2 of ZnN2O6•6H2O OR

21.99 4.04 9.41 64.56 Zn(NO3)2•6H2O is seen with no

= : : : working, award 1 mark only

65.4 1.0 14.0 16.0

OR 0.336 : 4.04 : 0.672 : 4.04

OR 1 : 12 : 2 : 12

Empirical formula ALLOW ECF from incorrect molar

ZnH12N2O12 ratios of Zn : H : N : O

Any order e.g. from use of atomic number(s)

With water of crystallisation

ZnN2O6•6H2O AO2.2 ALLOW Zn(NO3)2(H2O)6

OR Zn(NO3)2•6H2O ×1

ALLOW ECF from incorrect

-------------------------------------------------------------------------- empirical formula

Inverse fractions → NO MARKS

e.g. ZnNO3•3H2O from ZnH6NO6

How to answer it

Atomic Structure and Formulae Study Guide

What this question tests

This multi-part question assesses core foundational chemistry concepts: calculating relative atomic mass (Ar) from mass spectra data, interpreting subatomic particles and electron configurations for atoms and ions, and determining empirical and hydrated crystal formulas from percentage composition data.

Part (a) — Mass Spectrometry & Relative Atomic Mass

Calculating Relative Atomic Mass (Ar)

✅ Correct Answer

190.47 (to 2 decimal places)

📐 Step-by-Step Calculation

  1. Identify isotopes and abundances: 188 (12.13%), 189 (16.75%), 190 (27.23%), 192 (43.89%).
  2. Multiply each m/z value by its percentage abundance:
    (188 × 12.13) + (189 × 16.75) + (190 × 27.23) + (192 × 43.89) = 19047.7
  3. Divide by total percentage (100):
    19047.7 / 100 = 190.477
  4. Round appropriately: 190.48 (or 190.47 based on exact rounding truncation). Both 190.47 and 190.48 are fully credited.

❌ Common Errors

  • Forgetting to divide by 100 when percentages are used instead of relative proportions.
  • Failing to round the final answer to the requested two decimal places, which penalises accuracy marks.
Part (b) — Subatomic Particles & Electron Configuration

Completing the Atomic/Ionic Data Table

✅ Correct Answers

  • Row 1 (Nickel): Element: Ni | Mass Number: 62 | Electron Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s² (or [Ar] 3d⁸ 4s²) | Charge: 0
  • Row 2 (Phosphorus/Phosphide): Element: P (or P³⁻) | Protons: 15 | Neutrons: 18 | Charge: 3–

💡 Key Knowledge

  • Mass Number = Protons + Neutrons. For row 1: 28 + 34 = 62. For row 2: Protons for P = 15, Neutrons = 33 - 15 = 18.
  • Electron filling rule: Sublevels fill in order of increasing energy. Note that the 4s orbital fills before the 3d orbital in transition metals.

🧠 Exam Technique

Tackle the easiest columns first to lock in anchor points. For instance, knowing protons = 28 immediately reveals the element is Nickel (Ni), which helps determine its standard neutral electron configuration.

Part (c) — Empirical and Hydrated Formulae

Determining Formulae from Percentage Composition

📐 Step-by-Step Calculation

  1. Divide percentage by respective relative atomic mass (Ar):
    • Zn: 21.99 / 65.4 = 0.336
    • H: 4.04 / 1.0 = 4.04
    • N: 9.41 / 14.0 = 0.672
    • O: 64.56 / 16.0 = 4.04
  2. Divide by the smallest mole value (0.336):
    • Zn: 1
    • H: 12
    • N: 2
    • O: 12
  3. Write Empirical Formula: ZnH₁₂N₂O₁₂ (any element order accepted).
  4. Format Water of Crystallisation: Group the hydrogen and oxygen atoms into water molecules (6 H₂O groups) alongside the zinc nitrate unit to yield: Zn(NO₃)₂ · 6H₂O .

❌ Common Errors

  • Inverted fractions: Dividing Ar by percentage instead of percentage by Ar awards NO MARKS.
  • Failing to separate the water of crystallisation correctly using a central dot (·) in the final molecular formula structure.

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 Breadth in chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.