Atomic Structure & Mass Spectrometry (OCR AS Chemistry A): Revision Note

Exam code: H032

Richard Boole

Written by: Richard Boole

Reviewed by: Philippa Platt

Updated on

Isotopes & relative atomic mass

  • Isotopes are different atoms of the same element that contain:

    • The same number of protons (and electrons)

    • A different number of neutrons

  • Isotopes are also described as atoms of the same element but with different mass numbers

  • The mass of an element is given as relative atomic mass (Ar)

    • This is the weighted average mass of its isotopes

  • The relative atomic mass of an element can be calculated from its relative isotopic abundances, using the following steps:

    1. Multiply each isotope’s mass by its abundance

    2. Add the results together

    3. Divide by the total abundance (usually 100)

  • The equation for this is:

fraction numerator open parentheses r e l a t i v e space a b u n d a n c e subscript i s o t o p e space 1 end subscript space cross times m a s s subscript i s o t o p e space 1 end subscript close parentheses space plus open parentheses r e l a t i v e space a b u n d a n c e subscript i s o t o p e space 2 end subscript space cross times m a s s subscript i s o t o p e space 2 end subscript close parentheses space e t c over denominator 100 end fraction

  • Relative abundance values are either provided or can be read from a mass spectrum

Worked Example

Calculating the relative atomic mass of oxygen

A sample of oxygen contains the following isotopes:

Isotope

Percentage abundance

16O

99.76

17O

0.04

18O

0.20

What is the relative atomic mass of oxygen in this sample, to 2 decimal places?

A. 16.00

B. 17.18

C. 16.09

D. 17.00

Answer:

  • The correct answer option is A:

Arbegin mathsize 16px style fraction numerator open parentheses 99.76 space cross times space 16 close parentheses space plus open parentheses 0.04 space cross times space 17 close parentheses space plus open parentheses 0.20 space cross times 18 close parentheses over denominator 100 end fraction end style

Ar = 16.0044

Ar = 16.00

Mr from mass spectra

  • The percentage abundance of the isotopes in an element can be found using a mass spectrometer

  • In A-Level Chemistry, mass spectra are interpreted assuming all ions are singly charged (+1)

  • The basic processes of mass spectrometry are:

    1. The sample is vapourised

    2. The sample is ionised to form positive ions

    3. The ions are accelerated

      • Heavy ions move more slowly and are less deflected

    4. The ions are detected as a mass-to-charge ratio, written as bevelled m over z

      • Each ion produces a signal proportional to its abundance

      • So, a larger signal means a greater abundance

  • A mass spectrum can be used to calculate the relative atomic mass of an element

Worked Example

Calculating the relative atomic mass of boron

Calculate the relative atomic mass of boron using its mass spectrum, to 2dp:

Analytical Techniques Mass Spectrum Boron, downloadable AS & A Level Chemistry revision notes

Answer:

  • The relative atomic mass of boron, to 2dp, is 10.80

Arbegin mathsize 16px style fraction numerator open parentheses 19.9 space cross times space 10 close parentheses space plus space open parentheses 80.1 space cross times space 11 close parentheses over denominator 100 end fraction end style

Ar = 10.801

Ar = 10.80

Relative molecular and formula mass

  • These terms are closely related and both involve calculating mass from atomic contributions: 

    • Relative molecular mass refers to substances with simple molecules

      • For example, ethanol (C2H5OH) contains:

        • 2 carbon atoms

        • 6 hydrogen atoms

        • 1 oxygen atom

    • Relative formula mass refers to substances with giant structures

      • For example, calcium chloride (CaCl2) is an ionic lattice with a repeating ratio of:1 calcium ion to 2 chloride ions

  • These terms are often misused

    • Relative molecular mass is commonly and incorrectly applied to all compounds

  • Both the relative molecular mass and relative formula mass are calculated by adding up the relative atomic masses (Aᵣ) of all the atoms present

Worked Example

Calculate:

  1. The relative molecular mass of ethanol, C2H5OH

  2. The relative formula mass of calcium chloride, CaCl2

(Ar values: C = 12.0, H = 1.0, O = 16.0, Ca = 40.1, Cl = 35.5)

Answers :

  1. The relative molecular mass of ethanol, C2H5OH, is 46.0

Mr = (2 x C) + (5 x H) + O + H

Mr = (2 x 12.0) + (5 x 1.0) + 16.0 + 1.0

Mr = 46.0

  1. The relative formula mass of calcium chloride, CaCl2, is 111.1

Mr = Ca + (2 x Cl)

Mr = 40.1 + (2 x 35.5)

Mr = 111.1

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Richard Boole

Author: Richard Boole

Expertise: Chemistry Content Creator

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

Philippa Platt

Reviewer: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener