ASChemistryCIERevision Notes22. Analytical Techniques22.2 Mass SpectrometryIsotopic Abundance & Relative Atomic Mass

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Isotopic Abundance & Relative Atomic Mass (CIE AS Chemistry)

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Calculating Relative Atomic Mass

  • Isotopes are different atoms of the same element that contain the same number of protons and electrons but a different number of neutrons.
    • These are atoms of the same elements but with different mass numbers

  • Because of this, the mass of an element is given as relative atomic mass (Ar) by using the average mass of the isotopes
  • The relative atomic mass of an element can be calculated by using the relative abundance values
    • The relative abundance of an isotope is either given or can be read off the mass spectrum

  • A subscript r space equals space fraction numerator open parentheses relative space abundance subscript isotope space 1 end subscript mass subscript isotope space 1 end subscript close parentheses space plus open parentheses relative space abundance subscript isotope space 2 end subscript mass subscript isotope space 2 end subscript close parentheses space etc. over denominator 100 end fraction

Worked example

Calculate the relative atomic mass, Ar, of oxygen to 2 d.p.

Isotope Percentage abundance
16O 99.76
17O 0.04
18O 0.20

Answer:

  • Ar = fraction numerator open parentheses 99.76 space cross times 0.04 close parentheses space plus space open parentheses 0.04 space cross times space 17 close parentheses space plus space open parentheses 0.20 space cross times space 18 close parentheses over denominator 100 end fraction
  • Ar = 16.0044
  • Ar = 16.00 (to 2 d.p)

Worked example

Calculate the relative atomic mass of boron using its mass spectrum, to 2 d.p.

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

Answer:

  • A subscript r space equals space 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
  • Ar = 10.801
  • Ar = 10.80 (to 2 d.p)

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    Philippa

    Author: Philippa

    Expertise: Chemistry

    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.