1:17 be able to calculate the relative atomic mass of an element (Aᵣ) from isotopic abundances

1:17 be able to calculate the relative atomic mass of an element (Aᵣ) from isotopic abundances

75% of chlorine atoms are the type 35Cl (have a mass number of 35)

25% of chlorine atoms are of the type 37Cl (have a mass number of 37)

In order to calculate the relative atomic mass (Ar) of chlorine, the following steps are used:

  1. Multiply the mass of each isotope by its relative abundance
  2. Add those together
  3. Divide by the sum of the relative abundances (normally 100)

    \[ A_r = \frac{( (35 \times 75) + (37 \times 25) )}{100} \]

    \[ A_r = 35.5 \]

 

Example question:

A sample of bromine contained the two isotopes in the following proportions: bromine-79 = 50.7% and bromine-81 = 49.3%.

Calculate the relative atomic mass (Ar) of bromine.

    \[ A_r = \frac{( (79 \times 50.7) + (81 \times 49.3) )}{100} \]

    \[ A_r = 79.99 \]

 

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Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals