AP®ChemistryCollege BoardStudy GuidesUnit 1: Atomic Structure & PropertiesMass Spectroscopy of ElementsAverage Atomic Mass

Average Atomic Mass (College Board AP® Chemistry)

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Reviewed by: Stewart Hird

Average Atomic Mass

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

    • The same number of protons

    • A different number of neutrons

    • A different mass number

  • Most naturally occurring elements have more than one isotope

  • The percentage of each isotope in a sample is called relative abundance

  • The average atomic mass of an element is a weighted average of the masses of all of the naturally occurring isotopes of that element

 average space atomic space mass space equals space sum from blank to blank of open parentheses isotope space mass space cross times relative space abundance close parentheses subscript 1.. straight n end subscript

  • The relative abundance percentage must be converted to a decimal value by dividing by 100 before proceeding with an average atomic mass calculation

  • In decimal form, the sum of the relative abundances will equal 1

Worked Example

The average atomic mass of boron is 10.81 amu. The two naturally occurring isotopes of boron are listed in the table below.

Isotope

Mass (amu)

boron-10

10.01

boron-11

11.01

Using the information provided, calculate the percent abundance of each isotope.

Answer:

  • The sum of the relative abundances in decimal form is

    • x + y = 1

  • Rearranging the equation to solve for x

    • x = 1 – y

  • The average atomic mass of boron is

    • average atomic mass = (isotope mass × relative abundance)1 + (isotope mass × relative abundance)2 

    • 10.81 = 10.01x + 11.01y

  • Substituting the value of x

    • 10.81 = 10.01(1–y) + 11.01y

    • 10.81 = 10.01 – 10.01y + 11.01y

    • 10.81 = 10.01 + 1y

    • 0.80 = y

  • Substituting y into the equation for x

    • x = 1 – 0.80

    • x = 0.20

  • The percent abundance of boron-10 is 20% and the percent abundance of boron-11 is 80%

Worked Example

Sample Mass (amu)

Relative abundance (%)

125.91

51.80

127.90

48.20

 The results of the analysis of a pure sample of AgF using mass spectroscopy are shown in the table above. Fluorine only has one naturally occurring isotope. Based on this information, what is the average atomic mass of silver?

Answer:

  • The average atomic mass of fluorine is 19.00 amu

  • The mass of each silver isotope must be

    • isotope mass1 = 125.91 amu - 19.00 amu

    • isotope mass1 = 106.91 amu

    • isotope mass2 = 127.90 amu - 19.00 amu

    • isotope mass2 = 108.90 amu

  • The average atomic mass of silver is

    • average atomic mass = (isotope mass × relative abundance)1 + (isotope mass × relative abundance)2

    • average atomic mass = (106.91 amu × 0.5180) + (108.90 amu × 0.4820)

    • average atomic mass = 55.38 amu + 52.49 amu

    • average atomic mass = 107.87 amu

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    Fallon

    Author: Fallon

    Expertise: Chemistry Content Creator

    Fallon obtained a double major in chemistry and secondary education, and after graduating she taught Chemistry and Organic Chemistry for 7 years. Fallon’s passion for creating engaging classroom materials led her to pursue a career in content development. For over 3 years, Fallon has created videos, review materials, and practice questions for AP Chemistry, IGCSE, and other international exam boards.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.