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11 January, 16:43

Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 amu,

what is the natural abundance of each isotope? (expressed as a percentage)

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  1. 11 January, 18:55
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    The abundance of 107Ag is 51.5%. The abundance of 109Ag is 48.5%.

    Explanation:

    The average atomic mass of silver can be expressed as:

    107.87 = 106.90 * A1 + 108.90 * A2

    Where A1 is the abundance of 107Ag and A2 of 109Ag.

    Assuming those two isotopes are the only one stables, we can use the equation:

    A1 + A2 = 1.0

    So now we have a system of two equations with two unknowns, and what's left is algebra.

    First we use the second equation to express A1 in terms of A2:

    A1 = 1.0 - A2

    We replace A1 in the first equation:

    107.87 = 106.90 * A1 + 108.90 * A2

    107.87 = 106.90 * (1.0-A2) + 108.90 * A2

    107.87 = 106.90 - 106.90*A2 + 108.90*A2

    107.87 = 106.90 + 2*A2

    2*A2 = 0.97

    A2 = 0.485

    So the abundance of 109Ag is (0.485*100%) 48.5%.

    We use the value of A2 to calculate A1 in the second equation:

    A1 + A2 = 1.0

    A1 + 0.485 = 1.0

    A1 = 0.515

    So the abundance of 107Ag is 51.5%.
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