Chemistry H.

Indium has two naturally occurring isotopes and an atomic mass of 114.818 amu. In- 113 has a mass of 112.904 amu and an abundance of 4.3%. What is the identity and percent abundance of indium's other isotope. I am very confused on how do I set up this conversion problem?

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  1. atomic masses are weighted averages of all of the isotopes.
    % in 113 = 4.3% and since the isotopes must add to 100%, then 100 - 4.3 = 95.7% for the other isotope.
    Let x = to mass of the other isotope. Then set up the average.

    112.904(0.043) + (x)(0.957) = 114.818 and solve for x.
    95&% is part of the answer. x is the other part of the question. Check my numbers against typos. I make a typo now and then.

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  2. 5.07301149

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