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Today, 01:56

an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with an abundance of 24.10%. a third isotope weighs 206.9745 amu and is 22.10%, and the fourth isotope weighs 207.9766 amu and an abundance of 57.40. calculate the average atomic mass and identify the unknown element.

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  1. Today, 04:05
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    Answer is: the average atomic mass 217.606 amu.

    Ar₁ = 203.973 amu; the average atomic mass of isotope.

    Ar₂ = 205.9745 amu.

    Ar₃ = 206.9745 amu.

    Ar₄ = 207.9766 amu.

    ω₁ = 1.40% = 0.014; mass percentage of isotope.

    ω₂ = 24.10% = 0.241.

    ω₃ = 22.10% = 0.221.

    ω₄ = 57.40% = 0.574.

    Ar = Ar₁ · ω₁ + Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.

    Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

    Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

    Ar = 217.606 amu.

    But abundance of isotopes is greater than 100%.

    It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.
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