Silver has an average atomic mass of 107.9 u and is known to have only two naturally occurring isotopes. If 51.84% of Ag exists as Ag-107 (106.9051 u), what is the identity and the atomic mass of the other isotope?

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  1. Ag 107 = 0.5184 = 106.9051
    Ag x = 1-0.5184 = 0.4816
    0.5184*(106.9051) + 0.l4816*X = 107.9
    Solve for X.
    X will give you the atomic mass, round that to the nearest whole number for the mass number for that Ag isotope.

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