How do you find the relative abundance and the isotopic mass of an element? Zirconium has four naturally occuring isotopes.
90 = amu 89.905 and % 51.45
91 = amu 90.906 and % 11.22
92 = amu 91.905 and % 17.15
94 = amu ?? and % ??
I need to know the relative abundance and the isotopic mass and how to reach the answer.
I know that the equation is set up like this:
91.224 = 89.905(.5145) + 90.906(.1122) + 91.905(.17.15) + x(1-.5145 + .1122 + .1715).
I just don't understand how to do the math. I keep coming up with different answers every time. I haven't had algebra for two years so I'm not sure what to do with the x.
2007-01-25
15:20:32
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2 answers
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asked by
shane200388
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in
Science & Mathematics
➔ Chemistry