Homework Help: Chemistry

Posted by Samantha on Thursday, October 14, 2010 at 12:04am.

Congratulations! You have discovered a new element called Bhauzium (Bz). The average atomic mass of Bz was found to be 348.72 amu and its atomic number is 123.
a) if the atomic masses of three isotopes of Bhsuzium are 342.92 amu, 356.43 amu (6.52%), and 351.29 amu, what is the relative abundance (%) of each isotope?

Chemistry - DrBob222, Thursday, October 14, 2010 at 12:21am
Let X = fraction of 342.92
356.43 is 6.52% or 0.0652 as a fraction.
351.29 = (1-X-0.0652) = 0.9348-X

Sum those values = 348.72

342.92X + 356.43(0.0652) + 351.29(0.9348-x) = 348.72
Solve for x which is the fraction of the first isotope and 0.9348-x is the fraction of the third isotope listed. The percentages should add to 100.0%.

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