There are four naturally occurring isotopes of iron, Fe-54, Fe-56, Fe-57, and Fe-58. The iron block on the periodic table gives an atomic mass of 55.845 amu. If one were able to pick out a single iron atom at random, what is the probability that it will weigh 55.845 amu?

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It is true that there are four naturally occurring isotopes of iron: 5.845% of 54Fe, 91.754% of 56Fe, 2.119% of 57Fe and 0.282% of 58Fe.  However, the atomic mass listed for an element of the periodic table is not represented by any of these.  The atomic mass is a weighted average of all of the possible isotopes. 

The actual mass of each isotope is given by its mass number (ignoring the mass of the electrons).  So, Iron-54 has a mass of 54 amu, Iron-56 is 56 amu, Iron-57 is 57 amu, and iron-58 is 58 amu.  To get the atomic mass of iron as stated on the periodic table you would have to multiply each isotopes mass by its percent (as a decimal) and then add the results.  This will produce the weighted average of 55.845 amu found in the table.

So, if you were to randomly pick an iron atom out of a well mixed representative sample you would most likely pick iron-56 which is the most abundant isotope.  However, you will never pick an isotope with a mass of 55.845 amu because there is no such thing.

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