Question

Iron exists in nature as a mixture of predominantly three isotopes:

54 26fe (53.94 amu), 56 26fe (55.93 amu), and 57 26fe (56.94 amu). if the
most common isotope, 56 26fe, accounts for 91.75% of iron atoms, and
the average atomic mass of iron is 55.85 amu, what is the percent
abundance of the rarest of these three isotopes of iron?

Answer 1

Answer: 2.81%

Step-by-step explanation:

If the abundance of Fe-56 is 91.75%, then the other two isotopes have a combined abundance of 100%-91.75%=8.25%.

If we let the abundance of Fe-54 be x, then the abundance of Fe-57 is 0.0825-x. Substituting this into the atomic mass formula, we get:

`55.85=(55.93)(0.9175)+(53.94)(x)+(56.94)(0.0825-x)\\55.85=51.315775+53.94x+4.69755-56.94x\\55.85=56.0133-3x\\-0.1633=-3x\\x=(-0.1633)/(-3) \approx 0.0544`

If x=0.1633/3, this means that 0.0825-x=0.0281, so Fe-57 is the rarest isotope, with an abundance of 2.81%