Question

Iron has many isotopes but only 4 are found in significant amounts in naturally found mixtures. The amounts by mass percent are: 5.845% of 54 Fe (53.9396 amu) 91.754% of 56Fe (55.9349 amu), 2.119% of 57Fe (56.9354 amu) and 0.282% of 58Fe (57.9333 amu). What would you determine the average mass of iron to be

Answer 1

Answer:

The average mass of iron to be is 54.76 amu.

Explanation:

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

1)

Mass of  Fe-54 isotope = 53.9396 amu

Percentage abundance of Fe-54 = 5.845%

Fractional abundance of Fe-54 = 0.05845

2)

Mass of  Fe-56 isotope = 55.9349 amu

Percentage abundance of Fe-56 = 91.754 %

Fractional abundance of Fe-56= 0.91754

3)

Mass of  Fe-57 isotope = 56.9354 amu

Percentage abundance of Fe-57 = 2.119%

Fractional abundance of Fe-57 = 0.002119

4)

Mass of  Fe-58 isotope = 57.9333 amu

Percentage abundance of Fe-58 = 0.282%

Fractional abundance of Fe-58 = 0.00282

Average atomic mass of iron :

$=53.9396 amu\times 0.05845+55.9349 amu\times 0.91754 +56.9354 amu\times 0.002119 + 57.9333 amu\times 0.00282=54.76 amu$

The average mass of iron to be is 54.76 amu.