Respuesta :
Answer:
The average mass of iron to be is 54.76 amu.
Explanation:
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
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 :
[tex]=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[/tex]
The average mass of iron to be is 54.76 amu.
The average mass of the iron having four significant isotopes is 55.8452 amu
Let isotope Fe-54 be A
Let isotope Fe-56 be B
Let isotope Fe-57 be C
Let isotope Fe-58 be D
- From the question given above, the following data were obtained:
For A (isotope Fe-54):
Mass of A = 53.9396 amu
Abundance of A (A%) = 5.845%
For B (isotope Fe-56):
Mass of B = 55.9349 amu
Abundance of B (B%) = 91.754%
For C (isotope Fe-57):
Mass of C = 56.9354 amu
Abundance of C (C%) = 2.119%
For D (isotope Fe-58):
Mass of D = 57.9333 amu
Abundance of D (D%) = 0.282%
Average mass of iron =?
The average mass of iron can be obtained as follow:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100] + [(Mass of D × D%)/100]
= [(53.9396 × 5.845)/100] + [(55.9349 × 91.754)/100] + [(56.9354 × 2.119)/100] + [(57.9333 × 0.282)/100]
= 3.1528 + 51.3225 + 1.2065 + 0.1634
= 55.8452 amu
Therefore, the average mass of the iron is 55.8452 amu
Learn more: https://brainly.com/question/25104108
