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Suppose that a fictitious element, X, has two isotopes: 59X (59.015 amu) and 62X (62.011 amu). The lighter isotope has an abundance of 81.7%. Calculate the atomic mass of the element X.

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Answer:

Atomic Mass of Element = 59.57 amu

Given Data:

Element X Isotope 1 atomic weight = 59.015 amu

Element X Isotope 2 atomic weight = 62.011 amu

The abundance of lighter isotope = 81.7%

Atomic Mass of Element X = ?

Solution:

81.7% is the abundance of element that is lighter so it is the of the isotope having weight 59.015 amu

Then the remaining abundance = 100 -81.7 = 18.3 %

18.3 % abundance of the isotope weight = 62.011 amu

Formula:

Mass of the element = % of abundance (atomic mass of Isotope 1) + % of abundance (atomic mass of Isotope 2)

Put the values

Atomic mass of Element X = 81.7/100 (59.015) + 18.3 /100 (62.011)

Atomic mass of Element = 0.817 (59.015) + 0.183 (62.011)

Atomic mass of Element = 48.22 + 11.35

Atomic Mass of Element = 59.57 amu

User Ricardo Markiewicz
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