Question

An undiscovered element has three naturally occurring isotopes of X-55, X-57, and X-58. Isotope X-55 has an abundance of 27.80 % and isotope X-57 has an abundance of 44.39 %. What is the average mass of this element in amu?

Answer 1

Answer: The average atomic mass of this elements is 56.7221 amu.

Explanation: The average atomic mass is the sum of the masses of its isotopes each multiplied by their natural abundances.

$\text{average atomic mass}=\sum_{i=1}^{n}(mass)_i(\text{Fractional abundance})_i$            .....(1)

$\text{Fractional abundance}=\frac{\%\text{ abundance}}{100}$

We are given 3 isotopes of an element.

For Isotope $X^{55}$,

Mass = 55 amu

Fractional abundance = 0.2780

For isotope $X^{57}$,

Mass = 57 amu

Fractional abundance = 0.4439

Total Fractional abundance = 1

For isotope $X^{58}$,

Mass = 58 amu

Fractional abundance = Total abundance - abundances of the other isotopes

Fractional abundance = 1 - 0.7219

                                     = 0.2781

Now, putting all the values in equation 1, we get

$\text{Average atomic mass}= (55 amu\times 0.2780)+(57 amu\times 0.4439)+(58 amu\times 0.2781)$

Average atomic mass = 56.7221 amu.