Question

How do I calculate the average atomic mass of the element X?

Isotope	Abundance	Mass
${}^{221}X$	74.22	220.9
${}^{220}X$	12.78	220.0
${}^{218}X$	13.00	218.1

Answer 1

Hint: In the given table the natural abundance value and the masses of the isotope is provided. The average atomic mass is calculated by adding the masses of the isotopes of the chemical element where each mass is multiplied with the natural abundance.

Complete step by step answer:
The average atomic mass of the chemical element is the weighted average mass of the atoms in the naturally occurring sample of the chemical element.
The average atomic mass of the chemical element is defined as the sum of the masses of the isotopes of the chemical element each mass multiplied by the natural abundance value.
The natural abundance is defined as the decimal related with the percent of the atoms of that chemical element which are of the given isotope.
The formula used to calculate the average atomic mass of the chemical element is shown below.
$M{M_{Avg}} = \sum {{X_i}} {M_i}$
Where,
$M{M_{Avg}}$is the average atomic mass of a chemical element.
${X_i}$ is the fraction abundance of isotope i.
${M_i}$ is the atomic mass of isotope i.
First determine the contribution made by each isotope by multiplying the natural abundance with the mass.
For ${}^{221}X$
$ \Rightarrow {}^{221}X = 74.22 \times 220.9$
$ \Rightarrow {}^{221}X = 163.95$
${}^{220}X$
$ \Rightarrow {}^{220}X = 12.78 \times 220.0$
$ \Rightarrow {}^{221}X = 28.12$
${}^{218}X$
$ \Rightarrow {}^{218}X = 13.00 \times 218.1$
$ \Rightarrow {}^{221}X = 28.35$
Now, to determine the average atomic mass of the element, add all the contributions made by the isotope.
$ \Rightarrow 163.95 + 28.12 + 28.35$
$ \Rightarrow 220.4$

Therefore, the average atomic mass of the element X is 220.4.

Note: Don’t get confused by the term atomic weight, as the average atomic mass and atomic weight are the same. The average masses are expressed as unified atomic mass units (u). 1u is equal to exactly one twelfth mass of the neutral carbon-12 atom.