Question

The average molecular weight for element X is 59.97 g/mol. There are two known isotopes of element X, one weighing 59 g/mol, and the other weighing 61 g/mol. What is the relative abundance of each?

Possible Answers:
50% X-59, 50% X-61
75% X-59, 25% X-61
63% X-59, 37% X-61
51.5% X-59, 48.5% X-61
48.5% X-59, 51.5% X-61

Answer 1

The correct answer is:48.5% X-59, 51.5% X-61

Step-by-step explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

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

Let the fractional abundance of X-59 isotope be 'x'. So, fractional abundance of X-61 isotope will be '1 - x'

For X-59 isotope:

Mass of X-59 isotope =59 g/mol

Fractional abundance of X-59 isotope = x

For X-61 isotope:

Mass of X-61 isotope = 61 g/mol

Fractional abundance of X-61 isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

`59.97 g/mol=[(59 g/mol* x)+(61 g/mol* (1-x))]\\\\x=0.515`

Percentage abundance of X-59 isotope =
`0.515* 100=51.5\%`

Percentage abundance of X-61 isotope =
`(1-0.515)=0.485* 100=48.5\%`

Hence, the percentage abundance of both the isotopes X-59and X-61 are 51.5% and 48.5% respectively.