Question

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}Cl37 17 C l . Which isotope is the most abundant?

Answer 1

Answer: The percentage abundance of
`_(17)^(35)\textrm{Cl}` and
`_(17)^(37)\textrm{Cl}` isotopes are 77.5% and 22.5% respectively.

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
`_(17)^(35)\textrm{Cl}` isotope be 'x'. So, fractional abundance of
`_(17)^(37)\textrm{Cl}` isotope will be '1 - x'

• For
  `_(17)^(35)\textrm{Cl}` isotope:

Mass of
`_(17)^(35)\textrm{Cl}` isotope = 35 amu

Fractional abundance of
`_(17)^(35)\textrm{Cl}` isotope = x

• For
  `_(17)^(37)\textrm{Cl}` isotope:

Mass of
`_(17)^(37)\textrm{Cl}` isotope = 37 amu

Fractional abundance of
`_(17)^(37)\textrm{Cl}` isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

`35.45=[(35* x)+(37* (1-x))]\\\\x=0.775`

Percentage abundance of
`_(17)^(35)\textrm{Cl}` isotope =
`0.775* 100=77.5\%`

Percentage abundance of
`_(17)^(37)\textrm{Cl}` isotope =
`(1-0.775)=0.225* 100=22.5\%`

Hence, the percentage abundance of
`_(17)^(35)\textrm{Cl}` and
`_(17)^(37)\textrm{Cl}` isotopes are 77.5% and 22.5% respectively.