Question

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine has an atomic mass of 35.4527 arnu, what is the percent abundance of each isotope?

Answer 1

Answer: The percentage abundance of
`_(17)^(35)\textrm{Cl}` and
`_(17)^(37)\textrm{Cl}` isotopes are 75.77% and 24.23% 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 = 34.9689 amu

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

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

Mass of
`_(17)^(37)\textrm{Cl}` isotope = 36.9659 amu

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

• Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

`35.4527=[(34.9689* x)+(36.9659* (1-x))]\\\\x=0.7577`

Percentage abundance of
`_(17)^(35)\textrm{Cl}` isotope =
`0.7577* 100=75.77\%`

Percentage abundance of
`_(17)^(37)\textrm{Cl}` isotope =
`(1-0.7577)=0.2423* 100=24.23\%`

Hence, the percentage abundance of
`_(17)^(35)\textrm{Cl}` and
`_(17)^(37)\textrm{Cl}` isotopes are 75.77% and 24.23% respectively.