Question

Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is the natural abundance of Cl-35? (The atomic mass of chlorine is 35.45 amu)a) 75.95%b) 24.05%c) 50.00%d) 35.00%e) 37.00%

Answer 1

Answer:

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1  ⇒  X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

Answer 2

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope is 75.90 % respectively.

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:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$  .....(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 = 35.45 amu

Putting values in equation 1, we get:  

$35.45=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7590$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7590\times 100=75.90\%$

Percentage abundance of $_{29}^{65}\textrm{Cu}$ isotope = $(1-0.7590)=0.241\times 100=24.10\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope is 75.90 % respectively.