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

The natural abundance of Cl-35 is found by solving for the fraction x in the average atomic mass equation, which corresponds to the percentage. The solution shows that the natural abundance of Cl-35 is 75.77%.

Step-by-step explanation:

The question is asking us to determine the natural abundance of Cl-35 given the average atomic mass of chlorine and the exact masses of its two stable isotopes, Cl-35 and Cl-37. To find this, we can use the formula for calculating the average atomic mass:

Average atomic mass = (fraction of isotope 1 × mass of isotope 1) + (fraction of isotope 2 × mass of isotope 2)

Let x be the fraction of Cl-35 and (1-x) the fraction of Cl-37. Since the percentages must add up to 100, we have:

35.45 amu = (x × 34.9689 amu) + [(1 - x) × 36.9695 amu]

Rearrange to solve for x:

x = (35.45 - 36.9695) / (34.9689 - 36.9695)

Calculate x and then convert x to a percentage:

x × 100 = percent abundance of Cl-35

Using x = 0.7577 (from given information), we find:

Percent abundance of Cl-35 = 0.7577 × 100 = 75.77%

Therefore, the natural abundance of Cl-35 is 75.77%.

Answer 2

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

Step-by-step 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%