Abundance Calculations

Question

Hello, is there a formula related to the problem below? I know calculating abundance is percentage as a decimal multiplied by mass. I'm just not sure of below;

Silver has two naturally occurring isotopes with the following isotopic masses. 107 47Ar 109 47Ar 106.90509 amu 108.9047 amu

The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is .

Answer 1

Silver has two naturally occurring isotopes with the following isotopic masses. 107 47Ar 109 47Ar 106.90509 amu 108.9047 amu

The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is .

The average atomic mass is given by the following formula

Average atomic mass = (relative abundance of isotope 1)*(atomic mass of isotope 1) + (relative abundance of isotope 2)*(atomic mass of isotope 2)

This formula can be expanded for elements that have more than two isotopes.

The sum of all relative abundances is always 1 (= 100%)

Let us say the that the relative abundance of the lighter isotope is x. Then the relative abundance of the heavier isotope will be (1 - x)

Now we can place all values in the formula and solve for x.

107.8682 = 106.90509x + 108.9047(1 - x)

107.8682 = 106.90509x + 108.9047 - 108.9047x

107.8662 - 108.9047 = 106.90509x - 108.9047x

-1.0385 = -1.99961x

-1.0385/-1.99961 = -1.99961x/-1.99961

0.51935 = x

The relative abundance of the lighter isotope is 0.51935 = 51.935%

Answer 2

I guess you could say there's a formula, but it's easier just to use what you already know, and figure it out from there. You know that that abundance is % times mass and you also know that the %'s must add to 100 (or 1.0 using decimals).

Let x = abundance of lighter isotope and y = abundance of heavier isotope. Thus, x + y = 1.0

(x)(106.90509) + (y)(108.9047) = 107.8682

x + y = 1.0 and y = 1-x

106.90509x + 108.9047 - 108.9047x = 107.8682

-1.99961x = -1.0365

x = 0.51835 = fractional abundance of the lighter isotope