Question b220c

Sep 16, 2015

Here's why that is so.

Explanation:

I assume that you're referring to this video

The idea with finding the abundance of two isotopes that contribute to the relativ atomic mass of an element is that you know for sure that these abundances must add to 100%.

Since you're only dealing with two isotopes, you can say for sure that their abundances must add up to give 100%.

Moreover, if you take the fractional abundance of one of these two isotopes to be equal to $x$, then you know for a fact that the abundance of the second isotope must be $1 - x$.

The fractional abundances of the two isotopes must add up to give $1$, which is why you can say that

underbrace(""x"")_(color(blue)("abundance of first isotope")) + overbrace(1-x)^(color(green)("abundance of second isotope")) = 1

Now, in Tyler's first problem, he sets up the equation so that the fractional abundance of $\text{^35"Cl}$ is equal to $x$. This means that the fractional abundance of $\text{^37"Cl}$ must be $1 - x$.

You know that the relaatie atomic mass of an element is equal to the sum of the contributions each isotope brings to the table.

He sets up the equation and solves for $x$, which comes out to be $0.760$. This means that the fractional abundance of the second isotope must be

$1 - x = 1 - 0.760 = 0.240$

The exact same approach is true for the second problem. This time $x$ represents the fractional abundance of $\text{^6"Li}$.

When $x$ comes out to be equal to $0.075$, he calculates the fractional abundance of the $\text{^7"Li}$ isotope to be

$1 - x = 1 - 0.075 = 0.925$

Remember, fractional abundances are simply percent abundances divided by 100.

So if an isotope has an abundance of 50%#, its fractional abundance will be $\frac{50}{100} = 0.5$.