# Question 257f3

Sep 10, 2015

Yes (possibly).

#### Explanation:

Since your question is very vague, I'll assume that you're dealing with isotopes, abundances, and atomic masses.

The idea here is that when you have two isotopes, the lighter isotope will always contribute less than the heavier isotope to the average atomic mass of the element if the mass of the heavier isotope is closer to the average atomic mass.

So, for example, let's say that you have two isotopes, one of $\text{8 amu}$ and one of $\text{12 amu}$, and the average atomic mass of the element is $\text{10 amu}$.

The difference between the mass of the isotope and the average atomic mass is equal for both isotopes, which means that they will contribute equally to it, i.e. have equal abundances.

You can write

$8 \cdot x + 12 \cdot \left(1 - x\right) = 10$

$8 x + 12 - 12 x = 10$

$- 4 x = - 2 \implies x = \frac{\left(- 2\right)}{\left(- 4\right)} = \frac{1}{2}$

Each isotope's fractional abundance is $0.5$, which is equivalent to a percent abundance of 50%.

Now let's say that that the lighter isotope is $\text{7 amu}$ and the heavier one is $\text{12 amu}$, but that the atomic mass average is still $\text{10 amu}$.

$7 x + 12 \left(1 - x\right) = 10$

$- 5 x = - 2 \implies x = \frac{\left(- 2\right)}{\left(- 5\right)} = 0.4$

The lighter isotope contributes less, i.e. 40%, than the heavier one because you have

$| 10 - 7 | = 3 \to$ for the first isotope

$| 10 - 12 | = 2 \to$ for the second isotope

The mass of the heavier isotope is closer to the average atomic mass, which means that it must contribute more.

Now let's say that the lighter isorope is $\text{6 amu}$, the heavier one is $\text{12 amu}$, and the average atomic mass is $\text{10 amu}$.

$6 x + 12 \left(1 - x\right) = 10$

$- 6 x = - 2 \implies x = \frac{\left(- 2\right)}{\left(- 6\right)} = \frac{1}{3}$

The lighter isotope will once again contribute even less, i.e. 33%#, to the average atomic mass because you have

$| 10 - 6 | = 4 \to$ for the first isotope

$| 10 - 12 | = 2 \to$ for the second isotope

So, the idea is that when you have two isotopes, the lighter one will contribute less if the difference between its mass and the average atomic mass is bigger than the difference between the mass of the heavier isotope and the average atomic mass.