# Question #257f3

##### 1 Answer

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

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 * x + 12 * (1-x) = 10#

#8x + 12 - 12x = 10#

#-4x = -2 implies x = ((-2))/((-4)) = 1/2#

Each isotope's *fractional abundance* is

Now let's say that that the lighter isotope is

#7x + 12(1-x) = 10#

#-5x = -2 implies x = ((-2))/((-5)) = 0.4#

The lighter isotope contributes **less**, i.e.

#|10 - 7| = 3 -># for the first isotope

#|10 - 12| = 2 -># 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

#6x + 12(1-x) = 10#

#-6x = -2 implies x = ((-2))/((-6)) = 1/3#

The lighter isotope will once again contribute **even less**, i.e.

#|10 - 6| = 4 -># for the first isotope

#|10 - 12| = 2 -># 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.