**What is Percent Abundance**

The percentage of a specific isotope that exists in nature is the relative concept of percent abundance in chemistry. Abundance can be calculated by three ways: by mass fraction, by mole fraction and by volume fraction. The atomic weight (atomic mass) indicated on the periodic table for an element constitutes an average mass of all recognized isotopes.

The atomic nuclei only comprise protons and neutrons, each of these having a mass of around 1 atomic mass unit (amu). The atomic weight of each element should also be a whole number. Moreover, electron weights are considered negligible and are not involved in the atomic weight of an element. As we know that, when we look at the periodic table, it is mentioned that the most of the elements have a decimal fraction of their atomic weights. So, actually, the mass specified for each element is mean of all the isotopes that exist in nature which is. The abundance of each isotope of the element can be determined instantly if we know how often the isotopes have atomic weights.

The recognition of the element appears the same as the number of neutrons inside the nucleus varies. A modification in the number of neutrons in the nucleus indicates an isotope. There might be zero, one, two, more neutrons in the nuclei depending on the element. For example; Hydrogen has three isotopes. The nucleus of protium (11H) or simply hydrogen consists no neutron but only proton, deuterium (12H) has a neutron in the nucleus and tritium(13H) has two neutrons in the nucleus.

This article will show you the percent abundance formula, how to calculate it step by step and give you examples on how the formula can be utilized.

**Percent Abundance Formula**

Percent abundance is calculated by dividing the average atomic mass of the element by the summation of isotopic masses. The average atomic mass of an element is a weighted average. Isotopes are atoms with the same atomic number but different mass numbers due to a different number of neutrons. The formula can be written as;

Percent abundance=Massavg(Isotopic mass)

**How to calculate Percent Abundance**

The steps to calculate percent abundance are given below;

- Determine the average atomic mass
- Use the relative abundance concept
- Calculate the relative abundance of the unknown isotope
- Lastly, find the percent abundance

**Example**

There are two stable isotopes of chlorine: chlorine-35 and chlorine-37. The atomic mass of chlorine-35 is 34.97 and the atomic mass of chlorine-37 is 36.97 amu. If you have a glance in the periodic table, you will get that the mass of chlorine is mentioned as 35.45 amu. How did this happen?

Let’s have a deep dive regarding the average atomic chlorine mass. Why is the total atomic mass of chlorine-35 and chlorine-37 not only the average of these two values, respectively, if the atomic masses are 34.97 and 36.97 amu?

The reason is due to a differing distribution of various isotopes, meaning that some isotopes on Earth abound more naturally than others.

**Calculating the average atomic mass of chlorine. **

The following steps are followed to calculate the average atomic mass of chlorine.

**Step 1: Determine the average atomic mass **

Define the isotopic abundance challenge on the periodic table of the element. As, we have taken chlorine which has an average atomic mass of 35.45 amu.

**Step 2: Use the relative abundance concept**

Then, we will use the relative abundance formula for the given problem. As the given element consists only two isotopes, an equation can be set by the given equation below;

M1x+M21-x=MA…………(i)

Where,

M1 is the mass of one isotope, x is the relative abundance, M2 is the mass of other isotope and MA is the average atomic mass of the element

As we have, the atomic mass of chlorine-35 is 34.97 and the atomic mass of chlorine-37 is 36.97 amu. So, now let’s find the relative abundance.

Here, we need to solve for unknown x which is the relative abundance. One isotope is related as M1 and the other as M2.

We have;

M1=34.97 amu Chlorine-35

M2=36.97 amu [Chlorine-37]

MA=35.45 amu

Placing the data in the first equation, we get

34.97×x+36.97×1-x=35.45

**Step 3: Calculate the relative abundance of the unknown isotope**

Now, we will use the algebra to find the value of x. Firstly, property called distributive is used.

34.97×x+36.97-36.97×x=35.45

Then, like terms is combined given by;

-2x=-1.52

Finally, solving for x, we get

x=0.76

**Step 4: Find the percent abundance**

To get percent abundance, we will multiply the relative abundance value by 100 and put a percent (%) sign.

As we have now x= 0.76. Simply, multiply by 100 to get percent. So, chlorine-35 is 0.76*100=76%

Now,

For chlorine-37, we get

(1-x) = (1-0.76) = 0.24

Multiplying above by 100, we get 24%.

Therefore, the percent abundance of chlorine-35 is 76% and the percent abundance of chlorine-37 is 24%.

**What if elements have more than two isotopes? **

For example:

Oxygen has three naturally occurring isotopes 16O, 17O and 18O. The average atomic mass of oxygen is 15.9994 amu. Given, the atomic weight of 16O is 15.995 amu, 17O is 16.999 amu and 18O is 17.999 amu. Also, 17O has 0.037 percent in nature. What are the other isotopes percent abundances?

Firstly, we have the abundance of one isotope which is 0.0037. So, the abundances of the other two remaining isotopes is (1-0.00037) = 0.99963.

Let x be the unknown abundance of 16O and other isotope abundance of 18O be (0.99963-x).

Now, modifying equation (i), we get

(15.995) * (x) + (16.999) *(0.00037) + (17.999) * (0.99963 – x) = 15.9994

or 15.995x – 17.999x = 15.9994 – (16.999) • (0.00037) – (17.999) (0.99963)

or x = 0.9976

So, we get the abundance of 16O is 0.9976 and the abundance of 18O is (0.99963 – 0.9976) = 0.00203.

Hence, the percent abundance of three isotopes are given by;

16O=0.9976*100=99.76%

17O=0.00037*100=0.037% 18O=0.00203*100=0.203%