Animated Solution for Chemistry - Basic Concepts in Chemistry: The average molar mass of chlorine is $35.5{\text{ g mol}}^{-1}$. The ratio of ${}^{35}\text{Cl}$ to ${}^{37}\text{Cl}$ in naturally occurring chlorine is close to

Have you ever looked at the periodic table and wondered why the atomic mass of Chlorine is listed as $35.5$? Atoms are made of protons and neutrons, which have integer masses. So, how can an atom have a fractional mass?

The secret lies in isotopes. In nature, elements rarely exist as a single type of atom. Chlorine, for instance, is a mixture of two primary isotopes: ${}^{35}\text{Cl}$ (which has a mass of $35\text{ u}$) and ${}^{37}\text{Cl}$ (which has a mass of $37\text{ u}$). The value $35.5$ you see on the periodic table is not the mass of any single chlorine atom; it is the weighted average of these isotopes based on how abundantly they occur in nature.

To find out the exact ratio of these isotopes, we need to set up a mathematical model. Let's assume we have a large sample of naturally occurring chlorine.

Let the fractional abundance of ${}^{35}\text{Cl}$ be $x$.

Since the sum of all fractions in a whole must equal $1$, the fractional abundance of the heavier isotope, ${}^{37}\text{Cl}$, must be $(1-x)$.

The formula for the average molar mass ($M_{\text{avg}}$) is the sum of the mass of each isotope multiplied by its respective fractional abundance:

Now, we substitute our known values into this master equation. We know the average mass is $35.5$, the mass of the first isotope is $35$, and the mass of the second is $37$.

Let's carefully expand the brackets to avoid any silly mistakes:

Next, we group the like terms. Let's move the $x$ terms to the left side and the constant terms to the right side:

This simplifies beautifully to:

Isolating $x$, we get:

Converting this decimal into a fraction gives us $x=\frac{3}{4}$. This means that $75\%$ of naturally occurring chlorine is the lighter ${}^{35}\text{Cl}$ isotope.

If the fraction of ${}^{35}\text{Cl}$ is $\frac{3}{4}$, then the fraction of ${}^{37}\text{Cl}$ is simply $1-\frac{3}{4}=\frac{1}{4}$.

The question asks for the ratio of ${}^{35}\text{Cl}$ to ${}^{37}\text{Cl}$. We just need to divide their fractions:

So, the isotopes exist in a $3:1$ ratio. For every three atoms of Chlorine-35, there is one atom of Chlorine-37.

The Lever Rule (Alligation)

While the algebraic method is foolproof, competitive exams like JEE and NEET demand speed. There is a brilliant, visually intuitive shortcut known as the Rule of Alligation or the Lever Rule.

Imagine a number line with $35$ on the left, $37$ on the right, and the average $35.5$ sitting between them.

The distance from the average ($35.5$) to the heavier isotope ($37$) is $1.5$. The distance from the average ($35.5$) to the lighter isotope ($35$) is $0.5$.

According to the lever rule, the ratio of their abundances is exactly the inverse ratio of these distances!

In just five seconds, without writing a single algebraic equation, we arrive at the exact same $3:1$ ratio. The closer the average mass is to a particular isotope, the higher that isotope's abundance must be. Since $35.5$ is much closer to $35$, it makes perfect physical sense that ${}^{35}\text{Cl}$ dominates the mixture!