Isotopes

Isotopes

The chemical properties of an element depend on the number of protons and electrons inside the atom. So if a neutron or two is added or removed from the nucleus, then the chemical properties will not change. This means that such an atom would remain in the same place in the periodic table. For example, no matter how many neutrons we add or subtract from a nucleus with 6 protons, that element will always be called carbon and have the element symbol C (see the periodic table). Atoms which have the same number of protons (i.e. same atomic number Z), but a different number of neutrons (i.e. different N and therefore different mass number A), are called isotopes.

Did You Know?

In Greek, “same place” reads as ίσoςτόπoς (isos topos). This is why atoms which have the same number of protons, but different numbers of neutrons, are called isotopes. They are in the same place on the periodic table!

Definition: Isotope

Isotopes of an element have the same number of protons (same Z), but a different number of neutrons (different N).

The chemical properties of the different isotopes of an element are the same, but they might vary in how stable their nucleus is. We can also write elements as E–A where the E is the element symbol and the A is the atomic mass of that element. For example Cl–35 has an atomic mass of $$\text{35}$$ $$\text{u}$$ (17 protons and 18 neutrons), while Cl–37 has an atomic mass of $$\text{37}$$ $$\text{u}$$ (17 protons and 20 neutrons).

In nature the different isotopes occur in different percentages. For example Cl–35 might make up $$\text{75}\%$$ of all chlorine atoms on Earth, and Cl–37 makes up the remaining $$\text{25}\%$$. The following worked example will show you how to calculate the average atomic mass for these two isotopes:

Example: The Relative Atomic Mass of an Isotopic Element

Question

The element chlorine has two isotopes, chlorine–35 and chlorine–37. The abundance of these isotopes when they occur naturally is $$\text{75}\%$$ chlorine–35 and $$\text{25}\%$$ chlorine–37. Calculate the average relative atomic mass for chlorine.

Step 1: Calculate the mass contribution of chlorine–35 to the average relative atomic mass

$$\text{75}\%$$ of the chlorine atoms has a mass of of $$\text{35}$$ $$\text{u}$$.

Contribution of Cl–35 $$= (\frac{75}{100} \times 35 = \text{26.25}\text{ u})$$.

Step 2: Calculate the contribution of chlorine–37 to the average relative atomic mass

$$\text{25}\%$$ of the chlorine atoms has a mass of of $$\text{37}$$ $$\text{u}$$.

Contribution of Cl–37 $$= (\frac{25}{100} \times 37 = \text{9.25}\text{ u})$$.

Step 3: Add the two values to arrive at the average relative atomic mass of chlorine

Relative atomic mass of chlorine = $$\text{26.25}$$ $$\text{u}$$ + $$\text{9.25}$$ $$\text{u}$$ = $$\text{35.5}$$ $$\text{u}$$.

If you look on the periodic table (see figure below), the average relative atomic mass for chlorine is $$\text{35.5}$$ $$\text{u}$$.