# Question #41494

Jun 28, 2017

Average atomic mass of stable isotopes.

#### Explanation:

Many elements, like chlorine, exist in many isotopes. An isotope is a version of an element with different numbers of neutron. When it comes to isotopes, an average will be taken. This is done by doing ${A}_{r} = \frac{\Sigma \left(m \cdot p\right)}{\Sigma p}$ (in other words, an averag of the masses and the amount they appear), where m is the atomic mass of the isotope, and p is how often it occurs, either as a percentage or mass.

Another thing with ${A}_{r}$ is that only stable isotopes are used, as they can easily be measured in the substance, unstable isotooes will gradually decay into other isotopes, changing the composition if the substance.

The two stable isotopes of chlorine are $\text{^(35)"Cl}$ and $\text{^(37)"Cl}$. $\text{^(35)"Cl}$ appears roughly 75.78% of the time, and $\text{^(37)"Cl}$ appears roughly 24.22% of the time.

Using our equation from earlier we find, ${A}_{r} = \frac{\Sigma \left(m \cdot p\right)}{\Sigma p} = \frac{\left(35 \cdot 75.78\right) + \left(37 \cdot 24.22\right)}{100} = 35.4844 \approx 35.5$