# What is the weighted average mass of chlorine?

Apr 3, 2017

This is the weighted average of the individual chorine isotopes.....

#### Explanation:

This site reports that the isotopic abundance of elemental chlorine is 76% with respect to ""^35Cl, and 24% with respect to ""^37Cl. What has ""^37Cl got that ""^35Cl ain't got?

And if we take the weighted average of these isotopes:

(75%xx35+24%xx37) $\text{amu}$ $=$ $35.5 \cdot \text{amu}$

And thus when we calculate the mass of molecular chlorine, i.e. $C {l}_{2}$, we use a mass of $70.9 \cdot g \cdot m o {l}^{-} 1$.

Apr 3, 2017

The average mass of chlorine is $35.45$ amu.

#### Explanation:

The average atomic mass of some substance is measured with the percentages of how much they exist in nature. An average is always calculated by adding up the numbers and divide that by the amount of number you had, so, for example, the average for the numbers 1, 3, and two times 7 is:

$\frac{1 + 3 + 7 + 7}{4} = 4.5$

If we now look at the percentage of the numbers encountered we see that we encounter
25% the number 1
25% the number 3
50% the number 7

Following this calculation, the average can also be determined by multiplying the numbers with the percentages and add them up.
$\left(0.25 \cdot 1\right) + \left(0.25 \cdot 3\right) + \left(0.50 \cdot 7\right) = 4.5$

For calculating the average atomic mass this is no different.
For chlorine we now that
$C l$ with mass 34.9689 amu is encountered 75.77% in nature
$C l$ with mass 36.9659 amu is encountered 24.23% in nature

75.77% + 24.23%=100%

From the total of all chlorine, you will encounter 75.77% times the 34.9689 amu one, and 24.23% times the 36.9659 amu one.
We use exactly the same approach as in the example above:

$\left(0.7577 \cdot 34.9689\right) + \left(0.2423 \cdot 36.9659\right) = 35.45$amu