Weighted Averages

Calculating an average atmoic mass using relative abundances

There are two naturally occurring isotopes of bromine: $^{79} Br$ (mass of 78.91338 u) and $^{81}Br$ (mass of 80.916291 u). Bromine-79 accounts for 50.541% of all bromine found naturally, and bromine-81 has a percent isotopic abundance of 49.459%. What is the weighted average atomic mass of bromine?

Solution

To calculate a “regular” or mean average of n values, we would use: $$\text{mean value} = \frac{\text{value(1) + value(2) + … value(n)}}{n}$$ In this formula, each value “counts” equally in the sum, and we divide by the number of items as our total.

Since we want a weighted average in this calculation, we will modify this formula to account for the different weights, and the fact that our “total” = 100% (all the bromine in the world): $$\text{average mass} = \frac{(\%\, ^{79} Br)(\text{mass}\,^{79} Br)+(\%\, ^{81} Br)(\text{mass}\,^{81} Br)}{100\%} \\ \text{average mass} = \frac{(50.541\%)(78.91338 u)+(49.459\%)(80.916291 u)}{100\%} \\ \text{average mass} = \mathbf{79.904\, \text{u}}$$

This aligns with the mass that is reported on the periodic table (hooray!)

Check Your Learning

Silicon has three naturally occurring isotopes. Their masses and relative abundances are:

• $^{28}Si$: 27.97693 u; 92.223%
• $^{29}Si$: 28.97649 u; 4.685%
• $^{30}Si$: 29.97337 u; 3.092%

What is the average atomic mass of silicon?

Answer

28.08 u

Calculating a course grade

In a chemistry course, the grade breakdown (from the course outline) is:

• Labs – 25%
• Tutorials – 25%
• Midterm – 15%
• Final – 35%

If Jessie has earned an average grade of 82.0% in the labs, 94.0% in the tutorials, and 76.0% on the midterm exam, what is Jessie’s grade going into the final exam?

Solution

What we need to calculate here is the weighted average of all the grades except the final exam (which we don’t have yet). We can use a similar form to the way the bromine mass was calculated above: $$\text{average grade} = \frac{(\text{% for labs})(\text{grade on labs})+(\text{% for tutorials})(\text{grade on tutorials})+(\text{% for midterm})(\text{grade on midterm})}{{\text{ lab + tutorial + midterm}}} \\ \text{average grade} = \frac{(\text{25%})(\text{82})+(\text{25%})(\text{94})+(\text{15%})(\text{76})}{{\text{ 25 + 25 + 15}}} \\ \text{average grade} = 85.2% $$

Check Your Learning

Hollis, a student in the same chemistry class, would also like to predict their grade coming into the final. Hollis’ grades so far are: 67.0% in the labs, 81.0% in the tutorials, and 73.0% on the midterm exam.

Answer

73.8%

Check Your Learning

If Hollis wants to get a grade of A- on the course (a minimum overall grade of 82.0%), what minimum grade do they need on the final exam?
Hint: You will need to extend the formula to include the final exam. Try assigning a grade of “x” to the final, then solve for “x”.

Answer

97.3%