If 80.0% of an element has an atomic mass of 42.0 u and 20.0% has an atomic mass of 43.0 u, what is the average atomic mass of the mixture?

The average atomic mass of the mixture is 42.2 u.

Solution:
80.0% of an element has an atomic mass of 42.0 u
20.0% of an element has an atomic mass of 43.0 u
Average atomic mass of the mixture can be calculated by the formula:

$\frac{At.mass of isotope 1 \times \% of occurence + At. mass of isotope 2 \times \% of occurence}{100}$

Average atomic mass =

 $$\begin{gathered} \frac{ 42 \times 80 + 43 \times 20}{100} \\ =\frac{3360 +860}{100} \\ =\frac{4220}{100} \\ = 42.2 u \end{gathered}$$