If 7.4% of an element has an atomic mass of 6.015 u and 92.6% of an element has an atomic mass of 7.016 u, then what is the average mass of the element?

Given:
7.4% of an element has an atomic mass of 6.015 u
92.6% of an element has an atomic mass of 7.016 u
Formula:

The average atomic mass of the mixture can be calculated by the formula:

$\frac{At.mass of isotope 1 \times \% of occurence + At. mass of isotope 2 \times \% of occurence100}{100}$

Solution:

Average atomic mass =  $\frac{6.015X 7.4 +7.016 X92.6}{100}$

                                               $$\begin{gathered} =\frac{44.511 +649.68 }{100} \\ = \frac{694.19}{100} \\ = 6.941u \end{gathered}$$

The average atomic mass of the element is 6.941u