Question

A student wants to calculate the average atomic mass of neon in nature. Information about neon is in the table below.

Isotopes in Nature Atomic Mass (amu) Natural Abundance (%)
neon-20 19.992 90.48
neon-21 20.993 .27
neon-22 21.991 9.25
Which of the following is a correct way for the student to set up the calculation?



(1 point)

(19.992 amu×.0027)+(20.993 amu ×.9048)+(21.991 amu ×.0925)(19.992 amu×.0027)+(20.993 amu ×.9048)+(21.991 amu ×.0925)

(19.992 amu×.9048)+(20.993 amu ×.0027)(19.992 amu×.9048)+(20.993 amu ×.0027)

(19.992 amu×.9048)×(20.993 amu ×.0027)×(21.991 amu ×.0925)(19.992 amu×.9048)×(20.993 amu ×.0027)×(21.991 amu ×.0925)

(19.992 amu×.9048)+(20.993 amu ×.0027)+(21.991 amu ×.0925)(19.992 amu×.9048)+(20.993 amu ×.0027)+(21.991 amu ×.0925)

Answer 1

I don't like any of them. Since there are only 3 isotopes, it should be

(19.992 amu ×.9048) + (20.993 amu ×.0027) + (21.991 amu ×.0925)