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

Question

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)
Responses

(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×.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)

Answer 1

To calculate the average atomic mass of neon, the student needs to multiply the atomic mass of each isotope by its natural abundance (expressed as a decimal), and then sum these values.

The correct way to set up the calculation would be:

\((19.992 \text{ amu} \times 0.9048) + (20.993 \text{ amu} \times 0.0027) + (21.991 \text{ amu} \times 0.0925)\)

In terms of the available responses, the correct option matches:

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

This choice uses the correct abundances in decimal form and correctly sums the contributions from each isotope to get the average atomic mass.