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

Answer 1

To calculate the average atomic mass of neon in nature, you need to take the weighted average based on the natural abundance of each isotope. This means that for each isotope, you multiply its atomic mass by its natural abundance (expressed as a decimal) and then sum those values together.

Given the data:

- Neon-20: \(19.992 \, \text{amu} \times 0.9048\)
- Neon-21: \(20.993 \, \text{amu} \times 0.0027\)
- Neon-22: \(21.991 \, \text{amu} \times 0.0925\)

The correct setup for the calculation is:

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

So, the correct response is:

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