Question

students are provided a sample of an element with a table describing the composition. element sample’s properties
isotope | atomic mass (amu) | percent abundance
a | 27.976927 | 92.23
b | 28.976495 | 4.67
c | 29.973770 | 3.10
what is the average atomic mass of the substance to the nearest thousandth? record your answer.

Explanation:

Step1: Calculate contribution of Isotope A

Multiply atomic mass of A by its percent abundance (converted to decimal: \(92.23\% = 0.9223\)).
\(27.976927 \times 0.9223 \approx 25.803\)

Step2: Calculate contribution of Isotope B

Multiply atomic mass of B by its percent abundance (converted to decimal: \(4.67\% = 0.0467\)).
\(28.976495 \times 0.0467 \approx 1.353\)

Step3: Calculate contribution of Isotope C

Multiply atomic mass of C by its percent abundance (converted to decimal: \(3.10\% = 0.0310\)).
\(29.973770 \times 0.0310 \approx 0.929\)

Step4: Sum the contributions

Add the contributions from all isotopes: \(25.803 + 1.353 + 0.929 = 28.085\)

Answer:

28.085