Question

1. an element exists as 4 different isotopes. 4.35% have a mass of 49.9461 amu, 83.79% have a mass of 51.9405 amu, 9.50% have a mass of 52.9407 amu, and 2.36% have a mass of 53.9389 amu.

a. what is the average atomic mass of this element? round to 5sf.
b. what is the identity of this element?

Explanation:

Step1: Calculate the contribution of each isotope

For the first isotope: \(4.35\%\times49.9461 = 0.0435\times49.9461\approx2.172665\)
For the second isotope: \(83.79\%\times51.9405=0.8379\times51.9405\approx43.526055\)
For the third isotope: \(9.50\%\times52.9407 = 0.095\times52.9407\approx5.029367\)
For the fourth isotope: \(2.36\%\times53.9389=0.0236\times53.9389\approx1.272958\)

Step2: Sum up the contributions

\(2.172665 + 43.526055+5.029367 + 1.272958\approx51.99\)

Answer:

a. The average atomic mass is approximately \(51.99\) amu.
b. The element is Chromium (\(Cr\)) as its average atomic mass is close to the known average atomic mass of Chromium.