Q.22E - Chemistry (OpenStax)

Question

An element has the following natural abundance, and isotopic masses; 90.90% abundance with 19.99 amu, 0.26% abundance with 20.99 amu, and 8.82% abundance with 21.99 amu. Calculate the average atomic mass of this element.

Step-by-Step Solution

Answer

Average atomic mass = (0.9090×19.99)+(0.0026×20.99)+(0.082×21.99)        

 Average. atomic mass = (18.17+0.054+1.80) amu

                                        = 20.02 amu. 

Step 1 – Determine the above solution

Given, 90.90%, 0.26%, 8.82%

Change percentage into decimal

 90.90% = \(\frac{{90.90}}{{100}} = 0.9090\) 

0.26% = \(\frac{{0.26}}{{100}} = 0.0026\)

8.82% = \(\frac{{8.82}}{{100}} = 0.082\)

Step 2 – Determine the Formula of Average mass

Average mass = \(\sum\limits_i {{{(fractional\,\,abundance \times isotopic\,\,mass)}_i}} \)

Average atomic mass = (0.9090×19.99)+(0.0026×20.99)+(0.082×21.99)

 Avg. atomic mass = (18.17+0.054+1.80) amu

                                = 20.02 amu.