# Question #c4218

##### 1 Answer

#### Answer:

#### Explanation:

Right from the start, you can look at the atomic masses of the two **isotopes** and at the *average atomic mass* of the element and say that isotope **more abundant** than the other isotope.

An element's **average atomic mass** is calculated by taking the **weighted average** of the atomic masses of its stable isotopes. In other words, each isotope will contribute to the average mass of the element **in proportion** to its *abundance*.

#color(blue)( |bar( ul( color(white)(a/a)"avg. atomic mass" = sum_i "isotope"_i xx "abundance"_icolor(white)(a/a)|)))#

Since the average atomic mass of **closer** to the mass of the **more abundant** than the other.

If you take *decimal abundance* of the first isotope, you can say that

#y = 1 - x#

This happens because the decimal abundances of the two isotopes **must** be equal to

You can thus say that the average atomic mass of element

#"106.33" color(red)(cancel(color(black)(u))) = overbrace(x xx "105.95" color(red)(cancel(color(black)(u))))^(color(purple)("the contribution of" color(white)(a)""^106"X")) + overbrace((1-x) xx "108.95" color(red)(cancel(color(black)(u))))^(color(red)("the contribution of"color(white)(a)""^109"X"))#

Solve this equation for

#106.33 - 108.95 = x * (105.95 - 108.95)#

#x = 2.62/3 = 0.873333#

Since *lighter isotope*, you can say that the decimal abundance of the *heavier isotope* will be equal to

#1 - 0.873333 = 0.126667#

The **percent abundances** of the two isotopes, which are simply the decimal abundances multiplied by

#""^106"X: " color(green)(|bar(ul(color(white)(a/a)"87.333%"color(white)(a/a)|)))#

#""^109"X: " color(green)(|bar(ul(color(white)(a/a)"12.667%"color(white)(a/a)|)))#

As predicted, the lighter isotope contributes more to the average mass of the element because it is more abundant than the heavier isotope.