# Question #6aa96

##### 1 Answer

#### Answer:

#### Explanation:

The problem tells you that indium has two stable isotopes, so right from the start, you know that their percent abundances **must** add up to give

This means that the percent abundance of the indium-115 isotope is equal to

#100% - 4.305% = 95.695%#

Before moving forward, convert the two percent abundances to *decimal abundances* by dividing them by

You will have

#""^113"In: " (4.305 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.04305#

#""^115"In: " (95.695 color(red)(cancel(color(black)(%))))/(100color(red)(cancel(color(black)(%)))) = 0.95695#

Now, the **average atomic mass** of indium is given by the **weighted average** of the isotopic masses of its stable isotopes.

If you take

#overbrace(114.8 color(red)(cancel(color(black)("u"))))^(color(blue)("avg. atomic mass")) = overbrace(112.90 color(red)(cancel(color(black)("u"))) * 0.04305)^(color(blue)("contribution from"""^113"In")) + overbrace(xcolor(red)(cancel(color(black)("u"))) * 0.95695)^(color(blue)("contribution from"""^115"In"))#

Rearrange to solve for

#x = (114.8 - 112.90 * 0.04305)/0.95695 = 114.9#

You can thus say that indium-115 has an isotopic mass equal to

#color(darkgreen)(ul(color(black)("isotopic mass"color(white)(.)""^115"In = 114.9 u")))#

The answer is rounded to four **sig figs**, the number of sig figs you have for the average atomic mass of the element.