Question #6aa96

1 Answer
Jul 10, 2017

Answer:

#"114.9 u"#

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 #100%#.

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 #100%#.

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 #x# #"u"# to be the isotopic mass of indium-115, you can say that you have

#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#

#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.