# Question 6aa96

Jul 10, 2017

$\text{114.9 u}$

#### 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 $x$ $\text{u}$ 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 $x$

$x = \frac{114.8 - 112.90 \cdot 0.04305}{0.95695} = 114.9$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{isotopic mass"color(white)(.)""^115"In = 114.9 u}}}}$

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