Question #654a6

1 Answer
Sep 5, 2016

Here's what I got.


Well, you got the abundance of bromine-79 right, but that's about it.

When you set up the equation to find the atomic mass of bromine-79, i.e. #x#, you added the abundance of bromine-81 with the atomic mass of bromine-81 instead of multiplying these two values.

So instead of

#color(red)(cancel(color(black)(79.904 = 0.5069 * x + 0.4931 + 80.9163)#

you should have had

#79.904 = 0.5069 * x + 0.4931 xx 80.9163" "color(green)(sqrt())#

You know that the average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its stable isotopes.

In this case, you know that bromine has two stable isotopes, bromine-79 and bromine-81. You also know that for these two isotopes you have

#"For " ""^79"Br: " overbrace(50.69%)^(color(blue)("percent abundance")) = overbrace(0.5069)^(color(purple)("decimal abundance"))#

#"For " ""^81"Br: " overbrace(49.31%)^(color(blue)("percent abundance")) = overbrace(0.4931)^(color(purple)("decimal abundance"))#

You also know that bromine-81 has an atomic mass of #"80.9163 u"# and that the average atomic mass of bromine is equal to #"79.904 u"#.

So, what would be the contribution of bromine-81 to the average atomic mass of bromine?

To find that out, multiply the atomic mass of bromine-81 by the isotope's decimal abundance

#"For " ""^81"Br: " "80.9163 u" xx 0.4931 = "39.89983 u"#

What about the contribution of bromine-79?

Assuming that #x# is the atomic mass of bromine-79, you have

#"For " ""^79"Br: " = x xx0.5069 = (0.5069 * x)" u"#

These two contributions must add up to give you the average atomic mass of bromine, so

#79.904 color(red)(cancel(color(black)("u"))) = overbrace(39.89983color(red)(cancel(color(black)("u"))))^(color(brown)("from " ""^81"Br")) + overbrace((0.5069 * x)color(red)(cancel(color(black)("u"))))^(color(darkgreen)("from " ""^79"Br"))#

This means that you have

#x = (79.904 - 38.89983)/(0.5069) = 78.919#

Therefore, the atomic mass of bromine-79 is #"78.919 u"#.

I'll leave the answer rounded to five sig figs.