# Question #654a6

##### 1 Answer

Here's what I got.

#### Explanation:

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. * added* the abundance of bromine-81 with the atomic mass of bromine-81 instead of

*these two values.*

**multiplying**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 **average atomic mass** of bromine is equal to

*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

#"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

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