# Question #0f03b

##### 2 Answers

The answer is indeed **(b)**.

#### Explanation:

The thing to remember about **isotopes** is that they contribute to the *average atomic mass* of the element **in proportion** to their **abundance**.

That is, the average mass of the element will be **closer** to the atomic mass of the **most abundant isotope**.

So right from the start, you can look at the values given to you and say that **more abundant** than *closer* to the atomic mass of

I'll try to use a more intuitive approach here.

If the two isotopes were to have a

#"avg. atomic mass" = ("85 u " + " 87 u")/2 = "86 u"#

Now, notice that the difference between the atomic masses of the two isotopes is equal to

#Delta_"mass" = "87 u" - "85 u" = "2 u" = color(red)(4) * "0.5 u"#

This means that **for every**

#(100%)/color(red)(4) = 25%#

**increase in abundance** from **closer** to the average atomic mass of the more abundant isotope by a margin of

In your case, the average atomic mass is equal to

#"85.5 u" = overbrace("86 u")^(color(purple)("50% - 50% split")) - "0.5 u"#

So the average atomic mass is now **higher** than

This means that you have

#"Abundance """^85"Rb" = 50% + 25% = color(green)(75%)#

The abundance of

#"Abundance """^87"Rb" = 50% - 25% = color(green)(25%)#

Therefore, **three times as abundant as**

Alternative approach.

#### Explanation:

Here's another approach to use in order to find the abundances of the two isotopes.

As you know, each isotope will contribute to the *average atomic mass* of rubidium in proportion to their abundance.

#color(blue)(|bar(ul(color(white)(a/a)"avg. atomic mass" = sum_i i xx "abundance"_icolor(white)(a/a)|)))#

Here **atomic mass** of an isotope **decimal abundance**, which is simply *percent abundance* divided by

#color(blue)(|bar(ul(color(white)(a/a)"decimal abundance" = "percent abundance"/100color(white)(a/a)|)))#

For example, if an isotopes has a **percent abundance**, it will have a

#"decimal abundance" = 13/100 = 0.13#

So, you know that your element has two stable isotopes, **decimal abundance** of

This is the case because the abundances of the two elements must add up to give

You know that the average atomic mass of rubidium is

The equation will thus take the form

#85.5 color(red)(cancel(color(black)("u"))) = 85color(red)(cancel(color(black)("u"))) xx x + 87color(red)(cancel(color(black)("u"))) xx (1-x)#

#85.5 = 85x + 87 - 87x#

Rearrange to solve for

#87x - 85x = 87 - 85.5#

#2x = 1.5 implies x = 0.75#

So, the abundances of the two isotopes will be

#"For """^85"Rb: " overbrace(0.75)^(color(purple)("decimal abundance")) = overbrace(75%)^(color(green)("percent abundance"))#

#"For """^87"Rb: " 1 - 0.75 = overbrace(0.25)^(color(purple)("decimal abundance")) = overbrace(25%)^(color(green)("percent abundance"))#

Once again, this shows that **three times** as abundant as