Naturally occurring rubidium consists of just two isotopes. One of the isotopes consists of atoms having a mass of 84.912 amu; the other of 86.901 amu. What is the percent natural abundance of the heavier isotope

The percent natural abundance is 27.95% for the heavier isotope (86.901 amu)

Explanation:

The atomic mass of Rubidium (Rb) is 85.468.

To find the percent natural abundance, first we need to use the following equation:

84.912x + 86.901(1-x) = 85.468

x represents the percent natural abundance, in basic decimal form

In order to find x, we will use the distributive property, combine like terms and then move the constant to the other side to get x.

We are first finding the percent natural abundance of the first isotope to make this easier for us.

84.912x + 86.901(1-x) = 85.468

-> 84.912x + 86.901- 86.901x = 85.468

-> 86.901 - 1.989x  = 85.458      (like terms were combined)

-> Subtract both sides by 86.901

-> Divide both sides by-1.989x

x = 0.720462544

Now turn this into a percent by multiplying by 100.

--> 72.05% for the first isotope (the one that contains 84.912 amu)

To find the natural abundance of the other isotope (the denser, heavier isotope), you could just subtract 72.05 from 100 to save time, but the usually desired way (the way that your teacher would probably want you to do) is down below:

We plug this decimal for the x-value into the parentheses from earlier:

(1 -x) ---> ( 1 - 0.720462544 ) = 0.279537456

Now we just need to convert this to a percent by multiplying by 100

0.279537456 * 100 = 27.95% of the heavier isotope (86.901 amu)

answer: c.

explanation: