A sample of rhenium (Re) is #"37.40% Re-185"# and #"62.60% Re-187"#. What are the atomic masses of these isotopes? The sum of the masses of the two isotopes is #"371.9087 u"#. Find the mass of the individual isotopes.

The atomic weight of Rhenium (Re) as found on the periodic table of the elements is 186.207. Let x = the atomic weight of Re-185 and (371.9087 - x) = the atomic weight of Re-187.

Known/Given:
atomic weight of Re = 186.207 u
x = atomic weight of Re-185 in u
(371.9087 - x) = atomic weight of Re-187 in u
abundance of Re-185 = 37.40% = 0.3740
abundance of Re-187 = 62.60% = 0.6260

Unknown:
atomic masses of Re-185 and Re-187

Solution:
Note: In order to reduce rounding errors, I will include guard digits and round to the proper number of significant figures at the end.

1. Multiply x times the abundance of Re-185 and multiply (371.9087 - x) times the abundance of Re-187.

Re-185: (0.3740)(x) = 0.3740x

Re-187: (0.6260)(371.9087 - x) = 232.8148462 - 0.6260x

2. Add the results and set them equal to 186.207.

0.3740x + 232.8148462 - 0.6260x = 186.207

3. Solve for x by subtracting 232.8148462 from both sides and then divide both sides by -0.2520.

0.3740x + 232.8148462 - 0.6260x - 232.8148462 = 186.207 -
232.8148462

0.3740x - 0.6260x = -46.6078462

-0.2520x = -46.6078462

-0.2520x/-0.2520x = -46.6078462/-0.2520

x = 184.9517706 u

4. Atomic weights of Re-185 and Re-187.

x = 185.0 u = the atomic weight of Re-185 (rounded to four significant figures due to significant figures rule for multiplication and division.).

(371.9087 - 184.9517706) = 186.9569294 = 187.0 u = the atomic weight of Re-187 (rounded to one decimal place due to significant figures rule for addition and subtraction...Notice 184.9517706 was subtracted, not 185.0, because rounding is done after all calculations)

The atomic weight of Re-185 is 185.0 u, and the atomic weight of Re-187 is 187.0 u.