Rubidium has two principle isotopes, 85rb (84.911789 amu) and 87rb (86.909183 amu). what is the percentage abundance of the lower mass isotope?

(84.911789x) + (86.909183(1 - x)) = 85.468 --> Multiply out the brackets

84.911789x + 86.909183 - 86.909183x = 85.468 --> Subtract the bold part from both sides

84.911789x - 86.909183x = - 1.441183 --> Simplify the left side of the equation

- 1.997394x = - 1.441183 --> Divide both sides by the bold part to get x

x = 0.721532 --> Convert to percentage
= 72% (2.s.f.)

The teacher is mostly talking about the distillation of a mixture