How can I calculate the relative abundance of three isotopes? The letters have nothing to do with the periodic table. 22y = 22.341 amu, 23y =...

Denote the abundances in numbers (not percent) as `a_22,` `a_23` and `a_24.` Then if we assume that there are no other isotopes found in nature,

`a_22+a_23+a_24=1.`

Also, by the definition of relative atomic mass:

`23.045 = 22.341*a_22+23.041*a_23+24.941*a_24.`

There is no more information and we have only two linear equations for three unknowns. Usually this means infinitely many solutions.

From the first equation `a_22=1-a_23-a_24,` substitute it into the second equation:

`23.045 = 22.341*(1-a_23-a_24)+23.041*a_23+24.941*a_24,` or

`23.045- 22.341=(23.041-22.341)*a_23+(24.941-22.341)*a_24,`...

Denote the abundances in numbers (not percent) as `a_22,` `a_23` and `a_24.` Then if we assume that there are no other isotopes found in nature,

`a_22+a_23+a_24=1.`

Also, by the definition of relative atomic mass:

`23.045 = 22.341*a_22+23.041*a_23+24.941*a_24.`

There is no more information and we have only two linear equations for three unknowns. Usually this means infinitely many solutions.

From the first equation `a_22=1-a_23-a_24,` substitute it into the second equation:

`23.045 = 22.341*(1-a_23-a_24)+23.041*a_23+24.941*a_24,` or

`23.045- 22.341=(23.041-22.341)*a_23+(24.941-22.341)*a_24,` or

`0.704=0.7*a_23+2.6*a_24.`

Therefore `a_23=(0.704-2.6*a_24)/0.7 approx 1.006-3.714*a_24.` So if we would know `a_24` (or `a_23,` or `a_22`), we could find the other abundances.

To find the relative abundances in percent, we have to multiply `a_24,` `a_23` and `a_22` by `100.` The formal answer for your question is "not enough data." Also note that the atomic masses of isotopes are almost always integers, while your numbers aren't.