Question

How to sove this "percentage" problem?

From the average value of the relative atomic mass of potassium, 39,10, the relative atomic masses of the isotopes `K^39`, 38,9637; `K^40`, 39,9639; `K^41`, 40,9693 and the percentage of the least represented isotope `K^40` 0,0117%, calculate the percentage for other isotopes, `K^39`and `K^41`. How can I solve this?

Answer 1

The isotopic percentages are `""^39"K" = "93,50 %"` and `""^41"K" = "6,50 %"`

Explanation:

The relative atomic mass is a weighted average of the individual atomic masses.

That is, we multiply each isotopic mass by its relative importance (percent or fraction of the mixture).

If we let `f_i` represent the fraction of isotope `i`, we get two equations:

`f_39 + f_40 + f_41 = 1`

`f_39 + f_41 = 1 - f_40 = 1 - "0,000 117"`

`bb((1))color(white)(m)f_39 + f_41 = "0,999 883"`

The correct atomic masses are `""^40A_text(r) = "39,9640", ""^41A_text(r) = "40,9618"`, and `A_text(r)("Ca") = "39,0983"`

`f_39 × ^39A_text(r) + f_40 × ^40A_text(r)+ f_41 × ^41A_text(r) = A_text(r)("Ca")`

`f_39 × "38,9637" + "0,000 117" × "39,9640" + f_41 × "40,9618" = "39,0983"`

`"38,9637"f_39 + "40,9618"f_41 = "39,0983 - 0,004 676"`

`bb((2))color(white)(m)"38,9637"f_39 + "40,9618"f_41 = "39,0936"`

From Equation 1,

`bb((3))color(white)(m)"f_39 = 1 - f_41`

Substitute Equation 3 into Equation 2.

`"38,9637"(1-f_41) + "40,9618"f_41 = "39,0936"`

`"38,9637"- "38,9637"f_41 + "40,9618"f_41 = "39,0936"`

`"1,9981"f_41 = "0,1299"`

`f_41 = "0,1299"/"1,9981" = "0,0650"`
`f_39 = 1 - f_41 = "1 - 0,06501 = 0,9350"`

∴ The calculated percentages are `""^39"K" = "93,50 %"` and `""^41"K" = "6,50 %"`