Boron consists of two isotopes with masses of 10.00 and 11.00. The average atomic mass of boron is 10.81. What are the relative abundances of the two isotopes? To solve this and similar problems, most textbooks present the “X” method in which the fraction of one of the components is assumed to be “X” and that of the other is “1 - X” (1, 2). A new approach, named as the “Swap Method”, that is more “student-friendly” than the “X” method is presented here. In the “Swap Method”, the relative abundances of the components of a binary mixture are determined by following the steps below:

1. Determine the absolute differences between the weighted average value of the property under consideration and the values associated with pure forms of the components.
2. Swap the differences determined in step 1.
3. Divide the values obtained in step 2 by the difference in values of the property associated with pure forms of the two components. The answers are fractions of the two components.

For boron, the differences between average mass and the masses of the two isotopes (10 and 11) are 0.81 and 0.19 respectively. The percentage of boron-10 is 19 (= 0.19/[11-10]).

The mathematical rationale behind this approach is as follows. Let the fraction of the total that has the value (A) for a specific property in a mixture be X, the fraction with the value (B) is Y and the weighted average is C.

Students who normally freeze whenever they hear the statement, “Let us call the unknown X”, seem to like this approach. A majority of the students guestimate well and rarely miss the answer in a test.

Acknowledgment

The author appreciates the valuable inputs received from former student Farris Rushay, and Professor Viet Dung Nguyen at Ohio University Center in Zanesville, OH.

Literature Cited

1. Petrucci, R. H.; Harwood, W. H. General Chemistry: Principles and Modern Applications; Macmillan: New York, 1993; pp 47-48.
2. Kotz, J. C.; Treichel, P. Jr. Chemistry and Chemical Reactivity; Saunders: New York, 1999; pp 77-78.