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Stolyarov’s Formula of Comparative Abundance

by sumo nova

Let us begin this derivation with a hypothetical scenario. There exists an element with two isotopes, N and Q. N has a molar mass of 65.4 grams/mole and Q has a molar mass of 67.2 grams/mole. A sample of this element found in nature has a molar mass of 66.8 grams/mole. Given this data, it is desired to determine the fraction of the sample which is composed of Isotope N.

The conventional approach to this fairly simple problem would be brute, unwieldy force. The modern chemist and mathematician alike would designate x to be the desired fraction and set up the following equation:

65.4x + 67.2(1-x)= 66.8

Solving this is an irritant rather than a stumbling block, as it requires expanding parentheses and performing numerous acts of addition and subtraction that need only be done once! This approach is indeed logically correct with respect to solving the problem, but it amounts to reinventing the wheel. This is where Stolyarov’s Formula of Comparative Abundance can greatly condense the work required.

The hypothesis of Stolyarov’s Formula begins thus: In any given set with two elements, A and B, where a is the magnitude of element A and b is the magnitude of element B, while c is the mean magnitude of the set, the fraction x of element A can be determined as follows….

Since a and b are the sole elements of the set, (1-x) must be the fraction representing the relative abundance of element b. The equation which in this age is too often and mistakenly treated as the beginning of each individual problem is a mere starting point toward a far more elegant formulation.

a * x + b(1 -x) = c
a * x + b – b* x = c
a *x – b *x = c -b
x (a -b) = c -b
x= (c-b)/(a-b)– This is Stolyarov’s Formula of Comparative Abundance.

No more algebraic manipulation is necessary! One needs only insert known values of a, b, and c to determine x. A previously frustrating problem is condensed to a mere single step.

To see how this applies to the hypothetical problem given earlier, we note that a= the molar mass of N (65.4), b= the molar mass of Q (67.2), and c= the molar mass of the sample (66.8).

x= (66.8 – 67.2)/(65.4 – 67.2)= 2/9 or about 0.222. The fraction of Q can be easily determined therefrom as well, either by subtracting the fraction of N in the sample from 1 or by performing a Stolyarov’s Formula calculation with a= the molar mass of Q.

Another application of Stolyarov’s Formula involves alloys of two elements. Given the molar mass of a given sample of an alloy, one can determine the relative abundance of each of the elements within it. Similar situations can be conceived with regard to partial pressures of gases or even mixtures of two different foods that exhibit different prices individually (for example: if a mix almonds and walnuts costs $3.00 per unit volume, and almonds cost $3.56 per unit volume, while walnuts cost $2.78 per unit volume, determine the comparative abundance of almonds in the sample).

To review: In any given set with two elements, A and B, where a is the magnitude of element A and b is the magnitude of element B, while c is the mean magnitude of the set, the fraction x of element A can be determined by Stolyarov’s Formula of Comparative Abundance: x= (c-b)/(a-b).

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