Application of Diophantine to Problems in Chemistry
Roger Crocker
Queen Elizabeth College London University London England
T h e mathematical method of diophantine equations is shown to apply to two problems in chemistry. First, in the balancing of chemical equations, the diophantine approach, in addition to its intrinsic interest, has the advantages of mechanical application and of illustrating the variety of possible methods. Second, in determining the molecular formula of a compound, we are enabled to dispense with the percentage composition by weight (or the empirical formula), while conserving the advantage of rapid, mechanical solution. The conclusion briefly discusses some conditions a problem must satisfy to he approachable bv the method of dio~hantineeauations. These applications suggest that diophantine equations may prove useful in attacking other problems in chemistry. Suppose one has an equation in two or more unknowns each having integral coefficients and raised to a positive integral power. One then wishes to find just integral solutions (that is, solutions in which every unknown is an integer) to the equation. Furthermore, integral unknowns should make every term integral; in fact there should then be no non-integers anywhere within the equation. Then the equation is said to be a diophantine equation. If the equation is of the form ax by cz . . . = d where every unknown x, y, z, . . . and every coefficient a, b, c, . . . must be integral, then one has a linear diophantine equation. All the diophantine equations in this article are linear.
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Balancing Equations
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Given the chemical equation zA.,Bs,C,,. . . + YA~,B&,,.. . ... z'A.,'Bs'C, ,'... + ylA.,'B*'C,:'.
+
. + . . . (1) where a,, bl, cl, . . . , a b c , . . . are positive integers or 0,and where x, y, . . . , x', y', . . . are the unknown coefficients of the reactants and products. (The rest of the notation is obvious, the equation being written in standard form.) Then
Actually in the given applications,"integrd" may be replaced by "positive integral." While "positive" is not absolutely necessary in order that the problem be diophantine, "positive" is necessary to find solutions that make physical sense, as the unknowns must bephsitivein physical reality. * "Set of values (satisfying)" and "solution" are really the same thing. Actually there are an infinite numher of solutions; however one need find (in this application) only one solution provided it leads to positive (integral) vdues for all the unknowns in eqn. (2). Hence the indefinite article in "a solution" in some places in the text.
That is, there is one equation for each separate element (A, B, C . . .) in the reaction, expressing the conservation of the number of atoms of that element, and one unknown for each term (xA.,B&, . . . , . . .)-reactant or product-in the reaction. Each of these equations is a diophantine equation as x, y, . . . , x', y', . ., al, an, . . . , bl, b2, . . . , . . . are all integral.' Now if the number of terms n' > the number of elements n, one has a system of n diophantine equations in n' unknowns, and the number of unknowns is > the number of (diophantine) equations by n' - n. Thus, after elimination of the unknowns, one gets a single (linear) dio1 unknowns, which, phantine equation in n' - n since there are no constant terms (in this application), can be solvedZvery easily by simple rules or by trial and error. This situation (n' > n) occurs extremely often, particularly where n' - n = 1, a case in which the resulting diophantine equation (having two unknowns) can he solved immediately. n' - n = 2 is also a case in which the resulting diophantine equation can he solved easily (see Examples). I n ionic equations, one also gets an equation analogous to the equation for each element, expressing conservation of charge instead of atoms of mass. (Thus in an ionic equation, the number of equations n = number of separate elements 1.) If n' = n, one has n equations i n n unknowns, a system which is immediately completely determined, and of course no resulting diophantine equation occurs after elimination of unknowns. If n' < n, the same situation occurs (actually the system seems overdetermined but the extra equations will be consistent with the others in any cdrrectly analyzed physical system). Of course both these latter situations may be solved by simple elimination of unknowns. Given system (2) with n' > n, it is of course possible to obtain several different resulting diophantine equations, depending on what unknowns one chooses to eliminate. It does not always follow, however, that a solution of a resulting diophantine equation will immediately give an acceptable solution to (2), for several of the unknowns in (2) may turn out to be fractions (see Example 11). However, one can multiply the value for each unknown in (2) by the least common multiple (lowest common denominator) of the denominators of these fractions-immediately determined-to find a solution to (2); this can be done since there are no constant terms in (2). (This multiplying will subsequently be referred to as "clearing of fractions.") I n general, if two or more independent sets of values (sets that are not simple integral multiples of each other) satisfy (2), then they all balance eqn. (1) and are possible; in that case, eqn. (1) can be balanced in more than one independent way.
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Volume 45, Number 1 1, November 1968
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731
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Example I
xKMn01
+ yHnSOl + ZHZOI
dK2S04
y'MnSO4
+
+ z'HzO + ~ ' 0 % (3)
+ 4y + 22x == y47' + 4y' + z' + 2w'
42
for 0 for Mn for K for S for H
z = %'
221
+ 2zy == z22'' + y'
Example 111
22
or an equivalent diophantine equation in three variables. This is easily solved by inspection: x = 3,z = 1 , w' = S is one solution corresponding to which y' = 2, x' = 1,y = 3, z' = 4. Therefore,
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+ 3H.SOl + H2O1
KxSO,
+
=SO2+ YCIO~ zH20 x = 2'
This system is very easily reduced to
2KMnOd
into the form ax = by (a,b positive integers) which has as a solution x = b, y = a. This is the only independent solution from which one obtains (clearing of fractions if necessary) the only independent solution of eqns. (2) and (1).
+ 2MnSOa + 4H10 + 301
Other independent solutions to (3) found by inspection of (3s) arex = 2, z = 3, w' = 4, x' = 1, y' = 2, y = 3, z' = 6 and x = 2, z = 5, w' = 5 from which y' = 2, x' = 1, y = 3, z' = 8. I n fact x = 2, 2 = 1 29, where q is any non-negative integer, gives independent sets of solutions to (3a), all of which give independent sets of values which satisfy and balance (3). In this example n' - n = 2. Usually when this is the case, the resulting diophantine equation has been or can he put into the form of (3a), say m l by, = czl (where a, b, c, are positive integers and XI,y,, zl,represent an appropriate triple of the above unknowns). This always has as a solution XI = c, y, = c, zl = a b from which one obtains (clearing of fractions if necessary) a solution to (2) and (1). More generally, XI = cg, yl = ck, zl = ag b k ( g, k any positive integers) from which one can obtain an infinity of mutually independent solutions of (1).
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+
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22 = z' -y = - 2 2 ' + z 1 -
+22z
42 f 4~ f
++ y'2w'+ 38'
= 2' = z' 4%= 42'
+ 4 ~ +' 122' + w'
for K for Mn for Fe for S for H for 0
Thissystem is easily reduced to or an equivalent diophantine equation in two variables. x = 1, w' = 4 is a solution of eqn. (4a) from which x' = 1, y' = 1, z = 9/2, z' = 5/2, and y = 5; clearingof fractions (the lowest common denominator is 2), gives a solution for eqn. (4) of x = 2, w' = 8, x' = 2, y' = 2, z = 9, z' = 5, and y = 10. This is the only independent set of values satisfying eqn. (4). For if w' = 4x, then x' = x, y' = x, x = x, z = (9/2)x, z' = (5/2) x, and y = 5xsolving for the other unknowns in terms of x, and thus zKMnO*
+
+ SzFeSO, + 9/2xHaS04 xMnS01
zKHSOI
5/2zFel(SO&
+
+ 4xH20
from which, clearing of fractions, one gets the solution mentioned above, as x "divides out." Here n' - n = 1. In general, when n' - n = 1, the resulting diophantine equation always has or may be put 732 / Journal of Chemical Education
for CI far H (conservation of charge)
This system is very easily reduced to x = 3y or to an equivalent diophantine equation in two variables. Taking y = 1, it follows x = 3, z = 3, x' = 3, y' = 1, and z' = 6 which is a solution of (5). Here n' - n = 1 and again, there is only one independent set of values (the set just above) balancing eqn. (5). Finding the Molecular Formula
Suppose that a substance with a molecular weight M contains two elements A and B with atomic weights a and b. Then for a molecule of the substance containing x atoms of A and y atoms of B, az
+ by = M
(6)
wherex and y, the unknowns, are integral (and positive). To have a diophantine equation, a, b, and M should he integral. (If a and b are integral, then M will be.) a and b are practically integral for many common, important elements. I n fact, if I. and I, denote the integers nearest a and b, and if a - I, and b - I, are so small for these elements that - ' A < (a - I.)z (b - I ~ ) < Y '/z (6a) for x and y reasonably small enoughto be found in the molecule, then
+
=
IM
(7)
where I , is the integer nearest M. Equation (7) is a diophantine equation which may easily be solved3 for x and y. If several sets of (positive) values for x and y satisfy (7), one may substitute them in (6) and find which one will satisfy (6)-at least with minimum error (deviation from M). This set will be the correct set of values for x and y; that is, the solution to eqn. (6). (Also a set of values satisfying (7) may contain an x or y too large to reasonably expected for the size of the molecule, and hence can be eliminated on that account.) This reasoning extends to substances containing three or even more elements. Thus given a substance of molecular weight M containing elements A, B, C, . . . with atomic weights a, b, c, . . . ; if x, y, z, . . . represent the number of atoms of A, B, C, . . . ,respectively, in a molecule, az+by+cz+
where x, y, z, . . too). If -'/%
