CORRESPONDENCE BALANCING EQUATIONS ALGEBRAICALLY If we treat quantities to the right of the vertical line To the Editor as positive and those to the left as negative, we have DEARSIR: Our purpose in asking students to balance equations For 0 : + y-4=0 should surely be to teach chemical principles. The ionFor H : -2% + 2 y -3 = 0 ; electron method probably has the greatest instructional value. Those who nevertheless prefer the algebraical hence y = 4 and x = 5/2. We may then double the method ought to use it elegantly. A large number of whole equation, to clear of fractions. Whenever y represents molecules of water, as in the the substances appearing in any chemical equation may be balanced with respect to one another by simple in- example just given, y may be eliminated automatically spection. There probably never remain more than by calculating htke the atoms of oxygen minus the atoms of hydrogen, instead of calculating the atoms for these two others, to be treated as unknowns. By a group we shall mean a set of substances whose two elements separately. In the above example this porportions relative to one another can be determined would give by inspection, independently of the rest of the equation. In the example below, we assume l molecule of =no4, then see at once that there must be 1 whence x = 5 / 2 . molecule of KHSO4,l of MnSOa, and hence 2 of HzSOa. In the next example the determinant elements are These four substances form a group. The elements Cu and H, and the non-determinant elements, used (here K, Mn, S) which are considered in adjusting pro- in the final summation, are N and 0: portions within the group, and which do not occur outside the group, we shall call determinant elements. A Cu HNOs + Cu(NO& + N O H20 group of n substances will contain n - 1 determinant Molecules: 1 2% 1 Y Z Atoms of N : Y elements. 6 Y x Let us next assume x molecules of H2Cz04,whence we Atoms of 0 : see that there will be 2x molecules of Con. The element C is here the determinant element for a group of two substances. Finally, let the number of molecules of HzO be represented by y. Hence z = 4/3 and y = $3. Clear of fractions by Since there are two unknown quantities we must get multiplying the equation through by 3. We might two independent equations, by counting atoms for the have eliminated y automatically by calculating atoms two nun-determinant elements. 0 and H: of oxygen minus atoms of nitrogen, instead of calculating KMnO, HzS04 H G O I ---f atoms for these two elements separately. Molecules: 1 2
+
+
Atoms of 0 : Atoms of H :
4
+
+
8 4
4x 22
-
C
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