N is the numher of species. The total numher of moles of element k in the system is represented by R k , and M is the numher of elements. Equation (4) can he used to balance a chemical equation involving N species and M elements in the following way. We will use the M simultaneous equations to calculate the numbers of moles x i of N - 1 species required to make one mole of the remaining species. We can do this only if N = M 1. I n order to illustrate this process we will consider a specific example. Doris Kolh (15) has recently discussed halancing complex redox equations by inspection. Her most complicated redox equation provides a good example for use of a hand-held calculator. Without coefficients the reaction is
To put the equation in more familiar form we multiply through by five to clear of fractions and put the species with the negative roefficients on the right-hand side.
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There are 5 elements and so there are 5 element-halance equations of the form of eqn. (4). To calculate the numbers of moles of Ph(N312, C r ( M n 0 4 ) ~Cr20s, , MnOa, and NO required to make one mole of Pbj04, we have to solve the simultaneous linear equations
There is a row for each element and a column for each species, except for the species Ph304 which we are using to define the B column. The unknown numher of moles of Ph(Nd2, Cr(MnO&, Cr203, MnOz, and NO are represented by xl, xz, xa, x4, and zs,respectively. These equations may be written more conveniently in matrix notation (16,171.
The first matrix is referred to as the system formula matrix. The matrix on the rieht - is the element-abundance vector written as a column matrix. In order to understand how this problem is solved with a hand-held calculator we need to introduce some linear algebra. Matrices are represented with hold-face roman type, and so eqns. (4) and (7) may be written Ax=B
(8)
The same result is obtained no matter which species is put in the B rolumn. This formulation of the problem makes it easy to see several ways we can get into trouhle in trying to balance a chemical eqnatioh. 1) Too few species. If we eliminate any one of the species, we do not have a square matrix which is required for the inversion. 2) Too many species. If we add NO2 to the list of species, we will have six unknowns but only five linear equations. However. we can substitute NO? for NO in the statement of the problem. 3) Two elements in constant ratio. If, for example, N and O had occurred in these five species only as Nos, the five equations would not have been linearly independent, and we could have calculated the numbers of moles of only four species. The system effectively contains four "elements." 4) Ions. If ions are involvid, electrical charge has to be conserved throuah an additional linear eon. (14). Now let us return to the more general questions, given above, which are addressed hy Smith and Missen (14). If a closed chemical system contains S species and C "elements," the numher R of independent reactions required to represent all possible chemical changes in the system is given by In the example given above R = 1. If we add Pb(NO& to the list of species, R = 2. We can obtain the stoichiometric coefficients for the second chemical reaction by substituting Pb(NO3)2for Pbs04 in the above procedure. If we want to halance the eauation for the formation of PblN01Ig. . ". ", instead of Pb304, we simply multiply A-' given in eqn. (10) by the column matrix for P h i N o ~ ) ~ , . When S > C + 1,we h G e to be careful in selecting the suecies for the system formula matrix. These soecies have to ancing redox equations. They are not necessary for balancing equations but may he useful for other reasons.
Mass Spectral Analysis of Halogen Compounds David K. Holdsworth
If matrix A has an inverse A-I, then x = A-' B
University of Papua New Gu~nea BOX320 University P.O. Papua New Guinea
(9)
The program for the hand-held calculator can obtain the inverse A-' and multiply it hy B to obtain, in this example, the numbers x, of moles of Ph(NaI2, Cr(MnOa)z, Cr203, Mn02, and NO required to make a mole of Ph304. The result, expressed in the form of eqn. (9) is
The inverse of the svstem formula matrix is simwlv . . eiven " for interest; it is an intermediate step in obtaining the solution eiven in the column matrix on the left. The values of x riven Fn the column matrix on the left are the stoichiometriccoefficients in the balanced chemical eauation
Chemistry students can readily note the presence of one chlorine or one or two bromine atoms in an organic molecule by examining a mass spectrum and recognizing the characteristic molecular ion abundance patterns, spaced a t twomass-unit intervals, 3:1,1:1 and 1:2:1. Other halogen combinations give patterns that are more difficult to interpret. The abundances of some peaks may be less than 5% of the base peak and would not normally be plotted on a line diagram. A pocket calculator can be programmed to decide unequivocally and display the halogen combination in a rnolecular-ion or fragmentation-ion cluster. It can be seen from Table 1that it is not possihle to indicate particular halogen atoms in an ion-cluster by examination of the ( M 2)lM or (X 2llX percentage values alone, since certain halogen combinations give practically identical values. However,
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Volume 60
Number 2
February 1983
103
Table 1. Abundance Percentages of Isotopic Peaks in the Mass Spectra of Organic Halogen Molecules. (Calculator Conditional Limits Are Indicated.) Halogen Combination
% ( M + 2)IM or % ( X + 2)lX
% ( M + 4)/M or % ( X + 4)/X
Molecular Ion (Mt) and Fragmentation Ion (Xf) Isotopic Peaks
Table 2.
Mt(*)
or xt mass
Molecule ~n mass spectrometer
352' octachlorohexa-1,3,5-tr#ene 317 h e ~ a ~ h l o r ~ d l f l ~ ~ r ~ p e n t a d308' dene 1.1-dlfluoroperchloropropane 249 1 2.3.5-tetrachlorobenzene 214' pentabromofluoroethane 340 1-chloro-3 6dlbromobenzene 268' tr~bromofluoromethane 189 1.1.1-tr~chloro-3-bromopropene189 1,4.5-tr~chloroanthraq~1none I * 2-br0mo-4-chlor0phenoi 206' 2-bromo-l 2-d~chloroorooane 111
octachiorohena-l,3,5-tr1ene
% (M+ 2)lM. % (X+ 2)/X
% (M+ 4)IM. % (X+ 4)IX
273 229 192 161 127 395 228 195 160 97 130 64
312 225 155 101 62 578 157 96 73 29 30 10
Calculator d~splay Halogens in cluster CIS CII Cia CIS cI4 69 Br,CI Br2
BrCI, cI3 BrCl CI7
taining five and three chlorine atoms respectively at mle 199, 201, 203 and mle 117, 119,121 in hexachloroethane (M+ = 234). A compilation of mass spectral data (18) listing the ten most abundant peaks yielded suitable information on about 100 halogen compounds. The calculator made the correct decision in every example where the molecular isotopic peaks M, M 2, and M 4 or the fragment isotopic peaks X, X 2, X 4 were available. A selection of results is given in Table 2. (18, 19) Occasionally, the M 1peak was also listed enabling the C, H, N, 0 combinations for the rest of the ion to be predicted
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(20).
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Editor's Note
In the recent article by House (J. CHEM. EDUC.,59, 132 (1982)) a calculator program for carrying out Job's method was described. The program provides an alternative to the usual graphical method, and it can he used to find a break point in titration or other data. However, the method of intersecting tangents does not always work in determining composition of complexes. For a complete discussion see T. W. Gilbert, Jr., J Phys. Chem ,63,1788 (1959). Literature Cited
abundance data of the first and second isotopic only . peaks . ~i 4 i 1 ,I( 111 rs,r 111, prv,.~it 11:11u;~11 c t nil~ini~~icm I.)h1. d e f e r ~ l l i n r d .I I , I ' G ~ I I I ~L IW:IIW~LII ~I ~ I I I Y1, u r l h t . r i, " .L"." ~,"..-.-.
System,6. Lutus, United Softw& New
YO;^, NY
a,."
131 (a)Yeh,C., QCPE, 11, #332(1971).(blGnrdon,M.,Pople,.l., QCPE, 10, #136(1974). (
