Balancing Chemical Equations
Date: 09/21/2000 at 20:02:17 From: Dan Kneezel Subject: Balancing chemical equations I know this question has its roots in chemistry, but I was highly intrigued and figured I'd pass it along and see what the pros could do with it. Is it possible to create a finitely terminating algorithm to balance a chemical equation, i.e., find the coefficients to balance this example problem: _C8H8 + _O2 --> _H2O + _CO2? I know that both the introductory and AP level classes that I've taken suggest solely the trial and error method, but it seems highly likely that a more formal approach is both possible and time efficient. I thought about the question for a little while, though not with much concentration, I don't have time to. I suppose it could be solved with matrices and/or simultaneous equations. If there is a method for balancing the equations mathematically, please pass the information along. My math background should be extensive enough that I will at the very least understand the gist of what is being said. If it is not possible to create a finitely terminating algorithm, please explain why. Thank you in advance, Dan Kneezel
Date: 09/24/2000 at 10:18:35 From: Doctor Ian Subject: Re: Balancing chemical equations Hi Dan, You're right, simultaneous equations can be used to balance an equation. To take your example, _C8H8 + _O2 --> _H2O + _CO2 we can assign unknowns (variables) to the blanks, uC8H8 + vO2 --> xH2O + yCO2 which leads directly to a system of equations: C: 8u = y H: 8u = 2x O: 2v = x + 2y One thing you can say for sure is this: The solution won't be unique. (For example, you can always multiply all the unknowns by any integer, and get another 'solution'.) So you'll always have more unknowns than equations. One way to proceed is by eliminating variables. For example, we can substitute 8u for y wherever y occurs, giving us 8u = 2x 2v = x + 2(8u) = x + 16u Then we can substitute 4u for x wherever x occurs, giving us 2v = 4u + 16u = 20 u v = 10u This delineates the space of solutions. Once you choose a value for v, the other coefficients are determined. I hope this helps. Write back if you have more questions, about this or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.