Balance chemical equations

AP chem is the science course for my 11th year. In our chemistry text book, the only method is (intelligently) trial and error.
1. Trial and error
Pros:
Easy.
Cons:
From the start of learning stoichiometry. I was hoping to find some other way to avoid trials. Reasons include:

  1. Trial and error are only good when all the subscripts are small numbers or large numbers that have large GCDs
  2. Trial and error should not even be considered when all the subscripts are coprime to each other and fairly large(say, 2^37 and 3^11)
  3. Difficult to keep track of each trial
  4. Trial and error was never sexy

The most basic example
H2O -> O2 + H2
now you see there are one less O in the left, so add a 2 in front of H2O so there are 2 oxygen
2H2O -> O2 + H2
There are 4 hydrogen in the left and 2 in the right, add a 2 in front of H2 in the right.
The equation is balanced.
2H2O -> O2 + 2H2

2. Basic Algebra
Pros:
Hyperpwns trial and error. Hypopwns Matrices when equations are easy.
Cons:
calculators can't do much

To begain, we have to propose a mathematical model for a equation.
H2O -> O2 + H2
turns into
x (2(H)+O) = y 2O + z 2H
so x, y, z are the coefficients and H, O are variables with no relationship known.
Then, it's possible to list some relations for O and H
H: 2x = 2z
O: x = 2y
divide by 2 on both sides for H
H: x = z
we simply get
x=2y=z
Nice relationship.
now, assign 1 to the variable with the highest coefficient(which is of course, y). get the value for x and z.
Now, just put x, y, z back into the chemical equation
2H2O -> 1O2 + 2H2
eliminate the coefficient 1...
DONE!!!
you might say "what is the point, trial and error suppose to give me the answer like a LONG time ago.

Think again, what if you meet this chemical equation
C97H127O37 + O2 -> CO2 + H2O
try figure it out with the original method, please do.

Or, follow what I just did, find the relationships, suppose the coefficients are w,x,y,z
C: 97w = y
H: 127w = 2z
O: 37w + 2x = 2y + z
C and H are easy, O is a bit tough. But, anyone who is learning chem suppose to know how to eliminate variables.
substitute y with 97w and z with 63.5 w
37w + 2x= 194w + 63.5w
x = 110.25w
so the list goes
x = 110.25w
y = 97w
z = 63.5w
set w as 1
then
1 C97H127O37 + 110.25 O2 -> 97 CO2 + 63.5 H2O
balanced equation can't have decimals, so multiply each coefficient by some number(this case, 4) and behold... the balanced equation.
4 C97H127O37 + 441 O2 -> 388 CO2 + 254 H2O

3. Matrix
Pros:
Calculators does most of the work
Cons:
It's possible to output answers that fit the algebra equation but does not fit the chemical equation

Anyway, because the finding of relationships, it's also possible to work with Matrix for this problem.
For the same question, I will write them a new form. Note, this time, y will be a negative number since I'm only using pluses and no minuses. just remember to take the absolute value of it when the result come out.
C: 97w + 0x + 1y= 0z
H: 127w + 0x + 0y = 2z
O: 37w + 2x + 2y= 1z

If you are taking precalculus and using the Precalculus with limits: A graphing approach 4th edition[if you are in my pre-calc class, you have one]. Read page 489 to 546 to understand the reason behind each operation. Else, just look at some transformation that solve a system of equation

Take the coefficient for C, make it the 1st row of the matrix A
Take the coefficient for H, make it the 2st row of the matrix A
Take the coefficient for O, make it the 3st row of the matrix A
then take the coefficient of z and create a matrix with 3 rows 1 column and call it matrix B(because it only have 1 column, it's also a vector)

and the coefficient for w,x,y will be the matrix returned from this operation
The matrix A MUST have the same amount of rows and columns! matrix B MUST have the same amount of rows as matrix A and only 1 column.

A-1 * B
This was discussed in page 524, using inverse to solve system of equations.
but the return of the answer are seriously disturbing, it is not guaranteed they are integers.(all numbers are rounded)
w=0.0157480315
x=1.736220472
y=-1.527559055
That's why, introduce the determinant of matrix a
Update the former equation A-1 * B * |A|
Because I'm lazy so I directly quote from wikipedia.
what is a determinant?
"a determinant is a function depending on n that associates a scalar, det(A), to every n×n square matrix A. The fundamental geometric meaning of a determinant is as the scale factor for volume when A is regarded as a linear transformation."
After reading the above, it's obvious the previous answer multiply with the determinant will become integers.
w=4
x=441
y=-388
z= |A| = 254

Cheers~

Remember I said matrix A must have the same amount of rows and columns? What if there are a equation does not give you the power to do so? Like the exactly the same equation as the last one, but with one more O3.
C97H127O37 + O2 -> CO2 + H2O+O3
There will be 4 coefficients except the last one, but there are only 3 variables. Which means 3 rows and 4 columns.
C: 97w + 0x + 1y + 0z = 0a
H: 127w + 0x + 0y + 2z = 0a
O: 37w + 2x + 2y + 1z = 3a

Easy, make a new row.
w/e: 0w + 0x + 0y + 1z =0a
set the number for the new row in B, 0

so the matrices would be

Suppose matrix A is a identity matrix, then there would be a 1 in one block of the row just got expanded. then add 1 in that place.(page 511)

The answer would be
0, 381, 0, 0, 254
and yet, it is correct.

Calculation of the inverse, multiplication between matrix and matrix, matrix and scalar are all done by the calculator. All you have to do is input the matrix and input the equation. If you are using TI-83 plus and you don't know how to do all those operations, find your guidebook that comes with the calculator and read section 10.

Note:
If the determinant = 0, both the mathematical and the chemical equation can not be solved
It's possible to only have an solution to the mathematical equation but no solution to the chemical equation, for example
CH2O5 + O2 -> CO2 H2O
is a chemical equation that can not be balanced
but algebraical solution exist
4CH2O5 - 4O2 -> 4CO2+4H2O
be careful to check for those cases


Comments

Anonymous's picture

helsp

thx helps alot

Mgccl's picture

Glad you like it :)

Glad you like it :)

Anonymous's picture

chem

stepping stones for of mankind as wells for students..

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[...] Here are some other sources that will help you understand the question and formulate your answers: http://mgccl.com/2007/09/20/balance-chemical-equations http://www.science.uwaterloo.ca/~cchieh/cact/c120/reaction.html [...]

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Anonymous's picture

I Hope this works on my

I Hope this works on my midterm exams*_*

Anonymous's picture

NIce. It helped me to help

NIce. It helped me to help my daughter.

Anonymous's picture

taking up your challenge

First, it is not trial and error, you need a systematic approach, looking for elements in one compound on each side of the equation, and balancing them first, but leaving oxygen and htdrogen to the last

so you said balance this
C97H127O37 + O2 -> CO2 + H2O

Carbon in one substance on each side

C97H127O37 + O2 -> 97CO2 + H2O

hydrogen in one compound on each side

C97H127O37 + O2 -> 97CO2 + 63 1/2H2O

remove half by doubling

2C97H127O37 + 2O2 -> 194CO2 + 127H2O

Left with oxygen now

oxygen count to get the extra O2 required
lhs 2x37 + 2x2 = 74 + 4 = 78
rhs 2x194 + 127 = 388 + 127 =515
lhs deficient = 515-78 = 437
need to add218 1/2 more O2

2C97H127O37 + 220 1/2O2 -> 194CO2 + 127H2O

double to remove half

4C97H127O37 + 441O2 -> 388CO2 + 254H2O

check
Carbon lhs 4 x 97 = 388 = C on rhl = okay
hydrogen lhs 4 x 127 = 508 = 2 x 254 rhl = okay
oxygen lhs 4x37 + 441x2 = 148 + 882 = 1030
oxygen rhl 388x2 + 254 = 676 + 254 = 1030

This is explained in detail at this site
http://www.sky-web.net/science/balancing_chemical_equations.htm

Anonymous's picture

Second challenge - the very last "equation" you give

Challenge Two

CH2O5 + O2 -> CO2 H2O
C is currently balanced
H is currently balanced

Oxygens last
lhs = 5 + 2 = 7
rhl = 2 + 1 = 3
therefore original equation has an error, as need to add 4 oxygens to rhl

CH2O5 + O2 -> CO2 + H2O + 2O2

now correct for O2 being on each side by removing the same number of O2 from each side to get zero on the side with the least O2

CH2O5 -> CO2 + H2O + O2 which is the correct answer.

Note how this method even corrected the erroneous original equation.

The removal of species which exist on both sides is part of the standard method used in balancing half equations.

PS CH2O5 - not exactly likely to be a stable entity, is it?
Likely structure H-O-O-O-C=O(OH) or H-O-O-C=O(O-OH) So no need for O2 to make it decompose! Just say Boo! and it would fall apart.

PS forgot the html to give a working link to the site I mentioned.
how to balance chemical equations

PPS love that changing captcha test ;)

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