Balancing Chemical Equations (without Resorting To The Dark Arts)

Whenever I take on a new GCSE Chemistry student, invariably the topic they most want to learn is how to balance chemical equations.

They tell me that their school teacher often uses a hand-waving method akin to black magic, where balance only appears out of a fog of "doubling this coefficient, while acknowledging that this has a knock-on effect on something that you thought you had already bolted down, but then re-adjusting that causes another thing to go out of kilter etc etc". However, seemingly out of nowhere, you suddenly realise the equation has become balanced.

So here I am going to describe a simple method using algebra no more difficult than x + x = 2x that works 100% of the time without going on a wild goose chase.

First some basic ground rules:
Rule 1: You cannot change any chemical subscri pts, only coefficients you can drink H₂O but you wouldn`t want to drink H₂O₂ just for the sake of adding an extra Oxygen to the balance!

Rule 2: If one molecule of H₂O contains 2 Hydrogen atoms and 1 Oxygen atom, then x molecules of water contain 2x (two times x) Hydrogen atoms and x atoms of Oxygen

Rule 3: Chemists are lazy and whenever there should be a subscri pt of 1, such as in H₂O, the 1 is invisible

Rule 4: When a compound contains a bracket, every atom in the bracket adopts the bracket subscri pt

So, let s choose a really tricky equation to balance
Fe₂(SO₄)₃ + KOH ->& K₂SO₄ + Fe(OH)₃ & & & (unbalanced)

But what actually makes this so tricky? For a start, there are brackets to trip unwary atom-counters up, and then there are atoms of Oxygen in each compound.

Start by adding placeholder letters where the coefficients would normally go. You can choose a, b, c, d etc but I find the c can get mistaken for Carbon, so I often use w, x, y and z. Of course, if you re balancing equations involving Tungsten, Xenon, Yttrium or Zinc, think again!

wFe₂(SO₄)₃ + xKOH ->& yK₂SO₄ + zFe(OH)₃

Now, for each type of atom, count them up, but include the fact that you have w, x, y or z molecules. For example, in the reactants, there are 2 atoms of Fe in each molecule of Fe₂(SO₄)₃, and w molecules, so 2w Fe atoms in total& in the products, there are z atoms of Fe. Since no atoms can be created or destroyed during the reaction, 2w must equal z (ie whatever the value w turns out to be, z must be double it)

Hence for all the types of atoms we have:
Fe: 2w = z
S: 3w = y
K: x = 2y
H: x = 3z

The slightly more difficult one is Oxygen, because atoms must be counted in more than one compound in the reactants and in the products. Other than that, it s the same process.

O: 12w + x = 4y + 3z

Looks scary, but all this can be simplified by arbitrarily setting one of w, x, y or z & equal to 1 as a starting point, and see where it leads us:

So let`s assume w = 1 (in a minute, I`ll look at what happens if another starting letter is set to 1 instead.) If w = 1, the equations change to:
Fe: 2 = z
S: 3 = y
K: x = 2y
H: x = 3z
O: 12 + x = 4y + 3z

Straight away, using just the Fe, S and K equations, we see that z = 2, y = 3, and x = 6

Hence
Fe₂(SO₄)₃ + 6KOH ->& 3K₂SO₄ + 2Fe(OH)₃

So what would happen if you did choose something other than w as 1? Let s assume x = 1 instead. The equations change to:
Fe: 2w = z
S: 3w = y
K: 1 = 2y
H: 1 = 3z
O: 12w + 1 = 4y + 3z

From K: y = 1/2
From H: z = 1/3
From Fe: w = z/2 = 1/6

ie w = 1/6
x = 1
y = 1/2
z = 1/3

The ratios are the same as before, but we have to scale them all to be integers. To do this, multiply all of them by the largest denominator (ie 6)

Hence w = 1, x = 6, y = 3, and z = 2, exactly the same answer as before.

Always go back to the original equation with your proposed coefficients to double check everything tallies OK.

Dr Tony