Balancing chemical reaction equations by inspection
Balancing by inspection is the most basic method used. It works best for simple equations. More complicated ones require experience - if you don't have it and you are not afraid of relatively simple mathematics, you will probably find the algebraic method more effective.
How to balance an equation by inspection? You should follow these four basic rules:
- Start with the most complicated compound. Assume there is only one molecule of this compound taking part in the reaction.
- Start balancing atoms present in the lowest number of compounds. Don't be afraid of using fractions at this stage.
- Balance atoms present in most compounds at the end.
- Once all atoms and charges are balanced, remove fractions and find set of the smallest coefficients.
Let's try with ethylene glycol combustion:
C2H6O2 + O2 → CO2 + H2O
The most complicated molecule is that of ethylene glycol, so let's put 1 in front of it. Technically speaking if there is no coefficient, it means there is only one molecule of the compound, but we will put ones into the equations now to differentiate between molecules we have already tried to balance, and those that are not touched yet.
1C2H6O2 + O2 → CO2 + H2O
Both carbon and hydrogen are present in two molecules, oxygen is present in all four molecules. Let's start balancing with carbon and hydrogen. There are two atoms of carbon in ethylene glycol molecule, so let's add 2 in front of CO2.
1C2H6O2 + O2 → 2CO2 + H2O
Carbon is balanced, what about hydrogen? There are six hydrogen atoms on the left side of the reaction, so we need three water molecules on the right to balance hydrogen:
1C2H6O2 + O2 → 2CO2 + 3H2O
Time for oxygen balance. There are seven atoms of oxygen on the right - four in molecules of carbon dioxide and three in water molecules. To balance equation thus we need seven oxygen atoms on the left. Two are already present in the ethylene glycol, so we need five more - 2 1/2 oxygen molecule will do. Remember, we can use fractions at this stage:
1C2H6O2 + 2 1/2O2 → 2CO2 + 3H2O
Right now there are same numbers of atoms on both sides of the equation - but to finish balancing we have to remove the fraction. To do so let's multiply all coefficients by two:
2C2H6O2 + 5O2→ 4CO2 + 6H2O
And finally equation is balanced.
There is additional rule that says:
- Treat polyatomic groups that don't change in the reaction as if they were kind of large 'atoms' - instead of balancing atoms in the group individually, balance group as a whole.
We will use it to balance equation of reaction between calcium carbonate and phosphoric acid:
CaCO3 + H3PO4 → Ca3(PO4)2 + H2O + CO2
Obviously the most complicated molecule - one that we will start with - is that of calcium phosphate, so let's assume there is one such molecule in the balanced equation:
CaCO3 + H3PO4 → 1Ca3(PO4)2 + H2O + CO2
We will treat PO4 group (note that to be correct we should treat it as a ion, which we will neglect at the moment in notation) as one large entity (we will make it bold so that it will be easier to spot):
CaCO3 + H3PO4 → 1Ca3(PO4)2 + H2O + CO2
Now, instead of calculating individual atoms of phosphorus and oxygen we can balance PO4 as a whole - there are two such groups in calcium phosphate, thus we need two phosphoric acid molecules on the left side of the equation:
CaCO3 + 2H3PO4 → 1Ca3(PO4)2 + H2O + CO2
We also need three calcium atoms on the left:
3CaCO3 + 2H3PO4 → 1Ca3(PO4)2 + H2O + CO2
What is left to balance? Carbon, oxygen and hydrogen. There are three carbon atoms on the left, so we add 3 in front of CO2 on the right:
3CaCO3 + 2H3PO4 → 1Ca3(PO4)2 + H2O + 3CO2
There are 6 hydrogens on the left - once again we need to add 3 on the right side of the equation, this time in front of H2O:
3CaCO3 + 2H3PO4 → 1Ca3(PO4)2 + 3H2O + 3CO2
Last element to balance is oxygen. We can omit oxygen atoms in PO4 group as they are already balanced. Let's count - on the left side of equation we have 9 atoms of oxygen in calcium carbonate. On the right - there are three oxygen atoms in water molecules and 6 in carbon dioxide - nine together. Reaction equation is balanced.
If there are charged species present (for example in case of net ionic reactions) we have to balance charge as well:
Fe2+ + PO43- → Fe3(PO4)2
This one is relatively easy - most complicated molecule is that of the iron (II) phosphate on the right side. Assuming there is one molecule of Fe3(PO4)2 produced we have to add 3 in front of Fe2+ and 2 in front of PO43- - and equation is balanced.
3Fe2+ + 2PO43- → Fe3(PO4)2
It'is easy to check that charge is balanced too - charge on the right side is zero. On the left side we have 3×(+2) + 2×(-3) = 6-6 = 0 - so everything is correct.
But balancing by inspection have very severe limitations and is sometimes of no use. For example take a look at following reaction:
CH3ONa + NaClO2 + HCl → COCl2 + NaCl + H2O
Balancing by inspection leads nowhere. Using first two rules of starting compound selection we can start with balancing carbon and put 1 in front of CH3ONa which leads to conclusion that there is one molecule of COCl2 produced - and that's about all we can easily do. All other elements are present in several compounds, so whenever we try to put coefficient in front of one molecule, it changes at least two other coefficients. It makes us going into circles. Three hydrogen atoms from the CH3ONa form 1 1/2 water molecule - but we can't continue assuming there is 1 1/2 water molecule on the right, as for sure there will be more water on the right side since there is additional source of hydrogen on the left. No matter which element you select, you end with some circular reasoning.
(Thanks to Wilco for the last example).