# How to balance chemical equations

*Chemistry students often struggle with balancing chemical equations, but these tips will make this vital skill easier to master*

## Is there an easy way to balance chemical equations?

I get asked this *a lot* by students wanting help with this notoriously difficult skill. What students want to know is, is there are a set of rules or step-by-step process to follow that, once learned, will allow you to balance even the hardest equation?

The answer is *yes*, there is a method but unfortunately, it’s a mathematical method that uses algebra to find the balancing numbers, so it’s not that suitable for less mathematically-inclined chemistry students.

So, what are the other options?

If you’re a GCSE, A Level or IB student, you can learn the ‘**balancing by inspection**’ method, which is one of four methods used to balance chemical equations. (The other three methods – redox balancing, algebraic balancing, and simple algebraic balancing – I will cover in other blogs.)

## What is balancing chemical equations by inspection?

‘*By inspection*’ suggests some magical process where you stare long and hard at the equation and the balancing numbers mysteriously float into your head. That’s pretty much what happens when you’ve been balancing equations for a long time.

Before you get to that point, the good news is this method can be reduced to a set of rules that, once mastered and practised, will equip you with Jedi balancing skills allowing you to deal with **95% of chemical equations you are likely to meet** (the other 5% are either so difficult you can only use mathematical methods, or you have to use the redox balancing method).

## Principles behind balancing chemical equations

Let’s first state the key principles behind balancing chemical equations:

**1. You cannot destroy atoms**

The total number of atoms on each side of the equation **must be the same. **The number of atoms of each element must therefore be the same on each side of the equation.

**2. You cannot destroy charges**

The total number of charges **must be the same** on both sides of the equation. Charges are particles, so a different number on each side of the equation means some matter has been destroyed. Not allowed.

These first two principles are based on the Conservation of Matter. The other principle you mustn’t violate is:

**3. You cannot change chemical formulae**

Meaning, you may only balance equations by changing **coefficients** (*i.e.* the balancing numbers), not by changing **subscripts**.

Changing the subscripts changes the compound into something else – that would be alchemy, not chemistry.

The above approach is certainly cunning but there is a very big difference between H_{2}O and H_{2}O_{2}!

## Tricks to balance chemical equations

With those basics out of the way, let’s have a look at the rules that collectively make up the ‘balancing by inspection’ method. These rules are shown in action in the worked examples below.

### Rule 1: Look for an element that appears only once on the LHS and RHS

In other words, avoid elements present in more than one compound to begin with. Why? Because if an element appears in more than one compound, each compound will need a coefficient that will potentially unbalance other elements. It’s not a good idea to start out your balancing by creating imbalances.

If you’re unlucky and there isn’t an element that only appears once on each side, start with the one present in the fewest compounds

### Rule 2: Try subscripts as balancing numbers

Often, an element (or ion’s) subscript on one side of the equation will be its coefficient on the other side.

**This is the single most useful trick to remember**. After working through over 350 equations to see which rules are most used (I need to get out more!), I estimate you will use this trick to balance 90% of equations.

You can’t change subscripts so use them instead.

### Rule 3: Look for forced coefficients

If an element or ion appears only once on each side of the equation and is already balanced, its coefficient on both sides **must be the same**

This rule also applies to polyatomic ions (*e.g.* SO_{4}^{2-}, NO_{3}^{–}, CO_{3}^{2-}, PO_{4}^{3-} *etc*.)

### Rule 4: Don’t start with compounds containing many elements

Usually, you should try to balance compounds that contain **as few elements as possible first**

Changing a coefficient in front of a compound containing three elements will potentially balance one element and unbalance two. You are likely to be making life difficult for yourself. This rule is about keeping it simple. Of course, in some equations applying the earlier rules will need you to break this one.

### Rule 5: Remember you can use fractions

Coefficients can be fractions and they are useful for correcting odd/even imbalances and adding single elements

### Rule 6: Sum of charges LHS = sum of charges RHS

Correct negative charge imbalances using electrons.

### Rule 7: Balance water later

Unless the earlier rules force you to, **don’t start with water**

Water is a very common substance in a chemical reaction and oxygen and hydrogen are very common elements. Any number you put in front of water will change oxygen and hydrogen, forcing you to use another coefficient to balance those elements in other compounds.

The exception to this rule is when hydrogen or oxygen have a subscript on one side of the equation that you must use as the coefficient on the other side.

## Worked examples

#### Example 1

_Cl_{2} + _KBr → _KCl + _Br_{2}

- Don’t start with KBr or KCl (
**rule 4**) - Use the subscripts from Cl and Br (
**rule 2**) and it’s balanced:

Cl_{2} + 2KBr → 2KCl + Br_{2}

#### Example 2

_N_{2} + _H_{2} → _NH_{3}

- Equations where an element has an odd number of atoms on one side and an even number the other are common
- Here, there are 3 H’s on the RHS and 2 on the LHS
- Use
**rule 2**and swap the subscripts for H by putting 3 in front of H_{2}and 2 in front of NH_{3}

N_{2} + 3H_{2} → 2NH_{3}

#### Example 3

_C_{2}H_{5}OH + _O_{2} → _CO_{2} + _H_{2}O

- Don’t start with ethanol, C
_{2}H_{5}OH (**rule 4**) – change this, you are going to change the number of C, H and O atoms - Instead, use C’s subscript (
**rule 2**) and put it in front of CO_{2}.

C_{2}H_{5}OH +_O_{2} → 2CO_{2} + _H_{2}O

- The rest is just counting atoms. Notice there are 6 H’s in ethanol and only 2 in water? We can correct that with a 3 in front of water.

C_{2}H_{5}OH +_O_{2} → 2CO_{2} + 3H_{2}O

- Lastly, count and correct the oxygen atoms with a 3 in front of O
_{2}

C_{2}H_{5}OH +3O_{2} → 2CO_{2} + 3H_{2}O

#### Example 4

_OH^{–} → _H_{2}O + _O_{2} + _e^{–}

- First of all, use H’s subscript as the coefficient for OH
^{– }(**rule 2**):

2OH^{–} → H_{2}O + _O_{2} + _e^{–}

- We can sort out oxygen by halving O
_{2}(**rule 5**):

2OH^{–} → H_{2}O + 1/2O_{2} + _e^{–}

- Lastly, balance the charges (which are currently -2 on the LHS and -1 on the RHS) by putting a 2 in front of e
^{–}(**rule 6**):

2OH^{–} → H_{2}O + 1/2 O_{2} + 2e^{–}

- This is now balanced, but if you’re one of those students that doesn’t like fractions in chemical equations, or the question asks for whole number coefficients, just multiply by the denominator (
*i.e.*double everything):

4OH^{–} → 2H_{2}O + O_{2} + 4e^{–}

#### Example 5

_Fe_{2}O_{3} + _CO → _Fe + _CO_{2}

- If you don’t spot the trick you have to use for this one, you can end up going around in circles. The trouble with this one arises because you have 3 O’s on the LHS and 2 O’s on the right
*and*O in CO. - You certainly don’t want to start with Fe
_{2}O_{3}first (**rule 4**) - If it were not for CO containing oxygen, using
**rule 2**to swap oxygen’s coefficients (2Fe_{2}O_{3}and 3CO_{2}) would be a great way to balance oxygen, then you could just balance iron with a 4:

2Fe_{2}O_{3} + _CO → 4Fe + 3CO_{2}

- But now you’re stuck. You’d be forced to put a 3 in front of CO to balance carbon, unbalancing oxygen in the process.
- Instead, make the logical move and use
**rule 2**to balance iron:

Fe_{2}O_{3} + _CO → 2Fe + _CO_{2}

- Now use the ‘forced coefficients’
**rule 3**. - CO and CO
_{2}are the only compounds containing carbon, and carbon is already balanced (there is one atom in each compound containing it on each side). This means the coefficient in front of CO and CO_{2}*must be the same*. - To find it, you could use trial and error (
*e.g.*try 1, 2, 3 and so on until carbon and oxygen are balanced) or solve it with simple algebra - If we let
*x*= the coefficient for CO and CO_{2}, we can write a rule for the total number of oxygen atoms on the LHS and RHS:

Fe_{2}O_{3} + *x*CO → 2Fe + *x*CO_{2}

3 + *x* = 2*x*

Hence x = 3

Fe_{2}O_{3} + 3CO → 2Fe + 3CO_{2}

#### Example 6

_HNO_{3} + _CaCO_{3} → _Ca(NO_{3})_{2} + _CO_{2} + _H_{2}O

- Certainly I would not start with HNO
_{3}or Ca(NO_{3})_{2}because you’re going to upset too many elements changing those coefficients (**rule 4**) - Don’t start with H
_{2}O either because oxygen is present in every other compound (**rule 7**) - Instead, we’ll use the forced coefficient rule again (
**rule 3**), this time applied to a polyatomic ion, NO_{3}^{–} - NO
_{3}^{–}is a**spectator ion**in this equation, meaning it doesn’t participate in the reaction at all. All the NO3^{–}ions on the reactant side get transferred to the product side with no change. What that means is the N and O atoms it contains are stuck together and can be treated the same way you’d treat an atom - We can therefore use
**rule 2**and use the NO_{3}^{–}subscript from the RHS to balance HNO_{3 }on the LHS:

2HNO_{3} + _CaCO_{3} → _Ca(NO_{3})_{2} + _CO_{2} + _H_{2}O

- This now forces the coefficients for CaCO
_{3}, CO_{2}and H_{2}O to be 1. The equation is now balanced.

2HNO_{3} + CaCO_{3} → Ca(NO_{3})_{2} + CO_{2} + H_{2}O

#### Example 7

_Na_{2}SnO_{3} + _H_{2}S → _SnS_{2} + _NaOH + _H_{2}O

- This looks tricky but the above rules will help smash it
- Always scan an equation that has lots of elements to see if you can apply
**rule 1** - For this equation, we can apply it to Na, Sn and S as these are the elements that only appear in one compound on each side
- On the LHS, Sn is tied up with Na and O so changing Na
_{2}SnO_{3}‘s coefficient is a bad idea because you’ll change all those elements on the RHS too - The smart move is use
**rule 2**twice: use Na’s subscript on the LHS to balance NaOH (which sets the coefficient for Na_{2}SnO_{3 }as 1) and S’s subscript to balance H_{2}S on the LHS (which sets the coefficient for SnS_{2 }as also 1):

Na_{2}SnO_{3} + 2H_{2}S → SnS_{2} + 2NaOH + _H_{2}O

- Now all that remains is to correct H and O, which conveniently turned out to be already balanced:

Na_{2}SnO_{3} + 2H_{2}S → SnS_{2} + 2NaOH + H_{2}O

#### Example 8

_Hg(OH)_{2} + _H_{3}PO_{4} → _Hg_{3}(PO_{4})_{2} + _H_{2}O

- As ever, we avoid messing with compounds containing lots of elements unless we’re forced to (
**rule 4**) and hunt for an element that only appears once on each side (**rule 1**) - Turns out it’s Hg, and since it appears only once we have no choice other than to use its RHS subscript as its coefficient on the LHS (
**rule 2**)

3Hg(OH)_{2} + _H_{3}PO_{4} → Hg_{3}(PO_{4})_{2} + _H_{2}O

- We then apply
**rule 2**again, this time to the PO_{4}^{3-}ion:

3Hg(OH)_{2} + 2H_{3}PO_{4} → Hg_{3}(PO_{4})_{2} + _H_{2}O

- Counting up, there are 12 H’s on the LHS and none on the RHS currently, and we’re also short of 6 O’s on the RHS. A 6 in front of H
_{2}O now balances this equation (**rule 7**, using water last to to correct H’s and O’s):

3Hg(OH)_{2} + 2H_{3}PO_{4} → Hg_{3}(PO_{4})_{2} + 6H_{2}O

#### Example 9

_(NH_{4})_{3}AsS_{4} + _HCl → _As_{2}S_{5} + _H_{2}S + _NH_{4}Cl

- Well, this looks awful!
- We avoid the complicated-looking compounds first (
**rule 4**) and note that As appears only once on each side of the equation (**rule 1**) - Since we can’t change its subscript on the RHS, let’s use it on the LHS (
**rule 2**). In doing so, we will also set the coefficient for As_{2}S_{5}as 1:

2(NH_{4})_{3}AsS_{4} + _HCl → As_{2}S_{5} + _H_{2}S + _NH_{4}Cl

- Because S has a 4 subscript, that 2 coefficient has multiplied up the number of S atoms on the LHS to 8. We have 5 on the RHS in As
_{2}H_{5}, so we’re short 3. To fix that, put a 3 in front of H_{2}S:

2(NH_{4})_{3}AsS_{4} + _HCl → As_{2}S_{5} + 3H_{2}S + 6NH_{4}Cl

- NH
_{4}^{+}is a polyatomic ion and that 2 we just used as a coefficient has also created 6 of them on the LHS because of its 3 subscript. This forces us to put a 6 in front of NH_{4}Cl…

2(NH_{4})_{3}AsS_{4} + _HCl → As_{2}S_{5} + 3H_{2}S + 6NH_{4}Cl

- …and that 6 means we must now put a 6 in front of HCl to balance Cl and complete the equation:

2(NH_{4})_{3}AsS_{4} + 6HCl → As_{2}S_{5} + 3H_{2}S + 6NH_{4}Cl

#### Example 10

_MnO_{4}^{–} + _H^{+} + _Fe^{2+} → _Mn^{2+} + _Fe^{3+} + _H_{2}O

- Oh hell, this looks like a lot of fun. This one is firmly in A Level / IB Diploma territory and you’d ordinarily use the redox balancing method
- Let’s try to sort it out without that though…
- The first thing to note is that Fe is already balanced, because it only appears by itself once on each side of the equation. That means the coefficient for Fe
*must be the same*on both sides (**rule 3**). We don’t know what that number is yet, but it’s got to be the same for Fe^{2+}and Fe^{3+}. - The logical first move is use O’s subscript on the LHS to balance it on the RHS (
**rule 2**again, and we violate**rule 7**by changing H_{2}O first here but hey, this one’s tough…)

MnO_{4}^{–} + _H^{+} + _Fe^{2+} → Mn^{2+} + _Fe^{3+} + 4H_{2}O

- Note we’ve also set the coefficients for Mn and Mn
^{2+}to 1 by making that change, so we can’t now change those - The 4 in front of water means we have to put an 8 in front of H
^{+}, and now the equation is balanced in terms of elements**but not charge**

MnO_{4}^{–} + 8H^{+} + _Fe^{2+} → Mn^{2+} + _Fe^{3+} + 4H_{2}O

- We now need to use
**rule 6**to find the missing coefficient, which we can do using simple algebra (or trial-and-error if you hate maths): - Let
*x*= the coefficient for Fe^{2+}and Fe^{3+}

MnO_{4}^{–} + 8H^{+} + *x*Fe^{2+} → Mn^{2+} + *x*Fe^{3+} + 4H_{2}O

- Since total charges LHS = total charges RHS, we can write:

-1 + 8 + 2*x *= 2 + 3*x*

7 + 2*x* = 2 + 3*x*

*x* = 5

- Finally, it’s balanced:

MnO_{4}^{–} + 8H^{+} + 5Fe^{2+} → Mn^{2+} + 5Fe^{3+} + 4H_{2}O

## The importance of practise

As I’ve stressed repeatedly, balancing chemical equations is a skill that takes lots of **practise** to master.

Here are the rules summarised:

- Rule 1: Look for an element that appears only once on the LHS and RHS
- Rule 2: Try subscripts as balancing numbers
- Rule 3: Look for forced coefficients
- Rule 4: Don’t start with compounds containing many elements
- Rule 5: Remember you can use fractions
- Rule 6: Sum of charges LHS = sum of charges RHS
- Rule 7: Balance water later

The above set of rules is not a checklist; you are never likely to use all of these tricks to balance an equation and some won’t apply to certain equations. The only way you will get better at quickly selecting the right rule to apply is by *practising lots of balancing equations*.

Finally, be aware that appearances can be deceptive. Simple-looking equations can be fiendishly difficult. Ones with many elements and complicated-looking compounds can turn out to be easy.

With that in mind, here are some practise questions for KS3/GCSE level and A Level / IB Diploma, complete with answers. Some are extremely easy to balance and only need one rule, others are much more difficult.

Good luck!

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