Lesson 4: Formulae and Equations

8 05 2012

Click to download Lesson 4: Formulae and Equations (Word document) Formatting looks much better when you download it! When pasting from Word to this post some of my tables and colours were altered.

Empirical Formula is the simplest whole number ratio of each type of atom in a compound.


Chemical Molecular Formula Empirical Formula
Glucose C6H12O16 CH2O
Ethanoic Acid CH3CO2H CH2O

Looking at the molecular formula for Ethanoic Acid you can see there are

2 Carbon, 2 Oxygen and 4 Hydrogen atoms, this whole molecular formula has then been divided by a common multiple, in this case 2, so in the Empirical Formula you have:

1 Carbon, 1 Oxygen and 2 Hydrogen atoms.

In some questions you will be asked to calculate the Empirical Formula of a compound.

As an example, let’s work out H2SO4 by breaking it into its component parts (ratio H : S : O). If the question says we have 2.45g of H2SO4 containing 0.05g Hydrogen, 0.8g Sulfur and 1.6g Oxygen. Create yourself a little table, and work out the details from top to bottom:

In basic form, your working table could look like this:

Mass (g) 0.05 0.8 1.6
Molar mass (gmol-1) 1 32 16
Moles 0.05 0.025 0.1

Then divide all moles values by the smallest; you’re left with the result of the simplest ratio of how many of each atom makes up the compound.

Let’s try another Question…

Analysis shows that 0.6075g of Magnesium combines with 3.995g of Bromine to form a compound. Find the Empirical Formula of this compound.

Mg Br
Mass (g)÷Molar mass (gmol-1) 0.6075g ÷ 24gmol-1 3.995 ÷ 80gmol-1
Moles 0.025 ≈ 0.3mol 0.5mol
Divide mol values by the smallest 0.3 ÷ 0.3 0.5 ÷ 0.3
Simplest Ratio 1 1.6666… ≈ 2
Empirical Formula Mg Br2

The simplest ratio of Magnesium is 1 so just write Mg, which is one atom. For Bromine the ratio was 2, so you have to write Br2 two atoms being the simplest ratio. Simply it is a 1:2 ratio.

Molecular Formula shows the exact number of atoms in a molecule.

It is used for compounds that form covalent molecules (covalence is where elements share electrons in a chemical bond.) Remember that Noble gases are monoatomic (mono means one, single) and other elemental gases are diatomic (di means two or double)

Diatomic example: in molecular form Hydrogen gas is H2, Chlorine gas is Cl2, and Nitrogen gas is N2.

CH4 Methane is made up of 1 Carbon atom and 4 Hydrogen atoms.

We can use the ‘octet’ rule with ionic formula (which is explained in the next heading) to predict molecular formula, but only for s and p block elements not d block. (In Lesson 1 we discussed that the noble gases all have a ‘stable octet’ of 8 electrons in their outer shell making them very unreactive, because they don’t want to give away or accept electrons).

So if we wanted to predict the molecular formula of chlorine oxide…

Cl: 8 – 7 = 1 so Cl1-

O:  8 – 6 = 2 so O2+

The charges of each element must always balance, but currently we have 1- and 2+

You cannot subtract from the 2+ or add to the 1-, the charge does not change, but the amount of an element can change. In this case we need 2 atoms of Chlorine to balance out the 2+ charge of Oxygen. So we can predict the molecular formula to be:


So we have Cl 1- (x2) with O 2+, balanced together so that the overall charge always equals zero. This compound is called dichlorine monoxide.

Ionic Formula

This is the formula of an ionic compound whereby the charges of the elements are made to balance.

 Notice how in each equation Chlorine has a different charge, the ionic charge of the element itself has not changed (it is still 1- in each), but each time it has needed more atoms of itself to balance with the positive charge of the element it is bonding with. So sometimes it is just Cl, other times 2 atoms of Cl are needed, Cl2, and sometimes 3, Cl3.

We will go into more details on ionic compounds, and charges on the periodic table soon.

Structural Formula

The structural formula shows the arrangement of atoms and groups in a molecule.

Essential for organic compounds.



oxidised to

Ethanal (Acetaldehyde)




For organic reactions we often write unbalanced equations showing only the structural formula of the principle organic reactant and products. (reactant Þ products).

For inorganic reaction we usually write balanced equations showing all the reactants and products.

Examples of an inorganic equation:

1) Ordinary equation

Mg + H2SO4  => MgSO4 + H2

2) Ionic equation

Mg + 2H+ => Mg2+ + H2

Hydrogen gains electrons so is reduced. (Oxidation is loss of electrons, Reduction is gain of electrons. Remember OIL RIG”: Oxidation Is Loss, Reduction Is Gain.)

Half Equations

Half equations are used to show the loss and gain of electrons in a reaction, in other words to demonstrate the oxidation and reduction.

Oxidation Is Loss

Mg => Mg2+ + 2e

The 2e are the two electrons lost by Magnesium, electrons are negative so a loss of them makes the atom more positive, and hence Mg becomes Mg2+

Reduction is Gain

2H+ + 2e => H2

In this half equation Hydrogen is initially positive, until 2 electrons (2e) are added and it becomes negative. Remember, it must balance.