Lesson 4
Formulae and Equations
Empirical Formula is the simplest whole number ratio of each type of atom in a
compound.
Example:
Chemical
Glucose
Ethanoic Acid
Molecular Formula
C6H12O16
CH3CO2H
Empirical Formula
CH2O
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:
Mass (g)
Molar mass (gmol1)
Calculation to find
how many moles
Mol
Divide all mol
values by the
smallest number,
in this case 0.025
Simplest ratio of
each atom in the
compound
Drawn as…
H
0.05
1
S
0.8
32
O
1.6
16
0.05 ÷ 1
0.8 ÷ 32
1.6 ÷ 16
0.05
0.05 ÷ 0.025
0.025
0.025 ÷ 0.025
0.1
0.1 ÷ 0.025
2
1
4
H2
S
O4
By Ruth N. www.ruthlearns.wordpress.com
In basic form, your working table could look like this:
H
S
O
Mass (g)
0.05
0.8
1.6
Molar mass (gmol- 1
32
16
1)
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.
Mass (g)
÷
Molar mass (gmol-1)
Moles
Divide mol values by the
smallest
Simplest Ratio
Empirical Formula
Mg
0.6075g ÷ 24gmol-1
Br
3.995 ÷ 80gmol-1
0.025 ≈ 0.3mol
0.3 ÷ 0.3
0.5mol
0.5 ÷ 0.3
1
1.6666… ≈ 2
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).
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So if we wanted to predict the molecular formula of chlorine oxide…
Cl: 8 – 7 = 1 so Cl1O: 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:
Cl2O
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.
Example:
NaCl  Na+Cl-
This is 1+ 1-
Mg2+ + Cl1-  Mg2+Cl21Al3+ + Cl1-  Al3+Cl31-
This is 2+ 2This is 3+ 3-
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.
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Example
Ethanol
oxidised to
Ethanal (Acetaldehyde)
CH3CH2OH

CH3CHO
For organic reactions we often write unbalanced equations showing only the
structural formula of the principle organic reactant and products. (reactant 
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+ + 2eThe 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.
By Ruth N. www.ruthlearns.wordpress.com