1. Chemical Formulae
1. Each element is represented by its own chemical symbol.
2. In an element, the chemical formula represents the atoms in the
molecule.
3. The chemical formula tells us:
a. the types of atoms or ions in the compound,
b. the number of atoms or ionsin the compound,
4. For example, the chemical formula for ethene is C2H4. This shows that
ethene is the result of the combination of the elements carbon and
hydrogen, and there are 2 carbon atoms and 4 hydrogen atoms in each
molecule of ethene.
5. Generally, chemical formulacan be divided into
a. Empirical formula
b. Molecular formula
6. The empirical formula of a substance is the chemical formula that gives
the simplest whole-number ratioof atoms of each element in the
substance.
7. The molecular formula of a substance is the chemical formula that gives
the actual number of atomsof each element in the substance.

2. Empirical Formula
1. The empirical formula of a substance is the chemical formula that gives
the simplest whole-number ratio of atoms of each element in the
substance.
2. Empirical = information gained by means of observation, experience, or
experiment.
Example:
Chemical Substances Molecular Formula
Glucose
C6H12O6
Water
H2 O
Carbon Dioxide
CO2
Benzene
C6 H 6
Butane
C4 H 8
Empirical Formula
CH2O
H2 O
CO2
CH
CH2
Finding Empirical Formula
Steps to determine the empirical formula of a compound
STEP 1: Find the mass
STEP 2: Find the mole
STEP 3: Find the ratio of mole
STEP 4 : Find the simplest ratio of moles
Example 1
In a chemical reaction, 4.23g of iron reacts completely with 1.80g of oxygen
gas, producing iron oxide. Calculate the empirical formula of iron oxide.
[Relative atomic mass: Iron = 56; Oxygen = 16]

3. Answer:
Element
Fe
Mass, g
4.23g
Number of mole, mol 4.23/56=0.0755
Ratio of mole (÷ with 0.0755/0.0755=1
smallest n)
Simplest Ratio of
2
mole (in round
number/whole
number)
O
1.80g
1.80/16=0.1125
0.1125/0.0755=1.5
3
The empirical formula of iron oxide is Fe2O3
Example 2
Determine the empirical formula of a compound which has a percentage of
composition Mg: 20.2%, S: 26.6%, O: 53.2%. [Relative atomic mass: Mg = 24; S
= 32; O = 16]
Answer
Element
Percentage
Mass in 100g
Number of mole
Ratio of mole
Simplest ratio of
mole
Mg
S
O
20.2%
26.6%
53.2%
20.2g
26.6g
53.2g
20.2/24=0.8417mol 26.6/32=0.8313mol 53.2/16=3.325mol
0.8417/0.8313=1 0.8313/0.8313=1 3.325/0.8313=4
1
1
4
The empirical formula of the compound is MgSO4
Example 3
From an experiment, a scientist found that a hydrocarbon contains 85.7% of
carbon according to its mass. Find the empirical formula of the hydrocarbon.
[Relative atomic mass: Carbon = 12; Hydrogen = 1]
Answer:

4. Element
Percentage
Mass in 100g
Number of mole
Ratio of mole
Simplest ratio of
mole
C
85.7%
85.7g
85.7/12=7.142mol
7.142/7.142=1
1
H
14.3%
14.3g
14.3/1=14.3mol
14.3/7.142=2
2
The empirical formula of the hydrocarbon = CH2
Molecular Formula
1. The molecular formula of a substance is the chemical formula that gives
the actual number of atoms of each element in the substance.
2. A molecular formula is the same as or a multiple of the empirical
formula.
3. For example, the empirical of carbon dioxide is CO2 and the molecular
formula is also CO2.
4. Whereas, the empirical formula of ethane is CH3 while the molecular
formula of ethane is C2H6.
5. Molecular formula = (Empirical formula)n , n is integer
Finding Molecular Formula
Example 1
Given that the empirical formula of benzene is CH and its relative molecular
mass is 78. Find the molecular formula of benzene. [Relative Atomic Mass:
Carbon: 12; Hydrogen: 1]

5. Answer:
Let's say the molecular formula of benzene is CH.
The relative molecular mass of CH=(CH)n = 78
n(12+ 1) = 78
13n = 78
n = 78/13 = 6
Change n is 6 in molecular formula.
Therefore, the molecular formula of benzene
= C6 H6
Example 2
What is the mass of metal X that can combine with 14.4g of oxygen to form X
oxide with molecular formula X2O3. (RAM: O = 16; X = 56)
Answer:
Number of mole of oxygen=14.4/16=0.9 mol
From the molecular formula, we learn
that the ratio of element X to oxygenX:O = 2:3
Therefore, the number of mole of X=2/3 x 0.9=0.6 mol
Number of mole, n = mass/Molar mass
mass= 0.6 X 56
*mass=33.6g
The mass of element X = 33.6g
*Molar mass of a substance= Relative atomic mass of the substance
Percentage of Composition of a Compound
1. To find the percentage of composition of a substance means to find the
percentage of mass of each element in the molecule of the substance to
the mass of the molecule.
2. The percentage of mass of an element can be determined by the
following equation:

6. Percentage of mass of an element=Sum of the Relative Atomic Mass of the Ele
ment / Relative Molecular Mass of the substance×100%
Example
Calculate the percentage of composition of DDT (C14H9Cl5). [Relative atomic
mass: Carbon = 14; Hydrogen = 1; Chlorine = 35.5]
Answer:
Relative molecular mass of DDT=14(12)+9(1)+5(35.5)=354.5
Percentage of carbon=14(12) / 354.5×100%=47.4%
Percentage of hydrogen=9(1) / 354.5×100%=2.5%
Percentage of chlorine=5(35.5) / 354.5×100%=50.0%