Chemical formulae

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 Empirical Formula
Glucose C6H12O6 CH2O
Water H2O H2O
Carbon Dioxide CO2 CO2
Benzene C6H6 CH
Butane C4H8 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 O
Mass, g 4.23g 1.80g
Number of mole, mol 4.23/56=0.0755 1.80/16=0.1125
Ratio of mole (÷ with
smallest n)
0.0755/0.0755=1 0.1125/0.0755=1.5
Simplest Ratio of
mole (in round
number/whole
number)
2 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 Mg S O
Percentage 20.2% 26.6% 53.2%
Mass in 100g 20.2g 26.6g 53.2g
Number of mole 20.2/24=0.8417mol 26.6/32=0.8313mol 53.2/16=3.325mol
Ratio of mole 0.8417/0.8313=1 0.8313/0.8313=1 3.325/0.8313=4
Simplest ratio of
mole
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 C H
Percentage 85.7% 14.3%
Mass in 100g 85.7g 14.3g
Number of mole 85.7/12=7.142mol 14.3/1=14.3mol
Ratio of mole 7.142/7.142=1 14.3/7.142=2
Simplest ratio of
mole
1 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
= C6H6

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%