Term: 2^nd Term

Week: 5

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:

Subject: Chemistry

Topic:- Symbols, formulaes and equations II

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

State the law of constant composition
State the law of multiple proportions
Balance chemical equations

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

PRESENTATION	TEACHER’S ACTIVITY	STUDENT’S ACTIVITY
STEP 1 INTRODUCTION	The teacher reviews the previous lesson on symbols, formulae and equations	Students pay attention
STEP 2 EXPLANATION	He states and explains the laws of constant composition and multiple proportions	Students pay attention and participates
STEP 3 DEMONSTRATION	He outlines guidelines on how to balance chemical equations and solves some examples	Students pay attention and participate
STEP 4 NOTE TAKING	The teacher writes a summarized note on the board	The students copy the note in their books

NOTE

SYMBOLS, FORMULAES AND EQUATIONS

LAW OF CONSTANT COMPOSTION

It states that all pure samples of the same chemical compounds no matter their method of preparation will contain the same elements and will combine in the same proportion by mass.

LAW OF MULTIPLE PROPORTIONS

It states that when two elements A and B combined together to form more than one compound then the several masses of A which combined with a fixed mass of B are in simple ratio.

BALANCING CHEMICAL EQUATIONS
A chemical equation is a symbolic representation of a chemical reaction in which the reactants and products are denoted by their respective chemical formulae.

An example of a chemical equation is 2H₂ + O₂ → 2H₂O which describes the reaction between hydrogen and oxygen to form water

The reactant side is the part of the chemical equation to the left of the ‘→’ symbol whereas the product side is the part to the right of the arrow symbol.

STOICHIOMETRIC COEFFICIENT

A stoichiometric coefficient describes the total number of molecules of a chemical species that participate in a chemical reaction.

It provides a ratio between the reacting species and the products formed in the reaction.

In the reaction described by the equation CH₄ + 2O₂ → CO₂ + 2H₂O, the stoichiometric coefficient of O₂ and H₂O is 2 whereas that of CH₄ and CO₂ is 1.

The total number of atoms of an element present in a species (in a balanced chemical equation) is equal to the product of the stoichiometric coefficient and the number of atoms of the element in one molecule of the species.

For example, the total number of oxygen atoms in the reacting species ‘2O₂’ is 4.

While balancing chemical equations, stoichiometric coefficients are assigned in a manner that balances the total number of atoms of an element on the reactant and product side.

Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical equation must obey the law of conservation of mass and the law of constant proportions, i.e. the same number of atoms of each element must exist on the reactant side and the product side of the equation.

The Traditional Balancing Method

The first step that must be followed while balancing chemical equations is to obtain the complete unbalanced equation. In order to illustrate this method, the combustion reaction between propane and oxygen is taken as an example.

Step 1

The unbalanced equation must be obtained from the chemical formulae of the reactants and the products (if it is not already provided).
The chemical formula of propane is C₃H₈. It burns with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O)
The unbalanced chemical equation can be written as C₃H₈ + O₂ → CO₂ + H₂O

Step 2

The total number of atoms of each element on the reactant side and the product side must be compared. For this example, the number of atoms on each side can be tabulated as follows.

Chemical Equation: C₃H₈ + O₂ → CO₂ + H₂O
Reactant Side	Product Side
3 Carbon atoms from C₃H₈	1 Carbon atom from CO₂
8 Hydrogen atoms from C₃H₈	2 Hydrogen atoms from H₂O
2 Oxygen atoms from O₂	3 Oxygen atoms, 2 from CO₂ and 1 from H₂O

Step 3

Now, stoichiometric coefficients are added to molecules containing an element which has a different number of atoms in the reactant side and the product side.
The coefficient must balance the number of atoms on each side.
Generally, the stoichiometric coefficients are assigned to hydrogen and oxygen atoms last.
Now, the number of atoms of the elements on the reactant and product side must be updated.
It is important to note that the number of atoms of an element in one species must be obtained by multiplying the stoichiometric coefficient with the total number of atoms of that element present in 1 molecule of the species.
For example, when the coefficient 3 is assigned to the CO₂ molecule, the total number of oxygen atoms in CO₂ becomes 6. In this example, the coefficient is first assigned to carbon, as tabulated below.

Chemical Equation: C₃H₈ + O₂ → 3CO₂ + H₂O
Reactant Side	Product Side
3 Carbon atoms from C₃H₈	3 Carbon atoms from CO₂
8 Hydrogen atoms from C₃H₈	2 Hydrogen atoms from H₂O
2 Oxygen atoms from O₂	7 Oxygen atoms, 6 from CO₂ and 1 from H₂O

Step 4

Step 3 is repeated until all the number of atoms of the reacting elements are equal on the reactant and product side. In this example, hydrogen is balanced next. The chemical equation is transformed as follows.

Chemical Equation: C₃H₈ + O₂ → 3CO₂ + 4H₂O
Reactant Side	Product Side
3 Carbon atoms from C₃H₈	3 Carbon atoms from CO₂
8 Hydrogen atoms from C₃H₈	8 Hydrogen atoms from H₂O
2 Oxygen atoms from O₂	10 Oxygen atoms, 6 from CO₂ and 4 from H₂O

Now that the hydrogen atoms are balanced, the next element to be balanced is oxygen. There are 10 oxygen atoms on the product side, implying that the reactant side must also contain 10 oxygen atoms.

Each O₂ molecule contains 2 oxygen atoms. Therefore, the stoichiometric coefficient that must be assigned to the O₂ molecule is 5. The updated chemical equation is tabulated below.

Chemical Equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Reactant Side	Product Side
3 Carbon atoms from C₃H₈	3 Carbon atoms from CO₂
8 Hydrogen atoms from C₃H₈	8 Hydrogen atoms from H₂O
10 Oxygen atoms from O₂	10 Oxygen atoms, 6 from CO₂ and 4 from H₂O

Step 5

Once all the individual elements are balanced, the total number of atoms of each element on the reactant and product side are compared once again.
If there are no inequalities, the chemical equation is said to be balanced.
In this example, every element now has an equal number of atoms in the reactant and product side.
Therefore, the balanced chemical equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

EVALUATION: 1. State the law of constant composition

State the law of multiple proportions
Define chemical equations and describe its constituents

4. Balance these equations

Al + O₂→ Al₂O₃
N₂+ H₂ → NH₃
FeCl₃+ NaOH → NaCl + Fe(OH)₃
Zn + HCl → ZnCl₂+ H₂
P₂O₅+ H₂O → H₃PO₄
FeSO₄+ NaOH → Na₂SO₄ + Fe(OH)₂
Mg + HCl → MgCl₂+ H₂

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively

Lesson Notes By Weeks and Term - Senior Secondary 1