# Lesson Notes By Weeks and Term - Senior Secondary 1

Symbols, formulaes and equations II

Term: 2nd 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

1. State the law of constant composition
2. State the law of multiple proportions
3. 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 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

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

4. Balance these equations

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

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively