FSc Notes Chemistry Part 1 Chapter 1 Basic Concepts Lecture 4

FSc Notes Chemistry Part 1 Chapter 1 Basic Concepts

Stoichiometry:

The branch of chemistry which deals with the study of relationship between the quantities of reactants and products is called stoichiometry. It is based on two laws.

Law of Conservation of Mass:

This law states that during a chemical reaction the total mass of reactants is always equal to the total mass of the products.

Law of Definite Proportion

This law states that a particular compound consists of particular elements combined together in a fixed ratio, no matter from which source it is obtained.

E . g Water is a molecule and it always consists of H and O elements in 2:1 no matter how it is obtained .
Lets see how the stoichiometry depends upon these laws take the formation of H2O from H2 and O2
i.e.
H2 + O2 H2O
2 g 32g 18 g
34g 18g
Non-Stoichiometric
H2 + O2 H2O2
2 g 32g 2g + 32 g
34g 34 g
Non-Stoichiometric
Now
2H2 + O2 2H2O
4g 32g 36 g
36g 36 g

Stoichimetric equation

From the above discussion it is clear that a stoichiometric equation must follow both the above mentioned laws. If an equation follows one but not the other, then it will not be stoichiometric.
From the knowledge of stoichiometry came to know that during balancing of a chemical equation we should always put some digit as the co-efficient of any reactant a product , not in the formula.

Characteristics of a Stoichimetric equation

A stoichiometric equation must have the following characteristics.

We can study the following type of relationships from a stoichiometric (balanced) chemical equation.

Mass – Mass Relationship
From a stoichiometric equation we can easily find the mass of any reactant or product, if the mass of any reactant or product is driven.
For example.
How much k2 SO4 will be formed from 14g of KOH according to the following stoichiometric equation.
2KOH + H2 SO4------------------ K2SO4 + 2H2O

DATA
Mass of KOH = 14 g
Mass of K2SO4 = ?

Solution
According to the stoichiometric equation
2KOH + H2SO4 --------- K2SO4 + 2H2O
2 (39+16+1) 2*39+32+16*4
2*56 78+32+64
112g/mol 174 g/mol
As
112f of KOH give = 174g K2SO4
So 14 g of KOH will give = 174 *14 K2SO4 = 112 = 21.75g of K2SO4

Mole – Mole relationship
If no of moles of any reactant or product are given then the no of moles of any other reactant or product can easily be found from a stoichiometric equation.

Mass – Mole relationship
If the mass of any reactant or product is given then the no of moles of any other reactant or product can be calculated from stoichiometric equation

Mole – Mass Relationship
If the no of moles of any reactant or product are given then the mass of any other reactant or product can be calculated from the stoichiometric equation.

Limiting Reactant:

The reactant which gives the  no of moles of the product is called limiting reactant.
Or
The reactant which is consumed earlier is called limiting reactant.
Or
The reactant which stops a chemical reaction or which controls the formation of products is called limiting reactant .

Identification of Limiting Reactant:

For the identification of the limiting reactant, the following three steps are performed. First of all find the no of moles of each reactant if their masses are given. If no of moles of reactant are given directly, then this step is not required. Find out the no of moles of the product from all the reactants whose moles are given The reactant which gives or produces. The least no of moles of the products is selected as the limiting reactant.

Yield:

The amount of products obtained is known as yield.
Types:
There are two types of yield.

1. Theoretical yield
2. Actual Yield

Theoretical yield:
The amount of products calculated from a stoichiometric equation is known as theoretical yield.

Actual yield:
The amount of products obtained during chemical reaction or experiment is called actual yield.

The actual yield of a reaction is usually less then the theoretical yield.
It is because of the following reasons:

1. Due to the formation of by- produce is that product which unwanted whether it is indicated or not is the chemical equation.
2. Due to reversible reaction e . g N2 + 3H2----------------2NH3 Here some of the NH3 converts back into reactants before it is collected.
3. Due to mechanical loss during the process of weighing, drying, filtration etc.

Usually yield is indicated as percent yield to note the efficiency of the process. The percent yield is calculated as. % Y = A.Y/T.Y * 10

Written by: Asad Hussain