CHAPTER # 1

Significant Figures:

                        The most reliable digits in a number that are known with certainty.

Rules of Determining Significant Figures:

• Non zero digits are all significant; for example 363 has three significant figures and 0.68 has only two significant figures.
• Zero’s between non zero digits are significant; for example, 5004 has four significant figures and likewise 20.4 has three significant figures.
• Zero’s locating the decimal point in numbers less than one are not significant; for example, 0.062 has two significant figures and 0.001 has only one significant figure.
• Final zeros to the right of the decimal point are significant; fir example, 2.000 has four significant figures and 506.40 has five significant figures.
• Zeros that locate the decimal point in numbers larger than one are not necessarily significant; for example, 40 has one significant and 2360 has three significant figures.

Rounding off Data:

                        A process of reducing the digits in a number in order to get the desired number of digits is called rounding off data.

Rules for Rounding off Data:

• If the digits to be dropped is greater than 5, then add 1 to the last digit to be retained and drop all digits farther to the right. Thus 5.768 is rounded off to 5.77 to three significant digits (the last digit “9” dropped us greater than 5) and to 5.8 to two significant figures (the last digit “6” dropped is greater than 5).
• If the digit to be dropped is less than 5, then simply drop it and also all digits farther to the right. Thus 5.734 is rounded off to 5.73 to three significant digits (the last digit “4” dropped is less than 5) and 5.7 to two significant digits (the last digit “3” dropped is also less than 5).
• If the digit to be dropped is exactly 5, then two possible situation arise due to even and off nature of the last digit to be retained:

1. If the digit to be retained is even, then just drop the “5”. Thus 7.865 is rounded off to 7.86 to three significant digits (the digit to be retained “6” is even).
2. If the digit to be retained is off, then add 1 to it. Thus 6.75 is rounded off to 6.8 to two significant digits (the digit to be retained “7” is odd).

Atom:

            Atom is the Greek word. The smallest particle if an element that does not exist freely but can take part in chemical reaction is called an Atom

Molecule:

            Combination of two or more atoms which exist element independently.

Ion:

            The charged particle of an atom is called Ion

Cat ion(+ve) , An ion (-ve)

Atomic Number:

            The number of proton present in the nucleus of an atom is called Atomic Number.

It is represented by ‘Z’

Mass Number:

            The sum of number of proton and number of neutrons present in the nucleus of an atom is called Mass Number.

Atomic Mass:

            The mass of an atom of an element as compared to the mass of one atom like C¹² is called Atomic mass.

Formula Mass:

            The sum of atomic masses of an atoms of all elements presents in the formula of a compound is called formula mass.

e.g: NaCl (23+35.5 = 58.5 amu)

Mole:

            Gram atomic mass or Gram formula mass is called a mole

Avogadro Number:

            6.02 x 10²³

Formula:

1)      Number of moles = Mass / molar mass

2)      No. of particles = (Mass / molar mass) x Na

Emperical Formula:

            A formula which represent the least number of atoms is called Emperical formula.

e.g: Emp. Formula of Benzene is CH

Molecular Formula:

            A formula that expresses the actual numbers of each element in a molecule of the compound is called molecular formula.

e.g: Molecular formula of Benzene is C₆H₆

llustrated Example of Empirical Formula

Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.

Step I – Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.

Step II – Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm

Step III – Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%

Step IV – Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5

Step V – Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO

Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.

Illustration of Molecular Mass:

Example
The empirical formula of a compound is CH₂O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH₂O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH₂O)6
= C₆H₁₂O₆

Stoichometry:
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called Stoichometry.

1)      Mass – Mass relationship

2)      Mass – Volume relationship

3)      Volume – Volume relationship

1. Mass – Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.

Example
Calculate the mass of CO₂ that can be obtained by heating 50 gm of limestone.

Solution
Step I – Write a Balanced Equation
CaCO₃ —-> CaO + CO₂

Step II – Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO₃ —-> CaO + CO₂

Method I – MOLE METHOD
Number of moles of 50 gm of CaCO₃ = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO₃ gives 1 mole of CO₂
0.5 mole of CaCO₃ will give 0.5 mole of CO₂
Mass of CO₂ = Moles x Molecular Mass
= 0.5 x 44
= 22 gm

Method II – FACTOR METHOD
From equation we may write as
100 gm of CaCO₃ gives 44 gm of CO₂
1 gm of CaCO₃ will give 44/100 gm of CO₂
50 gm of CaCO₃ will give 50 x 44 / 100 gm of CO₂
= 22 gm of CO₂

2. Mass – Volume Relationship

The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm³ volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.

Example
Calculate the volume of CO₂ gas produced at S.T.P by combustion of 20 gm of CH₄.

Solution
Step I – Write a Balanced Equation
CH₄ + 2 O₂ —-> CO₂ + 2 H₂O

Step II – Write Down The Molecular Masses And Moles Of Reactant & Product
CH₄ + 2 O₂ —-> CO2 + 2 H₂O

Method I – MOLE METHOD
Convert the given mass of CH₄ in moles
Number of moles of CH₄ = Given Mass of CH₄ / Molar Mass of CH₄
From Equation
1 mole of CH₄ gives 1 moles of CO₂
1.25 mole of CH₄ will give 1.25 mole of CO₂
No. of moles of CO₂ obtained = 1.25
But 1 mole of CO₂ at S.T.P occupies 22.4 dm³
1.25 mole of CO₂ at S.T.P occupies 22.4 x 1.25
= 28 dm³

Method II – FACTOR METHOD
Molecular mass of CH₄ = 16
Molecular mass of CO₂ = 44
According to the equation
16 gm of CH₄ gives 44 gm of CO₂
1 gm of CH₄ will give 44/16 gm of CO₂
20 gm of CH₄ will give 20 x 44/16 gm of CO₂
= 55 gm of CO₂
44 gm of CO₂ at S.T.P occupy a volume 22.4 dm³
1 gm of CO₂ at S.T.P occupy a volume 22.4/44 dm³
55 gm of CO₂ at S.T.P occupy a volume 55 x 22.4/44
= 28 dm³

3. Volume – Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac’s law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH₄ + 2 O₂ —-> CO₂ + 2 H₂O
In this reaction one volume of CH₄ gas reacts with two volumes of oxygen gas to give one volume of CO₂ and two volumes of H₂O

Examples
What volume of O₂ at S.T.P is required to burn 500 litres (dm³) of C₂H₄ (ethylene)?

Solution
Step I – Write a Balanced Equation
C₂H₄ + 3 O₂ —-> 2 CO₂ + 2 H₂O

Step II – Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 —-> 2 CO2 + 2 H2O

According to Equation
1 dm³ of C₂H₄ requires 3 dm³ of O₂
500 dm³ of C₂H₄ requires 3 x 500 dm³ of O₂
= 1500 dm³ of O₂

Limiting Reactant

In Stoichometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the Stoichometry problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.

Problem
We are provided 50 gm of H2 and 50 gm of N₂. Calculate how many gm of NH₃ will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N₂ + 3 H₂ —-> 2 NH₃

Solution
In this problem moles of N₂ and H₂ are as follows
Moles of N₂ = Mass of N₂ / Mol. Mass of N₂
= 50 / 28
= 1.79
Moles of H₂ = Mass of H₂ / Mol. Mass of H₂
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H₂ are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N₂ is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N₂ + 3 H₂ —-> 2 NH₃

1 mole of N₂ gives 2 moles of NH₃
1.79 mole of N₂ gives 2 x 1.79 moles of NH₃
= 3.58 moles of NH₃

Mass of NH₃ = Moles of NH₃ x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH₃