Unit 1: Foundation and Fundamental of Chemistry | Class 11 Chemistry Notes

Unit 1: Foundation and Fundamental of Chemistry | Class 11 Chemistry Notes

Fundamental And Foundation of Chemistry

Fundamental And Foundation of Chemistry

Syllabus:

Unit Content
1. Foundation and Fundamentals (2 hr)

1.1 General introduction of chemistry

1.2 Importance and scope of chemistry

1.3 Basic concepts of chemistry (atoms, molecules, relative masses of atoms and molecules, atomic mass unit ( amu), radicals, molecular formula, empirical formula )

1.4 Percentage composition from molecular formula

Recognize the importance and scope of the chemistry:

The branch of science which deals with the study of chemical combination, chemical reaction, atomic and molecular structure, microscopic particles like electron, proton, neutron etc. is called chemistry.
Chemistry has a broad area of services. It is not only a practice of academic study and research,but also frequently plays a vital role in invention, discovery, manufacture, and medical services as well.
Most of the chemical substances are related to chemical combinations. Chemistry provides various services like cosmetics, foodstuffs, medicines, glass wares, soaps and detergents, synthetic clothes, rubber, nylon, plastic, etc.

Chemistry is ruling over all other branches of science in the modern era. Chemistry is further divided into the following branches:

  1. Inorganic chemistry
  2. Organic chemistry
  3. Physical chemistry
  4. Electrochemical chemistry
  5. Biochemistry chemistry
  6. Natural chemistry
  7. Analytical chemistry
  8. Nuclear chemistry
  9. Industrial chemistry
  10. Spectroscopic chemistry

Chemistry can serve in different fields of manufacturers and production such as:

  1. Food industry
  2. Distillery and beverage
  3. Soap and detergents
  4. Rubber, plastics, and paper industry
  5. Pharmaceutical
  6. Perfumes and sprays
  7. Analytical research
  8. Textile and tanning industry

Some basic terminology:

  1. Atom: Best Western purest particle of an element and compound which is self-existence and take part in a chemical reaction is called atom.
    Example: H, O, He
  2. Molecule: the smallest and purest particle of a substance is called molecule.
    Example: H2O, CO2
  3. Radicals: an atom or group of atoms that carries positive or negative charge is called radical. Radicals are of two types:
    1. Acidic Radicals: The radical which carries a negative charge and comes from a base is called a basic radical.
      Example: Na+, Mg+2, Ca+2
    2. Basic Radicals: The radical which carries a positive charge and comes from an acid is called a basic radical.
      Example: Cl-, CO3-2
  4. Valency: The combining capacity of an element with another element is called valency.
  5. Molecular Formula: the symbolic representation of a molecule which shows the actual number of atoms of a particular element present in a molecule is called molecular formula. Example:
    Compound Molecular Formula
    Glucose C6H12O6
    Ethane C2H6
    Benzene C6H6
  6. Empirical Formula: The simplest whole number ratio of atoms of different elements present in a molecule is called empirical formula.
    Example:
    Compound Empirical Formula
    Glucose CH2O
    Ethane CH3
    Benzene CH
  7. Structural Formula: the molecular formula which shows the arrangement of atoms in two-dimensional space is called formula. Example:CH4 Structural Formula of Methane
  8. Spatial Formula:The Molecular formula which shows the arrangements of atoms in a molecule in three-dimensional space is called spatial formula. Example:CH4 Spatial Structure of Methane

Calculation of percentage composition:

The number of parts by weight of an element present in 100 parts by weight of a molecule is called the percentage composition of the element.

Mathematically,

\[ \text{Percentage composition} = \frac{\text{Mass of Element}}{\text{Molecular Mass}} \times 100\% \]

Example: Calculate the percentage composition of oxygen and hydrogen in water.

Solution:

Molecular Mass of Water (\( \text{H}_2\text{O} \)) = \( 1 \times 2 + 16 = 18 \) gm

Mass of Oxygen = 16 gm

Mass of Hydrogen = 2 gm

Now,

Percentage composition of Oxygen = \( \frac{\text{Mass of Oxygen}}{\text{Molecular Mass of Water}} \times 100\% \)

\( = \frac{16}{18} \times 100\% \)

\( = 88.89\% \)

Percentage composition of Hydrogen = \( \frac{\text{Mass of Hydrogen}}{\text{Molecular Mass of Water}} \times 100\% \)

\( = \frac{2}{18} \times 100\% \)

\( = 11.11\% \)

Example for Students: Find the percentage composition of each element in sulfuric acid (\( \text{H}_2\text{SO}_4 \)).

Relationship between Molecular Formula and Empirical Formula:

Molecular Formula = \( n \times \) Empirical Formula

Example: Find the actual molecular formula of a compound whose empirical formula is CH and molecular mass is 78.

Given:

Molecular Formula Mass = 78

Empirical Formula Mass = 12 + 1 = 13

Now,

From the formula:

\( n = \frac{\text{Molecular Formula Mass}}{\text{Empirical Formula Mass}} = \frac{78}{13} = 6 \)

Thus, the Actual Molecular Formula is \( \text{C}_6\text{H}_6 \)

Relationship between Molecular Formula and Empirical Formula:

Molecular Formula = \( n \times \) Empirical Formula

Example: Find the actual molecular formula of a compound whose empirical formula is CH and molecular mass is 78.

Given:

Molecular Formula Mass = 78

Empirical Formula Mass = 12 + 1 = 13

Now,

From the formula:

\( n = \frac{\text{Molecular Formula Mass}}{\text{Empirical Formula Mass}} = \frac{78}{13} = 6 \)

Thus, the Actual Molecular Formula is \( \text{C}_6\text{H}_6 \)

Relative Atomic Mass:

An atom is a very tiny and invisible particle that can't be measured by any physical method. So, atomic mass is determined relative to the mass of an atom of other elements like hydrogen, carbon, and oxygen. That is why atomic mass is called relative atomic mass.

Definition:

The number of parts by weight which shows how many times an atom of an element is heavier than the mass of 1 atom of hydrogen or one by 1/16th of the mass of 1 atom of oxygen O16 or 1/12th of the mass of 1 atom of carbon C12 is called as relative atomic mass.

Formulas:

Relative Atomic Mass = \( \frac{\text{Mass of 1 atom of element}}{\text{Mass of 1 atom of Hydrogen}} \)

Relative Atomic Mass = \( \frac{\text{Mass of 1 atom of element}}{\text{Mass of 1 atom of Carbon}} \)

Relative Atomic Mass = \( \frac{\text{Mass of 1 atom of element}}{\text{Mass of 1 atom of Oxygen}} \)

Example: Atomic Mass of Na is 23. What does it mean?

Ans: It means one atom of sodium is 23 times heavier than the mass of 1 atom of hydrogen or 1/16th of the mass of oxygen (O16) or 1/12th of the mass of 1 atom of carbon (C12).

Relative Molecular Mass:

The number of parts by weight which shows how many times a molecule of a compound is heavier than 1/12th of the mass of 1 atom of carbon (C12) is called relative molecular mass.

Formula:

Relative Molecular Mass = \( \frac{\text{Mass of 1 mole of Compound}}{\left( \frac{1}{12} \right)^{\text{th}} \text{Mass of 1 atom of Carbon}} \)

Example: Molecular Mass of CO2 is 44. What does it mean?

Ans: It means one mole of CO2 is 44 times heavier than 1/12th of the mass of 1 atom of carbon (C12).

Relative Molecular Formula Mass:

It is defined as the sum of the relative atomic mass of all elements present in the molecules.

Example:

Relative molecular formula Mass of CaCO3 = 40 + 12 + 16 × 3 = 100 g/mol

Average Atomic Mass:

The average of different isotopes of an element as their percentage occurrence in nature its called average atomic mass.

Mathematically:

Average Atomic Mass = \( \frac{\text{Total % of Isotopes Occurrence}}{\text{Total no. of Percentage}} \)

Example: Cl35 and Cl37 are found in nature about 75% and 25% respectively. Then the average mass of Cl will be:

Average atomic mass of Cl = \( \frac{35 \times 75 + 37 \times 25}{75 + 25} = 35.5 \)

Questions

  1. Some of the elements don't have the whole number of atomic masses. Why?
    Ans: Some elements have different isotopes with different occurrence percentages in nature, and such elements' atomic mass is calculated by taking the average, which leads the atomic mass to be fractional.
  2. Why is the atomic mass of chlorine fractional?
    Ans: Chlorine exists with different isotopes. So, to maintain balance, we calculate the average of the existence of chlorine in nature, due to which chlorine's atomic mass value comes in fractional part.
  3. How is your atomic mass of chlorine 35.5?
    Ans: As we know, chlorine has various isotopes, such as 35 and 37. Due to this, we take the average of 35 and 37 according to their occurrence in nature. 75% of occurrence is for 35 and the rest for 37. So, \[ \text{Atomic Mass of Chlorine} = \frac{35 \times 75 + 37 \times 25}{75 + 25} = 35.5 \]

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2 comments

  1. Thank You, Sir.
  2. how do i dowload the pdf
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