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Atoms and Molecules Notes for class 9:

Atoms and Molecules Notes: Introduction

Are you aware that you can save a lot of time while preparing for exams by reading class 9 notes? In this section we have provided you with atoms and molecules notes. It consists of Law of chemical combination (law of conservation of mass, law of constant proportions), Dalton’s atomic Theory, atoms, molecules, atomicity, to write chemical formulae, molecular mass and the most important mole concept.

Introduction:

Law of chemical Combination:

There are two laws of chemical combination given by Lavoisier and L.Proust as follows:

Law of conservation of mass:

During a chemical reaction mass can neither be created nor destroyed.
As, A + B → C + D
Here, Mass of A and B = Mass of C and D

Law of constant proportion:

The elements in a chemical substance are always present in definite proportions by mass.
As, In the case of water (H2O), the ratio of masses of hydrogen and oxygen is always fixed i.e. 1:8.

Atomic Theories:

Many Scientists got involved in researching about the ultimate particles and tried to gave the explanation of the two chemical laws. John Dalton was the first among them who came forward to give his theory. Atoms and molecules notes considered all atomic facts. According to him all matter whether it is an element, a compound or a mixture consists of minute invisible particles called ‘atoms’.

Dalton’s Atomic Theory:

Atoms:

Substance Radii
Atoms of Hydrogen 10-10m
Molecule of water 10-9m
Molecule of Haemoglobin 10-8m
Grain of Sand 10-4m
Ant 10-2m
Watermelon 10-1m

 Atomic Symbols:

Elements Latin name Symbols
Hydrogen H
Helium He
Lithium Li
Beryllium Be
Boron B
Carbon C
Nitrogen N
Oxygen O
Fluorine F
Neon Ne
Sodium Natrium Na
Magnesium Mg
Aluminum Al
Potassium Kalium K
Silver Argentum Ag
Gold Aurum Au
Mercury Hydragyrum Hg
Iron Ferrum Fe
Copper Cuprum Cu

Atomic Mass:

Atomic masses of some elements:

Elements Atomic Mass
Hydrogen 1 u
Carbon 12 u
Nitrogen 14 u
Oxygen 16 u
Sodium 23 u
Magnesium 24 u
Sulphur 32 u
Chlorine 35.5 u
Calcium 40 u

Molecules:

Atomicity:

Molecules are classified into 4 categories based on the number of atoms present in it:

  1. Monoatomic: Molecules which contains only one atom is said to be monoatomic. Example: He, Ne, Ar etc.
  2. Diatomic: Molecules which contain two atoms are said to be diatomic. Example: O2, H2, Br­2 etc.
  3. Triatomic: Molecules which contain three atoms are said to be triatomic. Example: O3, NO2, CO2 etc.
  4. Tetra atomic:Molecules which contain four atoms are said to be tetra atomic. Example: P4, SO3 etc.

Molecules are further divided into two types:

  1. Homo-atomic Molecule: If the molecule is constituted with only single type of atoms then it is said to be Homo-atomic molecules. Learn and revise Atoms and Molecules Notes to score better.
Name Atomicity
Argon Mono-atomic
Helium Mono-atomic
Oxygen Di-atomic
Hydrogen Di-atomic
Nitrogen Di-atomic
Chlorine Di-atomic
Phosphorous Tetra-atomic
Sulphur Poly-atomic
Compound Combining Elements Atomicity
Water Hydrogen, Oxygen Tri-atomic
Ammonia Nitrogen, Hydrogen Poly-atomic
Carbon dioxide Carbon, Oxygen Tri-atomic

Ions:

Valency:

Remember: “All polyatomic ions with names starting with ‘S’ has valency 2 and CO32- has 2 and PO42- has valency of 2 otherwise each element has valency of 1.”

Revise atoms and molecules notes regularly.

Chemical Formulae:

Rules for writing Chemical formulae:

Chemical Formula of some simple compounds:

Molecular Mass:

Formula Unit Mass:

Mole Concept:

Q/A on atoms and molecules notes:

To understand it better here are some examples from NCERT class 9 chapter 3 notes. Let’s revise them as well:

Calculate the relative molecular mass of water (H2O).

Solution:
As we Know that,
Atomic mass of hydrogen (H) = 1u, oxygen(O) = 16 u
So the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is = 2 × 1+ 1×16 = 18 u

Calculate the formula unit mass of CaCl2.

Solution:
Formula unit mass of CaCl2 =
Atomic mass of Ca + (2 × atomic mass of Cl)
= 40 + 2 × 35.5 = 40 + 71 = 111 u

Calculate the mass of 0.5 mole of O2 gas.]

Solution:
Mass = molar mass × number of moles

⇒ 32 x ½ = 16 g

Calculate the mass of 6.022 × 1023 number of O2 molecules.

Solution:
As we know that

n = \(\frac{ given\:umber\: of\: particles}{Avogadro\: number}\) = \(\frac{N}{N_{A}}\)
n = \(\frac{6.022 × {10}^{23}}{6.022 × {10}^{23}}\)=1

Atoms and molecule notes for class 9 thus provides you with the actual curriculum what CBSE and other boards follow. Read, learn and revise it regularly.

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