Class - 9 Science (Chemistry) Chapter - 3 Atoms & Molecules Notes, NCERT Solutions & Frequently Asked Questions

                                               Class - 9

Science (Chemistry) 

Chapter - 3

Atoms & Molecules

Notes, NCERT Solutions & Frequently Asked Questions 

-- Notes --

The idea of divisibility by Indian philosophers

• Maharishi Kanad – He postulated that if we keep on dividing the matter (called ‘padarth’) we will get smaller and smaller particles. And soon we will achieve the smallest of particles (called as ‘parmanu’) which may not divide further.
• Pakudha Katyayama – He postulated that there are various forms of matter because the particles of matter exist together in combinations.

The idea of divisibility by Greek philosophers

• Democritus and Leucippus – They suggested that when we keep on dividing the matter there comes a time when no more division of particles can take place. Such particles are called atoms which means being invisible.

But all these ideas were not backed up by many experimental pieces of evidence until Antoine L. Lavoisier provided two laws of chemical combination.

Laws of Chemical Combination

1. Law of conservation of mass – mass can neither be created nor destroyed in a chemical reaction

2. Law of constant proportion/Law of definite proportion – the elements are always present in definite proportions by mass in a chemical substance

For example, Hydrogen and oxygen are present in water in a ratio of 1:8. So if we decompose 9g of water we will obtain 1g of hydrogen and 8g of oxygen.

 

The Atomic Theory

John Dalton proposed an atomic theory that acted as an explanation of the above two laws. As per the theory, all matter whether it is an element, a compound or a mixture consists of tiny invisible particles called ‘atoms’.

The postulates of the atomic theory by John Dalton

1. The matter is made up of tiny particles called atoms that cannot be divided.

2. Atoms are never formed or destroyed during a chemical reaction. 

 3. Atoms of an element exhibit the same nature. They have the same size, mass, and character.

4. Atoms of different elements exhibit variant nature. They do not have the same characteristics.

5. Atoms form compounds by combining them in a ratio of whole numbers.

6. A compound contains a constant number and kinds of atoms

 

Atoms

We can call atoms as the building blocks of matter. Just like bricks are the building blocks of a building.

What is the size of an atom?

Atoms are extremely small. Their size is measured in nanometers, where 1nm = 1/109 m.

Atomic radius is measured in nanometers 1/109 = 1nm 1m  = 109 nm
Relative Sizes
Radii (in m)	Example
10-10	Atom of hydrogen
10-9	Molecule of water
10-8	Molecule of Haemoglobin
10-4	Grain of Sand
10-2	Ant
10-1	Watermelon

 

Symbols for Atoms

The symbols for representing an atom are generated from the first two letters of the element’s name. The first letter is always in uppercase (capital letter) while the second letter is written in lowercase. Here are some examples –

 

The Atomic Mass

• Dalton’s Atomic Theory suggested that each element has a distinguishing atomic mass. With this theory, the law of constant proportions could be explained easily.
• But it is indeed difficult to evaluate the mass of an atom since the size of an atom is relatively small.
• Therefore scientists started evaluating the mass of an atom by comparing it with the mass of a standard atom.
• Earlier 1/16 of the mass of an oxygen atom was used as a standard for calculating the mass of other elements. Now, carbon - 12 is considered a standard atom for calculating the mass.
• Its atomic mass is 12u (12 atomic mass units). Thus we can say that one atomic mass unit is the mass of 1/12 the mass of a carbon-12 atom. Here is a list of atomic masses of a few elements.

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

 

Can atoms exist independently?

Atoms cannot survive independently. So, atoms join together and form molecules or ions. 

 

Molecule

• A molecule is a collection of various atoms that combine chemically with each other.
• These atoms are bound together by certain forces of attraction.
• Atoms of the same elements or different elements can bind together to form molecules.
• Therefore, a molecule is the smallest particle of a substance that can exist independently and shows all the properties of that substance.

Molecules of Elements

• The molecules of an element are formed by combinations of similar types of atoms. For example, Helium (He) is made up of only one atom while oxygen is made up of two atoms.
• Atomicity – the number of atoms in a molecule of an element is called its atomicity. For example, helium is monatomic and oxygen is diatomic.
• Monoatomic – when an element comprises a single atom. Example – all metals
• Diatomic – when an element comprises two atoms. Example – all gases
• Triatomic – when an element comprises of three atoms
• Tetra-atomic – when an element comprises of four atoms
• Poly-atomic – when an element comprises of more than two atoms

Here a few examples of atomicity of elements –

Atomicity of some Elements
Name	Atomicity	Formula
Argon	Monoatomic	Ar
Helium	Monoatomic	He
Oxygen	Diatomic	O2
Hydrogen	Diatomic	H2
Nitrogen	Diatomic	N2
Chlorine	Diatomic	Cl2
Phosphorous	Tetra – atomic	P4
Sulphur	Poly – atomic	S8

 

Molecules of Compounds

Molecules of compounds constitute atoms of different elements that combine together in a fixed proportion. For example, water comprises two atoms of hydrogen and one atom of oxygen.

 

Ions

• Compounds contain metals as well as non-metals. These elements include charged species which are known as ions.
• Thus, an ion is a particle that has a positive or negative charge.
• Anion – negatively charged ion
• Cation – positively charged ion
• There can be a single charged atom in an ion or there may be a group of charged atoms in an ion that have a net charge on the compound.
• When a group of atoms carries a charge in a compound it is called a polyatomic ion.

Chemical Formula

We use a chemical formula to represent the composition of a compound in the form of symbols. To write a chemical formula you must know two things –

1. Symbols of elements

2. Valency

 

Valency – it is also known as the combining capacity of an element. In other words, valency explains how atoms of one element will mix with atoms of another element. For example, the hydrogen ion is represented as H+ which means that its valency is 1. Similarly, the oxygen ion is represented as O2- which means that its valency is 2. Here is a list of valencies of various elements.

 

Rules of writing a Chemical Formula

• Valencies of on the ions must balance.
• In a case where both metal and non-metal substances are present in a compound, the name of the metal is always written first in the chemical formula. For example, Sodium Chloride is written as NaCl
• In the case of polyatomic ions, the ion is written in brackets before writing the number of ions associated with it. In the case of a single ion, there is no need to mention the ion in brackets

Writing the Formulae of Simple Compounds

 

Binary compounds – compounds that consist of two different elements

How to write a Formula of a Compound

• Write the symbols of the corresponding elements of the compound as explained above
• Write the valencies of the elements of the compound
• Crossover the valencies of the elements

Here are a few examples of writing the chemical formula

 

Molecular Mass and the Mole Concept

Molecular Mass – summation of all the atomic masses in a molecule

Molecular mass is expressed in atomic mass units (amu).

For example, the molecular mass of HNO3 can be calculated as:

Atomic mass of H =1u

Atomic mass of N =14u

Atomic mass of O =16u

Molecular mass of HNO3 = 1 + 14 + (16*3) = 63u

 

Formula Unit Mass

The sum of atomic masses of all atoms in a formula unit of a compound is called its formula unit mass.  The formula unit mass is used in the case of substances that constitute ions. For example, the formula unit mass of Sodium Chloride (NaCl) can be calculated as:    (1*23) + (1*35.5) = 58.5u

 

Mole Concept

How do we interpret a chemical equation?

• Suppose a chemical equation is: 2C + O2 = 2CO2
• We say that two molecules of carbon combine with one molecule of oxygen to form two molecules of carbon dioxide.
• We can also say that 24u of Carbon molecules combine with 32u of oxygen molecules to form 56u of carbon dioxide molecules.
• Therefore, we can characterise the quantity of a substance by its mass or by its number of molecules.
• A chemical equation directly indicates the number of molecules participating in the reaction. Thus, it is convenient for us to refer to the number of substances in a chemical reaction as numbers of molecules or atoms.

 

Mole

• Mole is a numerical quantity that has a mass equal to the atomic or molecular mass of species (atoms, molecules, ions or particles).
• 1 mole of any substance = 6.022 X 1023 number of particles (atoms, ions or molecules)
• This is called the Avogadro number or Avogadro Constant which is represented as N0
• The mass of 1 mole of a substance is the same as that of its atomic mass or molecular mass expressed in grams.
• The gram atomic mass of a substance – the atomic mass of a substance when expressed in grams is known as its gram atomic mass.
• The gram molecular mass of a substance – the molecular mass of a substance when expressed in grams is known as its gram molecular mass.
• For example, the atomic mass of Sulphur is 32u. The gram atomic mass of Sulphur is 32g.
• Also, 32u of Sulphur has 1 atom of Sulphur. 32g of Sulphur has 1-mole atoms, that is, 6.022 X 1023 atoms of Sulphur.
• Similarly, we can say that the gram molecular mass of Carbon Dioxide is 56g.
• But we know that in the case of the chemical equation mole is the measuring unit.
• Therefore, 1 mole = 6.022 × 1023 number = Relative mass in grams
• Wilhelm Ostwald introduced the word ‘mole’ which means a heap or a pile. Therefore, we consider a substance as a heap of atoms or molecules.

Consider these formulae –

A quick review of how mole, Avogadro number and Mass are related to each other – 

----- NCERT Solution -----

Exercise-3.1 Page: 32

Question 1.- In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide +  water

Solution:-

Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide +  water

5.3g                             6g                 8.2g     2.2g      0.9g

As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of  products.

As per the above reaction, L.H.S. = R.H.S.    i.e., 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g

Hence, the observations are in agreement with the law of conservation of mass.

Question 2.- Hydrogen and oxygen combine in a ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Solution:- We know hydrogen and water mix in a ratio 1: 8.

For every 1g of hydrogen, it is 8g of oxygen.

Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g

Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas.

Question 3.- Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Solution:- The relative number and types of atoms are constant in a given composition, says Dalton’s atomic theory, which is based on the rule of conservation of mass.

“Atoms cannot be created nor be destroyed in a chemical reaction.”

Question 4.- Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Solution:- The postulate of Dalton’s atomic theory that can explain the law of definite proportions is that the relative number and kinds of atoms are equal in given compounds.

Exercise-3.2 Page: 35

Question 1.- Define the atomic mass unit.

Solution:- An atomic mass unit is a unit of mass used to express the weights of atoms and molecules where one atomic mass is equal to 1/12th the mass of one carbon-12 atom.

Question 2.- Why is it not possible to see an atom with the naked eyes?

Solution:- Firstly, atoms are minuscule in nature, measured in nanometers. Secondly, except for atoms of noble gases, they do not exist independently. Hence, an atom cannot be visible to the naked eyes.

Exercise-3.3 Page: 39

Question 1.- Write down the formulae of

i) sodium oxide        ii) aluminium chloride      

iii) sodium sulphide        iv) magnesium hydroxide

Solution:- The following are the formulae:

i) sodium oxide – Na2O

ii) aluminium chloride – AlCl3

iii) sodium sulphide – Na2S

iv) magnesium hydroxide – Mg (OH)2

Question 2.- Write down the names of compounds represented by the following formulae:-

i) Al2(SO4)3        ii) CaCl2        iii) K2SO4        iv) KNO3        v) CaCO3

Solution:- Listed below are the names of the compounds for each of the following formulae:

i) Al2(SO4)3 – Alluminium sulphate

ii) CaCl2 – Calcium chloride

iii) K2SO4 – Potassium sulphate

iv) KNO3 – Potassium nitrate

v) CaCO3 – Calcium carbonate

Question 3.- What is meant by the term chemical formula?

Solution:- Chemical formulas are used to describe the different types of atoms and their numbers in a compound or element. Each element’s atoms are symbolized by one or two letters. A collection of chemical symbols that depicts the elements that make up a compound and their quantities.

For example, the chemical formula of hydrochloric acid is HCl.

Question 4.- How many atoms are present in a

i) H2S molecule and

ii) PO43- ion?

Solution:- The number of atoms present is as follows:

i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total.

ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.

Exercise-3.4 Page: 40

Question 1.- Calculate the molecular masses of H2 , O2 , Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Solution:- The following are the molecular masses:

The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u

The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u

The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u

The molecular mass of CO2 – atomic mass of C + 2 x atomic mass of O = 12 + ( 2×16)u = 44u

The molecular mass of CH4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u

The molecular mass of C2H6– 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) +

(6 x 1)u=24+6=30u

The molecular mass of C2H4– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) +

(4 x 1)u=24+4=28u

The molecular mass of NH3– atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u

The molecular mass of CH3OH – atomic mass of C + 3x atomic mass of H + atomic mass of O + atomic mass of H = (12 + 3×1+16+1)u=(12+3+17)u = 32u

Question 2.- Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u, Na = 23 u,  K=39u, C = 12u, and O=16u.

Solution:- Given:

The atomic mass of Zn = 65u

The atomic mass of Na = 23u

The atomic mass of K = 39u

The atomic mass of C = 12u

The atomic mass of O = 16u

The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u

The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u

The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u

Exercise-3.5 Page: 42

Question 1. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?

Solution:- Given:- 

1 mole of carbon weighs 12g

1 mole of carbon atoms = 6.022 x 1023

The molecular mass of carbon atoms = 12g = an atom of carbon mass

Hence, mass of 1 carbon atom = 12 / 6.022 x 1023  = 1.99 x 10-23g

Question 2.- Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given the atomic mass of Na = 23u, Fe = 56 u)?

Solution:-

a) In 100 grams of Na:

m = 100g, Molar mass of Na atom = 23g, N0 = 6.022 x 1023, N = ?

N = (Given mass x N0)/Molar mass

N = (100 x 6.022 x 1023)/ 23

N = 26.18 x 1023 atoms

b) In 100 grams of Fe:

m = 100 g, Molar mass of Fe atom = 56 g, N0 = 6.022 x 1023, N = ?

N = (Given mass x N0)/ Molar mass

N = (100 x 6.022 x 1023)/ 56

N = 10.75 x 1023 atoms

Therefore, the number of atoms is more in 100 g of Na than in 100 g of Fe.

Exercise Page: 43

Question 1.- A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.

Solution:- Given:-

 Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g

To calculate the percentage composition of the compound,

Percentage of boron = mass of boron / mass of the compound x 100

= 0.096g / 0.24g x 100  = 40%

Percentage of oxygen = 100 – percentage of boron

= 100 – 40 = 60%

Question 2.- When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen?  Which law of chemical combination will govern your answer?

Solution:- When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

Given that - 3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

Find out - We need to find out the mass of carbon dioxide that will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen.

Solution - First, let us write the reaction taking place here.

C + O2 → CO2

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

3g + 8g →11 g ( from the above reaction)

The total mass of reactants = mass of carbon + mass of oxygen

=3g+8g

=11g

The total mass of reactants = Total mass of products

Therefore, the law of conservation of mass is proved.

Then, it also depicts that carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

Thus, it further proves the law of constant proportions.

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

The remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of carbon dioxide will be formed

The above answer is governed by the law of constant proportions.

Question 3.- What are polyatomic ions? Give examples.

Solution:- Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.

Example: CO32-, H2PO4–

Question 4.- Write the chemical formula of the following.

a) Magnesium chloride        b) Calcium oxide        c) Copper nitrate

d) Alluminium chloride        e) Calcium carbonate

Solution:- The following are the chemical formula of the above-mentioned list:

a) Magnesium chloride – MgCl2                b) Calcium oxide – CaO

c) Copper nitrate – Cu(NO3)2                       d) Alluminium chloride – AlCl3

e) Calcium carbonate – CaCO3

Question 5.- Give the names of the elements present in the following compounds.

a) Quick lime        b) Hydrogen bromide        c) Baking powder      d) Potassium sulphate

Solution:- The following are the names of the elements present in the following compounds:

a) Quick lime – Calcium and oxygen (CaO)

b) Hydrogen bromide – Hydrogen and bromine (HBr)

c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)

d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)

Question 6.- Calculate the molar mass of the following substances.

a) Ethyne, C2H2                                    b) Sulphur molecule, S8

c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31)

d) Hydrochloric acid, HCl            e) Nitric acid, HNO3

Solution:- Listed below is the molar mass of the following substances:

a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g

b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8  x 32 = 256g

c) Molar mass of  Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g

d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g

e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+

3×16 = 63g

Question 7.- What is the mass of 

a) 1 mole of nitrogen atoms?

b) 4 moles of alluminium atoms (Atomic mass of alluminium =27)?

c) 10 moles of sodium sulphite (Na2SO3)?

Solution:- The mass of the above-mentioned list is as follows:

a) Atomic mass of nitrogen atoms = 14u

Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen atoms

Therefore, the mass of 1 mole of nitrogen atom is 14g.

b) Atomic mass of alluminium =27u

Mass of 1 mole of alluminium atoms = 27g

1 mole of aluminium atoms = 27g, 4 moles of alluminium atoms = 4 x 27 = 108g

c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O =  (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g

Therefore, mass of 10 moles of Na2SO3  = 10 x 126 = 1260g

Question 8.- Convert into a mole.

a) 12g of oxygen gas                b) 20g of water                c) 22g of carbon dioxide

Solution:- Conversion of the above-mentioned molecules into moles is as follows:

a) Given: Mass of oxygen gas=12g

Molar mass of oxygen gas = 2 Mass of Oxygen =  2 x 16 = 32g

Number of moles = Mass given / molar mass of oxygen gas = 12/32 =  0.375 moles

b) Given: Mass of water = 20g

Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g

Number of moles = Mass given / molar mass of water

= 20/18 = 1.11 moles

c) Given: Mass of carbon dioxide = 22g

Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g

Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles

Question 9.- What is the mass of:

a) 0.2 mole of oxygen atoms?                b) 0.5 mole of water molecules?

Solution:- The mass is as follows:-

a) Mass of 1 mole of oxygen atoms = 16u; hence, it weighs 16g.

Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2g

b) Mass of 1 mole of water molecules = 18u; hence, it weighs 18g.

Mass of 0.5 moles of water molecules = 0.5 x 18 = 9g

Question 10.- Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.

Solution:- To calculate the molecular mass of sulphur,

Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8×32 = 256g

Mass given = 16g

Number of moles = mass given/ molar mass of sulphur

= 16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur,

Number of molecules = Number of moles x Avogadro number

= 0.0625 x 6.022 x 10²³ molecules

= 3.763 x 1022 molecules

Question 11.- Calculate the number of alluminium ions present in 0.051g of alluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)

Solution:- To calculate the number of alluminium ions in 0.051g of alluminium oxide,

1 mole of alluminium oxide = 6.022 x 1023 molecules of alluminium oxide

1 mole of alluminium oxide (Al2O3) = 2 x Mass of alluminium + 3 x Mass of oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of alluminium oxide = 102g = 6.022 x 1023 molecules of alluminium oxide

Therefore, 0.051g of alluminium oxide has = 6.022 x 1023 / 102 x 0.051

     = 3.011 x 1020 molecules of alluminium oxide

One molecule of alluminium oxide has 2 alluminium ions; hence, the number of alluminium ions present in 0.051g of alluminium oxide = 2 x 3.011x 1020 molecules of alluminium oxide.

= 6.022 x 1020

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