Ch-3 Atoms & Molecules

LAWS OF CHEMICAL COMBINATION

Two laws of chemical combination:
1. Law of Conservation of Mass
2. Law of Constant Proportions

LAW OF CONSERVATION OF MASS

Established by Antoine L. Lavoisier, a French scientist. The law of conservation of mass
states, “Mass can neither be created nor destroyed in a chemical reaction”. In other
words, total mass of products is always equal to the total mass of reactants. Eg. Q1 Pg 32

Sodium carbonate + Acetic acid Sodium acetate + CO2 + H2O 5.3g + 6g

8.2g + 2.2g + 0.9g

11.3g 11.3g

Since total Mass of reactants = total mass of products, it follows Law of conservation of mass

LAW OF CONSTANT / DEFINITE PROPORTIONS

Law of Constant Proportion, given by Joseph Louis Proust, states that a chemical compound
is always made of same elements in exactly the same fixed/ definite proportion by
mass. This is irrespective of the method or the source from which it is obtained. If the ratio
of mass of constituent atoms will be altered, a new compound is formed. Eg. - The ratio of
masses of hydrogen and oxygen is always in 1:8 in water irrespective of source of water.

Nitrogen dioxide (NO2) is a compound, which is formed by the combination of nitrogen and
oxygen. The ratio of nitrogen and oxygen by mass in nitrogen dioxide is 7:16. Nitrous oxide
(N2O) is a compound which is also formed by the combination of nitrogen and oxygen. The
ratio of nitrogen and oxygen in nitrous oxide is 7:4.

Eg. Q2 Pg 33
1:8::3:x
x = 24g
24g of oxygen should combine with 3g of Hydrogen to form water

This law also helps us to calculate the percent of an element in the given compound by
using the formula:
% of element in a compound = Mass of element X 100
Mass of compound
Eg – Q1 Pg 43
Mass of the compound = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Mass of boron 0.096 g
Percentage of boron = Mass of boron x 100
Mass of compound

= 0.096 x 100 = 40%

 0.240

Percentage of oxygen = Mass of oxygen x 100

Mass of compound

= 0.144 x 100 = 60%

0.240

DALTON’S ATOMIC THEORY

John Dalton, a British chemist and scientist gave the Atomic Theory in 1808 on the basis of
Laws of Chemical combination. This theory is popularly known as Dalton’s Atomic Theory.
Its postulates are -
• All matter is made up of very small or tiny particles called atoms which participate in
chemical reactions.
• Atoms are indivisible and cannot be created or destroyed in a chemical reaction. • All

atoms of a given element are same in size, mass and chemical properties. • Atoms of

different elements are different in size, mass and chemical properties. • Atoms

combine in the ratio of a small whole number to form compounds. • The relative
number and kinds of atoms are constant in a given compound.

ATOM
On the basis of Dalton’s Atomic Theory, atom can be defined as the smallest particle of matter.

Characteristics of atoms:
• Atom is the smallest particle of matter.

• All elements are made of tiny particles called atom.

• Atoms are very small in size and cannot be seen through naked eyes.
• Atom does not exist in free-state in nature. But atom takes part in a chemical reaction. •
The properties of a matter depend upon the characteristics of atoms.
• Atoms are the building block of an element similar to a brick which combine together
to make a building.
• The size of atoms is indicated by its radius.

• In ancient time atoms was considered indivisible.

ATOMIC MASS
Mass of atom is called atomic mass. Since, atoms are very small, the actual mass of an atom
is very small. For example, the actual mass of one atom of hydrogen is equal to 1.673 x10-24
g. To deal with such small number is very difficult. Thus, for convenience relative atomic
mass is used.

The relative mass of all atoms are found with respect to an isotope of Carbon, C-
12. One atomic mass unit (1 amu) = 1/12 of the mass of one atom of C-12.
Thus atomic mass is the relative atomic mass of an atom with respect to 1/12 th of the mass of
carbon 12 atom. ‘amu’ is the abbreviation of Atomic mass unit, but now it is denoted just by
‘u’(unified mass)
 Mass of 1/12th of the C-12 atom = 1u
The atomic mass of carbon = 12 times 1/12th = 12u
The atomic mass of hydrogen atom = 1u.
This means one hydrogen atom is 1 time heavier than 1/12th of the carbon atom.

The atomic mass of oxygen is 16u, this means one atom of oxygen is 16 times heavier than1/12thof
carbon atom.

HOW DO ATOMS EXIST?

Atoms of most of the elements are not able to exist independently since they are quite
reactive. However, atoms of some elements, which are non-reactive, exist in free-state in
nature. Eg.- helium, neon, argon, etc.
Usually atoms exist in combined state with -
• Other atoms of same element (molecule of element)
• Atoms of other elements (molecule of compound)
Note – Atoms may also exist as ions

MOLECULE
A Molecule is a group of two or more atoms that are held together by some attractive forces
(chemical bond).
A molecule is the smallest particle of an element or a compound that is capable of an
independent existence and shows all the properties of that substance.
MOLECULE OF ELEMENT MOLECULE OF COMPOUND

It is formed by the combination of 2 It is formed when 2 or more atoms of

or more atoms of the same element different elements combine together
in a definite proportion by mass

Eg.- H2 , O2 , P4 , S8 Eg. – H2O , NH3

ATOMICITY
The no. of atoms of an element constituting a molecule of that element is called its Atomicity.

Monoatomic molecule:
When molecule is formed by single atom only.
Generally noble gas forms monoatomic molecules.
For example: Helium (He), Neon (Ne), Argon (Ar), Kr (Krypton), Xenon (Xe), Radon (Rn).

Diatomic
When molecule is formed by the combination of two atoms, it is called diatomic molecule. For
example: Hydrogen (H2), Oxygen (O2)Nitrogen (N2), Chlorine (Cl2), etc.

Triatomic
When molecule is formed by the combination of three atoms, it is called triatomic molecule.
For example: molecule of ozone (O3)

Tetra-atomic
When molecule is formed by the combination of four atoms it is called tetra-atomic
molecule. For example: Phosphorous molecule (P4)

Polyatomic
When molecule is formed by the combination of more than two atoms, it is called polyatomic
molecule. For example: Sulphur molecule (S8)
IONS
Atoms of several elements exist in the form of ion. Atoms or group of atoms which carry positive
or negative charge are called ions.

Cations:
Ions having positive charge.
For example: sodium ion (Na+), Magnesium ion (Mg2+), etc
Anions:
Ions having negative charge over them are called anions.
For example: Chloride ion (Cl–), Oxide ion (O2–), Sulphate ion (SO 42-)

Polyatomic ions:
Ions formed by two or more atoms are called polyatomic ions. These are group of atoms of
different elements which behave as single units, and are known as polyatomic ions. For
example: Ammonium ion (NH +), Hydroxide ion (OH–), etc
4

IONIC COMPOUNDS
The compounds consisting of cations and anions are called ionic
compounds. Eg. – NaCl, CaO

WRITING CHEMICAL FORMULA

Chemical formula of the compound is the symbolic representation of its composition. To write
chemical formula of a compound, symbols and valencies of constituent elements must be known.

Points to remember (Just explain- not to be written in nb)

➢ The symbols or formulas of the component radicals of the compound are written side by side.
➢ Positive radicals are written left and negative radicals on the right.
➢ The valencies of the radicals are written below the respective symbols. ➢ The criss-cross
method is applied to exchange the numerical value of valency of each radical. It is written as
subscript of the other radical.
➢ The radical is enclosed in a bracket and the subscript is placed outside the lower right corner.
➢ The common factor is removed.
➢ If the subscript of the radical is one, it is omitted.

The rules that you have to follow while writing a chemical formula are as
follows: ➢ The valencies or charges on the ion must balance.
➢ When a compound consists of a metal and a non-metal, the name or symbol of the metal is written
first. For example: calcium oxide (CaO), sodium chloride (NaCl), iron sulphide (FeS), copper
oxide (CuO) etc., where oxygen, chlorine, sulphur are non-metals and are written on the right,
whereas calcium, sodium, iron and copper are metals, and are writtenon the left.
 ➢ In compounds formed with polyatomic ions, the ion is enclosed in a bracket before writing the
number to indicate the ratio.
While writing the chemical formulae for compounds, we write the constituent elements and their
valencies as shown below. Then we must crossover the valencies of the combining atoms. The
formulae of ionic compounds are simply the whole number ratio of the positive to negative ions in
the structure. For magnesium chloride, we write the symbol of cation (Mg2+) first followed by the
symbol of anion (Cl–). Then their charges are criss-crossed to get the formula.

EXAMPLES

Formula of Sodium oxide Formula of Sodium hydroxide

Formula of Sodium chloride Formula of Calcium Chloride

Formula of Zinc chloride Formula of Aluminium hydroxide

Formula of Calcium
oxide Formula of Aluminium oxide

Formula of Silver oxide Formula of Silver chloride

Formula of Ammonium Chloride Formula of Ammonium

 carbonate
Formula of Silver Carbonate Formula of Silver Sulphate

Ag CO3 Ag SO4

+1 -2 +1 -2

Ag2CO3 Ag2SO4
Formula of Calcium hydroxide Formula of Ammonium sulphate

Ca OH NH4 SO4

+2 -1 +1 -2

Ca(OH)2 (NH4)2SO4

MOLECULAR MASS
Atomic mass: The atomic mass of an element is the mass of one atom of that element in u
(unified mass).
Atomic mass unit (amu): 1/12th of the mass of an atom of carbon-12 is called atomic mass
unit. It is a unit of mass used to express atomic masses and molecular masses. Molar mass:
The molar mass of an element is equal to the numerical value of the atomic mass. However,
in case of molar mass, the units change from ‘u’ to ‘g’. The molar mass of an atom is also
known as gram atomic mass.
For example, the atomic mass of carbon =12 u.
So, the gram atomic mass of carbon = 12 g.
Molecular mass of the molecule: The sum of the atomic masses of all the atoms in a
molecule of a substance is called the molecular mass of the molecule.
Example:
Molecular mass of Nitrogen gas (N2)
Atomic mass of Nitrogen = 14u
Molecular mass of N2 = 2 x 14 = 28u

Molecular mass of H2SO4

Atomic mass of Hydrogen = 1u
Atomic mass of sulphur = 32u
Atomic mass of oxygen = 16u
Molecular mass of H2SO4 = 2(Atomic mass of Hydrogen) + 1 (Atomic mass of sulphur) +
4(Atomic mass of oxygen)
= 2×1 + 32 + 4× 16 = 98 u.
Molecular mass of Hydrogen chloride:
Atomic mass of hydrogen + Atomic mass of chlorine = 1 + 35.5 = 36.5 u.
FORMULA UNIT MASS
The formula unit mass of a substance is the sum of the atomic masses of all atoms in a
formula unit of a compound. The term ‘formula unit’ is used for those substances which are
made up of ions.

Formula unit mass of NaCl:

=1 x Atomic mass of Na + 1 x Atomic mass of Cl
= 1 x 23 +1 x 35.5 = 58.5 u
Formula unit mass of ZnO:
= 1 x Atomic mass of Zn + 1 x Atomic mass O
= 1 x 65 + 1 x 16 = 81 u.
********
 Atoms and Molecules (NCERT solutions)

Exercise-3.1 Page: 27

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.
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.

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.
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.”
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: 30

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.
2. Why is it not possible to see an atom with the naked eyes?
Solution:
Firstly, atoms are minuscule in nature, measured in nanometres. Secondly, except for atoms of noble
gases, they do not exist independently. Hence, an atom cannot be visible to the naked eyes.

Exercise-3.3-3.4 Page: 34

1. Write down the formulae of

(i) sodium oxide---- Na2O
(ii) aluminium chloride---- AlCl3
(iii) sodium sulphide---- Na2S
(iv) magnesium hydroxide---- Mg (OH)2
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 – Aluminium sulphate
(ii) CaCl2 – Calcium chloride
(iii) K2SO4 – Potassium sulphate
(iv) KNO3 – Potassium nitrate
(v) CaCO3 – Calcium carbonate
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 symbolised 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.

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-3.5.1-3.5.2 Page: 34-35

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
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 Page: 36

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%
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:
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.
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–
4. Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium 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) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
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)
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