Mole

Concept
JEE • Grade 11 • Chemistry • Session No. 4
Average Atomic Mass

The atomic mass of an element is the weighted average of mass of all

the isotopes of that element.

If an element has three isotopes then

Isotopes (1) Isotopes (2) Isotopes (3)

Isotopes y1 y2 y3
Weights w1 w2 w3

Percentage
x1 x2 x3
Occurrence (by mole)

w1 x1 + w2 x2 + w3 x3
Average Atomic Mass =
100
Example

Cl35 Cl37
Percentage (by mole) 75 25

Ratio 3 1

35 × 3 + 37 × 1 142
Average Atomic Mass = = = 35.5
3+1 4
Q. An element exists in nature in two isotopic forms: X30 (90% by mole)
and X32 (10% by mole). What is the atomic mass of element?
Sol.

Ans. (30.2)
Q. Ag107, and Ag109 are two isotopes of silver if their atomic mass =
108.5. Find the % by mole of each isotope.
Sol.

Ans. (25, 75)

 Average Molecular Mass

The average molecular mass of a mixture is the average mass of all the
substances present in the mixture.

Substance (1) Substance (2) Substance (3)

Moles x1 x2 x3

Molecular
w1 w2 w3
Weights

w1 x1 + w2 x2 + w3 x3
Average Molecular Mass =
x1 + x2 + x3

Total mass of mixture

Mavg. =
Total mole
Q. A gaseous mixture contains 2 moles of He and 6 moles of H2 gas.
What is the average molecular mass of the mixture?
Sol.

Ans. (2.5)
Q. A gaseous mixture contains 40% H2 and 60% He by volume. What is the
average molecular mass of the mixture?
Sol.

Ans. (3.20)
Q. The percentage by mole of NO2 in a mixture of NO2(g) and NO(g)
having an average molecular mass of 34 is:
(A) 25% (B) 20% (C) 40% (D) 75%
Sol.

Ans. (A)
❑ NCERT REFERENCE
 Percentage Composition

wt. of that element

% of element in molecule = Molar mass of compound × 100

Example

Mass % of H, O in H2O is :

2
% of H in H2O = × 100 = 11.11%
18

16
% of O in H2O = × 100 = 88.89%
18
Q. Find the % composition of C in the following:
CH3COOH, HCHO

Sol.

Ans. 40%, 40%

Q. Find the minimum molar mass of the invertase enzyme, which contains
0.4% sulphur.

Sol.

Ans. 8000 gm/mol

Q. Find % of Fe2+ and Fe3+ present in nonstoichiometric compound Fe0.9O.

Sol.

Ans. Fe2+ – 77.77%

Fe3+ – 22.22%
❑ NCERT REFERENCE
Q. Number of hydrogen atoms per molecule of a hydrocarbon A having
85.8% carbon is _________
(Given : Molar mass of A = 84 g mol–1)
Sol. JEE Main 2023

Ans. (12.00)
Q. A metal M forms a metal carbonate M2CO3, if the carbonate contains
48% oxygen by mass. Then determine the atomic weight of metal.

Sol.

Ans. (20)
Q. Hemoglobin contains 0.25% of Fe. If molar mass of Hemoglobin is
89600 then calculate number of Fe atoms in 1 molecule of Hemoglobin.

Sol.

Ans. (4)
 Home-Work

In Allen Digital Module read MOLE CONCEPT theory from page no. 1-26.
Also practice illustrations given on these pages.

In Allen Digital Module Solve RACE- : Q No.

In Allen Digital Module Solve:

Exercise S : 13,14,15,16
Ideal Gas Equation

Under any condition of temperature and pressure, moles of gases may

be calculated using IDEAL GAS EQUATION :

PV = nRT

Where, P = Pressure applied on the Gas

V = Volume occupied by the Gas
n = Number of moles of gas
R = Universal Gas Constant
T = Temperature of Gas on Kelvin scale
Molar Volume

Volume of 1 mol gas at given temperature and pressure

NOTE

1. S.T.P condition (Standard Temperature and Pressure)

(P= 1 bar = 105 N/m2, T= 0˚C)
2. Old S.T.P condition (NTP)
(P= 1 atm, T= 0˚C)
1 mol of O2 gas = 32g = NA Molecules = 22.7 Lit. of volume at STP.
Mass in Gram Number of
× NA entities

÷ NA
÷ molar mass Number
of moles

× 22.7 l / 22.4 l

Vol (l) of gas

at STP/ NTP
Q. The number of molecules and moles in 2.8375 litres of O2 at STP are
respectively
(1) 7.527 × 1022 and 0.250 mol (2) 1.505 × 1023 and 0.250 mol
(3) 7.527 × 1023 and 0.125 mol (4) 7.527 × 1022 and 0.125 mol
Sol. JEE Main 2023

Ans. (4)
Q. Match List–I with List–II.
List – I List –II
A. 16g of CH4(g) I. Weighs 28 g
B. 1g of H2(g) II. 60.2 × 1023 electrons
C. 1 mole of N2(g) III. Weighs 32g
D. 0.5 mol of SO2(g) IV. Occupies 11.4 L volume at STP
Choose the correct answer from the options given below:
(1) A–I, B–III, C–II, D–IV (2) A–II, B–III, C–IV, D–I
(3) A–II, B–IV, C–III, D–I (4) A–II, B–IV, C–I, D–III
Sol. JEE Main 2023

Ans. (4)
❑ NCERT EXEMPLAR

Ans. (4)
❑ NCERT EXEMPLAR

Ans. (2)
❑ NCERT EXEMPLAR

Ans. (3)
❑ NCERT EXEMPLAR

Ans. (1-b, 2-c, 3-a, 4-e, 5-d)

 Home-Work

In Allen Digital Module read MOLE CONCEPT theory from page no. 1-21.
Also practice illustrations given on these pages.

In Allen Digital Module Solve RACE- :

In Allen Digital Module Solve:

Exercise O-1 : Q 13
Exercise (JEE Main PYQs) : Q No.
Exercise S : Q
Exercise 4 (JEE Advanced PYQs) : Q No.
 Density

Mass per unit volume of a substance is called density

Types of density

Absolute Density

Relative Density
For liquids and solids

Mass
Absolute density =
volume

Relative density/ Specific gravity density of the substance

=
(for liquids) density of water at 4° C (1 gmml−1 )
For gases

Mass of gas PM
Absolute density = =
volume of gas RT

Where

P = Pressure applied on Gas

M = Molecular Mass of Gas

R = Universal Gas constant

T = Absolute Temperature
Vapour Density

Vapour density is defined as the density of the gas with respect to

hydrogen gas at the same temperature and pressure.

Density of gas PMgas /RT

Vapour density = =
Density of H2 gas PMH2 /RT

Mgas Mgas
Vapour density = =
M H2 2

Mgas = 2 × V.D.
Q. A gas is found to have formula NOx its vapour density is 23, find x.

Sol.

Ans. NO2
Q. A metal chloride contains 55.0% of chlorine by weight. 100 mL
vapours of the metal chloride at STP weigh 0.57 g. The molecular
formula of the metal chloride is
(Given : Atomic mass of chlorine is 35.5u)
(1) MCl2 (2) MCl4 (3) MCl3 (4) MCl

Sol. JEE Main 2023

Ans. (1)
Law of Chemical Combination

Law of Conservation of Mass

According to this law, the mass can neither be created nor be destroyed in
all chemical and physical changes. This is called as “Law of Conservation
of mass”.
If the reactants are completely converted into products:

Total mass of reactants = Total mass of products

H2 + Cl2 ⎯⎯→ 2HCl

2 71 2 (1 + 35.5)

2+71=73g 73 g

NOTE

In nuclear reactions (Mass + energy) is conserved, not the mass separately.

 Law of Chemical Combination

Law of Definite Proportion / Law of Constant Composition

According to this law, a compound can be obtained from different sources.

But the ratio of each component by mass remain same. i.e. it does not
depend on the method of its preparation or the source from which it has
been obtained.

According to this law, a compound can be obtained from different sources.

But the ratio of each component by mass remain same. i.e. it does not
depend on the method of its preparation or the source from which it has
been obtained.
Example

Water can be obtained from different sources but the ratio of weight of H and
O remains same.
❑ NCERT REFERENCE
Law of Chemical Combination

Law of Multiple Proportion

It was given by John Dalton.

According to law of Multiple proportion if two elements combine to form

more than one compound then the different mass of one element which
combines with a fixed mass of other element bear a simple whole
number ratio to one another.
Example

Mass of oxygen which combine with the fixed mass of nitrogen in these
oxides are calculated as under :
Oxide Ratio of mass of nitrogen & oxygen
N2O 28 : 16
N2O2 28 : 32
N2O3 28 : 48
N2O4 28 : 64
N2O5 28 : 80
Number of parts by weight of oxygen which combine with 28 parts by mass
of nitrogen from the above are 16, 32, 48, 64, & 80 respectively. Their ratio is
1 : 2 : 3 : 4 : 5, which is a simple whole number ratio. Hence, the law is
illustrated.
❑ NCERT REFERENCE
Law of Chemical Combination

Law of Reciprocal Proportion

If two elements A and B combine with another element C, they will combine
with one another in the same ratio or a multiple or sub-multiple of the ratio
in which they combine with constant or fixed weight of C.
Example

Consider three compounds methane, carbon dioxide and water.

C

H O
H2O

Sr. No. Compounds Combining elements Combining weights

1 CH4 C H 12 4
2 CO2 C O 12 32

Hydrogen and oxygen react with carbon separately and forms methane and
carbon dioxide respectively.
The ratio of different weights of hydrogen and oxygen are combining
with fixed weight of carbon is 4:32 i.e. , 1:8.

Now hydrogen and oxygen combine to form water (H2O) in which the
ratio of the weight of hydrogen to that of oxygen is 2:16 i.e. , 1:8.

From this, the law of reciprocal proportion is illustrated, as the ratio is

same as that of the first ratio obtained.
 Law of Chemical Combination

Law of Gaseous Volume

It was given by Gay Lussac.

According to this law, in the gaseous reaction, the reactants are always
combined in a simple ratio by volume and form products, if gaseous, all
measurements being made under similar conditions of temperature and
pressure.
NOTE

This law is used only for gaseous reaction. It relate volume to mole or
molecules. But not relate with mass.

N2(g) + 3H2(g) → 2NH3(g)

 N2 + 3H2 → 2NH3

H2
NH3
N2

+
Example

1 volume of hydrogen combines with 1 volume of chlorine to produce 2

volumes of hydrogen chloride.

1 1 2 ⎯ Stoichiometry

H2(g) + Cl2(g) ⎯→ 2HCl(g)

1 Volume 1 Volume 2 Volume

 Law of Chemical Combination

Avogadro's law

Equal volume of all gases contain equal number of molecules (or moles)
at same temperature and pressure.

Example

Four, one litre flasks are separately filled with the gases H2, He, O2 and O3
at the same temperature and pressure. The number of molecules in the
flask are in the ratio 1:1:1:1, but the ratio of total number of atoms of
these gases present in different flask is 2:1:2:3.
❑ NCERT REFERENCE
❑ NCERT REFERENCE
Q. Under the same conditions, two gases have the same number of
molecules. They must
(A) be noble gases
(B) have equal volumes
(C) have a volume of 22.4 dm3 each
(D) have an equal number of atoms
Sol.

Ans. (B)
Dalton’s Atomic Theory

Introduction

Elements are made of extremely small particles called

atoms (Greek word “Atomos“ which means ‘indivisible’).

Maharishi Kaṇāda proposed that paramanu (atom)

is an indestructible particle of matter.
The salient features of Dalton’s Atomic Theory

Each element is composed of extremely small particles called atoms.

Atoms of a particular element are identical.

Atom of each element is an ultimate particle and it has a characteristic

mass but it is structure less.
Atoms are indestructible i.e. they can neither be created nor be destroyed.

Atoms of different elements take part in chemical reaction to form

molecule.

This theory is based on law of mass conservation and law of definite

proportions.
Limitations of Dalton’s Atomic theory:

It failed to explain the internal structure of atom.

It also failed to explain, how atom of different element differ from one
another.

Isotopes of hydrogen

Protium

Deuterium Tritium
❑ NCERT REFERENCE
Q. Choose the Incorrect Statement about Dalton's Atomic Theory
(1) Compounds are formed when atoms of different elements combine
in any ratio
(2) All the atoms of a given element have identical properties
including identical mass
(3) Matter consists of indivisible atoms
(4) Chemical reactions involve reorganization of atoms

Sol. [JEE Main 2024]

Ans. (1)
❑ NCERT EXEMPLAR

Ans. (1)
❑ NCERT EXEMPLAR

Ans. (2)
❑ NCERT EXEMPLAR

Ans. (1,4)
 Home-Work

In Allen Digital Module read MOLE CONCEPT theory from page no. 1-12.
Also practice illustrations given on these pages.

In Allen Digital Module Solve RACE-: Q No.

In Allen Digital Module Solve:

Exercise O-1 : Q 5-7
Exercise O -2: Q 2-3
Exercise (JEE Main PYQs) : Q No. ##,##,##,## /None /Complete
Exercise S : Q No. ##,##,##,## /None /Complete
EMPIRICAL
FORMULA
Empirical and Molecular Formula

Empirical formula

Formula which represents the constituent atoms in their simplest ratio.

Molecular formula

Formula which represents the actual number of atoms in one molecule

of the compound.

The molecular formula is generally an integral multiple of the empirical

formula.
Empirical and Molecular Formula

Structural formula

Formula which represents atoms along with bond present between

atoms.
 Molecular formula = n× Empirical formula

Compound Molecular formula Empirical formula

Formaldehyde CH2O CH2O

Acetic acid C2H4O2 CH2O

Glucose C6H12O6 CH2O

 Determination of Empirical Formula

To determine the empirical formula following steps should be followed :

(1) First of all, write down the % by mass of each element present in the
compound.

Compound C O H
Mass % 54 40 6

(2) The % by mass of each element is divided by its atomic mass. It gives
mole ratio of elements present in the compounds.

Mole % 54/12 = 4.5 40/16 = 2.5 6/1 = 6

 Determination of Empirical Formula

(3) If the number of moles of atom are not in simplest whole number ratio,
then divide or multiply it with a suitable constant, so as to obtain
simplest ratio.

Mole % 4.5 × 2 = 9 2.5 × 2 = 5 6 × 2 = 12

(4) Empirical formula is C9O5H12

❑ NCERT REFERENCE
❑ NCERT REFERENCE
Q. A moth repellent has the composition 49% C, 2.7% H and 48.3% Cl.
Its molecular weight is 147 gm. Determine its molecular formula.

Sol.

Ans. (C6H4Cl2)
Q. An organic compound contains 49.3% carbon, 6.84% hydrogen and its
molecular mass is 146. Molecular formula of compound is :-

Sol.

Ans. (C6H10O4)
Q. The empirical formula and molecular mass of a compound are CH2O
and 180 amu respectively. What will be the molecular formula of the
compound?
(A) C9H18O9 (B) CH2O (C) C6H12O6 (D) C2H4O2
Sol.

Ans. (C)
Q. The empirical formula of a compounds is CH2O. 0.25 mole of this
compound contains 1 gm hydrogen. The molecular formula of
compound is -

Sol.

Ans. (C2H4O2)
❑ NCERT EXEMPLAR

Ans. (3)
 Home-Work

In Allen Digital Module read MOLE CONCEPT theory from page no. 1-26.
Also practice illustrations given on these pages.

In Allen Digital Module Solve RACE- : Q No.

In Allen Digital Module Solve:

Exercise O-1 : Q 14-15