CHEMISTRY
MOLE CONCEPT
DALTON’S ATOMIC THEORY
By observing the laws of chemical combination, John Dalton proposed an atomic theory of matter. The
main points of Dalton’s atomic theory are as follows:
(i) Matter is made up of extremely small, indivisible particles called atoms.
(ii) Atom is the smallest particle that takes part in chemical reactions.
(iii) Atoms of same substance are identical in all respect i.e. they possess same size, shape, mass,
chemical properties etc.
(iv) Atoms of different substances are different in all respect.
(v) Atoms of different elements may combine with each other in a fixed, simple, whole number ratio to
form compound atoms.
(vi) Atom can neither be created nor destroyed.
Limitation of Dalton’s Atomic Theory
(i) It failed to explain how atoms of different elements differ from each other.
(ii) It failed to explain how and why atoms of elements combine with each other to form compound or
molecules.
(iii) It failed to explain the nature of forces that bind together different atoms in a molecule.
(iv) It did not make any distinction between ultimate particle of an element that takes part in reaction
(atoms) and the ultimate particle that has independent existence (molecules).
MOLE
The name ‘Mole’ comes from Latin word for “massive heap” Mole is the unit, chemists use to keep track
23
of large number of atoms, ions and molecules. One mole of object will always mean 6.023  10 of those
23
–1
objects. The number of object per mole, 6.023  10 mol is called Avogadro’s Number/Constant,
represented by NA.
–23
Experimentally we know that mass of one carbon12 atom is 1.9926  10
gm. It follows that the
number of atoms in exactly 12 gm of carbon–12 is
12
23
= 6.023  10
1.9926  10 23
This relationship of number of object to the atom is also true for ions or molecules.
amu
One atomic mass unit is defined as exactly 1/12 the mass of an atom of carbon12. We know
–23
experimentally that the mass of an atom of carbon12 is 1.9926  10 g, it follows that 1amu = 1.6605 
–24
10 g. The mass of any other type of atom in grams can be expressed in atomic mass unit by using this
relation as a conversion factor.
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GRAM ATOMIC MASS / GRAM MOLECULAR MASS
The mass of one mole atoms of any element is exactly equal to the atomic mass in grams of that
element. This is called as gram atomic mass or gram atoms. For example the atomic mass of oxygen
23
atom is 16 amu. One mole of oxygen atoms contains 6.023  10 oxygen atoms.
–24
Now, mass of one atom of oxygen = 16  1.66  10 g
–24
23
Mass of one mole oxygen atom = 16  1.66  10  6.023  10 = 16 grams
23
Similarly, the mass of 6.023  10 molecules (1 mole) of a substance is equal to its molecular mass in
grams or gram molecule.
–24
23
So, mass of one mole of oxygen molecule O2 = 32  1.66  10  6.023  10
or, 32 grams.
So, 32 grams is one gram molecular mass or one gram molecule.
As in atoms and molecules, mole concept is also applicable to ionic compound, which do not contain
molecules. In such cases, the formula of any ionic compound is representation of ratio between
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constituent ions. One mole of an ionic compound is represented by 6.023  10
formula units.
23
23
+
23
–
One mole of NaCl = 6.023  10 NaCl units = 6.023  10 units of Na + 6.023  10 units of Cl
Mass of one mole of NaCl = 23.0 g + 35.5 g = 58.5 gram NaCl

MOLE AND MOLAR VOLUME
From Avogadro’s law it is known that equal volumes of all gases contain equal number of molecules
under similar temperature and pressure conditions. Now, since one mole molecules of all gases contain
23
same number (6.023  10 ) of molecules therefore, they occupy same volume under similar conditions of
temperature and pressure. The volume occupied by one mole molecules of a gaseous substance is
called Molar volume. One mole molecules of all gases occupy 22.4 litres at 273 K and 760 mm pressure
(S.T.P.). Hence, molar volume of all gases at S.T.P. is 22.4 litres.
MOLE AT A GLANCE
For counting of articles, units like dozen or score is commonly used. Similarly in chemistry ‘mole’ is the
counting unit for chemical entities.

The amount of substance containing Avogadro’s number 6.023  1023

atom, molecule, ions, electrons
and protons, i.e.
(i)
Number of moles of molecule =
(ii)
Number of moles of atoms =
(iii) Number of moles of gases =
Weight in gram
Molecular weight
Weight in gram
Atomic weight
Volume at NTP
Standard molar volume (22.4 L)
(iv) Number of milli moles  Mole 1000
(v) For a compound PxQy
x moles of Q = y moles of P
(vi)
Number of entities

Avogadro's number 6.023×1023

 Number of moles
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PERCENTAGE COMPOSITION AND MOLECULAR FORMULA
Determination of chemical formula can be achieved by analysis the compound for the amounts of the
elements (moles) in a given mass of the compound.
eg. According to law of definite proportions 17.0 g of NH3 always contains 14.0 g of N and 3.0 g of H.
Mass % of N in NH3 
Mass of N in 1 mole of NH3
14
 100 
 100 = 82.35%
Mass of 1 mole of NH3
17
Mass % of H in NH3 
3
 100 = 17.65%
17
EMPIRICAL AND MOLECULAR FORMULA
If we know the percentage composition of the elements by chemical analysing we can calculate the
relative number of atoms of each element in the molecule of the compound. This gives us the empirical
formula of the compound. Further if the molecular mass is known then the molecular formula can easily
be determined.
Thus, while the empirical formula shows the relative number of atoms in the simplest ratio, the molecular
formula gives the actual number of atoms of each element in a molecule.
eg. A sample of hydrogen peroxide consists of 94.11% O and 5.89% H by mass. Therefore, the number
of moles of each element in the 100 g sample is
1
94.11
 5.88 mole of O
16
1
5.89   5.89 mole of H
1
Now, with help of the above data we can find out the number of moles of one element relative to the
other.
5.89 mole of H 1 mole of H
Thus, we have,

5.88 mole of O 1 mole of O
Showing that for hydrogen peroxide, the ratio is 1 mole of hydrogen to 1 mol of oxygen and the empirical
1
formula for hydrogen peroxide is HO. The molecular mass of hydrogen peroxide is 34 gmol  whereas the
1
empirical formula mass for hydrogen peroxide is 17 gmol , therefore molecular formula for hydrogen
34
peroxide is
 2 times, the empirical formula.
17
i.e., (HO)2 = H2O2
STOICHIOMETRY
A basic question raised in the chemical laboratory is, “How much product will be formed from specific
amounts of reactants?” Or in some cases we might ask the reverse question: “How much starting material
(reactant) must be used to obtain a specific amount of product?” To interpret a reaction quantitatively, we
need to apply our knowledge of molar masses and the mol concept.
For example,

CaCO3 
 CaO  g  CO2  g 
we see that CaCO3 on decomposition gives CaO and CO2.
The above equation is balanced. But if the equation is not balanced, then first balance the equation. For
above example, we can say that,
1 mol of CaCO3 decomposes to give 1 mole of CaO and 1 mol of CO2.
or we can also say that,
100 gms of CaCO3 gives 56 gm of CaO and 44 gms of CO2. So, it it is asked that how much CaO will be
formed from 20 gms of CaCO3.
So, we can apply unitary method and find out.
Here,
100 gms CaCO3 gives 56 gm of CaO.
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56
 20  11.2 gm
100
So, summarising the following points:
The mole method consists of the following steps
(i)
Write correct formula for all reactants and products, and balance the resulting equation.
20 gms CaCO3 gives
(ii)
Convert the quantities of some or all given or known substances (usually reactants) into mole.
(iii)
Use the coefficients in the balanced equation to calculate the number of mole of the unknown
quantities (usually products).
(iv)
Using the calculated number of mole and molar masses, convert the unknown quantities to
whatever units are required.
Mole of reactant  Mole of product
Mass of reactant  Mole of reactant  Mole of product
Mass of reactant  Mole of reactant  Mole of product  Mass of product
LIMITING REAGENT
A limiting reactant in a reaction is the species supplied in an amount smaller than that required by the
stoichiometric relation between the reactants.
A limiting reactant is like a part in an automobile factory, if there are 1000 headlights and 600 car bodies,
then the maximum member of automobile will be limited by the number of headlights. Because each body
requires two headlights and there are enough headlights are for only 500 cars. So the headlights play the
role of limiting reagents.
MOLE FRACTION
The mole fraction of any component in the solution is equal to the number of moles of that component
divided by the total number of moles of all the components. For a solution containing n 2 moles of the
solute dissolved in n1 mole of the solvent,
Mole fraction of solute in the solution
n
 x2   2
n1  n2
Mole fraction of solvent in the solution
n
 x1   1
n1  n2
x 1 + x2 = 1
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ASSIGNMENT PROBLEMS
31
1.
What is the mass of one mole of electrons? Mass of one electron is 9.11  10
2.
Simplest formula of the compound containing 60 % of element A (At. wt.10) and 40% of element
B (At.wt.20) is
3.
What volume of CCl4 having density 1.5 g/cc contains 1  10
4.
Calculate the number of atoms of oxygen present in 300 g of CaCO3.
5.
Calculate the mass of one atom of calcium, if atomic weight is 40 g mol .
6.
The number of atoms and molecules of nitrogen in 224 mL of nitrogen at STP are ‘x’ and ‘y’
respectively. Calculate the value of x and y.
7.
By heating 10 g CaCO3, 5.6 g CaO is formed. What is the weight of CO2 obtained in this
reaction?
8.
Find out empirical formula of an alkane (hydrocarbon) containing 82.76% carbon by weight?
9.
An inorganic salt gave the following percentage composition: Na = 29.11, S = 40.51 and O =
30.38. Calculate the empirical formula of the salt.
10.
Find out the mole of each component in 48 g of Na2CO3.
11.
How many atoms are there in 1 gm of CaCO3?
12.
The molecular formula of ferric sulphate is Fe2(SO4)3. How many moles are there in 40 g of ferric
sulphate? (Atomic mass: Fe = 56, S = 32, O = 16)
13.
If 200 g of calcium carbonate (formula weight = 100) occupies a volume of 67.0 ml, what is its
density?
14.
Calculate the molecular mass of (NH4)2SO4.FeSO4.6H2O.
15.
Calculate the number of moles present in 5 g of Ca.
16.
What will be the volume of 1 gram Helium at NTP?
17.
Calculate the number of molecules of sulphur (S8) in 16 g of solid sulphur.
18.
How many grams of neon will have the same number of atoms as are there in 4 g of Ca?
Atomic mass of Ne = 20; Ca = 40 amu]
19.
Calculate the number of moles of 52 g of He.
20.
Calculate the number of moles of 12.044  10
21.
Which of the following samples contains 2.0  10 atoms?
(a) 8.0 g O2
(b) 3.0 g Be
(c) 8.0 g C
(d) 19.0 g F2
25
kg.
chlorine atoms.
1
23
number of He atoms.
23
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22.
Which of the following samples contains the largest number of atoms?
(A) 1 g of Ni(s)
(B) 1 g of Ca(s)
(C) 1 g of N2(g)
(D) 1 g of B(s)
23.
Mass of single atom of X is 2.65  10
12
(A) C
16
(C) O
24.
Moles of He constitute by 9.033  10
25.
The volume of 2 moles of CO2 gas at NTP is equal to …….litres.
26.
Which of the following has maximum number of atoms?
(A) 24 g of C(12)
(B) 56 g of Fe(56)
(C) 27 g of Al(27)
(D) 108 g of Ag(108)
27.
If a mole were to contain 1  10
28.
1.12 mL of a gas is produced at STP by the action of 4.12 mg of alcohol, ROH with methyl
 


magnesium iodide. The molecular mass of alcohol is ROH  CH3MgI  CH4  R OMgI


29.
2 mol of H2S and 11.2 L SO2 at N.T.P. reacts to form x mol of sulphur; x is
SO2 + 2H2S  3S + 2H2O
30.
Hydrogen evolved at NTP on complete reaction of 27 gm of Al with excess of aq. NaOH would be
(Chemical reaction: 2Al + 2NaOH + 2H2O  2NaAlO2 + 3H2)
24
23
24
g. X should be
14
(B) N
1
(D) H
atoms of He is
particles, the mass of one mole of oxygen is (in g)
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ANSWERS
8
TO
ASSIGNMENT
1.
54.8422  10 kg
2.
A3B2
3.
0.426 litre
4.
5.42  10
5.
6.64  10
6.
1.210
7.
4.4 gm
8.
C2 H5
9.
Na2S2O3
10.
Na = 1, C = 0.5, O = 1.5
11.
3.012  10
12.
0.1
13.
2.99 g/ml
14.
392
15.
0.125
16.
5.6 litre
18.
2g
19.
13
20.
2
21.
b
22.
D
23.
C
24.
15
25.
44.8
26.
A
27.
53.2
28.
82.4
29.
1.5
30.
33.6 litre
23
22
g
22
24
atoms, 6.023  10
21
molecules
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