Chemistry
Session opener
Session Objectives
Session objectives
1. Atomic mass
2. Molecular mass
3. Avogadro’s law
4. Mole concept
5. Empirical and molecular formula
6. Limiting reagent
7. Chemical equation
Atomic mass or relative atomic
mass
How many times one atom
of an element is heavier than 1/12th
mass of one 12C carbon atom.
16 times
heavier
Ask yourself
Why chlorine has fractional
atomic mass?
Solution
Fractional atomic masses of elements are due to the
difference in abundance of isotopes.
e.g.,
37
2 isotopes of chlorine – 35
and
Cl
17 Cl have relative
17
abundance in the ratio of 3:1.




 Average atomic mass of chlorine atom
3  35  1 37

 35.5
3 1
Atomic Mass Unit (amu)
One atomic mass unit (amu) is equal
to 1/12th mass of an atom of
one carbon-12 atom
For example,
Atomic mass of H=1.008 amu
Atomic mass of O=16 amu
New symbol of amu is “u”
Atomic Mass
Statement:
Relative atomic mass states that how many times one
atom of a substance is heavier than 1/12th mass of
the one C-12 carbon atom.
Do you know
The term atom was proposed by John Dalton
The term molecule was proposed by Avogadro.
Illustrative example
A sample of strontium is found to
consist of three isotopes: 86Sr,
natural abundance 9.86%; 87Sr,
natural abundance 7.58% and 88Sr,
natural abundance 82.56%. Their relative
atomic masses are 85.91, 86.91 and 87.91 respectively.
Find the atomic mass of strontium.
Solution:
Atomic mass 
 Fractional abundance  mass
 Atomic mass of Sr  (Fractional abundance
86
Sr  mass86Sr)
 (Fractional abundance
87
Sr  mass87Sr)
 (Fractional abundance
88
Sr  mass88Sr)
9.86
7.58
82.56

 85.91 
 86.91 
 87.91  87.6364
100
100
100
Question
Illustrative Problem
What do you understand about
atomic mass of oxygen which is 16?
Solution:
It means oxygen atom is 16 times heavier than 1/12th
mass of one carbon-12 atom.
Gram atomic mass
When atomic mass is expressed
in gram, it is called gram atomic
mass.
For example,
Atomic mass of oxygen=16 amu
Gram atomic mass oxygen =16g
Molecular mass
Molecular mass of H2SO4
=2x1+32+4x16
= 98 amu
H2SO4 molecule is 98 times heavier
than 1/12th mass of one carbon-12
atom.
The molecular mass of a substance expressed in grams
is called Gram molecular mass.
Gram molecular mass of H2SO4=98g
Statement of Mole Concept
Mole is defined as the amount of the substance which
contains Avogadro’s number(6.023x1023)of atoms or
the molecules if the substance is atom and molecule
respectively
One Mole of X = 6.023 x 1023 X
Animation here
Questions
Illustrative example
Calculate the number of each
constituents present in 22g of
CO2.
Solution
Gram molecular Mass of CO2  12  2x16  44g
44g of CO2  6.023  1023 molecules of CO2
6.023  1023  22
 22g of CO2 
molecules of CO2
44
 3.011  1023 molecules of CO2
 3.011  1023 atoms of carbon
 6.023  1023 atoms of oxygen
Illustrative example
The volume of one atom of a metal
M is 1.66 x 10–23 c.c. Find the
gm-atomic weight of M
(Given density of M=2.7 g/c.c)
Solution:
Mass of one atom=Volume x density
1.66 x 10–23 x 2.7 g
 Mass of Avogadro's number of atoms=
1.66  10-23  2.7  6.023  1023 g
=27 g
Hence 1 gram atom of M=27 g
Illustrative example
How many atoms of carbon has a
young man has given to his bride
to be if the engagement ring contains
0.50 carat diamond?
There are 200 mg in a carat?
Solution:
1 carat of diamond =200 mg diamonds
0.5 carat of diamond =100 mg diamond
=0.1 g
Gram atomic mass of carbon (diamond)
=12 g
=1 mole
Solution
 12 g diamond contains carbon atoms
 6.023  1023
0.1g diamond contains carbon atoms
6.023  1023

 0.1
12
 5.02  1021 atoms of
12
C
Illustrative example
It has been estimated that the average
density of the universe is 1 x 10–30 g/cm3.
Assuming that this density is due to
hydrogen atoms, what is the average
number of hydrogen atoms in 1km3 of the universe.
Solution:
The mass of H in 1km3
 1  1030  (102  103 )3  1015 g / km3
Now 1g of H contains 6.023  1023 atoms
 1015 g of H contains
 1015  6.023  1023
 6.02  108 atoms
Empirical and molecular formula
Simplest whole number
ratio of the atoms in the molecule
For example,
Empirical formula of benzene=CH
Molecular formula of benezene=C6H6
Molecular formula is the true formula.
For example,
Empirical formula of hydrogen peroxide=HO
Molecular formula =H2O2
Empirical and molecular formula
Empirical formula of glucose=CH2O
Molecular formula =C6H12O6
Molecular formula=n x Empirical formula
n=1,2,3…..
n =
molecular mass
Empirical formula mass
Molecular mass=2 x Vapour density
Statement of Mole Concept
Mole is defined as the amount of the substance which
contains Avogadro’s number(6.023x1023)of atoms or
the molecules if the substance is atom and molecule
respectively
One Mole of X = 6.023 x 1023 X
Animation here
Questions
Illustrative example
Calculate the number of each
constituents present in 22g of
CO2.
Solution
Gram molecular Mass of CO2  12  2x16  44g
44g of CO2  6.023  1023 molecules of CO2
6.023  1023  22
 22g of CO2 
molecules of CO2
44
 3.011  1023 molecules of CO2
 3.011  1023 atoms of carbon
 6.023  1023 atoms of oxygen
Illustrative example
The volume of one atom of a metal
M is 1.66 x 10–23 c.c. Find the
gm-atomic weight of M
(Given density of M=2.7 g/c.c)
Solution:
Mass of one atom=Volume x density
1.66 x 10–23 x 2.7 g
 Mass of Avogadro's number of atoms=
1.66  10-23  2.7  6.023  1023 g
=27 g
Hence 1 gram atom of M=27 g
Illustrative example
How many atoms of carbon has a
young man has given to his bride
to be if the engagement ring contains
0.50 carat diamond?
There are 200 mg in a carat?
Solution:
1 carat of diamond =200 mg diamonds
0.5 carat of diamond =100 mg diamond
=0.1 g
Gram atomic mass of carbon (diamond)
=12 g
=1 mole
Solution
 12 g diamond contains carbon atoms
 6.023  1023
0.1g diamond contains carbon atoms
6.023  1023

 0.1
12
 5.02  1021 atoms of
12
C
Illustrative example
It has been estimated that the average
density of the universe is 1 x 10–30 g/cm3.
Assuming that this density is due to
hydrogen atoms, what is the average
number of hydrogen atoms in 1km3 of the universe.
Solution:
The mass of H in 1km3
 1  1030  (102  103 )3  1015 g / km3
Now 1g of H contains 6.023  1023 atoms
 1015 g of H contains
 1015  6.023  1023
 6.02  108 atoms
Empirical and molecular formula
Simplest whole number
ratio of the atoms in the molecule
For example,
Empirical formula of benzene=CH
Molecular formula of benezene=C6H6
Molecular formula is the true formula.
For example,
Empirical formula of hydrogen peroxide=HO
Molecular formula =H2O2
Empirical and molecular formula
Empirical formula of glucose=CH2O
Molecular formula =C6H12O6
Molecular formula=n x Empirical formula
n=1,2,3…..
n =
molecular mass
Empirical formula mass
Molecular mass=2 x Vapour density
Do you know
The present system of representing molecular formula
was introduced by Alfred Stock
Atomicity
The number of atoms present in one molecule
of a substance is called atomicity.
Question
Illustrative Problem
An inorganic salt gave the
following percentage composition
Na=29.11 S=40.51 O=30.38
Calculate the empirical and
molecular formula.
(Given vapour density of compound is 79.)
Solution:
Name of
element
% of
element
At.
mass
Relative no.
of atoms
Simplest
ratio
Na
29.11
23
29.11/23 =1.26
S
40.51
32
40.51/32
=1.26 1.26/1.26 =1
O
30.38
16
30.38/16
=1.89 1.89/1.26 =1.5
1.26/1.26=1
Solution
Simplest whole number atomic ratio
Na
S
O
2
2
3
Hence empirical formula=Na2S2O3
n
Molecular formula mass
Empirical formula mass
2 xVapour density

Empirical formula
2x79 158
n

1
158
158
Hence the molecular formula  Na2S2O3
Illustrative example
A crystalline salt on being rendered
anhydrous loses 45.6 % of its weight.
The percentage composition of the
anhydrous salt is:
Al = 10.5% ; K=15.1 %; S=24.96%; O=49.92 %
Find the simplest formula of the anhydrous and crystalline
salt.
Solution
Name of
element
% of
element
At.
mass
Relative no.
of atoms
Simplest
at. no.
ratio
K
15.1
39
15.1
 0.39
39
Al
10.5
27
10.50
 0.39
27
0.39
1
0.39
0.39
1
0.39
S
24.92
32
24.96
 0.78
27
0.78
2
0.39
O
49.92
16
49.92
 3.12
16
3.12
8
0.39
Simplest
whole no.
ratio
1
1
2
8
Solution
K : Al : S : O = 1 : 1 : 2 : 8
 Empirical formula of the anhydrous
salt is :
K1 Al1S2O8  KAlS2O8
Empirical formula of anhydrous salt=
1 x 39+1 x 27+2 x 32.0 + 8 x 16.0
=158 amu
Let the empirical formula mass of the hydrated salt be
=100 amu
Loss of weight due to hydration=45.6
Empirical formula mass of anhydrous salt
=100 – 45.6
=54.4 amu
Solution
Now, if the empirical formula mass
of the anhydrous salt is 54.4 that of
hydrated salt is 100.
If the empirical formula mass of the anhydrous salt
is 258, that of hydrated salt is
100

 258  474.3 amu
54.4
Total Loss in weight due to dehydration
=474.3–258=216.3 amu
Loss in weight due to one molecule of water =18.0 amu
 no. of molecules of water in the hydrated sample

216.3
 12
18
Solution
Hence, the no. of molecules of water
of crystallisation associated with the salt
KAlS2O8 is 12.
 Empirical formula of the crystalline salt  KAlS2O8.12H2O
Chemical Equation
It is symbolic representation of an
actual chemical change.
For example,
H2+Cl2
2HCl
Chemical Equation
(1) Should be arithematically balanced
N2  3H2  2NH3
(2) What is wrong with this equation
2HBr  H2  2Br
This reaction is not written in molecular form like
2HBr  H2  Br2 (correct)
Question
Illustrative Problem
Correct the following chemical
equations
KClO3
KCl + 3(O)
Cu(S) + 4HNO3
C2 +6H
Cu(NO3 )2+ (H2 O)2+2NO2
C2 H6
Solution:
2kClO3
Cu(S)+ 4HNO3
2C + 3H2
2KCl
+ 3O2
Cu(NO3)2 + 2H2O
C2H6
+2NO2
Limiting reagent
In the reaction
limiting reagent
consumed first.
Questions
Illustrative Problem
N2 + 3H2
2NH3
2 moles
2 moles 3 moles
Find out the limiting reagent?
Solution:
1 mole of N2 combine with 3 moles of H2
 2 mole of N2 combine with 3x2  6 moles of hydrogen
But actual moles of hydrogen in reaction are three.
Hence hydrogen is limiting reagent.
Illustrative Problem
Calculate limiting reagent in the
following reactions
H2
2 moles
+
I2
2HI
2 moles
2 moles
Solution:
1 mole of H2 combine with1mole of I2
 2 mole of H2 will combine with 2 mole of I2
Which is exactly same value of I2 given in the reaction
Similarly 2 moles of I2 required two moles of H2
Hence in this reaction there is no limiting reagent.
Class exercise
Class Exercise - 1
The largest number of molecules is in
(a) 54 g of nitrogen oxide
(b) 28 g of carbon dioxide
(c) 36 g of water
(d) 46 g of ethyl alcohol
Solution:
54
 1.23 mole of N2O
44
36
 2 mole of H2O
18
28
 0.64 mole of CO2
44
46
 1 mole of C2H5OH
46
Hence, answer is (c).
Class Exercise - 2
A compound has the empirical
formula C2H3O2, its vapour density
is 59. Its molecular formula will be
(a) C2H3O2
(c) C6H9O6
(b) C4H6O4
(d) C8H12O8
Solution
Empirical formula mass = 12 × 2 + 3 × 1 + 16 × 2
= 24 + 3 + 32 = 59
n
molar mass
empirical formula mass
Hence, answer is (b).
2  vapour density 2  59


2
59
59
Class Exercise - 3
One mole of CO2 contains
(a) 6.023 × 1023 atoms of C
(b) 6.023 × 1023 atoms of O
(c) 18.1 × 1023 molecule of CO2
(d) 3 g atoms of CO2
Solution
1 mol of CO2 contains 6.023 × 1023 molecule of CO2 and
6.023 × 1023 atoms of C atoms.
Hence, answer is (a).
Class Exercise - 4
The number of atoms in 4.25 g
of CH3 — O — CH3 is approximately
(a) 13.55 x 1022
(c) 5.0 x 1023
(b) 6.023 x 1023
(d) 12.23 x 1023
Solution
6.023  1023  4.25
No. of molecules =
46
= 5.56 x 1022
No. of atoms = 5.56 x 1022 x 9 = 5.0 x 1023
Hence, answer is (c).
Class Exercise - 5
The percentage of nitrogen in
urea (H2NCONH2) is about
(a) 46
(c) 18
(b) 85
(d) 28
Solution
NH2CONH2
Molecular formula = 28 + 4 + 12 + 16 = 60
28
% of N = 60  100  46.6%
Hence, answer is (a).
Class Exercise - 6
The number of gram molecules of
oxygen in 6.023 x 1024 CO molecules is
(a) 10 g molecules (b) 5 g molecules
(c) 1 g molecule
(d) 0.5 g molecule
Solution
6.023 x 1023 molecules = 1 g molecules of O2
 6.023 x 1024 molecules of CO
1 6.023  1024
 
 5 g molecules
23
2 6.023  10
Hence, answer is (b).
Class Exercise - 7
112 ml of hydrogen gas at STP contain
(a) 0.005 mol (b) 0.02 mole
(c) 1.02 g
(d) 6.011 x 1022 molecules
Solution
22,400 ml of H2 gas contains 1 mol.

1
112 ml of H2 gas contains 22,400  112 moles
= 0.005 moles
Hence, answer is (a).
Class Exercise - 8
60 g of a compound on analysis
gave 24 g C, 4 G H and 32 g O.
The empirical formula of the
compound is:
(a) C2H4O2
(b) C2H2O2
Solution:
24
% of C =  100  40%
60
4
% of H =  100  6.66
60
32
% of O =  100  53.33
60
(c) CH2O2
(d) CH2O
Class Exercise - 8
Element
C
H
O
Percentage Atomic ratio
Simplest ratio
3.33
1
3.33
40
40
 3.33
12
6.66
6.66
 6.66
1
6.66
2
3.33
53.33
53.33
 3.33
16
3.33
1
3.33
Hence empirical formula CH2O.
Hence, answer is (d).
Class Exercise - 9
Insulin contains 3.4% sulphur.
What will be the minimum
molecular weight of insulin?
(a) 941.17
(c) 941.76
(b) 1884
(d) 976
Solution:
3.4 g sulphur present in 100 g insulin
 32 g sulphur will be presenting in
Hence, answer is (a).
100
 32  941.17
3.4
Class Exercise 10
On analysis a certain compound
was found to contain iodine and
oxygen in the ratio of 254 g of iodine
(at. mass 127) and 80 g oxygen
(at. mass 16). What is the formula
of the compound?
(a) IO
(b) I2O
(c) I5O3
(d) I2O5
254
127  x  2
80
y 5
16
Hence formula of
compound is I2O5.
Solution:
Mass of iodine
x
atomic mass

mass of oxygen y
atomic mass
Hence, answer is (d).
Thank you