Stoichiometry JEE Notes | EduRev

JEE : Stoichiometry JEE Notes | EduRev

 Page 1


Stoichiometry – Nirmaan TYCRP
LEARNERS HABITAT EXPERTS Pvt. Ltd.: 97/1, IIIrd Floor, Near NCERT, Adchini, New Delhi, 011-32044009
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CLASSIFICATION OF MATTER
LAWS OF CHEMICAL COMBINATION
(a) Law of conservation of mass [Lavoisier]
In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to
mass after the reaction.
H
2
     +    1/2 O
2
   ? H
2
O ( ?)
 (g)            (g)
 1 mole 1/2 mole 1 mole
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm
mass after the reaction = 1 × 18 = 18 gm
Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two
substances react, releasing carbon dioxide gas to the atmosphere.  After reaction, the contents of the
reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction?
Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final
mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due
to the mass of released carbon dioxide gas.
Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm
(b) Law of constant composition [Proust]
All chemical compounds are found to have constant composition irrespective of their method of
prepration or sources.
? In H
2
O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap
water, river water or seawater or produced by any chemical reaction.
Ex. The following are results of analysis of two samples of the same or two different compounds of
phosphorus and chlorine. From these results, decide whether the two samples are from the
same or different compounds. Also state the law, which will be obeyed by the given samples.
S T O I C H I O M E T R Y
S T O I C H I O M E T R Y
Page 2


Stoichiometry – Nirmaan TYCRP
LEARNERS HABITAT EXPERTS Pvt. Ltd.: 97/1, IIIrd Floor, Near NCERT, Adchini, New Delhi, 011-32044009
1
www.learnershabitat.ac.in
CLASSIFICATION OF MATTER
LAWS OF CHEMICAL COMBINATION
(a) Law of conservation of mass [Lavoisier]
In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to
mass after the reaction.
H
2
     +    1/2 O
2
   ? H
2
O ( ?)
 (g)            (g)
 1 mole 1/2 mole 1 mole
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm
mass after the reaction = 1 × 18 = 18 gm
Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two
substances react, releasing carbon dioxide gas to the atmosphere.  After reaction, the contents of the
reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction?
Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final
mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due
to the mass of released carbon dioxide gas.
Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm
(b) Law of constant composition [Proust]
All chemical compounds are found to have constant composition irrespective of their method of
prepration or sources.
? In H
2
O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap
water, river water or seawater or produced by any chemical reaction.
Ex. The following are results of analysis of two samples of the same or two different compounds of
phosphorus and chlorine. From these results, decide whether the two samples are from the
same or different compounds. Also state the law, which will be obeyed by the given samples.
S T O I C H I O M E T R Y
S T O I C H I O M E T R Y
Stoichiometry – Nirmaan TYCRP
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Amount P Amount Cl
Compound A 1.156 gm 3.971 gm
Compound B 1.542 gm 5.297 gm
Sol. The mass ratio of phosphorus and chlorine in compound A,
m
P
 : m
Cl
 = 1.156:3.971 = 0.2911:1.000
The mass ratio of phosphorus and chlorine in compound B, m
P
 : m
Cl
 = 1.542:5.297 = 0.2911:1.000
As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion.
(c) Law of multiple proportions [Dalton]
When one element combines with the other element to form two or more different compounds, the mass of
one element, which combines with a constant mass of the other bear a simple ratio to one another.
? Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that
these figures shows the law of multiple proportion.
First oxide Second oxide
Carbon  42.9 % 27.3 %
Oxygen 57.1 % 72.7% Given
In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon.
1 part of oxygen will combine with 
42 9
571
.
.
 part of carbon = 0.751
Similarly in 2
nd
 oxide
1 part of oxygen will combine with 
27 3
727
.
.
 part of carbon = 0.376
The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1
This is a simple whole no ratio this means above data shows the law of multiple proportion.
Ex. Two oxide samples of lead were heated in the current of hydrogen and were reduced to the
metallic lead. The following data were obtained
(i)  Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm
(ii)  Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm.
Show that the data illustrates the law of multiple proportion.
Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the
loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the
hydrogen. Therefore,
the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm.
The mass ratio of lead and oxygen, r
1 
 = 
Pb
O
m 3.21 13.375
m 0.244 1.000
? ?
and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm.
The mas ratio of lead and oxygen, r
2
 = 
Pb
O
m 1.067 6.669
m 0.16 1.000
? ?
Now, r
1 
: r
2
 =
 
13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple proportion.
(d) Law of reciprocal proportions [Richter]
When two elements combine seperately with definite mass of a third element, then the ratio of their
masses in which they do so is either the same or some whole number multiple of the ratio in which they
combine with each other.
Page 3


Stoichiometry – Nirmaan TYCRP
LEARNERS HABITAT EXPERTS Pvt. Ltd.: 97/1, IIIrd Floor, Near NCERT, Adchini, New Delhi, 011-32044009
1
www.learnershabitat.ac.in
CLASSIFICATION OF MATTER
LAWS OF CHEMICAL COMBINATION
(a) Law of conservation of mass [Lavoisier]
In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to
mass after the reaction.
H
2
     +    1/2 O
2
   ? H
2
O ( ?)
 (g)            (g)
 1 mole 1/2 mole 1 mole
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm
mass after the reaction = 1 × 18 = 18 gm
Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two
substances react, releasing carbon dioxide gas to the atmosphere.  After reaction, the contents of the
reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction?
Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final
mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due
to the mass of released carbon dioxide gas.
Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm
(b) Law of constant composition [Proust]
All chemical compounds are found to have constant composition irrespective of their method of
prepration or sources.
? In H
2
O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap
water, river water or seawater or produced by any chemical reaction.
Ex. The following are results of analysis of two samples of the same or two different compounds of
phosphorus and chlorine. From these results, decide whether the two samples are from the
same or different compounds. Also state the law, which will be obeyed by the given samples.
S T O I C H I O M E T R Y
S T O I C H I O M E T R Y
Stoichiometry – Nirmaan TYCRP
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Amount P Amount Cl
Compound A 1.156 gm 3.971 gm
Compound B 1.542 gm 5.297 gm
Sol. The mass ratio of phosphorus and chlorine in compound A,
m
P
 : m
Cl
 = 1.156:3.971 = 0.2911:1.000
The mass ratio of phosphorus and chlorine in compound B, m
P
 : m
Cl
 = 1.542:5.297 = 0.2911:1.000
As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion.
(c) Law of multiple proportions [Dalton]
When one element combines with the other element to form two or more different compounds, the mass of
one element, which combines with a constant mass of the other bear a simple ratio to one another.
? Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that
these figures shows the law of multiple proportion.
First oxide Second oxide
Carbon  42.9 % 27.3 %
Oxygen 57.1 % 72.7% Given
In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon.
1 part of oxygen will combine with 
42 9
571
.
.
 part of carbon = 0.751
Similarly in 2
nd
 oxide
1 part of oxygen will combine with 
27 3
727
.
.
 part of carbon = 0.376
The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1
This is a simple whole no ratio this means above data shows the law of multiple proportion.
Ex. Two oxide samples of lead were heated in the current of hydrogen and were reduced to the
metallic lead. The following data were obtained
(i)  Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm
(ii)  Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm.
Show that the data illustrates the law of multiple proportion.
Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the
loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the
hydrogen. Therefore,
the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm.
The mass ratio of lead and oxygen, r
1 
 = 
Pb
O
m 3.21 13.375
m 0.244 1.000
? ?
and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm.
The mas ratio of lead and oxygen, r
2
 = 
Pb
O
m 1.067 6.669
m 0.16 1.000
? ?
Now, r
1 
: r
2
 =
 
13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple proportion.
(d) Law of reciprocal proportions [Richter]
When two elements combine seperately with definite mass of a third element, then the ratio of their
masses in which they do so is either the same or some whole number multiple of the ratio in which they
combine with each other.
Stoichiometry – Nirmaan TYCRP
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This law can be understood easily with the help of the following examples.
H S
O
H O
2
SO
2
H S
2
At. Mass 1 At. Mass 32
At. Mass 16
? Let us consider three elements – hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to
form H
2
O whereas sulphur combines with it to form SO
2
. Hydrogen and sulphur can also combine
together to form H
2
S. The formation of these compounds is shown in fig.
In H
2
O, the ratio of masses of H and O is 2 : 16.
In SO
2
, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S which
combines with a fixed mass of oxygen (say 32 parts) will be
4 : 32  i.e. 1 : 8 ...(1)
When H and S combine together, they form H
2
S in which the ratio of masses of H and S is
2 : 32  i.e., 1 : 16 ...(ii)
The two ratios (i) and (ii) are related to each other as
1
8
1
16
:  or   2  : 1
i.e., they are whole number multiples of each other.
Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole number
multiple of the ratio in which H and S combine together.
Ex. Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27
% carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89%
oxygen, by mass. Show that the data illustrates the law of reciprocal proportion.
S o l . Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third element. Now, let
the fixed mass of carbon = 1 gm. Then,
the mass of hydrogen combined with 1 gm carbon in methane = 
25 1
75 3
?
 gm
and the mass of oxygen combined with 1 gm carbon in carbon dioxide = 
72.73 8
27.27 3
?
 gm
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r
1
 = 
1 8 1
:
3 3 8
?
Now, the mass ratio of hydrogen and oxygen in water, r
2
 = 
11.11 1
88.89 8
?
As r
1
 and r
2
 are same , the data is according to the law of reciprocal proportion.
(e) Gay Lussac law of combining volumes :
When two or more gases react with one another, their volumes bear simple whole number ratio with
one another and to the volume of products (if they are also gases) provided all volumes are measured
under identical conditions of temperature and pressure.
? When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride
according to the following equation.
It has been observed experimentally that in this reaction, one volume of hydrogen always reacts with
Page 4


Stoichiometry – Nirmaan TYCRP
LEARNERS HABITAT EXPERTS Pvt. Ltd.: 97/1, IIIrd Floor, Near NCERT, Adchini, New Delhi, 011-32044009
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www.learnershabitat.ac.in
CLASSIFICATION OF MATTER
LAWS OF CHEMICAL COMBINATION
(a) Law of conservation of mass [Lavoisier]
In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to
mass after the reaction.
H
2
     +    1/2 O
2
   ? H
2
O ( ?)
 (g)            (g)
 1 mole 1/2 mole 1 mole
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm
mass after the reaction = 1 × 18 = 18 gm
Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two
substances react, releasing carbon dioxide gas to the atmosphere.  After reaction, the contents of the
reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction?
Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final
mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due
to the mass of released carbon dioxide gas.
Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm
(b) Law of constant composition [Proust]
All chemical compounds are found to have constant composition irrespective of their method of
prepration or sources.
? In H
2
O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap
water, river water or seawater or produced by any chemical reaction.
Ex. The following are results of analysis of two samples of the same or two different compounds of
phosphorus and chlorine. From these results, decide whether the two samples are from the
same or different compounds. Also state the law, which will be obeyed by the given samples.
S T O I C H I O M E T R Y
S T O I C H I O M E T R Y
Stoichiometry – Nirmaan TYCRP
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Amount P Amount Cl
Compound A 1.156 gm 3.971 gm
Compound B 1.542 gm 5.297 gm
Sol. The mass ratio of phosphorus and chlorine in compound A,
m
P
 : m
Cl
 = 1.156:3.971 = 0.2911:1.000
The mass ratio of phosphorus and chlorine in compound B, m
P
 : m
Cl
 = 1.542:5.297 = 0.2911:1.000
As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion.
(c) Law of multiple proportions [Dalton]
When one element combines with the other element to form two or more different compounds, the mass of
one element, which combines with a constant mass of the other bear a simple ratio to one another.
? Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that
these figures shows the law of multiple proportion.
First oxide Second oxide
Carbon  42.9 % 27.3 %
Oxygen 57.1 % 72.7% Given
In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon.
1 part of oxygen will combine with 
42 9
571
.
.
 part of carbon = 0.751
Similarly in 2
nd
 oxide
1 part of oxygen will combine with 
27 3
727
.
.
 part of carbon = 0.376
The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1
This is a simple whole no ratio this means above data shows the law of multiple proportion.
Ex. Two oxide samples of lead were heated in the current of hydrogen and were reduced to the
metallic lead. The following data were obtained
(i)  Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm
(ii)  Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm.
Show that the data illustrates the law of multiple proportion.
Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the
loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the
hydrogen. Therefore,
the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm.
The mass ratio of lead and oxygen, r
1 
 = 
Pb
O
m 3.21 13.375
m 0.244 1.000
? ?
and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm.
The mas ratio of lead and oxygen, r
2
 = 
Pb
O
m 1.067 6.669
m 0.16 1.000
? ?
Now, r
1 
: r
2
 =
 
13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple proportion.
(d) Law of reciprocal proportions [Richter]
When two elements combine seperately with definite mass of a third element, then the ratio of their
masses in which they do so is either the same or some whole number multiple of the ratio in which they
combine with each other.
Stoichiometry – Nirmaan TYCRP
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This law can be understood easily with the help of the following examples.
H S
O
H O
2
SO
2
H S
2
At. Mass 1 At. Mass 32
At. Mass 16
? Let us consider three elements – hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to
form H
2
O whereas sulphur combines with it to form SO
2
. Hydrogen and sulphur can also combine
together to form H
2
S. The formation of these compounds is shown in fig.
In H
2
O, the ratio of masses of H and O is 2 : 16.
In SO
2
, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S which
combines with a fixed mass of oxygen (say 32 parts) will be
4 : 32  i.e. 1 : 8 ...(1)
When H and S combine together, they form H
2
S in which the ratio of masses of H and S is
2 : 32  i.e., 1 : 16 ...(ii)
The two ratios (i) and (ii) are related to each other as
1
8
1
16
:  or   2  : 1
i.e., they are whole number multiples of each other.
Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole number
multiple of the ratio in which H and S combine together.
Ex. Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27
% carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89%
oxygen, by mass. Show that the data illustrates the law of reciprocal proportion.
S o l . Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third element. Now, let
the fixed mass of carbon = 1 gm. Then,
the mass of hydrogen combined with 1 gm carbon in methane = 
25 1
75 3
?
 gm
and the mass of oxygen combined with 1 gm carbon in carbon dioxide = 
72.73 8
27.27 3
?
 gm
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r
1
 = 
1 8 1
:
3 3 8
?
Now, the mass ratio of hydrogen and oxygen in water, r
2
 = 
11.11 1
88.89 8
?
As r
1
 and r
2
 are same , the data is according to the law of reciprocal proportion.
(e) Gay Lussac law of combining volumes :
When two or more gases react with one another, their volumes bear simple whole number ratio with
one another and to the volume of products (if they are also gases) provided all volumes are measured
under identical conditions of temperature and pressure.
? When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride
according to the following equation.
It has been observed experimentally that in this reaction, one volume of hydrogen always reacts with
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one volume of chlorine to form two volumes of gaseous hydrogen chloride. all reactants and products are
in gaseous state and their volumes bear a ratio of 1 : 1 : 2. This ratio is a simple whole number ratio.
“These are no longer useful in chemical calculations now but gives an idea of earlier methods of
analysing and relating compounds by mass.”
Ex. 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces
7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure
and temperature. Show that the data illustrates Gay Lussac’s law of volume combination.
Sol. V
hydrocarbon
 : V
oxygen
 : V
carbon dioxide
 : V
water vapou
r
  = 2.5 : 12.5 : 7.5 : 10.0
= 1 : 5 : 3 : 4 (simple ratio)
Hence, the data is according to the law of volume combination.
MOLE CONCEPT
Definition of mole : One mole is a collection of that many entities as there are number of atoms
exactly in 12 gm of C – 12 isotope.
or 1 mole = collection of 6.02 × 10
23
 species
6.02 × 10
23
 = N
A
 = Avogadro’s No.
1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion and 1 mole of
molecule termed as 1 gm-molecule.
METHODS OF CALCULATIONS OF MOLE
(a) If no. of some species is given, then no. of moles = 
Given no
N
A
.
(b) If weight of a given species is given, then no of moles = 
Given wt
Atomic wt
for atoms
.
( ),
or   = 
Given wt
Molecular wt
for molecules
.
.
( )
(c) If volume of a gas is given along with its temperature (T) and pressure (P)
use n = 
PV
RT
     where R = 0.0821 lit-atm/mol–K (when P is in atmosphere and V is in litre.)
1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litre.
1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litre.
Atom : Atom is smallest particle which can not be divided into its constituents.
Atomic weight : It is the weight of an atom relative to one twelvth of weight of 1 atom of C-12
RELATIONSHIP BETWEEN GRAM AND AMU
1 amu = 
1
12
 wt of one C - 12 atom.
for C 1 mole C = 12 gm = 6.023 × 10
23
 atoms
wt of 6.023 × 10
23
 atoms = 12 gm
wt of 1 atom of C = 
12
N
gm
A
(N
A
 ? Avogadro’s number = 6.23 × 10
23
)
1 amu = 
1
12
 wt of one C - 12 atom
= 
1
12
12
?
N
A
 gm
Page 5


Stoichiometry – Nirmaan TYCRP
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CLASSIFICATION OF MATTER
LAWS OF CHEMICAL COMBINATION
(a) Law of conservation of mass [Lavoisier]
In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to
mass after the reaction.
H
2
     +    1/2 O
2
   ? H
2
O ( ?)
 (g)            (g)
 1 mole 1/2 mole 1 mole
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm
mass after the reaction = 1 × 18 = 18 gm
Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two
substances react, releasing carbon dioxide gas to the atmosphere.  After reaction, the contents of the
reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction?
Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final
mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due
to the mass of released carbon dioxide gas.
Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm
(b) Law of constant composition [Proust]
All chemical compounds are found to have constant composition irrespective of their method of
prepration or sources.
? In H
2
O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap
water, river water or seawater or produced by any chemical reaction.
Ex. The following are results of analysis of two samples of the same or two different compounds of
phosphorus and chlorine. From these results, decide whether the two samples are from the
same or different compounds. Also state the law, which will be obeyed by the given samples.
S T O I C H I O M E T R Y
S T O I C H I O M E T R Y
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Amount P Amount Cl
Compound A 1.156 gm 3.971 gm
Compound B 1.542 gm 5.297 gm
Sol. The mass ratio of phosphorus and chlorine in compound A,
m
P
 : m
Cl
 = 1.156:3.971 = 0.2911:1.000
The mass ratio of phosphorus and chlorine in compound B, m
P
 : m
Cl
 = 1.542:5.297 = 0.2911:1.000
As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion.
(c) Law of multiple proportions [Dalton]
When one element combines with the other element to form two or more different compounds, the mass of
one element, which combines with a constant mass of the other bear a simple ratio to one another.
? Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that
these figures shows the law of multiple proportion.
First oxide Second oxide
Carbon  42.9 % 27.3 %
Oxygen 57.1 % 72.7% Given
In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon.
1 part of oxygen will combine with 
42 9
571
.
.
 part of carbon = 0.751
Similarly in 2
nd
 oxide
1 part of oxygen will combine with 
27 3
727
.
.
 part of carbon = 0.376
The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1
This is a simple whole no ratio this means above data shows the law of multiple proportion.
Ex. Two oxide samples of lead were heated in the current of hydrogen and were reduced to the
metallic lead. The following data were obtained
(i)  Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm
(ii)  Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm.
Show that the data illustrates the law of multiple proportion.
Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the
loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the
hydrogen. Therefore,
the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm.
The mass ratio of lead and oxygen, r
1 
 = 
Pb
O
m 3.21 13.375
m 0.244 1.000
? ?
and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm.
The mas ratio of lead and oxygen, r
2
 = 
Pb
O
m 1.067 6.669
m 0.16 1.000
? ?
Now, r
1 
: r
2
 =
 
13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple proportion.
(d) Law of reciprocal proportions [Richter]
When two elements combine seperately with definite mass of a third element, then the ratio of their
masses in which they do so is either the same or some whole number multiple of the ratio in which they
combine with each other.
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This law can be understood easily with the help of the following examples.
H S
O
H O
2
SO
2
H S
2
At. Mass 1 At. Mass 32
At. Mass 16
? Let us consider three elements – hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to
form H
2
O whereas sulphur combines with it to form SO
2
. Hydrogen and sulphur can also combine
together to form H
2
S. The formation of these compounds is shown in fig.
In H
2
O, the ratio of masses of H and O is 2 : 16.
In SO
2
, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S which
combines with a fixed mass of oxygen (say 32 parts) will be
4 : 32  i.e. 1 : 8 ...(1)
When H and S combine together, they form H
2
S in which the ratio of masses of H and S is
2 : 32  i.e., 1 : 16 ...(ii)
The two ratios (i) and (ii) are related to each other as
1
8
1
16
:  or   2  : 1
i.e., they are whole number multiples of each other.
Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole number
multiple of the ratio in which H and S combine together.
Ex. Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27
% carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89%
oxygen, by mass. Show that the data illustrates the law of reciprocal proportion.
S o l . Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third element. Now, let
the fixed mass of carbon = 1 gm. Then,
the mass of hydrogen combined with 1 gm carbon in methane = 
25 1
75 3
?
 gm
and the mass of oxygen combined with 1 gm carbon in carbon dioxide = 
72.73 8
27.27 3
?
 gm
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r
1
 = 
1 8 1
:
3 3 8
?
Now, the mass ratio of hydrogen and oxygen in water, r
2
 = 
11.11 1
88.89 8
?
As r
1
 and r
2
 are same , the data is according to the law of reciprocal proportion.
(e) Gay Lussac law of combining volumes :
When two or more gases react with one another, their volumes bear simple whole number ratio with
one another and to the volume of products (if they are also gases) provided all volumes are measured
under identical conditions of temperature and pressure.
? When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride
according to the following equation.
It has been observed experimentally that in this reaction, one volume of hydrogen always reacts with
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one volume of chlorine to form two volumes of gaseous hydrogen chloride. all reactants and products are
in gaseous state and their volumes bear a ratio of 1 : 1 : 2. This ratio is a simple whole number ratio.
“These are no longer useful in chemical calculations now but gives an idea of earlier methods of
analysing and relating compounds by mass.”
Ex. 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces
7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure
and temperature. Show that the data illustrates Gay Lussac’s law of volume combination.
Sol. V
hydrocarbon
 : V
oxygen
 : V
carbon dioxide
 : V
water vapou
r
  = 2.5 : 12.5 : 7.5 : 10.0
= 1 : 5 : 3 : 4 (simple ratio)
Hence, the data is according to the law of volume combination.
MOLE CONCEPT
Definition of mole : One mole is a collection of that many entities as there are number of atoms
exactly in 12 gm of C – 12 isotope.
or 1 mole = collection of 6.02 × 10
23
 species
6.02 × 10
23
 = N
A
 = Avogadro’s No.
1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion and 1 mole of
molecule termed as 1 gm-molecule.
METHODS OF CALCULATIONS OF MOLE
(a) If no. of some species is given, then no. of moles = 
Given no
N
A
.
(b) If weight of a given species is given, then no of moles = 
Given wt
Atomic wt
for atoms
.
( ),
or   = 
Given wt
Molecular wt
for molecules
.
.
( )
(c) If volume of a gas is given along with its temperature (T) and pressure (P)
use n = 
PV
RT
     where R = 0.0821 lit-atm/mol–K (when P is in atmosphere and V is in litre.)
1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litre.
1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litre.
Atom : Atom is smallest particle which can not be divided into its constituents.
Atomic weight : It is the weight of an atom relative to one twelvth of weight of 1 atom of C-12
RELATIONSHIP BETWEEN GRAM AND AMU
1 amu = 
1
12
 wt of one C - 12 atom.
for C 1 mole C = 12 gm = 6.023 × 10
23
 atoms
wt of 6.023 × 10
23
 atoms = 12 gm
wt of 1 atom of C = 
12
N
gm
A
(N
A
 ? Avogadro’s number = 6.23 × 10
23
)
1 amu = 
1
12
 wt of one C - 12 atom
= 
1
12
12
?
N
A
 gm
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1 amu = 
g m
N
1
A
ELEMENTAL ANALYSIS
For n mole of a compound (C
3
H
7
O
2
)
Moles of C   = 3n
Moles of H   = 7n
Moles of O   = 2n
Ex. Find the wt of water present in 1.61 g of  Na
2
SO
4
. 10H
2
O
Sol. Moles of Na
2
SO
4
. 10H
2
O = 
wt. in gram
molecular wt
 = 
1.61
322
 = 0.005 moles
Moles of water = 10 × moles of Na
2
SO
4
. 10H
2
O
= 10 × 0.05 = 0.05
wt of water = 0.5 × 18 = 0.9 gm Ans.
AVERAGE ATOMIC WEIGHT
= ? % of isotope X molar mass of isotope.
The % obtained by above expression (used in above expression) is by number (i.e. its a mole%)
MOLECULAR WEIGHT
It is the sum of the atomic weight of all the constituent atom.
(a) Average  molecular weight = 
nM
n
i i
i
?
?
where n
i
 = no. of moles of any compound and m
i
 = molecular mass of any compound.
Make yourselves clear in the difference between mole % & mass % in question related to above.
Shortcut for % determination if average atomic weight is given for X having isotopes X
A
 & X
B
.
100
X & X of  weight in difference
X of wt – ght atomic wei Average
X of %
B A
B
A
? ?
Try working out of such a shortcut for X
A
, X
B
, X
C
EMPIRICAL FORMULA, MOLECULAR FORMULA
Empirical formula : Formula depicting constituent atom in their simplest ratio.
Molecular formula : Formula depicting actual number of atoms in one molecule of the compound.
Relation between the two : Molecular formula = Empirical formula × n
n ?
Molecular mass
Empirical Formula mass
Check out the importance of each step involved in calculations of empirical formula.
Ex. A molecule of a compound have 13 carbon atoms, 12 hydrogen atom, 3 oxygen atoms and
3.02 × 10
–23
 gm of other element. Find the molecular wt. of compound.
Sol. wt. of the 1 molecule of a compound = 13 × 
12
N
A
 + 12 × 
1
N
A
 + 3 × 
16
N
A
 + 3.02 × 10
–23
      
A
A
23 –
N
N 10 02 . 3 48 12 156 ? ? ? ?
?
 = 234.18 / N
A
 = 234 amu.  Ans.
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