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Stoichiometry refers to calculations
based upon chemical equations.
A balanced chemical equation provides
the following information of mathematical significance:
(i) molecular and molar relationship between reactants and products
(ii) weight relationship between reactants and products
(iii) volume relationship between gaseous reactants and gaseous
products
Example:
2CO + O2 -->2CO2
The following information is provided
by the above chemical equation:
| (a) |
2 molecules of carbon monoxide will combine with
1 molecule of oxygen to form 2 molecules of carbon dioxide |
| (b) |
2 moles of carbon monoxide will combine with
1 mole of oxygen to form 2 moles of carbon dioxide |
| (c) |
2 x 28 = 56 parts by weight of carbon monoxide
will combine with 32 parts by weight of oxygen to form 2 x 44
= 88 parts by weight of carbon dioxide |
| (d) |
2 volumes of carbon monoxide will combine with
1 volume of oxygen to form 2 volumes of carbon dioxide |
Stoichiometry deals with calculations
based upon the information, provided by mass-mass, mass-volume and
volume-volume relationships in a balanced chemical equation.
| Laws of Chemical Combination
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Law of conservation of mass
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This law was proposed by Lavoisier and states
that during all physical and chemical changes, the total mass
of a system is conserved, i.e., the total mass of the products
is always equal to the total mass of the reactants. |
(i) Law
of constant composition or law of definite proportion
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This law was proposed by Proust and states that
a chemical compound is always made up of same elements combined
together in the same fixed proportion by mass. |
(ii)
Law of multiple proportions
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This law was proposed by Dalton
and states that when two elements combine to form two or more
chemical compounds, then the mass of one of the elements which
combines with a fixed mass of the other, bears a simple whole
number ratio.
Example:
Hydrogen combines with oxygen to form two compounds namely
water (H2O) and hydrogen peroxide (H2O2).
In water, 1 part by weight of hydrogen combines with 8 parts
by weight of oxygen while in hydrogen peroxide 1 part by weight
of hydrogen combines with 16 parts by weight of oxygen. The
ratio in which oxygen combines with a fixed weight of hydrogen
in water and H2O2, is 8 : 16 or 1 :
2, which is a simple whole number ratio.
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(iii)
Law of reciprocal proportion
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This law was proposed by Ritcher
and states that when two elements combine separately with
a fixed mass of a third element then the ratio in which they
do so is either the same or a whole number multiple of the
ratio in which they combine directly with each other.
Example:
Carbon and sulphur both combine with a third element oxygen
to form CO2 and SO2 respectively while
they combine together to form CS2.
In CO2, 12 parts by weight of carbon combine with
32 parts by weight of oxygen while in SO2, 32 parts
by weight of sulphur combine with 32 parts by weight of oxygen.
Therefore, the ratio of weights of carbon and sulphur which
combine with a fixed weight of oxygen (say 32 parts by weight)
is 12 : 32 or 3 : 8.
The ratio in which carbon and sulphur combine directly to
form CS2 is 12 : 64 or 3 : 16. The two ratios are
related to each other as
i.e. they are simple multiples
of each other.
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(iv)
Gay Lussac's Law of combining volumes
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Under a given condition of temperature and pressure,
when different gases combine together they always do so in volumes
which bear a simple ratio to one another and also to the volumes
of gaseous products. |
Avogadro's Law
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Equal volumes of all gases under
similar conditions of temperature and pressure contain equal
number of molecules.
Applications of Avogadro's Law
(a) In calculating atomicity
of elementary gases.
(b) In deriving the molecular formula of a gas.
(c) In deriving the relation, Molecular formula of a gas =
2 x vapour density of the gas
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Dulong and Petit's Law
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This law is applicable to metals only and states that
Atomic weight x specific heat is
 6.4
where specific heat is expressed in calorie/g.
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Atomic mass: The atomic mass
of an element is a number, which indicates how many times an atom
of that element is heavier than the 
of the mass of an atom of a carbon-12 isotope.
Unit of atomic mass
Since atomic mass of an element is
a simple ratio, it has no units but is expressed in amu (atomic
mass unit) and
Gram atomic mass
The atomic mass of an element expressed
in grams is called gram atomic mass.
| Example |
Atomic mass of nitrogen = 14 amu
Gram atomic mass of nitrogen = 14 g |
Molecular mass: The molecular
mass of a substance is a number, which indicates how many times
a molecule of that element is heavier than the 
of the mass of an atom of a carbon-12 isotope. The molecular mass
of a substance can be calculated by adding the atomic masses of
all the atoms present in one molecule of the substance.
Example:
Molecular mass of NaOH = 23 amu + 16 amu + 1 amu = 40 amu
Gram molecular mass
The molecular mass of a substance
expressed in grams is called its gram molecular mass or one gram
molecule.
| Example |
Molecular mass of NaOH = 40 amu
Gram molecular mass of NaOH = 40 gram = 40 g (or 1 gram molecule
of NaOH) |
Mole
One mole is a collection of 6.023
x 1023 particles and this number (6.023 x 1023)
is also known as Avogadro's number.
Thus
1 mole of atom = 6.023 x 1023
atoms
1 mole of molecule = 6.023 x 1023 molecules
1 mole of ion = 6.023 x 1023 ions
1 mole of electron = 6.023 x 1023 electrons
Relationship
of mole with mass
The gram atomic mass of any substance contains one mole of it, i.e.
6.023 x 1023 atoms.
Thus
the gram atomic mass of oxygen = 16
g
Now 16 gram of oxygen will contain
6.023 x 1023 atoms of oxygen, i.e
mass of 6.023 x 1023 atoms
of oxygen = 16 g
Similarly,
1 mole of carbon atoms = 6.023 x 1023
atoms of carbon
mass of 1 mole of carbon atoms = 12 g
or mass of 6.023 x 1023 carbon atoms = 12 g
Similarly, a gram molecular mass of
any substance contains 6.023 x 1023 molecules of it.
Thus
gram molecular mass of oxygen = 32
g
Now 32 g of oxygen will contain 6.023
x 1023 molecules,i.e.
mass of 6.023 x 1023 molecules
of oxygen = 32 g
Similarly,
1 mole of NaOH = 6.023 x 1023
molecules of NaOH
mass of 1 mole of NaOH = 40 g
or mass of 6.023 x 1023 molecules of NaOH = 40 g
The mass of one mole of a substance
is often called its molar mass. Thus the molar mass is equal to
the atomic mass or molecular mass expressed in grams depending upon
whether the substance is atomic or molecular.
Relationship
of mole with volume of a gaseous substance
For gases, the volume occupied by
one mole of any gas at STP (273 K and 1 atm. pressure) is 22.4 litre.
Example
1 mole of oxygen = 32 g
\Volume occupied
by one mole of oxygen or 32 g of oxygen at STP = 22.4 litre
Since one mole of oxygen or 32 g of oxygen
contain 6.023 x 1023 molecules of oxygen.
Thus volume of 6.023 x 1023
molecules of oxygen at STP will be 22.4 litres.
Similarly 1 mole of ammonia (NH3) or
17g of ammonia or 6.023 x1023 molecules of ammonia will
occupy 22.4 litre at STP.
Thus
and mass of 11.2 litres of any gas
at STP = vapour density of the gas
Equivalent weight
That part by weight of a substance,
which combines with or displaces 1.008 parts by weight of hydrogen
or 8.0 parts by weight of oxygen or 35.5 parts by weight of chlorine,is
called its equivalent weight. When the equivalent weight of a substance
is expressed in grams it is called the gram equivalent weight or
gram equivalent of the substance.
Calculation
of Equivalent weight of different substances:
| Chemical Calculations in
Solutions |
Standard Solution
A solution of known strength is known
as a standard solution.
Strength of a Solution
The strength of a solution refers
to the amount of solute present in a definite amount of the solvent
or solution.
It is expressed in several ways:
| (i) |
Per
cent Strength |
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(a) |
Percent
strength by Mass
It refers to amount of solute, in grams present in 100 g of
the solution. |
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(b) |
Percent
strength by volume
It refers to amount of solute, in grams present in 100 ml of
the solution. |
| (ii) |
Molarity
Number of gram moles of solute present in 1000 ml (1 litre)
of solution is called the molarity (M) of the solution. |
| (iii) |
Molality
Number of gram moles of solute present in 1000 g (1 kg) of solvent
is called the molality (m) of the solution. |
| (iv) |
Normality
Number of gram equivalents of solute present in 1000 ml of solution
is called the normality (N) of the solution. |
| (v) |
Formality
Number of formula weights of solute present in 1000 ml of the
solution is known as the formality (F) of the solution. |
| (vi) |
Mole
Fraction
It is expressed as the ratio of number of moles of solute
to the total number of moles present in the solution.
Thus if n is the nuber of moles
of solute and N that of solvent then
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Molarity equation
Molarity of a solution is inversely
proportional to its volume,i.e.
i.e.
or MV = constant
Thus if M1 and V1
are the initial molarity and volume of a solution, respectively,
and if this solution is diluted to volume V2 so that
its new molarity becomes M2, then
M1V1 = M2V2
This equation is known as the Molarity
equation.
Normality equation
If a solution with normality N1
and volume V1 is diluted to volume V2 so that
its normality changes to N2, then
N1V1 = N2V2
This equation is known as the Normality
equation.
Important
Relations to Remember
(ii) Number of gram equivalents 
(iii) Mass of one atom of an element
(iv) Mass of one molecule of a substance
For
Solutions
(v) Number of millimoles of solute =
Molarity x volume in ml
(vi) Number of milliequivalents of solute
= Normality x volume in ml
(vii) Normality |
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where
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W = weight of solute in grams
E = equivalent weight of solute
V = volume of solution in ml |
(vii) Molarity M 
where M = molar mass of the solute
(ix) Normality = x
x Molarity
where x
may be
(a) acidity of a base
(b) basicity of an acid
(c) number of electrons gained or lost by a redox reagent
(d) valency of a carbon
(x) Molality (m) 
where X is the weight in gram of solvent.
(xi) Weight per cent of an element in
a compound is given by
(xii) 
(xiii) Molecular weight = 2 x Vapour
density
Solved
Examples of Stoichiometry
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