STOICHIOMETRY SYLLABUS

SYLLABUS
STOICHIOMETRY
• Dalton’s atomic theory and its postulate
• Laws of stoichiometry
• Avogadro’s law and some deductions-relation between mass and vapour
density and molecular mass and number of particles.
• Mole and its relation with mass, volume and number of particles.
• Calculation based on mole concept( stoichiometric calculation)
• Limiting reactant and excess reactant (calculate the maximum amount
of products produced)
• Theoretical yield, experimental (actual ) yield and % yield (calculation)
• Calculation of empirical and molecular formula from % composition

Dalton’s Atomic Theory
 Who Is John Dalton?
John Dalton ( 1766 – 1844)
was an English chemist,
meteorologist and physicist.
He is best known for his
pioneering work in the
development of modern atomic
theory, so he is also one of
fathers of modern chemistry

 What Is DALTON'S ATOMIC THEORY?
The idea of atoms had been proposed much as earlier as
5th century B.C. by Demecritus, a Greek philosopher.
According to him, matter is composed of tiny particles.
These tiny particles were named ‘ATOMOS’ (means
indivisible or uncuttable ,in Greek). John Dalton adopted
this idea and in 1808, put forward the first theory about
the structure of matter which is known as Dalton’s
atomic theory.

All matters are composed of extremely small, indivisible or
microscopic particles called "Atom".
POSTULATES OF DALTON’S
ATOMIC THEORY
Postulate 1
Postulate 2
Atoms can neither be created nor can be destroyed. Since
they are indestructible, this is also called principle of
indestructibility of matter.

Atoms of same element are identical in all respects & are
different from the atoms of all other elements.
Postulate 3
Postulate 4
Atoms combine with each other in simple whole number
ratio to give a compound atom ( nowadays called molecule).

1. With the discovery of still smaller particles, sub-atomic
particles, e.g., electrons, neutrons and protons, the atom
can no longer be the smallest particles and indivisible.
2. The atoms of same elements may not be all alike . They
may vary in mass and density. For example: chlorine
has two isotopes having mass numbers 35 a.m.u and 37
a.m.u. These are known as Isotopes.
The theory remained undisputed up to the 19th century.
But the theory was discarded due to following reasons:

3. Atoms of different elements are not different in all
respects. Atoms of different elements that have the same
atomic mass are called Isobar.eg. Ca and Ar, both have
same mass no. 40 but different atomic number,20 and 18
respectively.
4. The ratio in which atoms combine may not be
simple in complex organic compounds like
sugar C12H22O11 (ratio 1.09 : 2 : 1 ) & Borax,
disodium tetra borate(Na2B4O7; 1:2:3.5)
etc.There are large no. of such compounds
available.

There are certain radioactive elements occurring in
nature the atoms of which change into other atoms by
emission of radiation. Example Uranium(92),
Thorium(90) etc. some elements have been
synthesized as for example, Americium, Curium,
Berkelium, Californium, etc., are the man made
elements.
5. Atoms can be altered.eg.
-α -α
92U 90Th 88 Ra

Laws of chemical
combination are based
on experimental facts ,
that is these laws are
empirical laws.
The calculation which deals with the mass and volume
relationship among reactants and products is termed as
STOICHIOMETRY (in Greek: stoichiometry meaning to
measure an element ). A chemical equation is the basis for such
calculations.
STOICHIOMETRY

1. LAW OF CONSERVATION OF MASS OR MATTER
2. LAW OF CONSTANT COMPOSITION/ LAW OF
DEFINITE PROPORTIONS
3. LAW OF MULTIPLE PROPORTIONS
4. LAW OF RECIPROCAL OR EQUIVALENT
PROPORTIONS
There are 5 basic laws of stoichiometry which govern the chemical
reactions as well as the combination of element to form a compound.
These laws are :

5. THE LAW OF GASEOUS VOLUME ( OR GAY
LUSSAC’S LAW OF GASEOUS VOLUME )
In first 4 laws, combination of substances by
weight takes place & in fifth law, combination of
gases by volumes takes place.

Enunciated by French
Chemist ANTOINE
LAVOISIER

“ In any chemical reaction, the initial weight of
reacting substances is equal to the final weight of the
product.”
OR
“Mass is neither gained nor lost during a chemical
reactions.”
OR
“The total mass of the substance taking part in a
chemical change remains the same through out the
change”

This law is also known as the law of indestructibility of matter.
•This law is in accordance with Dalton’s atomic theory.
• According to Dalton’s atomic theory, atoms neither can be
created nor can be destroyed.
•During a chemical reaction, simply rearrangement of atoms
takes place.
•Since an atom has a fixed mass, total mass during a reaction
is conserved.

The mass of the reactants (starting materials)
equals the mass of the products
2Mg (s) + O2 (g) → 2MgO
(s)
48.6 g 32.0 g 80.6
g
For example:

 The law of conservation of mass can be
demonstrated by the union of hydrogen and
oxygen. If the H2 and O2 are weighed before they
unite, it will be found that there combined weight
is equal to the weight of water formed.

EXPERIMENTAL VERIFICATION
The practical verification of this law was given
by a German Chemist H. Landolt.

The tube with NaCl & AgNO3 solutions before reaction is weighed, then
the tube is weighed after reaction. No change in weight is observed. Thus
this experiment verifies the law of conservation of mass.
In 1905, Albert Einstein enunciated the mass–energy relationship,
E=mc2
, where c=velocity of light & E=Energy corresponding to mass, m.
The mass and energy are interconverible to each other.
Therefore, there should be decrease in mass in case of reactions which
are accompanied by liberation of energy.
Modification of the law of conservation of mass

Hence the law of conservation of mass can also be stated as:
“There is no detectable gain or loss of mass in a chemical
reaction.”
However, the mass ,(m=E/C2 ) changing into energy is
extremely small because the value of C is very large.
OR
“ Total mass and energy during a chemical reaction
remains always constant.

In the sun, matter is changed into
energy and thrown out in the form of
heat, light etc. and probably
somewhere in the universe this
energy may be utilized in the
formation of matter

ESTABLISHED BY FRENCH
CHEMIST JOSEPH LOUIS
PROUST

“A chemical compound always
consists of the same elements
combined together in the same
proportions by weight irrespective
to the source & methods of
preparation.

OR,
“A chemical compound contains the same
elements in exactly the same proportions
(ratios) by mass regardless of the size of the
sample or source of the compound.”
OR
“The percentage composition of an element
in a compound is always fixed.”

For example, regardless its source, water is always
composed of hydrogen and oxygen atoms combined
together in the fixed ratio of 1:8 by weight or it is always
89 percent oxygen by mass and 11 percent hydrogen by
mass.
% of H=2/18 × 100 = 11.11%
% of O= 16/18 × 100= 88.88%

Every sample of
pure water,
though prepared
in the laboratory
or obtained from
rain, river or
water pump,
contains 1 part
hydrogen and 8
parts oxygen by
mass.

EXPERIMENTAL VERIFICATION OF THE LAW OF
CONSTANT COMPOSITION
Let us take 3 samples of copper oxide prepared by the
following methods :
a. ∆
2Cu(NO3)2 2CuO + 4NO2 +
O2
b.
∆
Cu(OH)2 CuO + H2O
c.
∆
CuCO CuO + CO

In all the three samples, the % of Cu & O2 by
weight has been found to be the same. Thus
the law is experimentally verified.

Berzelius heated 10g lead (Pb) with various amounts of
sulphur (S). But every time he got exactly 11.56g of Lead
sulphide, and the excess of sulphur was left over.
This shows the significance of law of constant
composition.

Given by: John Dalton in theyear 1804
“When two elements combine to form two or
more different compounds, then the weights
of one of the elements which combines with a
fixed weight of the other, bear a simple ratio
to one another."
33
STATEMENT

For Example carbon forms 2 stable compounds with
oxygen:
In Carbon monoxide (CO): 12 parts by weight of carbon
combine with 16 parts by weight of oxygen.
In Carbon dioxide (CO2): 12 parts by weight of carbon
combines with 32 parts by weight of oxygen.
 Ratio of the weights of oxygen that combines with a
fixed weight of carbon (12 parts) is 16: 32 or 1: 2 which is
a simple ratio. Thus the law is illustrated.

 Another illustration of this law is the
formation of Water and Hydrogen Peroxide.

Compounds
Elements
Nitrous
Oxide
Nitric
Oxide
Nitrogen
Trioxide
Nitrogen
Tetraoxide
Nitrogen
Pentoxide
Formula N2O NO N2O3 N2O4 N2O5
Wt. of
Nitrogen
28 14 28 28 28
Wt. of
Oxygen
16 16 48 64 80
When N is
taken 14
14 14 14 14 14
Then O is 8 16 24 32 40
ONE MORE EXAMPLE: Nitrogen and Oxygen combine with each
other to form compounds; N2O, NO, N2O3, N2O4, N2O5.

The ratio of Oxygen which combines with fixed weight of
Nitrogen (14 in this example) is 8:16:24:32:40 or 1:2:3:4:5,
which is a simple ratio. Thus the law is illustrated.
EXPERIMENTAL VERIFICATION OF THE LAW
Let us take pure samples of two oxides of same element, copper;
CuO & Cu2O.
These two samples are weighed & then subjected to reduction by
heating in the current of hydrogen.
CuO + H2 Cu + H2O
Cu2O + H2 2Cu + H2O

After complete of reduction, copper left behind in each sample is
weighed separately. Then wt. of oxygen in each oxide is calculated by
using formula, Wt. of O2=Wt. of oxide-wt. of copper
Now the weights of Cu that combine with the same wt. of oxygen is
calculated by unitary method.
The ratio of these weights of Cu is always found to be 1:2 Which is a
simple ratio.
Thus the law is experimentally verified.

The ratio of the weights of two different elements ( say A and
B) which combine separately with a constant weight of the
third element (say C) is either the same, or simple multiple of
the ratio of the their ( A and B) weights in their actual chemical
combination.
To understand this better, let us assume that A is Carbon, B is
Hydrogen and C is Oxygen .
STATEMENT

A=C, B=H AND C= O
C O
H
CH4
CO2
H2O
In CO2, 12 parts by weight of carbon combine
with 32 parts by weight of oxygen.
In H2O, 2 parts by weight of hydrogen combine
with 16 parts by weight of oxygen.
Or in H2O, 4 parts by weight of hydrogen combine with 32 parts by
weight of oxygen.
Hence the ratio of weights of carbon and hydrogen which combine with
fix weight of oxygen(32) is 12:4 Or 3:1 ……………(i)

In CH4, 12 parts by weight of carbon combine with 4 parts by weight
of hydrogen.
Here, in actual combination of C and H, The ratio of weights of C & H is
12:4 Or 3:1 ………………….(ii)
The ratios (i) & (ii) are found to be same , thus the law is illustrated.
THUS THE LAW IS ILLUSTRATED

ANOTHER EXAMPLE
C
S O
CS2
CO2
SO2
12 parts by weight of carbon combine with 32
parts by weihgt of oxygen.
In CO2
In SO2
32 parts by weight of sulphur combine with 32 parts by weight of
oxygen
From the above two compounds, the ratio of wts. Of carbon and
sulphur which combine with fixed wt. of oxygen is 12:32 or
3:8……………………………………(i)
In CS2 by actual combination
12 parts by weightof carbon combine with 64 parts by weight of sulphur.

The ratio of wts. of carbon and sulphur in their actual combination is
12:64 or 3:16………(ii)
Now, the ratios (i) & (ii) are related to each other as
3/8 : 3/16 or 3/8 ×16/3 or 2:1
This shows ratios (i) & (ii) are whole no. multiple of each other.
THUS THE LAW IS ILLUSTRATED.

This law is completely dependent on the combining
equivalents of elements i.e. 1.008 part by weight of
hydrogen, 8 parts by weight of oxygen & 35.5 parts by
weight of chlorine. So, this law is also called the law
of chemical equivalents or the law of equivalent
proportion. According to this “ Elements combine with
each other in the ratio of their chemical equivalents or
simple multiple of the equivalent weight.”

THE LAW OF GASEOUS VOLUME
ESTABLISHED
BY
JOSEPH-LOUIS GAY -
LUSSAC

STATEMENT
“When two different gases combine together
under similar conditions of temperature and
pressure to give gaseous products, the
volumes of the reactants as well as products
bear a simple whole number ratio.”
ILLUSTRATIONS
(1) H2(g) + Cl2(g) 2HCl(g)
1 Vol. 1 Vol.
2 Vol.
(Ratio 1:1:2)
(2) 2CO(g)+ O2(g) 2CO2(g)
2 Vol. 1 Vol. 2 Vol. (Ratio 2:1:2)
(3) N2(g) + 3H2(g) ↔ 2NH3(g)
1 Vol. 3 Vol. 2 Vol. (Ratio 1:3:2)

The ratios in all above three reactions are simple
whole number ratio.
Thus the law is illustrated.

MOLE CONCEPT
Mole has derived from molecular mass. Mole is the full form
& mol. is the abbreviated form of mole.
1 mol. equals to gram atomic weight or gram molecular
weight.
EXAMPLE, Atomic weight of sodium= 23 amu
Gram atomic weight of sodium=23g
Therefore, 1 Mole of sodium =23g
Similarly, Molecular weight of water=18 amu
and Gram molecular weight=18g
Therefore 1 Mole of water=18g
Mole is expressed in terms of-
1) Gram 2) Volume 3) Number
Mole in terms of
gram

Mole in terms of Volume
1 Mole of any gas at NTP occupies 22.4L of
volume.
Mole in terms of number
“ A mole is a collection of particles of anything
(atoms, molecules, ions, electrons, protons, etc.)
equal in number to the number of atoms present
in 12g ( 1g atom) of C-12 isotope.
It has been found that 1g atom of C-12 isotope contains
6.023×1023 carbon atoms. This number is called
Avogadro’s number.

1 Dozen= 12 pcs , 1 Gross= 144pcs
Similarly, 1 Mole=6.023×1023
1 Dozen banana= 12 pieces
1 Gross banana= 144 pieces
Similarly, 1 Mole banana=6.023×1023 pieces.

Therefore,
1 mole of H atoms =
6.023×1023 H atoms=1.008g
1 mole 0f SO4
-- ions= 6.023×1023 SO4
--ions=96g
1 mole of KCl = 6.023×1023 formula units=74.6g
1 mole of H2O molecules =
6.023×1023 H2O molecules =18g
1 mole of H2 molecules =
6.023×1023 H2 molecules=2.016g

IN CONCLUSION, WE CAN WRITE
1 Mole =6.023×1023 particles
= one gram atomic weight
or one gram molecular weight or
one gram formula weight in case of
ionic compounds or one gram ionic
weight or 22.4L at NTP in case of
gases, etc.

CALCULATION OF NO. OF MOLES
Weight in gram
1. For atoms, No. of moles=
Atomic weight
Given no. of atoms
=
NA
Weight in gram
2. For molecules, no. of moles=
Molecular weight
Given no. of molecules
=
NA

Weight in gram
3. For ionic compounds, no. of moles =
Formula weight
Weight in gram
For ions, no of moles =
Ionic weight
Given volume of gas in litres at NTP
For gases at NTP, no. of moles
22.4

Most commonly, following relations are involved during
chemical calculations;
1. a. Mole and mole relation
b. Mole and volume relation
c. Mole and number relation
d. Mole and mass relation
2. a. Mass and mass relation
b. Mass and volume relation
c. Mass and number relation
3. a. Volume and volume relation
b. Volume and number relation
4. Number and number relation

FEW NUMERICALS ON MOLE
CONCEPT
Q. Calculate the absolute mass of 1 amu.
Solution,
1 mole of carbon-12 isotope=6.023×1023 C-12 atoms
= 12g
6.023×1023 C-12 atoms= 12g
12g
1 atom =
6.023×1023
1 amu= 1/12 of wt. of 1 atom of C-12 isotope =
12 1
×
6.023×1023 12
g

Therefore 1 amu= 1.66×10-24g
= 1.66×10-24g
Q. Calculate the no. of moles of each of the following,
i. 5g of CaCO3
ii. 1.6g of SO4– - ions
iii. 2.24L of CO2 at NTP
iv. 3.011×1026 electrons
(Ans 0.05 moles)
(Ans 0.1 mole)
(Ans 500 moles)
(Ans 0.0167)
Q. Calculate the no. of moles and no. of molecules present
in o.85g of NH3 gasAns. 0.05mole,3.011×1022 molecules

Solution:
Given mass of NH3=0.85g, Molecular mass of ammonia=17
0.85
No. of moles of NH3 = = 0.05 moles
17
1 mole of NH3 gas = 6.023×1023 NH3 molecules
0.05…………………… = 0.05×6.023×1023 molecules
= 3.011×1022 molecules
Q. How many electrons are required to discharge 3 moles of
calcium cation? (Ans. 3.6138×1024 electrons)

Solution:
Ca++ + 2e- Ca
1 mole 2 moles 1 mole
3 moles 6 moles
6 moles of electrons are required to discharge
3 moles of Ca++
We have,
1 mole of electrons = 6.023×1023 electrons
6 moles of electrons = 6×6.023×1023 electrons
= 3.6138×1024 Ans.

1. The cost of table sugar is Rs. 40 per kg. Calculate its cost per mole.
2. Find the no. of electrons present in 1.6g of methane?
3. 16g of a gas at STP occupies 5.6L. What is the molecular mass of the
gas?
4. What volume of CO2 gas is produced by heating 50g of 50% pure
CaCO3?
5. Calculate the no. of atoms of carbon present in 25g CaCO3.
Questions
Answers
1. Rs.13.68
2. 6.023×1023 electrons
3. 64amu
4. 5.6L
5. 1.505×1023 carbon atoms

Q. Pure water sample having a mass of 1.8g is taken, answer the
following questions:
a. How many moles of water are present?
b. How many molecules of water are present?
c. How many atoms of hydrogen are present?
d. How many gram molecules of water are present?
e. How many gram atoms of hydrogen are present?
Answers:
a. 0.1mol.
b. 6.023×1022 molecules
c. 12.044×1022 atoms
d. 0.1 gram molecule of water
e. 0.2 gram atoms.

AVOGADRO’S HYPOTHESIS
STATEMENT
“The equal volume of all gases and vapours in
the same conditions of temperature & pressure
contain equal number of molecules (OR
MOLES).”
According to this hypothesis if we take a litre of
hydrogen or nitrogen or oxygen at NTP, then all of three
gases will contain same number of molecules.

Avogadro’s hypothesis helps:
Applications of Avogadro’s hypothesis
1. to harmonize Dalton’s atomic theory and the law of gaseous
volume.
2. in deduction of atomicity of elementary gases.
3. In the deduction of relationship between molecular mass
and vapour density of gas.
4.in deduction of relationship between molecular mass and
number of particles.

1. Harmonization of Dalton’s atomic theory and the law of
gaseous volume.
For example,
Hydrogen gas + Chlorine gas Hydrogen chloride gas
1 vol. 1 vol. 2 vol. (From expt.)
Thus this reaction follows the law of gaseous volume since
the ratio is simple.
Let 1 vol. of hydrogen gas contains ‘n’ no. of hydrogen
molecules.
(Volumes are measured under similar conditions)

Then by applying Avogadro’s hypothesis, 1 vol. of
chlorine also contains ‘n’ no. of chlorine molecules.
Similarly, 2 volumes of hydrogen chloride contain
‘2n’ no. of molecules of hydrogen chloride.
Hence we have,
‘n’ molecules + ‘n’ molecules ‘2n’ molecules
Therefore,
1 molecule + 1 molecule 2 molecules
½ molecule + ½ molecule 1 molecule
Therefore, 1molecule of hydrogen chloride contains ½
molecule of hydrogen & ½ molecule of chlorine.

We know that 1 molecule of any compound contains at
least 1 atom of its constituent elements.
Therefore, ½ molecule of hydrogen as well as ½
molecule of chlorine should contain at least 1 atom from
each of the corresponding elements, because, according
to Dalton’s atomic theory, atoms cannot be divided.
However, molecules can be divided. It is not against
Dalton’s atomic theory.
Hence, the law of gaseous volume is supported by
Dalton’s atomic theory.

Deduction of relationship between molecular mass and
vapour density of gas
Wt. of certain vol. of gas
vapour Density(V.D.) =
Wt. of same vol. of hydrogen gas
(Under similar conditions of temperature and pressure)
Let certain volume of gas contains ‘n’ molecules and then
applying Avogadro’s hypothesis:
Wt. of n molecules of the gas
V.D. =
Wt. of n molecules of hydrogen gas

If n=1,
Wt. of 1 molecule of the gas
V.D. =
Wt. of 1 molecule of hydrogen gas
Wt. of 1 molecule of the gas
=
2 × Wt. of 1 atom of hydrogen gas
Molecular mass of the gas
=
2
Hence, Molecular mass of a gas = 2 × V. D.

DEDUCTION OF RELATIONSHIP BETWEEN
MOLECULAR MASS AND NUMBER OF
PARTICLES
MOLECULAR MASS EXPRESSED IN GRAM IS
KNOWN AS GRAM MOLECULAR MASS OR MOLE.
1 Mole of anything = 6.023×1023 particles
We know,
1 mole of any gas at NTP = 22.4L
That is, 22.4L of any gas at NTP = 6.023×1023 gaseous
molecules (Avogadro’s hypothesis)
Therefore, wt. of 6.023×1023 gaseous molecules=1 mole
of gas

Limiting reactant and excess reactant
one of the reactants is present in larger amount than
the other.
While performing a chemical reaction, the reactants are
not always taken as stoichiometric equation required.
The reactant which is in deficit amount is consumed
first and thus decides the amount of product formation.
Hence, the chemical reactant which is in deficit
amount and finished first after completion of reaction,
is called LIMITING REACTANT OR LIMITING
REAGENT.

It is called so as it limits or controls the amount of product
formation in a chemical reaction.
The remaining reactant left over unreacted in the reaction
mixture is called EXCESS REACTANT.
For example,
HCl reacts with NaOH to form NaCl & H2O
Let us mix 38.5g HCl with 40g NaOH.
The balanced chemical equation is :
HCl + NaOH NaCl + H2O
1 mole 1 mole 1 mole 1 mole
36.5g 40g

The stoichiometric amount of HCl is 36.5 & that of NaOH
is 40g. for complete neutralization.
Here, mass of HCl is 38.5g which is more than required by
the balanced chemical equation.
Hence, HCl is excess reagent and NaOH is limiting reagent.
Q. 4g of methane is mixed with 20g of oxygen to
form CO2 & H2O.
i. Which one is the limiting reagent and why?
ii. Calculate the mass of water produced.
iii. What is the volume of CO2 gas produced at NTP?
Solution,

i. The balanced chemical equation is,
CH4 + 2O2 CO2 + 2H2O
1 mole 2 mole
16g 2×32g=64g
For 4g 64/16 ×4=16g
Hence, 4g of CH4 requires 16g of oxygen for complete combustion.
But, the mass of oxygen taken is 20g which is in excess.
Thus methane is limiting reagent as it is finished first.
ii. The balanced chemical equation is,
CH4 + 2O2 CO2 + 2H2O
16g 2×18g=36g

4g 36/16 ×4=9g of water Ans.
iii. The balanced chemical equation is,
CH4 + 2O2 CO2 + 2H2O
4g 22.4/16 × 4= 5.6L at NTP Ans.
16g 22.4L at NTP
Q. Given,
CaCO3 + 2HCl CaCl2 + H2O + CO2
If 10g 0f pure CaCO3 are added in a solution containing 7.665g HCl.
i. Find the limiting reagent.
ii. Calculate the no. of moles of excess reactant left over unreacted.
iii. Calculate the volume of CO2 gas produced at NTP.
iv. Calculate the no. of gram of NaOH required to absorb whole of the CO2
gas as Na2CO3.

Solution:
i. The given balanced chemical equation is :
CaCO3 + 2HCl CaCl2 + H2O + CO2
1 mole 2 mole
=100g =2×36.5
=73g
That is, 100g Of CaCO3 consumes 73g of HCl
Given, wt. of CaCO3= 10g
So, 10g of CaCO3 consumes 73/100 ×10 g HCl
= 7.3g(required amount)
But, given wt. of HCl = 7.665g, which is greater amount
than required amount.
Hence, CaCO3 is finished first, so CaCO3 is the limiting
reagent.

ii. Excess amount of HCl= Given weight – Required wt.
= 7.665 – 7.300
= 0.365g
o.365
Therefore, no of moles of HCl left = = 0.01 mole Ans
36.5
iii. CaCO3 + 2HCl CaCl2 + H2O + CO2
1 mole produces 1 mole
= 100g =22.4L at NTP
Therefore,
10g…………………………………...22.4/100 × 10
= 2.24L of CO2 gas at NTP
Answer

iv. The balanced chemical equation between NaOH
& CO2 is ( to be remembered) :
2NaOH + CO2 Na2CO3 + H2O
2 mole 1 mole
=2×40 = 22.4L at NTP
=80g
That is, 22.4L of CO2 at NTP are absorbed by 80g of NaOH.
But, volume of CO2 gas produced at NTP=2.24L
2.24L of CO2 at NTP are absorbed by 80/22.4 ×2.24
= 8g of NaOH Ans

Q. 20g of marble containing 90% CaCO3 is
allowed to react with 50g of concentrated acid
containing 36.5% of HCl by mass. Find the
i. Limiting reagent
ii. Number of moles of excess reactant left over
unreacted
iii. Mass of CaCl2 formed
iv. Volume of CO2 gas produced at NTP
v. No. of moles of H2O produced
Answers
i. CaCO3 ii) 0.14 mole
iii) 19.98g iv) 4.032L
v) 0.18 mole

Theoretical yield, experimental (actual )
yield and % yield (calculation)
Theoretical yield:-
For a given reaction, the amount of product to be
formed is calculated with the help of its stoichiometric
equation. This amount of product calculated from
stoichiometric equation is called theoretical yield or
stoichiometric yield. This is theoretical value.
Therefore, theoretical yield is the function of limiting
reactant.
More frequently, the actual amount of product
formed is different than the calculated or theoretical
amount.

The following are the reasons for the formation of
different amount of products than theoretical
amount:
i. Purity of chemicals (reactants)
ii. Completion of reaction
iii. Reversibility of reaction
iv. Processing of compounds (handling during process)
v. Formation of side products, etc.
Experimental (actual ) yield:-
The amount of product obtained from experiment is
called actual or experimental yield. This is
experimental value.

The experimental amount of product is usually
somewhat different from its theoretical amount.
Percentage yield:-
It is the ratio of experimental yield to the theoretical yield
expressed in percent.
Mathematically,
Experimental yield
Percent yield = ×100%
Theoretical yield
Basically(practically), the percentage yield is always less
than 100%.

Because, experimental yield is generally always
less than theoretical yield.
Q. 1 gram of HAuCl4.3H2O ( mol. Mass=393.833, and
atomic mass of gold is 199.966 )was reduced to gold
nanoparticles using sodium citrate. After subsequent
purification and drying , the mass of gold nanoparticles
was 0.325g. Calculate the percentage yield of gold
nanoparticles.
Solutio
n,
We know, 393.833g of HAuCl4 contain 199.966g Au, then
1g……………..contains 199.966/393.833
=0.507g
Therefore, theoretical yield=o.507g
Given, experimental yield=0.325g

Hence,
Experimental yield
Percent yield = × 100%
Theoretical yield
0.325
= × 100
0.507
= 64.10% Ans.
Q. 4.6g of pure sodium was reacted with excess of
Chlorine to produce sodium chloride. If the amount of
sodium chloride obtained was 10g, what is the percentage
yield of the reaction? Ans. 85.47%

Calculation of empirical and molecular
formula from % composition
EMPIRICAL FORMULA
It is the simplest form of formula of a compound. It indicates the
simplest ratio of the atoms of the constituent element present in one
molecule of the compound.
MOLECULAR FORMULA :
Chemical substances are written in abbreviated form by using
symbols, called chemical formula or molecular formula. It is
symbolic expression for a molecule of a substance. It shows actual
number of atoms of constituent element/s in one molecule of the
substance. Thus a molecular formula is the shorthand expression of
the composition of a molecule.

Molecular formula of a compound is determined with the help
of its empirical formula and empirical formula, in turn is
determined with the help of percentage composition of the
compound. And percentage composition of the compound is
determined with the help of a suitable chemical analysis.
DETERMINATION OF EMPIRICAL FORMULA
FROM PERCENTAGE COMPOSITION:

RULES:
1. The % composition of given element in the given
compound is determined by a suitable chemical
analysis.
2. If the total % composition of elements is not equal
to 100%, the remaining % will that of oxygen atom.
3.% composition of each of the element is divided by
its respective atomic weight to get relative no. of moles
of the element.

4. Each relative moles of element is divided by the
smallest relative number of moles to get the simplest
whole number ratio. (If it is not in whole no., the ratio is
multiplied by certain whole no. to convert into whole no.
ratio.)
5. The empirical formula is written from the simplest
whole no. ratio.
DETERMINATION OF MOLECULAR FORMULA FROM
EMPIRICAL FORMULA
The molecular formula is determined by using the relation;
Molecular formula=(Empirical formula)n
Molecular mass
Where, n(common factor) =
Empirical formula weight

Molecular mass of compound is determined by a
suitable chemical analysis.
Empirical formula weight is determined by adding
total relative atomic weights of elements from
empirical formula.
Thus n can be calculated and hence molecular formula is
determined.

Q. An organic compound contains 32% carbon and 4%
hydrogen. Determine empirical formula of the compound.
Solution:
S.N. Element %
composition
Atomic
weight
Number
of moles of atoms
Simplest ratio
of moles of
atoms
Simple-
st whole
no. ratio
1 C 32 12 32/12=2.66 2.66/2.66=1 2
2 H 4 1 4/1=4 4/2.66=1.5 3
3 O 64 16 64/16=4 4/2.66=1.5 3
Therefore, empirical formula of compound is C2H3O3.

Q. Find the molecular formula of an organic compound which gave
the following percentage composition. C=26.6%, H=2.22%. The
vapour density of the compound is 45 amu.
Solution:
S.N. Element %
Composition
Atomic
-weight
Number
of moles of
atoms
Simplest ratio of
moles of atoms
Simplest
whole no.
ratio
1 C 26.6 12 26.6/12=
2.21
2.21/2.21=1 1
2 H 2.22 1 2.22/1=
2.22
2.22/2.21=1 1
3 O 71.18 16 71.18/16=
4.4
4.4/2.21=2 2

Therefore, empirical formula of compound is CHO2.
Empirical formula weight= 12+1+16×2=45 amu
Vapour density of compound(given)=45 amu
Therefore, Mol.wt. of the compound=2×V.D.
=2×45=90amu
Mol.wt. 90
Therefore, n= = = 2
Emp.formula wt. 45
Therefore, molecular formula of compound =(CHO2)n
=(CHO2 )×2=C2H2O4 Ans.

Q. An organic compound was found to have carbon
65.73%, hydrogen 15.06% and nitrogen 19.21%. It’s V.D.
was found to be 37. Calculate the molecular formula of
the organic compound. Ans. C4H11N

STOICHIOMETRY SYLLABUS

Recommended

Recommended

More Related Content

Similar to STOICHIOMETRY SYLLABUS

Similar to STOICHIOMETRY SYLLABUS (20)

Recently uploaded

Recently uploaded (20)

STOICHIOMETRY SYLLABUS

Editor's Notes