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Representative Sample Chemistry:
Mole Concept and Stoichiometry
MOLE CONCEPT AND STOICHIOMETRY
1. Introduction
As we have learned about the units and various conversions, it’s time that we start using them. Before
starting with the core calculations, we need to know the laws which have been formulated with the help
of the facts known from experimental evidences. So now, get ready to play.
2. Dalton’s Atomic Theory
Postulates of Daltons theory:
Mole Concept and Stoichiometry
16 PLANCESS CONCEPTS
Many unexplained chemical phenomena were quickly explained by Dalton with his theory.
Dalton’s theory quickly became the theoretical foundation in chemistry.
Vipul Singh
AIR 1, NSTSE 2009
2.1 Law of Conservation of Mass
A. Lavoisier in 1789 introduced the concept of mass conservation. He stated that “In all physical changes
and chemical reactions, the total mass of the products is the same as the total mass of the reactants.” He
is considered as “Father of Modern Chemistry”.
Modification to the Law of Conservation of mass
The total sum of mass and energy of the system remains constant.
PLANCESS CONCEPTS
The above law will remain constant over time, provided the system is isolated.
A similar statement is that mass cannot be created/destroyed, although it may be rearranged
in space, and changed into different types of particles.
Sir Lavoisier was the one who coined the word “oxygen” meaning “acid former” and came up
with the combustion process through his experiment.
Vipul Singh
AIR 1, NCO
2.2 Law of Definite Proportions or Constant Composition
Law of Constant Composition was given by Sir Joseph Proust in 1799, The law states that “A pure
chemical compound always consists of the same elements combined together in a fixed (definite)
proportion by weight.”
Illustration 1: The mass of copper oxide obtained by heating 2.16 g of metallic copper with nitric acid
and subsequent ignition was found to be 2.7 g. In another experiment, 1.15 g of copper oxide on
reduction yielded 0.92 g of copper. Show that the data illustrates the law of constant composition.
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Mole Concept and Stoichiometry
17
Sol:
In the first experiment
In the second experiment
Weight of copper = 2.16 g
Weight of copper oxide = 1.15 g
Weight of copper oxide= 2.70 g
Weight of oxygen = 0.23 g
Weight of oxygen = 2.70 – 2.16 = 0.54 =
g
% copper
% copper =
=
Weight of copper
× 100
Weight of copper oxide
(2.16g)
× 100
= 80%
(2.70g)
% oxygen =
Weight of oxygen
× 100
Weight of copper oxide
=
(0.54 g)
× 100
= 20%
(2.70g)
Weight of copper
× 100
Weight of copper oxide
=
(0.92g)
× 100
= 80%
(1.15g)
% oxygen =
Weight of oxygen
× 100
Weight of copper oxide
=
(0.23g)
× 100
= 20%
(1.15g)
Since the ratio by weights of copper and oxygen in the two compounds remains the same, the law of
definite composition is illustrated.
2.3 Law of Multiple Proportions
In 1803 Sir John Dalton discovered the Law of Multiple Proportions. According to this law, “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, bears a simple ratio to one another.”
Simple ratio means the ratio between small natural numbers, such as 1:1, 1:2, 1:3.
It is one of the basic law of stoichiometry and it is used by Dalton to establish the atomic theory.
Mole Concept and Stoichiometry
18 Illustration 2: Carbon is found to form two oxides, which contain 42.8% and 27.27% of carbon
respectively. Show that these figures illustrate the law of multiple proportions.
Sol:
For the first oxide:
For the second oxide:
% of carbon in first oxide = 42.8
% of carbon in second oxide = 27.27
% of oxygen in first oxide = 100 – 42.8 = 57.2
Mass of oxygen in grams that
% of oxygen in second oxide =100 – 27.27=
72.73
combines with 42.8 g of carbon = 57.2
Mass of oxygen in grams that
∴ Mass of oxygen that combines with 1 g of
combines with 27.27g of carbon =72.73
carbon =
57.2
= 1.34 g
42.8
∴ Mass of oxygen that combines with 1 g of
carbon =
72.73
= 2.68 g
27.27
Ratio between the masses of oxygen that combine with a fixed mass (1 g) of carbon to the two oxides =
1.34:2.68 or 1:2 which is a simple ratio. Hence, this illustrates the law of multiple proportion.
Illustration 3: Carbon and hydrogen combine to form three compounds A, B and C. The percentage of
hydrogen in these compounds are: 25, 14.3 and 7.7 respectively. Show that the data illustrates the Law
of Multiple Proportions.
Sol. Let us fix 1g of hydrogen as the fixed weight in the three compounds.
In the first compound
In the second compound
In the third compound
Weight of hydrogen=25.0 g
Weight of hydrogen = 14.3 g
Weight of hydrogen = 7.7g
Weight of carbon
Weight of carbon
Weight of carbon
=100 –25 = 75.0g
= 100 – 14.3 = 85.7g
= 100 – 7.7 = 92.3g
25.0 g of hydrogen have
14.3g of hydrogen have
7.7g of hydrogen have
combined with carbon=75.0 g
combined with carbon
combined with carbon = 92.3g
1.0 g of hydrogen has with
= 85.7 g
1.0 g of hydrogen has
1.0g of hydrogen has
combined with carbon
combined with carbon
= 92.3/7.7 =12 g
carbon =
75.0
g = 3g
25.0
= 85.7/14.3 = 6 g
Ratio of weight of carbon which combine with 1g of hydrogen in the three compounds is 3 : 6 : 12 or 1
: 2 : 4. As the ratio is simple whole number in nature, the Law of Multiple Proportions is proved.
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Mole Concept and Stoichiometry
PLANCESS CONCEPTS
Dalton became the first scientist to explain the behavior of atoms in terms of the measurement
of weight.
Vipul Singh
AIR 1, NSO
2.4 Law of Reciprocal Proportions
Jeremiasa Richter German scientist in 1972 proposed a law known as “Law of Reciprocal Proportions “
The Law States that , “The ratio of the weights of two elements, A and B which combine separately with a
fixed weight of the third element C is either the same or some simple multiple of the ratio of the weights
in which A and B combine directly with each other.”
He introduced the term “Stoichiometry “
The elements C and O combine separately with the third element H to form CH4 and H2O and they
combine directly with each other to form CO2.
In CH4, 12 parts by weight of carbon combine with 4 parts by weight of hydrogen. In H2O, 2 parts by
weight of hydrogen combine with 16 parts by weight of oxygen. Thus, the weight of C and O which
combine with fixed weight of hydrogen (say 4 parts by weight) and 12 and 32 i.e., they are in the ratio
12:32 or 3:8. Now in CO2, 12 parts by weight of carbon combine directly with 32 parts by weight of
oxygen i.e., they combine directly in the ratio 12:32 or 3: 8 which is the same as the first ratio.
Illustration 4: Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contain 88.90%
of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of
nitrogen. Show that this data illustrates the law of reciprocal proportions.
Sol:
In NH3, 17.65 g of H combine with 82.35 g N
In N2O3, ratio of weights of N and O
∴ 1 g of H combine with N =
which combine with each other =36.85: 63.15
82.35
g = 4.67 g
17.65
In H2O, 11.10 g of H combine with 88.90 g O
∴ 1 g H combine with O =
= 1: 1.71
88.90
g = 8.01 g
11.10
∴ Ratio of the weights of N and O which
combine with fixed weight (1g) of H
= 4.67: 8.01 = 1: 1.72
Thus, the two ratios are the same. Hence, it illustrates the law of reciprocal proportions.
Mole Concept and Stoichiometry
20 Illustration 5: Phosphorus trichloride (PCl3) contains 22.57% of phosphorus, phosphine (PH3) contains
91.18% of phosphorus and Hydrochloric acid (HCl) contains 97.23% of chlorine. Show that the data is
according to the Law of Reciprocal Proportions.
Sol. Let us fix 1g of phosphorus (P) as the fixed weight.
In phosphorus trichloride (PCl3)
In phosphine (PH3)
Weight of phosphorus = 22.57g
Weight of phosphorus = 91.18g
Weight of chlorine = 100 – 22.57 = 77.43g
Weight of hydrogen = 100 – 91.18
22.57g of phosphorus have combined
= 8.82g
with chlorine = 77.43g
91.18g of phosphorus have combined
1.0g of phosphorus has combined
with hydrogen = 8.82g
with chlorine =
77.43
g = 3.43g
22.57
1.0g of phosphorus has combined
with hydrogen =
8.82
g = 0.097g
91.18
Thus, the ratios by weights of chlorine and hydrogen combining with a fixed weight of phosphorus in
the two compounds is 3.43:0.097 (This ratio is not the same) ....(i)
In hydrochloric acid gas (HCl), Weight of chlorine = 97.23g
Weight of hydrogen = 100 – 97.23 = 2.77g
Thus, chlorine and hydrogen have combined in the ratio 97.23:2.77
Let us compare the two ratios.
These are related to each other as
....(ii)
3.43 97.23
:
= 35.37:35.10 or 1:1 (approximately)
0.097 2.77
2.5 Gay Lussac’s Law of combining Volumes
Sir Gay Lussac found the relationship existing between the volumes of the gaseous reactants and their
products. In 1808, he put forward a generalization known as the Gay Lussac’s Law of combining volumes.
According to this Law “When gases react together, they always do so in volumes which bear a simple
ratio to one another and to the volumes of the product. If these are also gases, provided all measurements
of volumes are done under similar conditions of temperature and pressure.”
It implies that one molecule of hydrogen chloride gas is made up of 1 atom of hydrogen and 1atom of
chlorine.
2.6 Avogadro’s Hypothesis
According to this hypothesis “Equal volumes of all gases under similar conditions of temperature and
pressure contain equal number of molecules.”
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Mole Concept and Stoichiometry
21
This hypothesis explains elegantly all the gaseous reactions and is now widely recognized as a law or a
principle known as Avogadro’s Law or Avogadro’s principle.
The reaction between hydrogen and chlorine can be explained on the basis of Avogadro’s Law as follows:
Applying Avogadro’s hypothesis:
Hydrogen 2n
No. of molecules - +
Oxygen Water Vapour
n 2n
1
Thus, 1 molecule of water contains molecule of oxygen. But, 1 molecule of water contains 1 atom of
2
oxygen. Hence, 1 molecule of oxygen = 2 atoms of oxygen i.e., atomicity of oxygen = 2
2.6.1 Applications of Avogadro’s hypothesis
(A) In the calculation of atomicity of elementary gases
E.g. 2 volumes of hydrogen combine with 1 volume of oxygen to form two volumes of water
vapour.
Word equation for this reaction can be written as
Hydrogen +
Oxygen 1 vol. 2 vol.
Water Vapour
2 vol.
Example: Combination between hydrogen and chlorine to form hydrogen chloride gas. One
volume of hydrogen and one volume of chlorine always combine to form two volumes of
hydrochloric acid gas.
Word equation for this reaction can be written as
H2 (g) 1 vol. +
Cl2(g) ⟶
1 vol. 2HCl (g)
2 vol.
Thus proved, the ratio between the volume of the reactants and the product in this reaction is
simple, i.e., 1: 1: 2. Hence it illustrates the law of combining volumes.
(B) To find the relationship between molecular mass and vapour density of a gas
Density of gas
Massof acertainvolumeof thegas
=
Density of hydrogen
Massof thesame volumeof hydrogenat
thesametemp,andpressure
Vapour density (VD) = If n molecules are present in the given volume of a gas and hydrogen under similar conditions of
temperature and pressure.
VD =
Massofnmoleculesofthegas
Massof1moleculeofthegas
=
Massofnmoleculesofhydrogen Massof1moleculeofhydrogen
Mole Concept and Stoichiometry
=
22 Massof1moleculeofthegas
Molecularmass
=
Massof1moleculeofhydrogen
2
(Since molecular mass of hydrogen is 2). Hence, Molecular mass = 2 × vapour density
In a single reactant reaction, the calculations are arrived out with only that amount of the reactant
which has converted into the product.
In the reactions, where more than one reactant is involved, one has to first identify the limiting
reactant, i.e., the reactant which is completely consumed. All calculations are to be carried out
with the amount of the limiting reactant only. For this, limiting reactant should be identified.
The following example would help us.
A
+
2B ⟶4C
Initially — 5 moles
12 moles 0 moles
If A is the limiting reactant, moles of C produced = 20
If B is the limiting reactant, moles of C produced = 24
The reactant producing the least number of moles of the product is the limiting reactant and
hence A is the limiting reactant.
Thus,
A
+
2B ⟶4C
Initially — 5 moles
12 moles 0 moles
Finally — 0 moles
2 moles 20 moles
PLANCESS CONCEPTS
In a chemical reaction, not all reactants are necessarily consumed. One of the reactants may
be in excess and the other may be limited. The reactant that is completely consumed is called
limiting reactant, whereas unreacted reactants are called excess reactants.
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23
Mole Concept and Stoichiometry
Amounts of substances produced are called yields. The amounts calculated according to
stoichiometry are called theoretical yields whereas the actual amounts are called actual yields.
The actual yields are often expressed in percentage, and they are often called percent yields.
Neeraj Toshniwal
Gold Medalist, INChO
3. Main Drawbacks Of Dalton’s Atomic Theory
(contd.)
Mole Concept and Stoichiometry
24
SOLVED EXAMPLES
Example 1: Find the percentage of water of CRYSTALLISATION in the following
(a) Copper sulphate crystals CuSO4.5H2O (b) Washing soda crystals Na2CO3.10H2O
Sol.
Copper sulphate crystals CuSO4.5H2O
Washing soda crystals Na2CO3.10H2O
Gram molecular weight of compound
Gram-molecular weight of compound
= Cu + S + 4(O) + 5 (H2O)
= 63 + 32 + 4(16) + 5(18) = 249 g
Na2CO3.10H2O
= 2(Na) +C +3(O)+10(H2O)
Weight of water in the compound
= 2 ×23 + 12+3 ×16 +10 × 18 =286 g
Wt. of water in the compound
= 5 ×18 = 90g
% of water of crystallisation in
CuSO4.5H2O
=
90
× 100 =
36.14%
249
=10×18= 180g % of water of crystallisation in the
compound Na2CO3.10H2O
=
180
× 100 =
62.94%
286
Example 2: Calculate the volume of oxygen at S.T.P. required for the complete combustion of 100 l of
Carbon monoxide at the same temperature and pressure.
Sol.
2CO (l) + O2 (g) 2CO2 (g)
2 (22.4l) 22.4l
44.8 l of CO requires 22.4 l of O2 for combustion.
100 l of CO will require O2 =
22.4
× 100 l =
50 l
44.8
Example 3: How many litres of ammonia are present in 3.4 kg of it?
Sol.
At S.T.P., Molar mass has volume equal to 22.4
Molar mass of ammonia (NH3) = 17g
17g of NH3 has volume = 22.4L 3.4 kg or 3400 g will have volume =
22.4
× 3400 =
4480 l
17
(contd.)
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Mole Concept and Stoichiometry
Exercise 1— For school Examinations
Fill in the Blanks
Q.1
During a chemical reaction, the sum of
remains unchanged.
the of the reactants and products
Q.2 In a pure chemical compound, elements are always present in a
by mass.
Q.3 Clusters of atoms that act as an ion are called
proportion
ions.
Q.4 In ionic compounds, the charge on each ion is used to determine the
the compound.
Q.5 The Avogadro constant carbon-12.
of
is defined as the number of atoms in exactly of
Q.6 Mass of 1 mole of a substance is called its
Q.7 The abbreviation used for lengthy names of elements are termed as their
Q.8 A chemical formula is also known as a
Q.9 Those ions which are formed from single atoms are called
Q.10 Ionic compounds are formed by the combination between
Q.11 The valency of an ion is
Q.12 Mole is a link between the
and
to the charge on the ion.
and
Q.13 The SI unit of amount of a substance is
True / False
Q.14 Formula mass of Na2O is 62 amu.
True False
Q.15 Those particles which have more or less electrons than the normal atoms are called ions
True False
Q.16 Formula for sulphur dioxide is SO3.
True False
Q.17 Molar mass of ethyne (C2H2) is 26 g/mol.
True False
(contd.)
Mole Concept and Stoichiometry
26
Exercise 2 – For Competitive examinations
Multiple choice Questions
Q.1 Q.2 Q.3 Q.4 Which of the following has largest number of particles?
(a) 8g of CH4
(b) 4.4g of CO2
(c) 34.2g of C12H22O11
(d) 2g of H2
The number of molecules in 16.0g of oxygen is
(a) 6.02 x 1023
(b) 6.02 x 10-23
(c) 3.01 x 10-23
(d) 3.01 x 1023
The percentage of hydrogen in H2O is (a) 8.88
(b) 11.12
(c) 20.60
(d) 80.0
Find the mass of oxygen contained in 1 kg of potassium nitrate (KNO3)
(a) 475.5 g
(b) 485.5 g
(c) 475.24 g
(d) 485.2 g
Q.5 How many moles of electron weight one kilogram?
(a) 6.023 x 1023(b)
(c)
6.023
× 1054
9.108
(d)
1
× 1031
9.108
or=
y
=
75 5 3 cm
Q.6 25.4g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl3.
Calculate the ratio of moles of ICI and ICl3
Q.7 Q.8 (a) 1 : 1
(b) 2 : 1
(c) 3 : 1
(d) 1 : 2
The mass of sodium in 11.7 g of sodium chloride is
(a) 2.3 g
(b) 4.32 g
(c) 6.9 g
(d) 7.1 g
The formula of a chloride of a metal M is MCl3, the formula of the phosphate of metal M will be
(a) MPO4
(b) M2PO4
(c) M3PO4
(d) M2(PO4)3
(contd.)
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Mole Concept and Stoichiometry
27
SOLUTIONS
EXERCISE 1
Fill in the Blanks
Sol.1 masses
Sol.2 definite
Sol.3 polyatomic
Sol.4 chemical formula
Sol.5 6.022 x 1023, 12g
Sol.6 molar mass
Sol.7 symbol
Sol.8 molecular formula.
Sol.9 simple ions.
Sol.10 metal and non-metals. Sol.11 equal.
Sol.12 mass of atoms & number of atoms.
Sol.13 mole
True/False
Sol.14 True
Sol.15 True
Sol.16 False
Sol.17 True
Sol.18 False
Sol.20 False Sol.21 True
Sol.22 True
Sol.23 False
Sol.24 True
Sol.19 True
Match the Column
Sol.25 A ->(r), B -> (s), C -> (p), D->(q),
Sol.26 A->(q), B-> (p), C -> (r), D -> (s)
Very Short Answer Questions
Sol.27 In one gram atomic weight of any element 6.02 x 1023 atoms of that element are present. This quantity
of the element is known as one mole.
Sol. 28 One gram atomic mass of any element contains the same number of atom of that element as there are
carbon atoms in exactly 12 g of carbon-12. This number is Avogadro’s number i.e., 6.023 x 1023.
Sol.29 The atomic mass of an element when expressed in grams is known as gram atomic mass or simply
as gram atom (g-atom). Thus one g-atom of carbon (C-12) weighs 12.0 g. The number of gramatoms of any element in its certain weight is given by:
Number of gram − atom =
Mass of the element in grams
Atomic mass
(contd.)
Mole Concept and Stoichiometry
28
SOLUTIONS
EXERCISE 2
Multiple Choice Questions
Sol.1 (d)
Sol.2 (d)
Sol.3 =
(b) Percentage of hydrogen
Sol.4 (c)
Sol.5 (d)
Massofhydrogeninwater 2
× 100 = × 100 =11.12%
Molarmassofwater
18
Sol.6 (a)
Sol.7 (b)
Sol.8 (a) MCl3 → M3+ + 3Cl− ; M3+ + PO 43− → MPO 4
Sol.9 (b)
Sol.10 (d)
Sol.11 (c)
Sol.12 (a) Sol.13 (b)
Sol.14 (c)
Sol.15 (c)
Sol.16 (b)
Sol.17 (c)
Sol.18 (b)
Sol.19 (b)
Sol.20 (a)
Sol.21 (a)
Sol.22 (b)
Sol.23 (b)
Sol.24 (a)
Sol.25 (a)
Sol.26 (d)
Sol.27 (d)
Sol.28 (c)
Sol.29 (b)
Sol.30 (d)
Sol.31 (c)
Sol.32 (d)
Sol.33 (d)
Sol.34 (c) Law of multiple proportions.
Sol.35 (a) C
onstant proportions according to which a pure chemical compound always contains
same elements combined together in the same definite proportion of weight.
Sol.36 (b) 1 atom of Cu + 1 atom of sulphur + 9 atoms of oxygen + 10 atoms of hydrogen. Total number
of atoms in compound is 21.
Sol.37 (b) Law of multiple proportion. As the ratio of oxygen which combine with fix weights of 1 g
of nitrogen bears a simple whole number ratio
0.57: 1 : 12 : 1.7031 : 2 : 3
Sol.38 (a) Molecular weight of C60H122
= 12 x 60 + 122 x 1 = 720 + 122 = 842
 6 x 1023 molecule C60H122 has mass = 842 gm.
∴ 1 molecule C60H 122 has mass
842
= =
140.333 × 10−23 gm =
1.4 × 10−21 gm
23
6 × 10
(contd.)
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