# stoichiometry.pdf

September 3, 2017 | Author: Gadde Gopala Krishna | Category: Mole (Unit), Molecular Mass, Molecules, Stoichiometry, Chemical Compounds

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Synopsis... PART - I (Laws of chemical combination, amu, Atomic and molecular masses, average atomic mass, gm. atom & gm.molecule) Laws of Chemical combination : 1. Law of conservation of Mass (Lavoisier, 1744) “Mass can not be created or destroyed. In physical or chemical process, the total mass of the system remain conserved.”  This law can not be applied to the nuclear process where mass and energy are interconversable. (Einstein’s equation : E  m. C 2 )  On the basis of this law, we may conclude that for a reaction, if the reaction is 100% completed then, total mass of reactants before reaction = Total mass of products after reaction.  For incomplete reactions : Total mass of reactants before reation = Total mass of products formed + mass of unreacted reactants left. CAPS - 1 : .3.4g of AgNO3 in 100g water, when mixed with 1.17g of NaCl in 100g water, 2.87g AgCl and 1.70g NaNO3 were obtained. Verify law of conservation of mass. Sol : Total mass of substance before reaction = 3.4g AgNO3 + 100g H2O + 1.17g NaCl + 100g H2O = 204.57g of reactant  Total mass of substance after reaction 2.87g AgCl + 1.70g NaNO3 + 200g H2O = 204.57g of Products  The result proves law of conservation of mass. Law of constant (or definite) proportions (Proust, 1799) “A chemical compound always contains the same element combined together in the same proportions by mass.”  i.e, the composition of a compound always remain fixed and it is independent to the source from which the compound is obtained.  Eg : Compound CO2 can be formed by either of these process. i) by heating CaCO3  CaCO3   CaO  CO2

ii) by heating NaHCO3  2NaHCO3   Na2CO3  H 2O  CO2

iii) by burning in O2  C  O2   CO2

iv) by reaction of CaCO3 with HCl.

CaCO3  2 HCl   CaCl2  H 2O  CO2

 CO2 obtained by all these methods contains C:O ratio 12 : 32 by mass.  This law can not be applied to the compound, obtained by using different isotopes of the elements as the isotopes have different atomic masses. Eg : CO2 using C12 isotope has C : O :: 12 : 32 CO2 using C14 isotope has C : O :: 14 : 32  The elements combining in the same ratio of their masses may give different compounds under different Sr. INTER

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experimental conditions. Eg : Combination of ‘C’, ‘H’ and ‘O’ in the ratio 12 : 3 : 8 may give C2H5OH or CH3OCH3 under different experimental conditions. CAPS - 2 : .A sample of 1.375g cupric oxide when reduced in a steam of hydrogen gave 1.098g CU. On the other hand, a sample of 1.179g pure CU gave 1.476g cupric oxide when CU was in HNO3 and nitrate formed was heated strongly to get cupric oxide. Show that these data prove law of definite proportions. Sol : According to Ist experiment, the composition of cupric oxide is 1.375g cupric oxide = 1.098g CU + Oxygen (0.277g)

 % of CU in the sample =

1.098  100  79.85% 1.375

 In 2nd experiment, the composition of cupric oxide is : 1.476g cupric oxide = 1.179g CU + oxygen (0.297g)

 % of CU in the sample =

1.179 100  79.87% 1.476

 Since, in both the experiments % CU and % O are constant which validate law of constant proportions Law of Multiple proportions (Dalton) “If two elements combine to form more than one compound, then for the fixed mass of one element, the mass of other element combined will be in simple ratio”  Examples of law of multiple proportions : i) Combination of C and O may form CO and CO2 In CO ratio of C : O is 12 : 16 In CO2 ratio of C : O is 12 : 32 Thus ratio ‘O’ in CO and CO2 is 16 : 32 or 1 : 2 i.e, a whole number ratio. ii) N and O form five stable oxides, N2O, NO, N2O3, N2O4 and N2O5. In these oxides, amount of oxygen, which react with 28g N2 are in the ratio 16 : 32 : 48 : 64 : 80. i.e, 1 : 2 : 3 : 4 : 5. iii) In H2O and H2O2, 2g H combines with 16g and 32 g oxygen respectively and the ratio is 16 : 32 or 1 : 2.  However, the discovery of isotopes led to some disperencies in this law also. CAPS-3 : Two compounds each containing only tin and oxygen had the following composition. Mass Mass % of tin % of oxygen Compound A 78.77 21.23 Compound B 88.12 11.88 Verify law of multiple proportions for this data. Sol : In compound A 21.23 parts of oxygen combine with 78.77 parts of tin 1 part of oxygen combines with 78.77 / 21.23 = 3.7 parts of tin  In compound B 11.88 parts of oxygen combine with 88.12 parts of tin 1 part of oxygen combines with 88.13 / 11.88 = 7.4 parts of tin  Thus, the mass of tin in compound A and B which combine with a fixed mass of oxygen are in the ratio of 3.7 : 7.4 or 1 : 2  The data illustrates the law of multiple proportions Law of Reciprocal proportions (or) law of equivalent proportions (Richter 1792 - 94) “ If two elements combine separately with a third element, the mass ratio of the first two elements combined with a fixed mass of the third element will be equal to or in simple ratio to the mass ratio of first two elements in a compound formed by their direct combiantion”. For example : i) Hydrogen combines with Sodium and chlorine to form two compounds. NaH and HCl respectively. Sr. INTER

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In NaH : Sodium 23 parts ; Hydrogen one part In HCl : Chlorine 35.5 parts ; Hydrogen one part In NaCl : 23 parts of sodium ; 35.5 parts of chlorine. These are the same parts which combine with one part of hydrogen in NaH and HCl respectively. Hydrogen combines with sulphur and oxygen to form compounds H2S and H2O respectively. In H2S : Hydrogen 2 parts + Sulphur 32 parts In H2O : Hydrogen 2 parts + Oxygen 16 parts In SO2 : Sulphur 32 parts + oxygen 32 parts In SO2 the ratio of S and O by mass is 32 : 32. Which is double of the ratio of masses of these elements which combine with 2g of hydrogen.

2g H H2S

H2O

S 32g

O 16g

CAPS-4 : The % composition of NH3 , H2O and N2O3 is given below : NH3   82.35% N and 17.65% H H2O   88.90% O and 11.10% H N2O3   63.15% O and 36.85% N On the basis of above data prove law of reciprocal proportions. Sol : for NH3 : 1 part of H reacts with =

82.35  4.67 part N. 17.62

for H2O :

88.90  8.01 part O. 11.10 Thus, the ratio N : O :: 4.67 : 8.01  0.58.  As the two ratio are same, thus law of reciprocal proportions are correct. 1 part H reacts with =

Law of Gaseous volumes (Gay Lussac, 1808)  This law is applied to the reactions containing at least two gaseous components.  This law states that : “At the same temperature and pressure, the volumes of gaseous reactants reacted and the volume of gaseous products formed bear a simple ratio” Examples : i)

N 2 g   3H 2 g    2 NH 3 g  1vol

3vol

2vol

i.e, One volume of N2 reacts with three volumes of H2 to give two volumes of NH3.

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CO g   1 O2 g    CO2 g  2 1vol 1 vol 1vol 2

iii)

H 2  g   Cl2 g    2 HCl  g  1vol

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2vol

CAPS-5 : How much volume of oxygen will be required for complete combustion of 40ml of acetylene (C2H2) and how much volume of CO2 will be formed ? All volumes are measured at NTP ? Sol : 2C2 H 2  5O2   4CO2  2 H 2O 2vol

5vol

4vol

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5 x40mL 4 x40mL 2 2

40mL 100mL 80mL So, for complete combustion of 40mL of acetylene, 100mL of oxygen is required and 80mL of CO2 is formed. Daltons Atomic Theory :  On the basis of the laws of chemical combiantion and the work of Greek Philosophers, John Dalton in 1803 - 1807 proposed his atomic theory.  The basic postulates of Daltons atomic theory are as follows : i) Each element is composed of extremely small particles called atoms. ii) All atoms of a given element are identical i.e, atoms of a particular element are all alike but, differ from atoms of different other elements. iii) Atoms of different elements posses different properties including different masses. iv) Atoms are indestructible i.e, atoms are netiehr created nor destroyed in chemical reactions. v) Atoms of elements take part to form molecules i.e, compounds are formed when atoms of more than one element combine. vi) In a given compound, the relative number and kind of atoms are constant. Advantages :  Daltons theory supports laws of chemical combination and provides us a conceptual picture of matter. Limitations :  It could not explain why do atoms combine to form a molecule.  It could not explain the nature of forces which hold the atoms and molecules in solids, liquids and gaseous state.  It could not explain the GayLussac’s law of combining volume.  It could not explain, that why should atoms of an element differ in their masses. If one examines the Dalton’s atomic theory in the light of recent developments in science, then  The atom is no longer supposed to be indivisible the atom is not a simple particle but a complex one.  Atoms of the element may not necessarily posses the same mass but, possess the same atomic number and show similar chemical properties (Discovery of isotopes)  Atoms of the different elements may posses the same mass but, they always have different atomic numbers and differ in chemical properties (Discovery of isobars)  Atoms of one element can be transformed into atoms of other element (Discovery of artificial transmutation)  There are number of compounds which do not follow the law of constant proportions. Such, compounds are called non - stoichiometric compounds. Atoms and Molecules :  Atom may be defined as the smallest particle of an element which does not exist free in nature but takes Sr. INTER

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part directly in chemical combinations. Atoms of any element can be represented by the symbol of that element. Molecule may be defined as the smallest particle of an element or compound, which exist free in nature but, does not participate directly in chemical combinations. Molecules are represented by the molecular formula, which tells the exact number of atoms or same or different elements present in each molecule. For example, Water molecules are represented as H2O. It tells that each water molecule contains two atoms of hydrogen and one atom of oxygen. Atoms of inert gases exist free in nature. The term molecule should not be used for ionic compounds. The perfect term for them is formula unit. The smallest particles of metals are always atoms not molecules.

ATOMIC AND MOLECULAR MASS    

As atoms are very tiny particles, their absolute masses are difficult to measure. However, it is possible to determine the relative masses of different atoms, if small unit of mass is taken as a standard. For this purpose, in 1961, the international union of chemists selected a new unit for expressing the atomic masses.

 

12 They accepted the stable isotope of carbon C with mass number of 12 as the standard.

ATOMIC MASS UNIT (amu)

“The quantity,

1 12 mass of an atom of carbon -12  C  ” is known as the atomic mass unit and is 12

abbreviated as amu.

The actual mass of one atom of carbon - 12 is 1.9924  10 23 g (or) 1.9924  1026 kg

 1 amu =

1 1.9924 1023  1.66 1024 g  1.66 1027 kg 12

ATOMIC MASS Atomic mass of an element can be defined as “The number, which indicates how many times, the  mass of one atom of the element is heavier in comparision to

1 th part of the mass of one atom of 12

C-12”.

 Atomic mass of an element

Mass of one atom of the element 1  Mass of one atom of C  12 12

 Atomic mass of an element 

Mass of one atom of the element 1amu

The atomic masses of some elements on the basis of C-12 are given below:  Hydrogen : 1.008 amu 2. Oxygen : 16 amu  Chlorine : 35.5 amu 4. Magnesium : 24 amu  Copper : 63.5 amu 6. Iron : 55.847 amu  Sodium : 22.989 amu 8. Silver : 107.868 amu  Nitrogen : 14 amu 10. Sulphur : 32 amu  ACTUAL MASS OF AN ATOM The actual mass of an atom (absolute mass) of an element =  The atomic mass of an element in amu  1.66 1024 g Sr. INTER

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eg:- 1) The actual mass of H-atom = 1.008  1.66  10

24

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 1.6736  10

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2) The actual mass of O-atom = 16  1.66  1024  2.656  1023 g

AVERAGE ATOMIC MASS Most of the elements occur in nature as a mixture of isotopes. [isotopes-the atoms of the same 

 

element having different atomic masses] For example : chlorine is found in nature as a mixture containing two isopopoes Cl-35 and Cl-37 These are found in the ratio of 75% (Cl-35) and 25% (Cl-37)

 The average relative mass of chlorine is calculated as 35 

75 25  37   35.5amu 100 100

So, based on the average mass, atomic mass of chlorine is 35.5 amu

Average isotopic mass 

Here, x,y are percentage aboundance of the two isotopes with atomic masses a and b. [y=100-x] The average atomic masses of various elements are determined by multiplying the atomic mass of each isotope by it’s fractional abundance and adding the values thus obtained.

x y a  b 100 100

 Average atomic mass 

m a  nb mn

here, a, b are atomic masses of isotopes in the ratio m:n. CAP-6 Carbon occurs in nature as a mixture of Carbon - 12 and Carbon - 13. If the percentage abundance of C-12 is 98.9 then find the average atomic mass of carbon ? Solution : % abundance of C  12  98.9

 % abundance of C  13  1.1  Average atomic mass = 12 

98.9 1.1  13  100 100

 12.011amu CAP-7 Boron has two isotopes Boron - 10 and Boron - 11 whose average atomic mass is 10.8 amu. Then, what are percentage abundance of B-10 and B-11. Solution : Let x be the % abundance of B-10  (100-x) will be the % abundance of B  11

 10.8  10 

100  x  x  11 100 100

 x  20%; y  80% MOLECULAR MASS Similar to atomic mass, molecular mass is also expressed as a relative mass with respect to the mass 

of the standard substance an atom of carbon - 12. The molecular mass of a substance, may be defined as “A number which indicates how many times one molecule of a substance is heavier in comparision to

1 th of the mass of one atom of carbon 12. 12

Mass of one molecule of the substance  Molecular mass = 1 th mass of one atom of carbon  12 12 Sr. INTER

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CHEMISTRYv v v v v v v v v v v v v v v v v v v v v v v v STOICHIOMETRY the mass of a molecule is equal to sum of the mass of the atoms present in a molecule.  Example : 1) One molecule of water consists of 2 atoms of hydrogen and one atom of oxygen.

 Molecular mass of water   2 1.088  16  18.016amu

One molecule of H 2 SO4 consists of 2 atoms of hydrogen, one atom of sulphur and four atoms of oxygen.

 Molecular mass of H 2 SO4   2 1.008   32   4 16 

 98.016  98amu

Similarly molecular masses of some more molecules can be calculated as follows. ACTUAL MASS (OR) ABSOLUTE MASS OF A MOLECULE The mass of one molecule of substance is known as it’s actual mass.  For examples: The actual mass of one molecule of oxygen is 32  1.66  1024 g  5.32  1023 g GRAM-ATOMIC MASS (OR) GRAM ATOM When numerical value of atomic mass of an element is expressed in grams, the value becomes gram atomic mass or gram atom. For example : The atomic mass of oxygen is 16, while gram atomic mass (or) gram atom of oxygen is  16g. Similary the gram atomic masses of Hydrogen, Chlorine and Nitrogen are 1.008 g, 35.5g and 14.0g  respectively. One gram atom of every element consists of same number of atoms. This number is called Avogadro  number. AVAGADRO NUMBER “The number of atoms in 12 g of carbon - 12 has been found experimentally to be 6.023  1023 . This

number is known as “Avogadro’s number”  N A  , named in the honour of Amedeo Avogadro (17761856) Note : Mass of one atom of C-12 isotope  1.9924 1023 g

 Number of atoms present in 12g of C-12 

12 23 1.9924  1023  6.023  10

So, one gram atom of every element consists of Avogadro number of atoms.  GRAM - MOLECULAR MASS (OR) GRAM MOLECULE “A quantity of substance whose mass in grams is numerically equal to it’s molecular mass is called gram molecular mass”. In other words, molecular mass of a substance expressed in grams is called gram molecular mass or  gram molecule.

For Example : i)

The molecular mass of chlorine is 71 amu and therefore, it’s gram - molecular mass (or) gram molecule is 71g.

ii)

Molecular mass of O2 is 32 amu  gram molecular mass of O2 = 32g.

Molecular mass of nitric acid ( HNO3 ) = 1  14  (3  16)  63 amu  Gram - molecular mass of nitric acid = 63g Note : Gram - molecular mass should not be confused with the mass of one molecule of substance in gms. For example: Gram - molecular mass of oxygen is 32 gms where as mass of one molecule of oxygen  iii)

is 32  1.66  1029 g Sr. INTER

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CHEMISTRYv v v v v v v v v v v v v v v v v v v v v v v v STOICHIOMETRY Gram molecular mass (or) one gram molecule of any substance contains 6.023  1023 molecules.  Calculating

Given weight of the element gram atomic mass

a.

Number of gram atoms: Number of gram atoms 

b.

Number of atoms present in the given amount of the element

Given weight of the element  N A = no.of gm atoms  gram atomic mass

N

A

Given weight of the substance gram atomic mass

c.

Mass of an element 

e.

Number of gm. molecules 

f.

Mass of a substance = No.of gm. molecules  gm. molecular mass.

Given weight of the substance  NA gram atomic mass

CAPS-8 1. Calculate number of gram atoms in: a. 48 gm of O2 Solution: a. one gm. atom of oxygen - 16 gms ? - 48 gms 

 No.of gm atoms =

b. 8 gm of H2

48 =3 gm. atoms 16

b.

1gm of H2 - 1 gm. atom. 8 gm of H2 - ? 8 gm atoms. 2. Calculate the number of atoms present in a. 6g of Mg b. 8 gm of O2 Solution: a. 24gm of Mg contains - NA of atoms (1 gm atom)  6 gm of Mg contains - ? atoms

 b.

6 1  N A   6  1023  1.5  1023 24 4

16gm of oxygen contains - NA of atoms (1 gm atom)

 4 gm of oxygen contains - ?

4 1  N A   6  1023  3  1023 atoms 8 2 3. Calculate the mass of a. 1.5 gm of atom of Cl2 b. 2 gm atoms of sodium Solution: a. 1 gm atom of Cl2 - 35.5 g  1.5 gm atom of Cl2 = 1.5  35.5  50.45gm b. 1 gm atom of Sodium - 23g

 2 gm atoms of sodium = 23  2  46gm Sr. INTER

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 b.

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90  0.5 gm molecule. 180

71gm of Cl2 - 1 gm. molecule.  142 gm of Cl2 - 2 gm. molecules

5. Calculate number of molecules in: a. 8 gm of O2 b. 88 gm of CO2 Solution: a. 32 gm of O2 contains - NA of molecules

 8 gm of O2 contains -

8  NA of molecules 32

1   6  1023  1.5  1023 molecules 4 b.

44 gm of CO2 contains - NA of molecules (1 gm molecule) 88 gm of CO2 contains -

88  N A of molecules 44

1   6  10 23  3  1023 molecules 2 6. Calculate the mass of a. 2.5 gm molecules of CH4 Solution: a. 1 gm molecule of CH4 - 16 gm  2.5 gm molecules of CH4 - 2.5 16 b. 1 gm molecule of C12H22O11 - 342 gm  1.2 gm molecule - 12  342 = 410.4 gm.

b. 1.2 gm molecules of C12H22O11

PART - 2 (Mole concept) Mole:  Just like for the counting of articles, the unit dozen is commonly used irrespective of their nature, similarly chemists use the unit “mole” for counting of atoms, molecules, ions etc.  The mole was introduced by ostwald in 1896.  This is the Latin word “moles” meaning heap or pile.  A mole is defined as the number of atoms in 12.00g of carbon - 12.

 The number of atoms in 12g of C - 12 has been found experimentally to be 6.023  1023 .  Number of atoms in 12g of C - 12 

12 Mass of an atom of C  12

12 23 1.9924  10 23  6.023 10

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CHEMISTRYv v v v v v v v v v v v v v v v v v v v v v v v STOICHIOMETRY  This number is also known as Avagadro’s number named in honour of Amedeo Avogadro (1776-1856).  Thus, a mole contains 6.023  1023 units. These units can be atoms, molecules, ions, electrons or anything else. Examples: i) 1 mole of H-atoms means 6.023  1023 H-atoms. ii) 1 mole of CO2 molecules means 6.023  1023 CO2 molecules. iii) 1 mole of electrons means, 6.023  1023 electrons. iv) 1 mole of ions means, 6.023  1023 ions.  Definition of mole in terms of mass is: “One mole is the amount of substance that contains as many particles or entities as there are atoms in exactly 12.00g of the C - 12 isotope.”  Gram molar volume (GMV) : Volume occupied by 1 mole of any gas at STP is 22.4 Lt and this volume is called Gram Molar Volume. [STP conditions: T = 273 k, P = 1 atm]  “The mass of one mole atoms of any element is exactly equal to the atomic mass in grams (gram atomic mass or gram atom) of that element.” For Example : atomic mass of Aluminium is 27 amu. But, 1 amu = 1.66 1024 g .

 Mass of one mole Aluminium = 27  1.66  1024  6.023  1023  27g 27g is the atomic mass of Al in gms or it is one gram atomic mass or one gram atom of Aluminium.  “Similarly, the mass of 6.023  1023 molecules of a substance is equal to its molecular mass in grams or gram - molecular mass or gram molecule.” For Example : Molecular mass of CO2 is 44 amu 24 23  Mass of one mole of CO2  44  1.66  10  6.023  10 g  44 g  44 g is the molecular mass of CO2 in grams or one gram molecular mass or one gram

molecule.  Calculation of number of moles (n): a) When mass is given :

Mass of substance  W in g  Number of moles (n) = M  g-atomic or g-molecular mass  M = atomic mass of element or formula mass of ionic solids or molar mass of compound. Eg: nNaCl 

W W W ; nH 2O  ; nCH 4  58.5 18 16

b) When number of entities are given:

number of entities present Number of moles (n) = 6.023×1023 i.e Avagadro constant, N .  A Eg: n CO2 =

number of CO 2 molecules number of Na atoms ; n Na = NA NA

c) For gaseous substances: Number of moles (n) =

Volume of gas at S.T.P in Lt 22.4 Sr. INTER

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Volume of CH 4 at S.T.P in Lt 22.4

d) Calculation of number of entities: Number of entities present = n  N A . Eg: Number of Na - atoms in 0.5 mole Na  0.5  N A

 Usually mass of one mole of a substance is equal to it’s gm. molecular mass Eg: 1 mole of 1 mole of H 2  2g

O2  32 g

SO2  64 g

1 mole of

CO2  44 g

1 mole of

Cl2  71g

1 mole of

CH 4  16 g

1 mole of

H 2O  18 g

1 mole of

CaCO3  100 g

1 mole of

C12 H 22O11  342 g

1 mole of

1 mole of

C6 H12O6  180 g

1 mole of

NaOH  40 g

 2g of H 2 , 32 g of O2 , 64 g of SO2 , 44 g of CO2 , 71 g of Cl2 , 28 g of N 2 , 18 g water vapour at S.T.P occupies a volume of 22.4 Lt. CAPS-9 1. Calculate number of atoms of each kind present in 90 gms of Glucose? Sol. Number of moles of Glucose =

90 180  0.5mol.

1mol of C6 H12O6 contains  6  N A of ‘C’ - atoms 12  N A of ‘H’ - atoms 6  N A of ‘O’ atoms.  Number of ‘C’ - atoms  6  6.023  10 23  3.6 1024 Number of ‘H’ - atoms  12  6.023  10 23  7.2  1024 Number of ‘O’ - atoms  6  6.023  10 23  3.6  10 24 2. Calculate number of electrons present in 4.5 g of H 2O ? Sol. Number of moles 

4.5  0.25mol. 18

18g of H 2O contains ______10  N A of electrons 4.5g of H 2O contains ______ ?

4.5  10  N A of electrons 18 =2.5  6  1023 

 1.5 10 24 electrons. 3. Calculate the charge present on 1 mole of electrons. Sol. Charge on one electron = 1.609  10 19 coloumbs.

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 Charge on one mole of electrons  1.609  10 19  6.023  10 23

 96,500 coloumbs =1 Faraday. 4. Calculate the number of ions present in 1.11 g of CaCl2 Sol : CaCl2  Ca 2  2Cl 

1mol

1mol

111g

NA

1mol

2 NA

 111 g of CaCl2 contains - 3 NA of ions  1.11 g of CaCl2 contains - 0.03  N A of ions

 3  10 2  6  1023  1.8  1022 ions. 5. Calculate number of molecules present in a drop of water of volume 0.01mL? (density of H 2O is 1g/ mL) sol : Mass of 1 drop of H 2O  0.001 1

 0.01g  18 g of H 2O contains - N A of molecules. 2

 10 g of H 2O contains - ?

102  6  1023  3.3  1020 molecules. 18

6. Suppose, chlorophyll contains 2% of Mg by weight. Then, what should be the minimum molecular weight chlorophyll? Sol : 100 g of Mg contains - 2g of Mg So, chlorophyll should contain minimum 1g atom of Mg i.e 24 g of Mg

 24 g of Mg is present in -

24  100 2

 1200g of chlorophyll

 Minimum Mol. Wt = 1200 g

PART-3 (Percentage composition of elements in a compound - Emperical formula and Molecular Formula) Percentage composition of compounds :  The composition of any compound represents the relative amount of all the constituent elements by weight.

 Percentage of an element =

ZA  100 M

where, Z = No. of atoms of that element in each molecule A = atomic weight of the element M = molecular weight of the compound  The percent analysis of a substance is useful to determine the formula of unknown compound. Eg. i) In H2O : M H 2O = 18 Sr. INTER

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% H

2 100  11.12% ; 18

% O

16  100  88.89% 18

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ii) In CaCO3 : M CaCO3  100 Ca % = 40% ; C % = 12% ; O% = 48% CAPS - 10 : If insulin contains 3.4% sulphur, what will be the minimum molecular mass of insulin ? Sol : For minimum mol mass, insulin must contain at least one ‘s’ atom in it’s one molecule.  100g of insulin contains ........ 3.4g of sulphur ........ 32g of sulphur

100  32  941.176 g 3.4 CAPS - 11 : Chlorophyll contains 2.68% of Mg by mass. Calculate the number of Mg atoms in 2g of chlorophyll ? Sol : 100g of chlorophyll contains .......... 2.68g of Mg

  2g of chlorophyll contains .

2.68 mol of Mg 24

2.682 mol of Mg 24100

= 2.23  103 Mole of Mg  No. of Mg atoms =

2.23  103  6.023  1023 = 1.345  1021 atoms of Mg CAPS - 12 : Calculate the percentage composition of : MgSO4 . 7H2O Sol : MgSO4 . 7H2O =  24  96  7 18 = 246. i) % of Mg = ii) % of S =

24  100  9.75% 246

32 100  13.00% 246

iii) % of O =

176  100  71.54% 246

iv) % of H =

14  100  5.69% 246

v) % of H2O =

126  100  51.21% 246 Sr. INTER

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Determination of chemical formula from percentage compsotion : Emperical formula (E.F)

Types of chemical formulae

Molecular formula (M.F)

Structural formula (S.F)

 Molecular formula of any compound represents the exact number of atoms of different elements present in each molecule of the compound.

 Emperical formula of any compound represents the simplest atomic ratio of the different elemetns present in the compound. Compound

Molecular Formula

Empirical formula

Ethane

C2H6

CH3

Ethanol

C2H6O

C2H6O

Acetic acid

C2H4O 2

CH2O

Glucose

C6H12O6

CH2O

Benzene

C6H6

CH

 Two different compounds can have the same emperical formula.  Two different compounds can have the same molecular formula (isomers).  For most of the ionic compounds, the formula represented are their emperical or simplest formula. Determination of E. F from % composition :

Element

Write the symbols of elements present in the molecule

Mole ratio = % Simple mole % composition atomic mass ratio Write % composition here. If sum% given is not 100, then subtract this sum from 100 and use it as oxygen %

Divide % of element by atomic weight

Divide all the mole ratio by smallest mole ratio

Conversion into whole number

E. F

If simple mole ratio are Write the whole fractional, thus no.s obtained as multiply all with subscript with same number to the symbols of convert into elements whole no. or nearly whole no

Example : An oxide of iron contains 69.94% Fe and 30.06% O. Determine its E.F (Fe : 55.85 ; O = 16.0) Solution : Element Fe O

% Moleratio composition

Simplemole Whole E. F ratio no. ratio

69.94

69.94 1.25 55.85

1.25 12 1.25

2

30.06

30.06 1.88 16

1.88 1.52 1.25

3

Fe2O3

CAPS - 13 : 1. Calculate the M.F of a hydrocarbon which contains 85.7% carbon and has molecular mass 84  Sol :% C = 85.7 and % H = 14.3 This data gives E. F compound to be CH2 Sr. INTER

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CHEMISTRYv v v v v v v v v v v v v v v  M.F = E .F x n  n = 6   M. F = C6H12

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2.

Hydrated Barium iodide molecule contains x molecules of water (BaI2 . xH2O). A sample of it, weighing 10.407g when heated strongly gives off water completely and anhydrous residue left weighs 9.520g. Find the value of x. Sol :Mass of water escaped = 10.407 - 9.520 = 0.887 g

BaI 2 .xH 2O (10.407 g ) BaI 2 (9.520 g )

xH 2O(0.887 g )

Moleof H 2O  x (or) 0.887/18  x  x  2 Moleof BaI 2 9.520/ 391 Quantitative estimations : Determination of % analysis of elements in organic compounds or quantitative estimation of elements in organic compounds. I. Estimation of C and H : A known weight of organic compound is burnt in oxygen and the products H2O and CO2 formed are absorbed in conc. H2SO4 and KOH(aq) respectively. The % of C and H are obtained using the formulae. % H 

2 weight of H 2O   100 18 weight of organic compound

w eight of C O 2 12 % C  14  w eight of organic com pound  100 II. Estimation of Nitrogen : i) Duma’s method :

28  volume of N 2 at STP % N  22, 400  weight of organic compound  100 ii)

Kjeldhal’s method :

1 .4 N  V w e ig h t o f o rg a n ic c o m p o u n d

% N 

iii)

where, (N x V), is Meq. of NH3 given out during Kjeldahl’s method measured in terms of Meq. of acid used for NH3 neutralisation. Estimation of Sulphur :

weight of BaSO 4 32 % S  233  weight of organic compound  100 iv)

Estimation of halogens (Carius method) % Cl =

35.5 weight of AgC l   100 143.5 weight of organic compound

% Br 

80 w eight of AgBr   100 188 weight of organic compound

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CHEMISTRYv % I v)

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127 weight of AgI   100 235 weight of organic compound

Estimation of Phosphorus : % P

weight of Mg 2 P2 O7 62   100 222 weight of organic compound

vi)

Estimation of oxygen : There is no direct method for estimation of oxygen in organic compounds. It is usually calculated by the difference. % O = 100  [% of all other elements in organic compounds] CAP-14 0.236g of an organic compound yielded 0.528g CO2 and 0.324g H2O on combustion. 0.295g of the compound gave 56mL N2 at STP. If the molecular weight of the compound is 59, then find the molecular formula of the compound.

12

weight of CO

2 Sol : % of C  44  weight of organic compound  100

12 0.528   100  61.07% 44 0.236

=

% of H 

=

2 weight of H 2O  100 18 weight of organic compound

2 0.324   100  15.25% 18 0.236

% of N 

Volume of N 2 at STP 28   100 22, 400 weight of organic compound

28 56   100  23.73% 22, 400 0.295

Element

Percentage

C H N

61.07 15.25 23.73

Relative no. of atoms 5.09 15.25 1.695

Simplest ratio 3 9 1

 Empirical formula is C3H9N and E. F weight = 59  M. F = E. F = C3H9N CAPS-15 1. An organic compound contains 49.3% carbon, 6.84% hydrogen and its V. D is 73. Then find the M. F of the compound.

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CHEMISTRYv Sol : C 

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% Mol .Mass 49.3 146    6 100 At .mass 100 12

H

% Mol .Mass 6.84 146     10 100 At .mass 100 1

O

% Mol.Mass 43.86 146    4 100 At.mass 100 16

 Molecular formula = C6H10O4 Molecular mass = 12  6  10 1  16  4  146 2.

An organic compound containing C and H has 92.3% carbon and V. D 39. Then find the M. F of the compound.

Sol : Element

%

At Mass

Carbon Hydrogen

92.30 7.70

12 1

Relative no of atoms 7.69 7.70

Simplest ratio 1 1

 E.F  CH

PART - 4 (Chemical Stoichiometry : Calculations based on chemical equations) Introduction: “Chemical stoichiometry describes the quantitative realtionships that exist between substances undergoing chemical changes.”  A chemical equation represents an actual chemical change in terms of symbols / formula of reactants and products.  If the number of atoms in reactant side and product side are not same then reaction is said to be unbalanced. Eg : H 2  O2  H 2O  If the number of atoms in both sides are equal then reaction is said to be balanced.  The numbers written before atoms / molecules in a balanced chemical equation are known as “Stoichiometric coefficients”.  All the chemical equations can be treated as algebraic equations and coefficient 1 is not written.  HCl Eg : 1 2 H 2  1 2 Cl2  

(or) H 2  Cl2   2 HCl (or) 2 H 2  2Cl2   4 HCl  Thus, a balanced chemical equation shows conservation of mass and atoms only. Significance of chemical equations :  A chemical equation provides qualitative and quantitative details of a chemical reaction. A balanced chemical equation gives following informations about ratio of the reactants and products. i) Mole ratio ii) Molecules ratio iii) Mass ratio iv) Volume ratio  Eg : Consider a balanced chemical equation as

aA  bB  cC  dD Sr. INTER

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This equation gives information about i) Mole ratio : a mole A  b mole B  c mole C  d mole D Here symbol  signifies “Stoichiometryically equivalent to”. ii) Molecules ratio : a x NA molecules of A  b x NA molecules of B  c x NA molecules of C  d xNA molecules of D iii) Mass Ratio :

 a  M A  g of

A   b  M B  g of B   c  M C  g of C

  d  M D  g of D MA , MB , MC , MD are molar masses of A, B, C and D respectively. iv) Volume ratio : When all reactants and products are gases. a vol A  b vol B  c vol C  d vol D Consider for example, the reaction represented by a balanced chemical equation

N2 g   3H2 g    2NH3 g  Mole ratio Molecule ratio Mass ratio Volume ratio

1mol

3mol

2mol

1xNA 2xNA 1 molecule 3 molecules 28g 6g 1vol

3vol

2xNA 2 molecules 34g 2vol

 It is thus evident that the coefficients in a balanced chemical equation can be interpreted as the relative no. of moles, molecules or volume involved in the reaction.

 The stoichiometric relation can be used to given conversion factors for relating quantities of reactants and products in a chemical reaction.

 The chemical stoichiometric problems may be classified on the following relationship. a) Weight - Weight relationship - Gravimetric Analysis b) Weight - volume relationship c) Volume - volume relationship i) For gases - gas analysis (or) Eudiometry ii) For solutions - Volumetric analysis (or) titration NOTE : Eudiometry and volumetric analysis will be discussed separately. Methods of solving stoichiometric problems : 1. Mole Method (Based on mole concept) Step - I : Write the complete and balanced reaction concerned. Step - II : The stoichiometric coefficients in the balance reaction represents the relative number of moles of the different reaction components concerned, with the help of mole. Step - III : Use unitary method to get required amount.

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CAP-16. Calculate the weight of CO2 formed by complete combustion of 1.5 gm ethane.

 2CO2  3H 2O Sol : C2 H 6  7 2 O2  1 mol 2 mol = 30 gm 2x44 gm 30gm C H produces 2x44gm CO2 on complete combustion.  2 6

 1.5gm C2H6 will produce 2.

2  44  1.5  4.4 gm CO2 30

Principle of atomic conservation (POAC) method :

 As the reactions are balanced by conserving the atoms of each element, the mole method may be applied to such calculations, without balancing the reaction.

 The solution of above problem may be done by POAC method as : Moles of C - atoms in C2H6 = moles of C-atoms in (or) 2 x mole of C2H6 = 1 x mole of CO2

 2

1.5 w  1  w  4.4 gm 30 44

3.

Based on Equivalent concept (Equivalent method) The number of gm equivalents of each reactants reacted will remain the same and the same number of gm equivalents of each product will form.  The solution of above problem may be done by equivalent concept as : No.of gm equivalents of C2H6 = No. of gm equivalents of CO2

1.5 w   w  4.4 gm 30 44 14 7

  

[Note : Equivalent method based problems are done thoroughly further in equivalent weights concept]. Problem solving based on mole concept : I. Calculations based on mass - mass relationship : In such calculations, masses of reactants are given and mass of the product is required and vice - versa. CAP-17. Calculate the number of grams of MgCl2 that could be obtained from 17.0g of HCl, when HCl is reacted with an excess of magnesium oxide. Sol : Balanced equation, M gO  2 H C l    M gC l 2  H 2 O 1m ol

2 m ol

1m ol

(2  36.5) g  73 g

1m ol

(24  71) g  95 g

73g HCl produce   95g MgCl2 17g HCl produce  ? 

95  17  22.12 g 73

CAP-18. How many grams of oxygen are required to burn completely 570g of octane ? Sol : Balanced equation

2C8 H18  25O2  16CO2  18 H 2O 2mol 2x114g 570g

25mol 25x32g ? Sr. INTER

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CHEMISTRYv

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STOICHIOMETRY

570  25  32  2000 g 2  114

Calculations involving mass - volume relationship :  In such calculations, masses of reactants are given and volume of the products is required and vice - versa.  1 mole of a gas occupies 22.4 lit volume at STP mass of a gas can be related to volume according to the following gas equation. PV = nRT ; PV = w/m RT CAP-19. What volume of NH 3 g  at 27°C and 1atm pressure will be obtained by thermal decomposition of 26.25g NH4Cl ? Sol :

NH 4Cl S    NH 3 g   HCl g  1mol

1mol

53.5 g

1mol

 53.5g NH4Cl will give

1  26.25 mole NH 3 = 0.5 mole 53.5 From, PV = nRT

1 V  0.5  0.0821 300 V = 12.315lit CAP-20. Calculate the volume of H2 liberated at 27°C and 760mm pressure by treating 1.2g of Mg with excess of HCl.

 MgCl2  H 2 Sol : Mg  2 HCl  1mol 24 g

1mol 22.4 L at STP

24g Mg liberates   22.4 L of H2

22.4

1.2g Mg will liberate   24 1.2  1.12L  Volume of H2 under given conditions can be calculated by applying :

PV PV 1 1  2 2 T1 T2 P1 = 760mm T1 = 273K V1 = 1.12 L V2 =

P2 = 760mm T2 = 27 + 273 = 300 K V2 = ?

760  1.12 300  =1.2308 L 273 760

III. Calculations based on volume - volume relationship These calculations are based on two laws : i) Avagadro’s law ii) Gay - Lussac’s law For example :

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N 2 g   3H 2 g    2 NH 3 g  1mol

3mol

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STOICHIOMETRY

2mol

1  22.4 L 3  22.4 L

2  22.4 L

Under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes.

N 2 g   3H 2 g    2 NH 3 g  1vol 3vol

2vol

Under similar conditions ratio of coefficients by mole is equal to ratio of coefficient by volume. CAP-21.What volume of air containing 21% oxygen by volume is required to completely burn 1kg of carbon containing 100% combustible substances ?

 CO2 g  Sol : C s   O2 g   1mol

1mol

12 g

22.4 L

 1000g ?

1000  22.4lit 12

100L of air contains   21L of O2 ?

 

1000  22.4 L of O2 12

1000 22.4   100  8888.5 L of air 12 21 CAP-22. What volume of O2 gas at NTP is necessary for complete combustion of 20lit of propane measured at 27°C and 760mm pressure ?

 3CO 2 +4H 2 O Sol : C 3 H 8 +SO 2   1vol 1L

5 vol 5L

1L of propane requires   5L of O2 20L of propane will require  

5  20  100 L of O2 at 760mm and 27°C.

 This volume will be converted to STP conditions. Given conditions P1 =760mm V1=100lit T1 = 27 + 273 = 300K

STP conditions P2=760mm V2 = ? T2 = 273

PV PV 760  100 273 1 1  2 2  V2    91L T1 T2 300 760 Sr. INTER

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Different kind of problem profile on chemical stoichiometry : I. Simple problems : In such problems, the amount of any one of the reaction component will be required against the given amount of some other reaction component. CAP-23. Calculate the amount of NaOH needed for complete neutralisation of 2.45gm H2SO4.

 Na 2 SO4  2 H 2 O Sol : 2 NaOH  H 2 SO4  2 mol 1mol 2  40 gm 98 gm 98gm of H2SO4 requires   2x40gm

NaOH for complete neutralisation

 2.45 gm H2SO4 requires   2.0gm CAP-24. Calculate the volume of O2 gas liberated at STP by complete decomposition of 4.9 gm of KClO3. Sol : 2 KClO3   2 KCl  3O2 2mol 3mol 2x122.5gm 3x22.4L at STP ?  4.9 gm

4.9  3  22.4  1.344 L 2  122.3

II. Limiting Reagent based problems :  In such problems, the amount of any one of the product will be required against the given amount of more than one reactants.  In solving such problems, first we have to determine the reactant which will react completely, i.e, limiting reagent.  Limiting reagent, limits the amount of product to be formed. So, the amount of product formed will be according to the limiting reagent.  “Limiting reactant or reagent is the reactant i.e. entirely consumed, when a reaction goes to completion”. (or) “The reactant which gives least amount of product on being completely consumed” is called limiting reactant. CAP-25. How many grams of NH3 will form by combination of 3gm H2 and 7gm N2 ? Sol : N 2  3H 2   2 NH 3 1mol 3mol 2mol 28gm 3x2=6gm 2x17=34gm  28gm N2 requires   6gm H2 for complete r x n 7 gm N requires < 3.0gm  2 Hence, N2 is the limiting reagent. Now, 28gm N2 will produce 2x17gm NH3 8.5gm NH3  7 gm N2 will produce CAP-26. What volume of Cl2 gas will liberate at STP, when 2.61gm MnO2 is reacted with 2.92gm HCl ? Sol : MnO2  4 HCl   MnCl2  Cl2  2 H 2O 1mol 4mol 1mol 879gm 4x36.5g 22.4L 87gm MnO2 requires   4  36.5g HCl for complete reaction.  HCl is the L. R Now, 4  36.5g HCl will produce  C 2 at STP  22.4 L Cl Sr. INTER

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 2.92 gm HCl will produce   0.448L Cl2 at STP III. Reactions in succession based problems : In such problems, the amount of any one of the reaction component belonging from a reaction is to be related with the amount of some other reaction component belonging from other reaction with the help of some common components. CAP-27. How many grams of ethylene may be burnt completely by the oxygen gas produced by complete decomposition of 49gm KClO3 ? Method - 1 : Solve each concerned reaction separately. i) 2 KClO3   2 KCl  3O2 2mol 2x122.5g

 49g

3mol 3x32gm

3  32  49  19.2 gm 2  122.5

ii) C2 H 4  3O2   2CO2  2 H 2O 1mol 3mol 28g 3x32g ? 1.92g

19.2  28  5.6 gm C2H4 3  32

Method - 2 : Add or subtract the concerned reactions properly such that the common compound, by which the reactions are related, cancels out. The reaction, thus obtained, may be a hypothetical reaction but it will give the true molar relation. Method - 3 : Relate the moles of component of given amount with the component of required amount with the help of common compound.

2mol KClO3  3mol O2  1mol C2 H 4 2 x 122.5g

 49g

28gm

28  49  5.6 g 2  122.5

Method - 4 : Equivalent method (Discussed in volumetric analysis) CAP-28. One Lt of an aqueous solution contains 31.6gm KMnO4. The solution is to be decolourised by passing SO2 gas through it. Calculate the amount of iron pyrite, FeS2, which should be roasted to provide the necessary amount of SO2 (K = 39 , Mn = 55 , Fe = 56). The concerned reactions are :

2KMnO4  5SO2  2H2O   K2SO4  2MnSO4  2H2 SO4

4 FeS 2  11O2   2 Fe2O3  8SO2 Sol:

2 mole . KMnO4  5 mol SO2  5 mol FeS 2 2

 2x158g KMnO4

=

 31.6g KMnO4  

5  120 gm =300gm 2 5 60  31.6  30gmFeS2 2 158

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IV. Percentage based problems : A) Percentage purity calculations :  Depending upon the mass of the product, the equivalent amount of the reactant present can be determined with the help of a chemcial equation.  Knowing the actual amount of the reactant taken and the amount calculated with the help of a chemical equation, the percentage purity can be determined. CAP-29. When 1.25gm of a sample of chalk is strongly heated, 0.44gm CO2 gas is produce. Determine the % of pure CaCO3 in the chalk sample. Sol : CaCO3   CaO  CO2 1mol 1mol 100g 44g ? 0.44g 

100  0.44  1 gm 44

Hence, % of pure CaCO3 in the sample 

1.0  100  80% 1.25

CAP-30. Calculate the amount of 80% pure NaOH sample required to react completely with 21.3gm Cl2 in hot condition. Sol : 6 NaOH  6 mol 6 x 40 g 6  40  21.3 3  71

3Cl2   5NaCl  NaClO3  3H 2O 3mol 3 x 71 g 21.3g

 24g 100

 The weight of NaOH sample required = 24  80  30 g B.

Percentage yield determination :  In general, when a reaction is carried out in laboratory, we do not obtain actually the theoretical amount of the product.  The amount of the product that is actually obtained is called the actual yield.  Knowing the actual yield and theoretical yield, the percent yield can be calculated by the given formula :

Percent yield 

Actual yield  100 theoretical yield

CAP-31. When 3.9g Al(OH)3 is reacted with excess of HCl, 6.50gm AlCl3 is formed. Determine the percentage yield of product ? Sol : Al(OH)3 + 3HCl   AlCl3 + 3H2O 1mol 1mol 78gm 133.5g 3.9gm

133.5  3.9  6.675 gm AlCl3 78

But, the amount formed is only 6.5 gm

 Percentage yield =

6.5  100  97.38% 6.675 Sr. INTER

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CAP-32. When methyl chloride is treated with sodium in ether, ethane is produced along with some otehr products. If the percentage yield of ethane is 80%, calculate the weight of ethane formed by 2.02 kg methyl chloride. Sol : 2CH3Cl + 2Na   H3C - CH3 + 2NaCl 2mol 1mol 2x50.5g 30g

30  2.02  0.6kg C2 H 6 2  50.5

2.02 kg

But, the yield of ethane is only 80% and hence, the amount of ethane produced is 0.6 

80  0.48 kg 100

C) Percentage composition of mixture calcualtion Analysis of mixtures :  In such problems, one of the components is supposed to be x g and the other will be the difference from the total.  Balanced chemical equations for the reactions of both the components are now written and the total amount of the common product produced by the components of the mixture is calculated.  It is equated with the data given and the unknown factors are, thus, worked out. CAP-33. When 4gm of a mixture of NaHCO3 and NaCl is heated strongly, 0.66gm CO2 gas is evolved. Determine the percentage composition of the original mixture. Sol : 2NaHCO3   Na2CO3 + CO2 + H2O 2mol 1mol 2x84g 44g  44gm CO2 is evolved from 2x84g NaHCO3  0.66gm CO2 will evolve from 2  84  0.66  2.52 g NaHCO 3 44

Hence, % composition of the mixture is : N aH C O 3 

2.52  100  63% 4

NaCl  100  63  37% CAP-34. 2g of a mixture of CaCO3 and MgCO3 requires 2g of H2SO4 for complete reaction. Determine the percentage composition of the original mixture.

xg

Sol :

(2  x) g

CaCO3

MgCO3

}

2g

CaCO3  H 2 SO4   CaSO4  CO2  H 2 O

1mol 100g xg

1mol 98g 98 x 100

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MgCO3  H 2 SO4   MgSO4  CO2  H 2O 1mol 1mol 84g 98g

2  x g

98  2  x g 84

hence, the total weight of H2SO4 required =

98 98 x  (2  x )  2 g 100 84

 x = 1.78 1%  % CaCO3 = x/2  100 = 89%  MgCO3 = 11%

PART - 5 (EUDIOMETRY)

1.

Eudiometry or gas analysis involves the calculations based on gaseous reactions in which the amounts of gases are represented in volumes, measured at the same pressure and temeprature. Some basic assumptions related with calculations are : Gay - Lussac’s law of volume combination holds good. According to this law, the volumes of gaseous reactants reacted and the volumes of gaseous products formed, all measured at the same temperature and presume, bears a simple ratio.

N 2g   3 H 2g     2 N H 3 g 

2.

3.

1vol 3vol 2vol Problem may be solved directly in terms of volume, in place of mole. For non-reacting gaseous mixture, Amagat’s law holds good. According to this law, the total volume of a non-reacting gaseous mixture will be equal to sum of partial volumes of all the component gases. V = V1 + V2 + .............. The volume of solids or liquids is considered to be negligible in comparison to the volume of gas.

2 H 2g   O 2g     2 H 2 O l 

4. 5.

2mole 1mole 2mole 2vol 1vol 0vol Air is considered as a mixture of oxygen and nitrogen gases only. Nitrogen gas is considered as an unreative gas. The problems will be of two profiles 1) Amount determination 2) Molecular formula determination

AMOUNT DETERMINATION This type of problems will be as similar as Stoichiometry. CAP-35. What volume of oxygen is required for complete combustion of 40ml acetylene?

5 O2   2CO2  H 2O 2 5 1vol. vol. 2 5  1vol C2H2 requires vol. of O2 for complete combustion. 2

Sol : C2 H 2 

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5 x40 = 100ml O2 2

CAP-36. 20ml of C2H4 and 80ml of O2 gases are mixed in an eudiometer tube and fired. Determine the final composition of gases. Sol : C2 H 4  3O2   2CO2  2 H 2O 1vol

3vol 2vol 0vol 60ml 40ml 0 Final composition : O2 remained = 80 - 60 = 20ml CO2 formed = 40ml CAP-37. When 600ml of CO2 gas is passes through red hot charcoal, the volume become 800ml. If all the volumes are at the same temperature and pressure, determine the composition of final volume.

 20 ml

Sol : CO 2(g)  C(s)  2CO (g ) 1vol 2vol 1200 ml  600 ml but the final volume is not 1200 ml. It means that all CO2 is not reacted with carbon. Let only x ml CO2 has reacted with carbon.

C O 2g   C s   2C O g  1vol

2vol 2x ml Now, the final volume = vol.of CO2 remained + vol. of CO formed Or, 800 = (600 - x) + 2x  x = 200 Hence, the final composition : CO2 remained = 600 - x = 400 ml And CO formed = 2x = 400 ml CAP-38. When 40ml of pure ozone is heated, the volume increases by 10ml. Determine the percentage, by volume, of ozone, decomposed into oxygen. Sol : Let only x ml of ozone decomposed

 x ml

2O3   3O2 2vol

3vol 3/2x ml Increase in volume = vol. of product  vol. of reactant

 x ml

Or, 10 

3 xx 2

 x = 20 Hence, percentage decomposition of

O3 =

20 ×100=50% 40

CAP-39. 30ml of CH4 gas and 90ml of O2 gas are mixed and fired. The resulting gases are passed through excess of caustic soda solution. Determine the composition of gases before and after passing through caustic soda solution. Sol : CH 4 +2O 2   CO 2 +2H 2 O 1vol 2vol  30ml 60ml

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Composition of gases before passing through solution : O2 unreacted = 90 - 60 = 30ml CO2 formed = 30ml When these gases passed through NaOH solution, CO2 gas will be absorbed in the soluton. Hence, after passing through solution, the composition will be O2 gas, 30ml Note: Ab sorb en t Gases absor bed 1 NaOH sol, KO H so l, et c H aloge ns, CO 2, SO 2, SO 3, etc 2 Alkalin e pyro gallo l so l. O2 3 Tur pe ntin e oil O3 4 Co n c. H 2 SO 4, P 2O 5 etc M o istu re 5 H eated m e tal like Pd , Pt, etc H 2 CAP-40. 40ml of a mixture of C2H2 and CO is mixed with 100 ml of O2 gas and the mixture is exploded. The residual gases occupied 104 ml and when these are passed through KOH solution, the volume become 48 ml. All the volume are at same the temperature and pressure. Determine the composition of original mixture. Sol : Let the mixture contains x ml C2H2 . The volume of CO becomes (40 - x) ml.

5 C2 H 2 + O 2   2CO 2 +H 2 O 2 1vol.

 x ml

5 vol 2

2vol

0vol

5 x ml 2

2x ml

0

2 C O + O2   2CO 2 2vol

 (40 - x)ml

1vol

2vol

1 (40 - x) (40 - x)ml 2

In this problem, unknown is only one but number of informations are more. Hence, the problem may be solved by different methods. Some of them are : Method - I : Residual volume = vol. of O2 unreacted + vol. of CO2 formed 

5 

1

 

 10.4  100   2 x  2  40  x      2 x   40  x   

10.4  100 

5 1 x 2 2

 x = 16 Method - II : When the residual gases are passed through KOH solution, the volume contracted from 104 ml to 48 ml. It is due to absorption CO2 gas formed. Hence, Volume of CO2 formed = (104 - 48) ml 2x + (40 - x) = 56  x = 16 Sr. INTER

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Method - III : After passing through KOH solution, the volume of gas remained is 48 ml and definitely it is unreacted O2 gas. Hence, the vol. of O2 gas reacted = (100 - 48) ml or,

5 1 x   40  x   52 2 2

 x = 16 Composition of original mixture : C2H2 = x ml = 16 ml CO = (40 - x) = 24ml CAP-41. A 20 cm³ mxiture of CO, CH4 and He gases is mixed with excess of O2 and fired. A contraction of 13 cm³ observed. A further contraction of 14 cm³ is noticed when the residual gases are passed through aqueous KOH solution. What was the composition of original mixture? All the volume are at the same temeprature and pressure. Sol : Let the original mxiture contains x cm³ CO and y cm³ CH4.

CO 

1 O2    CO 2 2

1vol 1/2vol

1 vol

x cm³  x cm³ x/2 cm³ Contraction in volume = vol. of reactants  vol of products 

 

= x

x x 3   x  cm 2 2

CH 4 +2O 2   CO 2 +2H 2 O 1vol 2vol  y cm³ 2y cm³

1vol ycm³

0 vol 0

 Contraction in volume =  y  2 y   y  2 y cm

3

From question, total contraction in volume = 13 cm³

x  2 y  13 ........... 1 2 When the residual gases are passed through KOH solution, a contraction of 14 cm³ takes place. It is due to the absorption of CO2 gas formed. Hence,

x  y  14 ...........  2  From (1) and (2) ; x = 10, y = 4 Hence, composition of original mixture is CO = x cm³ =10 cm³ CH4 = y cm³ = 4 cm³ He = 20  (x + y) = 6 cm3 MOLECULAR FORMULA DETERMINATION In such problems, first assume the formula of the substance in terms of unknowns as ammonia, NxHy ; hydrocarbon CxHy etc. Then determine the values of unknowns with the help of given reactions and amounts. CAP-42. 20ml of a gaseous hydrocarbon requires 100ml of O2 for complete combustion and produces 60ml of CO2 gas. Determine the molecular formula of the hydrocarbon. Sol : Let the molecualr formula of the hydrocarbon is CxHy.

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y y  Cx H y   x   O2   x CO2  H 2O 4 2 

y   x   vol 4 

1 vol

x vol

0 vol

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y  20  x    100 4   y=8 Hence, the hydrocarbon is C3H8 CAP-43. When a gaseous, olefinic hydrocarbon is burnt completely in excess of oxygen, a contraction in volume equal to double the volume of hydrocarbon is noticed. Identify the hydrocarbon. Sol : Let the molecular formula of olefinic hydrocarbon is CxH2y and its V ml is burnt completely.

Cx H 2 y 

3x O2   xCO2  xH 2O 2

1vol

3x vol 2

x vol

V ml

3x .Vml 2

x. Vml

0 vol 0

The contraction in volume is

3x    x  V  .V    x.V   V 1   ml 2    2

 

From question, V  1 

x   2V 2

 x=2 Hence, the hydrocarbon is C2H4 (Ethene) CAP-44. At high temperature, gaseous S4N4 decomposes into nitrogen gas and sulphur vapours. It is found that for each volume of S4N4 decomposed, 2.5 volume of total gaseous products are formed. Determine the molecular formula of sulphur vapour. Sol : Let the molecular formula of sulphur vapour is Sx

xS 4 N 4    4S x  2 x N 2 

x vol. 1vol.

4 vol. 2x vol 4/x vol. 2 vol.

4   2  vol. x 

Total Vol of products = 

But, from question, it should be 2.5 vol.

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4 + 2 = 2.5 x

 x=8 Hence, the molecular formula of sulphur vapour is S8. CAP-45. When a certain volume of pure oxygen gas is passed through silent electric discharge, a contraction of 5 ml in volume is observed. A further contraction of 10 ml takes place when the resulting gases are passed through turpentine oil. Determine the volume of ozone formed, using these date. Sol : Let the volume of oxygen gas converted into ozone is V ml.

xO2   2O3 x vol

 V ml

2vol

2 V ml 3

Now, contraction in volume,

V

2V  5........ 1 3

The further contraction in volume is due to absorption of ozone in turpentine oil. Hence, volume of ozone formed.

2V  10........  2  3 From (1) and (2) ; V = 15

PART-6 (Oxidation states and Redox Reactions) OXIDATION - REDUCTION: OLD CONCEPT  Oxidation is the process involving addition of oxygen or removal of hydrogen or both. Emamples:  FeO Fe (Addition of oxygen)

CH 3CH 2OH  CH 3CHO

(removal of hydrogen)

CH 3CH 2OH  CH 3COOH

(both)

 Reduction is just reverse process. It involves addition of hydrogen or removal of oxygen or both. (removal of oxygen) ZnO  Zn C2 H 4

 C2 H 6 (addition of hydrogen)

CH 3COOH  CH 3CH 2OH

(both)

 This concept have the following drawbacks a) It includes only the reactions involing oxygen and hydrogen atoms. For example

Fe  FeO and Fe  FeCI 2 Both the processes are similar because iron is forming a compound in which its valency is two. The first process is oxidation but second process is not, it does not involve oxygen or hydrogen atoms. b) There may be some process involving oxygen and hydrogen atoms but can not be classified as oxidation or reduction. Sr. INTER

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CH 2  CH 2  CH 2  CH 2 OH

OH

The process involves addition of oxygen as well as hydrogen and hence it cannot be classified as oxidation or reduction by this definition.

 The old definition may be modified by replcing oxygen atom with electronegative atom or group and replacing hydrogen atom with electropositive atom or group.

Fe

FeCI 2

CH 2  CH 2  CH 2  CH 2 (Oxidation, addition of -OH groups)

OH

OH

reduction

 Oxidation and reducation are always simultaneous process.

Fe  Cl2  FeCl2 Oxidation

MODERN CONCEPT  In terms of electrons : Oxidation is the process involving loss of electrons and reduction is the process involving gain of electrons. Examples Zn  Zn 2  2e

(Oxidation)

2Cl-

Cl2  2e

(Oxidation)

Cu 2   2e

Cu

(Reduction)

Mn 2   4 H 2O

(Reduction)

MnO4  8H   5e 

(Oxidation) 2ClO3  6 H 2O  10e Cl2  12OH    Interms of oxidation state: Oxidation is the process involving increase in oxidation state of element while reduction is the process involving decrease in oxidation state of element.  Oxidation state or oxidation number: Oxidation state of any atom in any molecule or ion may be defined as arbitrary charge assigned to that atom according to some well defined rules.  Rules for determination of oxidation number: a) If the bonded atoms are identical, distribute the shared electron equally between them and if the bonded atoms are different, count the shared electron pair for more eletronegative atom. b) Determine the net charge developed on the atoms after re-distribution of shared electrons. It will represent the oxidation states of the atoms. Example

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(i) HCl molecule

1p 17 p H Cl 16e Hence, O.S. of H = + 1 Electrons except shared electrons 0e O.S. of Cl  1 2e Shared electrons 2e Electorns redistribution Net charge 1 1 (ii) Cl2 molecule

17 p 17 p Cl Cl Electrons except shared electrons 16e 16e Hence, O.S. of Cl= 0 Shared electrons 2e 1e Electrons redistribution 1e 0 Net charge 0 (iii) H2O2 molecule 1p H

Electrons except shared electrons 0e Shared electrons Electorns redistribution Net charge

8p O

8p O

6e

+1

6e 2e

2e

1p H 2e

2e 1e

2e 1e

-1

-1

0e

Hence, O.S. of H = +1 O.S. of O = –1

+1

(iv) CO2 molecule 8p O

2e

6e Shared electron 4e Electron redistribution 4e Net charge

-2

8p O

6p C

6e Hence O.S.of C = + 4 O.S.of O = -2

4e 4e +4

-2

The re-distribution of electron may be made simply by assuming the covalent bonds, ionic and assigning the charge on atoms on the basis of electronegativities and bond order. (v)H2SO4 molecule -2

O +1 H

-1 -1

O

+2 +1 S +1 +2

O

-1

-1

O

+1 H

O.S.of H = + 1 S = +6 O = -2

-2

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Mathematical method for O.S. determination: Rule I: The O.S. of any atom in its elemental state is zero. Rule II: The maximum O.S. of any atom will be equal to (+group number) and minimum O.S. will be equal to (group number-8), where group numbers are in roman numerals. Example: S is a member of group VIA and hence its max. O.S is +6 and min, is -2

Cu(IB) : 1,  2

Exception:

Au(IB) : 1,  3 Xe(0) : 2,  4,  6,  8, etc Rule III: The sum of O.S. of all the atoms in a molecule is zero and for ions, it is equal to the ionic charge. Rule IV: The O.S. of some elements are fixed in all of their compounds. +1: alkali metals, Ag +2: alkaline earth metals, Zn +3: Al -1: F Rule V: O.S. of hydrogen is +1 in all of its compounds except in metal hydrides, where it is -1. Rule VI: O.S. of oxygen is -2 in all of its componds except in a) Peroxides, where it is -1 b) Superoxides, where it is -1/2 c) Oxide of fluorine, where it is in positive states. (+1, +2) 2

1

O F2 , O 2 F2 etc Rule VII: The charges on different ions commonly used, should be known.

CO32 

HCO3

 Hydrogen carbonate ion or bicarbonate ion

SiO32  Silicate ion

PO34

 Phosphate ion

HPO 24   Hydrogen phosphate ion

H 2 PO 4  Dihydrogen phosphate ion

HPO32   Phosphite ion

NO 3

 Nitrate ion

NO 2

 Nitrite ion

SO 24 

 Sulphate ion

SO32 

 Sulphite ion

S2 

 Sulphide ion

S22

 Pyrite ion

S2 O 72 

 Disulphate ion

S2 O32   Thiosulphate ion

S2 O82 

 Perdisulphate ion

ClO 

 Hypochlorite ion

ClO3

 Chlorate ion

ClO 2

 Chlorite ion

ClO 4

 Perchlorate ion

 Carbonate ion

Rule VIII: In the complex compounds, the overall charge on ligand should be considered in place of considering individual atoms. Neutral ligands: H 2 O, NH 3 , CO, NO, pyridine (Py), ethylenediamine (en), triphenylphosphine ( Ph 3 P ) Negative ligands: X  , OH  , NH 2 , NO3 , NO2 , C 2O 24 (ox), O 2 , O 22 ,SO 24 ,S2O32 , CNO  , etc Positive ligands:

NO  , NO 2 , etc .

Sr. INTER

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CAP-46. Determine the oxidation state of the underlined atoms in the following compounds or ions. (a)

NH3

(b) C 2 H 4

x  3( 1)  0  x  3 (c)

2x  4( 1)  0  x  2

CH 3

(d) H 2 SO 4

x  3( 1)  1

2(1)  x  4(2)  0

 x  6 (f) NaH 2 PO 2

 x  2 (e) H 3 PO4

3( 1)  x  4( 2)  0

(1)  2( 1)  x  2( 2)  0

 x  5 (g) K 2 Cr 2 O 7

 x  1 (h) Mg2 P2O7

2( 1)  2x  7( 2)  0

2(2)  2x  7(2)  0

 x  6

(i)

 x  5 (j) H 2 S2 O8

H 2 SO5 2( 1)  x  5( 2)  0

2( 1)  2x  8( 2)  0

x  8 (not possible)

 x  7 (not possible)

The O.S of S may be upto +6 only. Let us see the structural formula of this compound: O

O

O

||

||

||

H  O  S O OH 

H  O  S O  O  S O  H 

Peroxide O linkage The molecule is containing 3 oxygen atoms in -2 state and 2 in -1 state.

Peroxide O linkage O The molecule is containing 6 oxygen atom in -2 state and 2 in -1 state.

Hence, 2( 1)  x  3( 2)  2( 1)  0

Hence, 2( 1)  2x  6( 2)  2( 1)  0

x  6

x  6

Pb(BrO3 ) 2

(l) As2 S3 As and S, both exhibits variable

||

(k)

The O.S. of S may be upto +6 only. Let us see the structural formula of this compound:

Pb and Br, both exhibits variable oxidation states. As the anion is BrO3 ion, the O.S of Pb may be determined as

||

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oxidation states. As the binary

x  2( 1)  0 x  2

compounds of sulphur are called sulphides O.S. of S = -2 now for arsenic

and the O.S. of Br may be determined as

2x  3( 2)  0

y  3( 2)  1

 x  3

 y  5

(m) K 4 [Fe(CN)6 ]

(n) Ni(CO) 4

CN  is a negative ligand 4( 1)  x  6( 1)  0

CO is neutral ligand

x  2

x  0

 x  4 0  0 Sr. INTER

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Example 2: Classify the following process as oxidation or reduction. Process Answer a) Zn   Zn 2

Zn   Zn 2 O.S = 0 O.S. = +2 Increase in O.S., hence Oxidation 0

b) Cl2   ClO3

5

C l 2   C lO 3 Increase in O.S., hence Oxidation 6

c) Cr2 O 72   Cr 3

Cr 2 O 72   Cr 3 Decrease in O.S., hence Reduction 5

d) NO3   NH 4

3

N O3   N H 4

Decrease in O.S, hence Reduction TYPES OF REACTIONS Redox reactions: These are the reactions involving oxidation and reduction. Types: I. Displacement reactions: Eg. Zn  2HCl  ZnCl 2  H 2 II. Inter molecular redox reactions: One molecule of reactant is oxidised where as molecule of other reactant is reduced. Eg. KMnO4  H 2SO 4  H 2S  K 2SO 4  MnSO4  S . ( KMnO 4 undergoes reduction and H 2S undergoes oxidation) III. Disproportionation reaction: The redox reactions in which the atoms of same element belonging from the same molecule or ion are oxidised as well as reuced are called diporportionation reaction. Such reactions are also called auto-redox or self-redox reaction. Examples: 0

1

5

(i) 3 C l 2  6NaOH  5Na Cl  Na Cl O 3  3H 2 O Cl-atoms are oxidised and reduced, both 4

7

(ii) 2 M nO 2  Mn O 4  Mn 2  Mn-atoms are oxidised and reduced, both 3

3

0

(iii) N H 4 N O 2  N 2  2H 2 O N-atoms are oxidised and reduced, both. But it is not disportionation becuase both N-atoms are belonging from different ions. IV. Comproportionation reaction: These are reverse of disproportionation reactions. Examples: 4

0

2H 2S2  SO 2  3S 2H 2 O Intramolecular redox: One atom of a molecule is oxidised and other atom of same molecule is reduced. Eg: 2Mn 2 O 7  4MnO 2  3O 2 Non Redox reactions: These are the reactions in which neither oxidation nor reduction take place. i) All neutralization (Acid -Base) reactions are non-redox. Examples

NaOH  HCl  NaCl  H 2 O 2KOH  H 2SO 4  K 2SO 4  2H 2 O ii) Almost all precipitation reactions are non-redox. Examples

AgNO3  NaCl  AgCl   NaNO3 Na 2SO 4  BaCl 2  BaSO 4  2NaCl Sr. INTER

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OXIDISING AGENT: Elements or compounds which oxidises the other compound and themselves undergo reduction in the reactions are called oxidising agents. Common oxidising agents are KMnO 4 , K 2 Cr2 O 7 , O 3 , F2 , H 2SO 4 etc. 7

1

2

0

Eg: 2K Mn O 4  3H 2SO 4  10H C l  K 2SO 4  2 Mn SO 4  8H 2 O  5C l 2 O.A. 0

6

2

4

 M g  2H 2 SO 4   Mg SO 4  S O 2  2H 2 O

Common oxidising agents and their reduced forms in the reactions:

MnO 4  Mn 2  (in acid medium)

HNO3  NO 2 or NO

MnO 4  MnO 2 (in neutral or weak basic medium)

O3  O 2

MnO 4  MnO 42  (in strong basic medium)

F2  F

Cr2 O 72   Cr 3 (Acidic medium)

H 2SO 4  SO 2

Reducing Agent: Elements or compounds which reduces the other compound and themsevles undergo oxidation in the reactions are called reducing agents. Common reducing agents are metals,

HI, H 2S, LiAIH 4 , Na 2S2 O3 etc. 1.

2

0

2.5

1

2Na 2 S 2 O3  I 2  Na 2 S 4 O 6  2N aI R.A.

2.

2

4

0

2H 2 S  S O 2  2S H 2 O

R.A. Common reducing agents and their oxidised forms in the reactions: I2 S2 O32   S4 O 62

Cl2 (or )Br2 S2 O32   SO 42

(with I2)

H 2S

S

I

 I2

(with Cl2 or Br2)

Substances in which, the central atom is neither in minimum nor in maximum O.S. can behave like both O.A. and R.A., both. Some examples are

as O.A

2

C

2

H 2 O 2 acts as O.A. 1

H 2 O2

CO

as R.A. H 2O

4

CO 2

H 2O2 acts as R.A.

as O.A

0

O2

2

S

4

SO 2

as R.A. Sr. INTER

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METHODS OF BALANCING CHEMICAL REACTIONS: A balanced chemical reaction should follow the conservation of (i) atoms of each elements (ii) mass (iii) charge Oxidation Number Method: This method is based on the fact that the total change in O.S in oxidation and reduction should be equal for a balanced reaction. A redox reaction may be balanced by this method using the following steps: Step I: Select the species undergoing oxidation and reduction and write them separately. Step II: Balance the atoms of responsible elements (elements responsible for change in O.S.) by simple counting. Step III: Determine the changes in O.S of both the process, due to the total number of atoms of responsible elements. Step IV: Make the changes in O.S. of both process, equal by multiplying with suitable numbers. Add both the processes after multipliation. Step V: If some reaction components are left, write them in proper side and balance them by simple counting. Step VI: If the reaction is not balancing at step IV or V, add some molecule or ion in the proper side. The species added should be according to the reaction and it should not create a new change in O.S. CAP-47. Balance the following reaction by O.N. method. a) KMnO 4  H 2SO 4  HCl  K 2SO 4  MnSO 4  H 2 O  Cl 2 Sol: Step I: Oxidation: Reduction: Step II: Oxidation: Reduction:

1

0

H Cl  Cl2 7

2

K Mn O 4  Mn SO 4

2HCl  Cl 2 KMnO 4  MnSO 4 Change in O.S.

Step III: Oxidation: Reduction: Step IV: Oxidation:

1

0

2

2H Cl  Cl2 7

2

5

K Mn O 4  Mn SO 4

[2HCl  Cl2 ]  5

[KMnO 4  MnSO 4 ]  2 ______________________________________ Reduction:

2KMnO4  10HCl  2MnSO 4  5Cl 2 Step V: 2KMnO 4  10HCl  3H 2SO4  2MnSO4  5Cl2  K 2SO 4  8H 2 O is the balanced equation b) Cl2  NaOH  NaCl  NaClO3  H 2 O Sol: Step I: Oxidation: Reduction: StepII: Oxidation: Reduction:

0

5

Cl2  Na Cl O3 0

1

Cl 2  Na Cl

Cl2  2NaClO3 Cl2  2NaCl Sr. INTER

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Change in O.S. StepIII: Oxidation: Reduction: StepIV: Oxidation:

0

5

10

2 Cl2  2Na Cl O3 0

1

2

Cl 2  2Na Cl

[Cl2  2NaClO3 ]  1

[Cl2  2NaCl]  5 ________________________________________ Reduction:

6Cl2  2NaClO3  10NaCl or, 3Cl 2  NaClO3  5NaCl StepV: 3Cl 2  6NaOH  NaClO3  5NaCl  3H 2 O is the balanced reaction c)

Zn  HNO3  Zn(NO3 ) 2  NO  H 2 O

Sol: Step I: Oxidation : Reduction: Step II: Oxidation: Reduction:

0

2

Zn  Zn(NO3 ) 2 5

2

HN O3  N O 0

2

Zn  Zn(NO3 ) 2 5

2

HN O3  N O Change O.S.

Step III: Oxidation: Reduction: Step IV: Oxidation: Reduction:

0

2

Zn  Zn(NO3 ) 2 5

2

H N O3  N O

2 3

[Zn  Zn(NO3 )2 ] 3 [HNO3  NO]  2 ________________________________

3Zn  2HNO3  3Zn(NO3 )2  2NO Step V:

3Zn  2HNO3  3Zn(NO3 ) 2  2NO  H 2 O

The reaction cannot be balanced without changing the stoichiometric coefficient of HNO3 . Step VI: Add six molecules of HNO3 in the left side to balance number of nitrogen atoms.

3Zn  2HNO3  6HNO3  3Zn(NO3 ) 2  2NO  H 2 O or, 3Zn  8HNO3  3Zn(NO3 ) 2  2NO  4H 2 O is the balanced reaction. Ion Electron Method: This method is based on the fact that the total loss of electrons in oxidation and total gain of electrons in reduction, should be equal for a balanced reaction. A redox reaction may be balanced by this method using the following steps: Step I: If the reaction is given in molecular form, convert it in the ionic form. For it, write acids, strong bases and salts in ionic form and then cancel out the spectator ions. Sr. INTER

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Step II: Select the species undergoing oxidation and reduction and write them separately. Step III: Balance the atoms of responsible element by simple counting. Step IV: Balance the atoms of other elements by adding some molecule or ion in the proper side. The species added should be according to the reaction and it should not creat a new change in O.S. In most of the reactions, the other elements remained will be hydrogen or oxygen. They are balanced according to medium of the reaction. In acidic medium: Add one water molecule in the opposite side for each excess of oxygen atom. Add one H  ion in the opposite side for each excess of hydrogen atom. In basic medium: * Add one water molecule on the same side and two OH  ions on opposite side for each excess of oxygen atom. * Add one OH  ion on the same side and one water molecule on the opposite side for each excess of hydrogen atom. * The atoms may also be balanced by balancing them first in acidic medium and then replacing the H  ions suitably by OH  ions. Step V: Balance the charges in both process by adding proper number of electrons on the proper side. Step VI: Make the total number of electrons lost and gained equal by multiplying with suitable numbers. Add both the processes. It should be balanced reaction in ionic form. Step VII: If the original reaction was in molecular form, convert it in molecular form. CAP-48. Balance of the following reactions by ion electron method. a) KMnO 4  H 2SO 4  HCl  K 2SO 4  MnSO 4  Cl2  H 2 O Sol: Step I: Ionic form of the given reaction is

K   MnO4  2H   SO 24  H   Cl  2K   SO 24  Mn 2  SO 24  Cl2  H 2 O Step II: Oxidation: Reduction: Step III: Oxidation: Reduction: Step IV: Oxidation:

0

Cl  C l 2 7

Mn O 4  Mn 2 2Cl   Cl2 MnO 4  Mn 2  2Cl   Cl2 MnO 4  8H   Mn 2   4H 2 O

Step V: Oxidation: Reduction: Step VI: Oxidation: Reduction:

2Cl  Cl 2  2e MnO 4  8H   5e  Mn 2  4H 2 O [2Cl  Cl 2  2e]  5 [MnO 4  8H   5e  Mn 2  4H 2 O]  2

______________________________________________________

2MnO4  16H   10Cl  2Mn 2  8H 2 O  5Cl2 It is balanced reaction in ionic form.

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Step VII: 2KMnO4  10HCl  3H 2SO 4  2MnSO 4  8H 2 O  5Cl2  K 2SO 4 is the balanced reaction. b)

Cl2  OH   Cl  ClO3  H 2O

Step I: Oxidation: Reduction: Step II: Oxidation: Reduction: Step III: Oxidation: Reduction: Step IV: Oxidation: Reduction: Step V: Oxidation: Reduction:

Cl 2  ClO3 Cl 2  Cl  Cl 2  2ClO3 Cl 2  2Cl  Cl2  12OH   2ClO3  6H 2O Cl 2  2Cl  Cl2  12OH  2ClO3  6H 2O  10e Cl2  2e  2Cl Cl2  12OH  2ClO3  6H 2O  10e [Cl2  2e  2Cl ]  5

________________________________________________

6Cl2  12OH  2ClO3  6H2O  10Cl or,

3Cl2  6OH   ClO3  5Cl  3H 2 O is the balanced reaction

Step III onwards may be replaced as Step III: Oxidation: Reduction: Step IV: Oxidation: Reduction: Step V: Oxidation: Reduction:

Cl 2  6H 2 O  2ClO3  12H 

Cl 2  2Cl  Cl2  6H 2O  2ClO3  12H   10e Cl2  2e  2Cl [Cl2  6H 2 O  2ClO3  12H   10e]  1 [Cl 2  2e  2Cl  ]  5

______________________________________________

6Cl2  6H 2 O  2ClO3  10Cl  12H  or,

3Cl2  3H 2 O  ClO3  5Cl   6H 

To remove H  ion, add equal number of OH  ions on both sides.

3Cl2  3H 2O  6OH   ClO3  5Cl  6H   6OH or,

3Cl2  3H 2O  6OH   ClO3  5Cl  6H 2 O

or,

3Cl2  6OH   ClO3  5Cl  3H 2 O is the balanced reaction.

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PART - 7 (EQUIVALENT WEIGHTS)  Molecular weight or molar mass of a substance is constant in any type of reaction. But, equivalent weight of a substance depends upon the nature of chemical reaction in which substance is taking part.  i.e Molar mass refers to amount of substance, while equivalent mass refers to mass that reacts with definite mass of another substance.  Concept of equivalent mass gives the amounts of substances that react with each other giving similar equivalents of products.  “Equivalent mass of a substance is the mass of it which react with 1.008g hydrogen, 8.0g oxygen or 35.5g chlorine”. Calculation of equivalent masses for different systems: I. Equivalent mass of an element: (or) E Element 

Atomic mass of element valence of element

It is the number of gm. of the element which combines with or displaces 1gm hydrogen or 8gm of oxygen or 35.5 gm of chlorine. Example:

E Ca 

M Ca 40   20 (In bivalent compounds) 2 2

E Na 

M Na 23   23 1 1

E Cl 

M Cl 35.5   35.5 1 1

E Al 

M Al 27   9 (In trivalent compounds). 3 3

CAP-49. Determin the equivalent weight of underlined elements in the following compounds. a)

PH 3

: EP 

31  10.33 3

b)

AlCl3

: E Al 

27 9 3

c)

FeO

: E Fe 

56  28 2

d)

32  n  factor  1 : E S  6  5.33

CAP-50. An oxide of a metal contains 40% oxygen by weight. What is the equivalent weight of the metal? Sol: Let we have 100g of metal oxide.  It will contain 40g oxygen and 60g metal  40gms oxygen combines with 60g metal.

 8gms oxygen will combine with

60  8  12 g of metal 40

 Eq.wt of metal = 12 Sr. INTER

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CAP-51. Calculate E03 in the following reaction: 2O3  3O2 Sol: 96g oxygen is obtained from 96g ozone.  8g oxygen obtained from 8g ozone.

 E03  8 i.e the amount which reacts with or releases 8g oxygen

 II

48 M  .M 8 6 6

Equivalent weight of an ion: It is the number of gm of the ion which can be discharged by loss or gain of one mole of electron. Hence.

E ion 

Formula Mass of ion Magnitude of charge on ion

Eg: i) E Na   ii) E Al3 

M Na  1 M Al3 3

27 9 3

2

M Cl

v) E SO 2  4

vi) E PO 3  4

vii) E NO  III

23  23. 1

M Na 2

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M SO 2 4

2 M PO 3 4

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M NO 1

40  20 2

35.5  35.5 1 

96  48 2

95  31.67 3

30  30 1

Equivalent weight of a salt:

 It is the number of gms of the salt formed by transfer of 1 mole electron. Hence. E Salt 

 

Molecular weight of the salt no.of e  transferred in salt formation

Molecular weight no.of metal atoms  Its valency

Formula mass of salt Total  ve(or)  ve chargeon cationic(or)anionic part

 Equivalent weight of any salt may also be given as the sum of equivalent weights of cation and anion present in it. i.e. Equivalent mass of salt = Equivalent mass of cationic part + Equivalent mass of anionic part. Sr. INTER

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* This relation should not be used for hydrated salts.  Examples:

58.5  58.5 (or) E  E Na   E Cl  23  35.5  58.5 1 1

i)

E NaCl 

ii)

E MgCl2 

95  47.5 (or) E  E Mg2  E Cl  12  35.5  47.5 2 1

iii)

E K 2SO4 

174  87 (or) E  E K   ESO2  39  48  87 4 2 1

iv)

E AlCl3 

v)

E Na 2CO3 

vi)

E CaCO3 

vii)

E Ca 3 ( PO 4 )2 

133.5  44.5 (or) E  E Al3  E Cl  9  35.5  44.5 3 1 106  53 (or) E  E Na   E CO2  23  30  53 3 2

100  50 (or) E  E Ca 2  E CO2  20  30 3 2

viii) E Al2 (SO4 )3 

310  51.67 (or) E  E Ca 2  E PO3  20  31.67  51.67 4 3 2

342  57 (or) E  E Al3  ESO2  9  48  57 4 3 2

 The second method should not be applied for hydrated salts. Example: i)

E BaCl2 .2H 2 O 

ii)

E Alum 

M Salt 137  71  36   122 2 2

M 958   119.375 8 8

[eg: K 2SO 4 .Al2 (SO 4 )3 .24H 2 O ]  Some times, when reaction is given, the equivalent mass of a substance is calculated by dividing its formula mass by number of moles of cationic or anionic charge replaced per mole of salt. For example: In the reaction, i)

Na 2SO 4  HCl  NaHSO4  NaCl E Na 2SO4 

ii)

M Na 2SO4 1

In the reaction,

2Na[Ag(CN) 2 ]  Zn  Na 2 [Zn(CN)4 ]  2Ag E N a [ A g ( C N )2 ] 

M [ N a Ag ( C N ) 2 ] 1

[ 1 mole Ag  is replaced per mole of Na[Ag(CN) 2 ] .

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IV Equivalent mass of an acid:  It is the number of gms of the acid which will furnish one mole of H  ion in water. Hence, equivalent weight of acid

Molecular weight basicity

 Where basicity = no.of replaceable H  ions present per molecule of an acid.(or) = moles of monovalent base required for complete neutralization of one mole acid. of acids may also be given as (1+ equivalent weight of anion).

Equivalent weight

36.5  36.5  1  ECl  1  35.5 1

i)

E HCl 

ii)

E H 2SO4 

98  49  1  E SO2  1  48 (when both the ‘H’s are replaced) 4 2

iii)

E H2SO4 

98  98  1  E HSO2  1  97 (where one ‘H’ is replaced) 4 1

iv)

E CH3COOH 

v)

E HNO3 

60  60  1  E CH COO  1  59 3 1

63  63  1  E NO  1  62 3 1

Note:  In oxy acids only those hydrogens are acidic which are attached to oxygen directly.

 Oxyacids are having at least one ‘= 0’ group and one OH group with the central atom Eg: H 3 PO 4 .  If reaction is not given, equivalent mass of an acid is defined for complete neutralization. vi) E H3PO4 

M 98 95   32.66  1  E PO3  1  4 3 3 3

 1  31.66 O   ||  HO  P  OH :Maximum 3 replaceable H 's in H PO  3 4  | OH   vii) E H3PO3 

M 82   41  1  E HPO2  1  40 3 2 2

O ||

[ In H 3 PO 3 : HO  P|  OH : Maximum two replaceable H’s] s] H

viii) E H3PO2 

M  66  1  E H PO  1  65 2 2 1 O ||

[ In H 3 PO 2 : H  P|  OH : only one replaceable hydrogen] H Sr. INTER IIT-JEE MATERIAL

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CHEMISTRYv ix)

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E H3BO3 

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( in water)

[Actually H3 BO3 is monobasic acid and its aqueous solution contains H+ obtained from dissociation of

H 2 O , as given below: B(OH)3  H 2 O  B(OH) 4  H  ] V

Equivalent mass of a base:

 It is the number of gms of base which will furnish one mole of OH  ions in water. Hence, Equivalent weight of a base =

Molecular weight . acidity

 Where acidity = No.of moles of OH  ions furnished by one mole of base in water. (or) = no.of moles of H  ions needed for neturalization of one mole of the base.  Equivalent weight of base (metal hydroxides) may also be given as (equivalent weight of cation +17)  Equivalent weight of a base (metal oxides) may also be given as (equivalent weight of cation +8). Ex: Determine the equivalent weight of following bases: a) b) c) d)

40  40  E Na   17  23  17 1 58 E Mg(OH)2   29  E Mg 2  17  12  17 2 78 E Al(OH)3   26  E Al3  17  9  17 3 62 E Na 2O   31  E Na   8  23  8 2 [ Na 2 O  H 2 O  2NaOH  2Na   2OH  (or) Na 2 O  2HCl  2NaCl  H 2 O . Hence, acidity of E NaOH 

Na 2 O is 2] e)

E Al2O3 

102  17  E Al3  8  9  8 6

[ Al 2 O 3  3H 2 O  2Al(OH)3  2Al3  6OH 

(or) Al 2 O3  6HCl  2AlCl3  3H 2 O

Hence, acidity of Al 2 O3 is 6]. f)

E NH3 

17  17 1

[acidity of NH3 is 1:

NH 3  H 2 O  NH 4 OH  NH 4  OH 

NH 3  HCl  NH 4 Cl ]. VI Equivalent mass of an acidic salt:  An acid salt is that which contains replaceable hydrogen atoms, eg.

NaHSO4 , NaHCO3 , NaH 2 PO4 , Na 2 HPO4 etc. Molecular weight  E acid salt  replaceable ' H 'left in the salt Sr. INTER

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Examples: i)

E NaHSO4 

ii)

E NaHCO3 

iii)

E NaH 2PO4 

M NaHSO4 1 M NaHCO3 1 M NaH 2PO4 2

VII Equivalent weight of basic salt:

E

Molecular weight of basic salt Re placeable OH gp in basic salt E Al(OH)2 Cl 

Eg:

M 2

VIII Equivalent weight of an oxidising agent:  [An oxidising agent undegoes reduction, i.e gains electrons (or) decreases its oxidation state]  So, equivalent wt. of an oxidsing agent can be defined as: “It is the number of gms of that substance involved in the gain of one mole of electrons” (or) the amount of substance that gives 8gms of nacesent oxygen” Equivalent wt. of O.A = (or) 

Molecular weight no.of moles of e s gain by1mole of O.A

Molecular weight change in O.St per mole of O.A

 NOTE: The number of e  s lost or gained or change in O.St should be according to the number of atoms of responsible element present in the respective molecule or ion of the O.A. CAP-52. Determine the equivalent weight of the following oxidising agents: I

KMnO 4 (reacting in acidic medium) MnO 4 converts to Mn 2 in acidic medium.

i)

O.St Method: 7

H Mn O 4   Mn 2

 Change in O.St  2  7  5  E MnO4  ii)

M 158   31.6 5 5

e  method:

MnO 4  8H   5e   Mn 2  4H 2 O  1 mole of O.A gains 5 moles of e  s  n.factor  5  E MnO4 

M 158   31.6 5 5

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CHEMISTRYv iii)

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Nascent oxygens method:

2KMnO4  3H2SO4  K2SO4  2MnSO4  3H2O  5(O) 2 mol

5x16g

2 mole 10 i.e. II

8g

1 1 mol   158 5 5

KMnO 4 (reacting in neutral medium or weak basic medium) MnO4 converts to MnO2 in neutral medium.

i)

O.St Method: 7

4

Mn O4  Mn O2  E 

ii)

M 158   52.67 3 3

Change in O.St = 4  7  3 . e  mehtod:

MnO4  2 H 2O  3e   MnO2  4OH 

 1 mole of O.A gains 3 moles of e  S .  EMnO  4

iii)

158  52.67 3

Nascent oxygens method:

4 KMnO4  2 H 2O  4 MnO2  4 KOH  6(O ) 4 158g

7 6  16g 8g

?

III

8  4  158 6  16

158  52.67 g 3

KMnO4 (in strong alkaline medium) MnO4 converts into MnO42 in strong alkaline medium.

i)

O.St method: 7

6

Mn O4  Mn O42   change in O.St = 6  7  1  EMnO4  ii)

M  158 . 1

  2 e  method: MnO4  e  MnO4  1 mole of O.A gains 1mole of e  s  n  factor  1

 EMnO4 

M  158 1 Sr. INTER

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Nascent oxygen Method:

2 KMnO4  2 KOH  2 K 2 MnO4  H 2O  (O) 2 158g

16g

158g

8g

 E  158 g IV

K 2Cr2O7 (in acidic medium): changes to Cr 3

a)

O.St method: C r 2 O 7 2  2 C r  3  Change in O.St = 2(3) - 2(6) = 6.  n-factor = 6  E  M / 6

b)

2   3 e  method: Cr2O7 14H  6e  2Cr  7H2O

6

 no.of moles of e  s gained per mole of O.A=6.

294  E  M / 6 = 6  49 c)

Nascent oxygen method:

K2Cr2O7  4H2 SO4  K2 SO4  Cr2 (SO4 )3  4H2O  3(O)

3 16g

1 mole M/6

8g

V

H 2O2 (as O.A): H 2O2 reduces to H 2O .

a)

H 2 O 2  H 2 O  change in O.St

1

2

 2(2)  ( 1)2

n2  E b)

34  17 2

H 2O2  H 2O  (O) 34g 17g

16g 8g

 EH 2O2  17 g c)

In acidic medium H 2O2  2 H   2e   2 H 2O  O2   no.of moles of e s gained = 2.

Equivalent weight of a reductant: Can be defined as  It is the no.of gms of that substance involved in the loss of one mole of electrons. (or)  The numbe of moles of electrons lost by the one mole of reductant (or) (or)  The decrease in O.St per molecule of reductant  The amount of the substance that takes 8g of nascent oxygen in the redox reaction. Sr. INTER

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 Eq uivalent weight of a R.A 

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STOICHIOMETRY

Molecular weight Molecular weight   No.of moles of e s lost permole of R.A Change in O.St

Examples: i) Mohr’s salt (in acidic medium)

FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O in acidic medium acts as a strong R.A.

 Mohr’s salt get oxidises from Fe 2 to Fe 3 a)

Fe 2  Fe 3  e  : Change in O.St = 1 (or) no.of moles of electrons lost per mole of R.A.=1  E  M  392

ii)

Oxalic acid (in acidic medium):

H 2 C 2 O 4 .2H 2 O get oxidises to CO 2 in acidic medium. a)

C 2 O 4 2  4H   2e   2CO 2  2H 2 O  no.of moles of e  s gained per 1 mole of R.A=1.  E C2O42 

b)

3

M 126   63 2 2 4

C 2 O 42  2C O 2 Change in O.St = 2(4)-2(3)=2  E 

M 126   63 2 2

Equivalent weight of a substance that undegoes disproportion: Eg (i) Equivalent weight of Cl2 : With cold & dilute alkali Cl2 disproportionates as: Change in O.St=2

Cl  ClO 

Cl2 Change in O.St=2

 ECl 2  E1  E2 

M M   M  71g 2 2

(ii) Equivalent weight of Cl2 with hot & conc.alkali

n2  10 Cl2 n1  2

Cl  ClO3

M M M M 3M  ECl2  E1  E2  n  n  2  10  5 1 2

3   71  42.6 g 5

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II Equivalent weight of a substance that is comproportionated Eg: Equivalent weight of sulphur in the following rxn.

n2  4

S2 SO 2

S

n1  2  E S  E1  E 2 

M M 3    32  24g 2 4 4

III Equivalent weight of a substance that is thermally decomposed:

change in O.St  6 oxidation 5

1

2K C lO32 2KC l  3O 2 Eg(i) change in O.St  6 Re duction

 Consider the n-factor, either w.r. to Cl2 or w.r. to oxygen. [which is same in both the cases]

 n  factor  6

 E KClO3 

M 122.5   20.4179 6 6

oxidation Change in O.St  2 5 1

Eg(ii)

0

3

PCl 5

PC l3  Cl 2

Re duction change in O.St  3

 

[From PCl5 only two Cl atoms are oxidised to Cl2 ] Change in O.St w.r to ‘P’ = change in O.St w.r to ‘Cl’=2 n-factor = 2

E PCl5 

M 208.5   104.25g 2 2

IV Equivalent weight of a substance in which two different elements either both oxidised (or) both reduced: Eg:(i) FeC 2 O 4 in acidic medium; Fe 2 oxidises to Fe 3 and C 2 O 4 2 to CO 2

n2  2 2  3

FeC 2 O 4

H

4

Fe3  CO 2

n1  1

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STOICHIOMETRY

 Total n-factor = n1  n 2  1  2  3

M  E FeC2O4  n  n 1 2 

M 142   47.33 3 3

Eg(ii) Equivalent weight of FeS2 in the half reaction,

Re duction

n 2  10 2 1

3

FeS2

4

Fe2 O3  SO 2 oxidation n1  1

M M 118  E FeS2  n  n  11  11  10.73 1 2 Equivalent volume: It is “volume occupied by the gaseous substance at STP corresponding to its gm equivalent weight”. Equivalent volume =

22.4 L . n  factor

 Where n-factor is the factor by which molecular weight of the substance is divided to get its equivalent weight. Examples: Determine the equivalent volume of the following gaseous compounds. a) H 2 gas

b) O2 gas

c) Cl 2 gas

2  1g  Eq.vol of 2

H 2 gas 

22.4 = 11.2 lt 2

a)

H 2 gas : E.wt.of

b)

Cl 2 gas : E.wt of Cl2 

71  35.5g  Eq.vol of 2

c)

O2 gas : E.wt of O 2 

22.4 32  5.6 lt  8g  Eq.Vol of O 2 gas  4 4

H2 

Cl 2 gas 

22.4  11.2 lt 2

CAP-53. Find the eq. volume of SO 2 gas in the reaction : SO 2  SO 42 n-factor = 2.

 Eq. vol of SO 2 

22.4  11.2 lt 2

Principle of gm equivalence: “Number of g-equivalents of all reactants reacted will be equal and the same number of g-equivalents of each product will form.” CAPS-54. 1. A metal oxide contains 47% oxygen by weight. Determine the equivalent weight of the metal. Sol: If we take 100gm of oxide, it contains 47gm oxygen & hence rest, 53gm metal.  no.of g.equ. of metal = no.of g.eg.of oxygen. Sr. INTER

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w w 53 47      E  9.02   E  E  Oxygen E 8 Metal 2.

5.04 gms of metal carbonate produces 2.4gm metal oxide on complete decomposition. What is the equivalent weight of the metal. Sol: no.of g-eq metal carbonate = no.of g-eq.of metal oxide.

 3.

5.04 2.4   E  12 E  30 E  8

5.55 gm of a base neturalises completely 1.68 lt SO 2 gas at STP. Calculate the equivalent weight of a base.

Sol: no.of g.eq. of base = no.of g.eq. of SO 2

 4.

5.55 1.68   E  37 E 11.2

How many gms of oxygen is needed for complete combustion of 6gm CH 4 ? 4

0

4  2

Sol: C H 4  O 2  CO 2  H 2 O

 E.wt.of CH 4  

16 32  2 Eq.wt of O 2  8 8 4

No.of g.eq. of CH 4 = no.of g.eq.of O2

6 w   w  24gms 2 8

Experimental Methods to determine atomic weight of elements:

(at.wt.  percent abundance) 100

a)

Isotope Method: Average atomic weight =

b)

Dulong and Petit’s Law: The product of atomic weight and specific heat of an element is about 6.4

atomic weight  specific heat  6.4 * This law is applicable for solid elements only. * It gives better result with heavy elements. The exact atomic weight of elements may be determined by using the following steps: (i)

atomic weight (approx) =

6.4 sp.heat

(ii) Equivalent weight determination from experiment (iii) Valency  = fractional, but close to an integer. Make it integer. It will give the correct ‘v’ of the element. (iv) Atomic weight (exact) = Eq. wt  valency CAP-55. The specific heat of a metal is 0.115 cal/oC-gm. The oxide of this metal contains 22.2g oxygen by weight. Determine the exact atomic weight of the metal. Sol: at.wt(approx) =

6.4 6.4   55.65 sp.heat 0.115

If we take 100gm of metal oxide, it contains 22.22 gm oxygen and remaining 77.78 gm will be metal. Sr. INTER

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No.of g-eq. of metal = no.of g-eq. of oxygen 

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STOICHIOMETRY

77.78 22.22  E 8

 E = 28 Now, valency  c)

at.wt.(approx) 55.65   1.9875  2 eq.wt. 28

Hence, exact at .wt. of metal = eq.wt  valency = 28  2=56 Vapour density of volatile chloride method: Let an element, M, of valency v, form a volatile chloride, MClV. Then, mol.wt. of MClV=1  at.wt.of M+v  at.wt.of Cl  2  V.D  V  E  V  35.5  V(E  35.5)

 V

2  V.D E  35.5

Hence, with the known V.D. of volatile chloride and equivalent weight of element, we may determine the valency of element.  at.wt. of element = eq.wt  valency.. CAP-56. The vapour density of a volatile chloride of a metal is 95. The chloride contains 37.368gm chlorine by weight. Determine the atomic weight of metal. Sol: Let we take 100gm of metal chloride. It will contain 37.368gm chlorine and remaining 62.632 gm metal. Now,

62.632 37.368  E 35.5 E = 59.5  Now, valency of metal, V 

d)

2  V.D 2  95  2 E  35.5 59.5  35.5

Hence, atomic weight of metal = E  V  59.5  2  119 Law of Isomorphism: Isomorphous compounds are the compounds having similar molecular formula as well as crystalline geometry.

Example: Cu 2S and Ag 2S

K 2SO 4 , K 2SeO 4 , K 2 WO 4 , K 2 CrO 4 MgSO4  7H2 O, ZnSO4  7H 2 O and FeSO4  7H 2 O ; etc Isomorphous compounds differs only in a single element or ion and the elements differing should definitely have the same valency. According to the law of isomorphism, the ratio of atomic weights of the element differing in isomorphous compounds will be equal to their mass ratio combined with the same mass of other elements.

At.wt.of A wt.of A  At.wt of B wt.of B combined with the same weight of other elements. CAP-57. Cu 2S and Ag 2S are isomorphous compounds. Cu 2S contains 20.14% sulphur and Ag 2S contains 12.90% sulphur, by weight. If the atomic weight of copper is 63.5, determine the atomic weight of silver. Sol: Cu 2S contains 20.14% sulphur and hence remaining 79.86% is copper. Hence, 1 gm sulphur will combine with

79.86  3.965gm copper 20.14 Sr. INTER

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Ag 2S contains 12.90% sulphur and hence remaining 87.10% is silver. Hence, 1 gm sulphur will combine with

87.10  6.752gm silver 12.90

Now, from the law of isomorphism,

At.wt of Ag wt.of Ag  At.wt of Cu wt.of Cu combined with same amount of sulphur

or,

At.wt of Ag 6.752  63.5 3.965

 At.wt. of Ag = 108.134

PART - 8 (CONCENTRATION TERMS AND VOLUMETRIC ANALYSIS) Solution : It may be defined as the homogeneous mixture of two or more substances. The substance present in larger amount is called solvent and the other substances present in smaller amounts are called solutes. Concentration or strength of solution : It normally represents the amount of solute present in the given amount of solution. Depending on the unit of amount, different methods are used to express concentration. Some of them are : i) Percentage method ii) gm / lt iii) PPM iv) Molarity (M) v) Molality (m) vi) Normality (N) vii) Formality (F) viii) Mole fraction (XA) ix) Volume strength I. Percentage method : It represents the amount of solute present in 100 amount of solution. The amount may be expressed in gm or ml resulting the following four kind of percentage methods. i) w/w (weight by weight method) : It represents the amount of solute (in gm) present in 100gm of solution. ii) w/v (weight by volume method) : It represents the amount of solute (in gm) present in 100ml of solution. iii) v/w (volume by weight) : It represents the amount of solute (in ml) present in 100gm of solution. iv) v/v (volume by volume) method : It represents the amount of solute (in ml) present in 100ml of solution. CAPS-58. Example - 1 : 250gm of a solution contains 20g solute. Determine the strength of solution in w/w percent. Sol : 250g solution contains 20g solute.  100g solution will contain

20  100  8 g solute 250 Hence, strength = 8% (w/w). Example - 2 : 120 ml of a solution contains 0.02 mole of solute of molar mass, 60gm. Express the strength of solution (w/v) percent. Sol : Weight of solute = mole x molecular weight = 0.02  60  1.2gm Now,  120ml of solution contains 1.2gm solute Sr. INTER

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1.2

 100ml of solution will contain 120  100  1 gm solute Hence, strength = 1% (w/v) Example - 3 : The strength of a solution of density 1.2 gm/ml is 10% (w/w). Express the strength in (w/v) percent. Sol : Strength of solution is 10% (w/w), means if we take 100 gm 

100 ml solution, it will contain 10gm solute, 1.2

100 ml solution contains 10gm solute 1.2

 100ml solution will contain

100 100  12 gm solute  100     1.2 

Hence, strength = 12% (w/v) II. Parts per million (PPM)  It represents the amount of solute (in gm) present in 106 gm of solution. It is normally used when strength of solution is very low.

PPM 

Mass of solute  106 Mass of solution

CAPS-59. Example - 1 : The strength of a solution is 0.002% (w/w). Express it in ppm. Sol : 0.002% (w/w) means 100gm of solution will contain 0.002gm solute. Hence, Strength =

0.002  106  20 ppm 100

Example - 2 : The fluoride concentration in a sample of toothpaste is 500ppm. What is the mass of fluoride present in a tube of toothpaste containing 200gm paste ? Sol : 500 ppm means 106 gm paste will contain 500g fluoride

 Mass of fluoride =

500  200  0.1gm 106

III. Strength (or) concentration (or) gm/lit : It represents “the amount of the solute in gms present in one litre of the solution”. Massof soluteingms

 Strength of solution = volumeof solutioninml 1000 CAPS-60 Example - 1 : 10 gm urea is dissolved in 70gm water to get a solution of density 1.25gm/ml. Express the concentration of solution in gm/lit. Sol : Weight solution prepared = 10 + 70 = 80 gm

 Volume of solution =

w 80   64ml d 1.25

 64ml solution containing 10 gm solute,  1000ml solution will contain

10  1000  156.25 gm solute. 64  Strength = 156.25 gm/lit. Sr. INTER

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Molarity (M) : It represents “The number of moles of solute present per litre of the solution”. Molarity =

No.of moles of solute vol.of solution in litres

M

wA M A  VLt

wA 1000  M A Vml

( wA = weight of the solute ; MA = molecular weight of the solute Vml = volume of solution in ml.) CAPS-61 Example - 1 : 4.9 gm H2SO4 is dissolved in water to get 250ml solution. What is the molarity of solution. Sol : M 

4.9  1000  0.2 M 98  250

Example - 2 : Calculate the molarity of 10% (w/v) NaOH solution. Sol : 10% (w/v) solution means 100ml of solution contains 10gm solute, NaOH. Hence,

M

10  1000  2.5M 40 100

Example - 3 : Calculate the molarity of 10% (w/v) NaOH solution of density 1.1 gm/ml. Sol : 10% (w/w) solution means

M

100 ml solution contains 10gm NaOH. Hence, 1.1

10  1000  2.75M 100 40  1.1

Example - 4 : How many moles of solute are present in 250ml of 0.8M KMnO4 solution. Sol : No. of moles = Molarity x vol. of solution in litres.

0.8 

250  0.02 1000

Dilution : Addition of excess solvent to the solution, without changing the amount of solute. Dilution is reciprocal of concentration i.e, Dilution =

1 concentration

Law of dilution : “The number of moles of solute present before and after dilution is constant”. No. of moles =

weight of the solute mol.weight of the solute

(or) No. of moles = Molarity x VLt Suppose M1 , M2 are the molarities and V1 , V2 are volumes of solution in lts before and after dilution respectively. Sr. INTER

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Then, according to law of dilution Before dilution After dilution M1 x V 1 = M2 x V 2 CAPS-62 Example - 1 : 40ml of 2M-HCl solution is diluted to 100ml. What is the molarity of diluted solution. Sol : 40 x 2 = 100 x M2  M2 = 0.8 M NOTE :  As the dilution is reciprocal that of concentration it should be remembered that : * If dilution increases by 10times molarity (concentration) decreases by 10times. Eg : 10ml of 0.1M solution is diluted to 100ml

 conc  by 10 times  final conc. 0.01 * If dilution increases by 5 times, conc  by 5 times Eg : 20ml of 0.2M solution is diluted to 100ml.  dilution is by 5 times  conc. should decrease by 5 times

 find molarity =

0.2  0.04 5

Eg : 40 ml of 0.5 M solution is diluted 160ml.  dilution is by 4 times

 conc. should  by 4 times 

0.5 = 0.125 is the final molarity 4

Eg : If 50ml of 0.8 M solution is diluted to 400ml.  dilution is by 8 times

 conc. should  by 8 times 

0.8  0.1 M is the final conc. 8

Example - 2 : 20ml of 0.5 M solution is diluted to 100ml. 25ml of this solution is taken and further diluted to 100ml. Then find the final conc. of the solution. i) 1st dilution : 20ml is diluted to 100ml  5 times dilution  conc.  by 5 times

0.5  0.1 5

ii) 2nd dilution : This 0.1M solution of 25ml is diluted to 100ml  4 times dilution  conc.  by 4 times.

0.1  0.025 M is final conc. 4

Example - 4 : 25ml of HCl solution is diluted to 100ml. 50ml of this solution is taken and diluted further to get 400ml of 0.1M HCl. Then, what is initial conc. of HCl. Sol : i) 25  M1  100   2 Sr. INTER

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25×M1 M1 = 100 4

M1  50  400  0.1 4  M1 = 3.2 M

Resultant Molarity of mixture of strong acids (or) strong bases : If two strong acids (or) strong bases with concentrations M1 and M2 and volumes V1 and V2 are mixed with each other then the resultant molarity. M Mix 

M 1V1  M 2V 2 V1  V2

 If there is a further dilution, the formula becomes M 1V1  M 2V2

M = final volume CAP-63. 20ml of 2M-NaOH solution is mixed with 40ml of M-NaOH solution and the mixture is diluted to 100mL. Determine the molarity of the diluted solution. Sol : The total no. of moles of solute in final mixture should be the sum of moles of solute in all the solution mixed, if the solutes are non - reacting. Hence, M =

20  2  40  1  0.8 M 100

Formality (F) : It represents “The number of gram formula units of solute present in one litre of solution”.  The term formality should be preferred in expressing the concentration of ionic compound, because their smallest particles are called formula units in place of molecules. Example : A 250ml solution contains 3.51 gm NaCl. Calculate the formality of the solution Formality=

3.51  1000  0.24F 58.5  250

Normality (N) : It represents “Number of gm equivalents of solute in one litre of solution”.

 Normality=  w

No.of gm equivalents of solute Volume of solution in litres

wA E A  Vlit 1000

A  N= E A × Vml

wA = weight of the solute in gms Vml = volume of solution in mL. EA = gm. equivalent weight of the solute Molecular weight n  factor Example : Calculate the normality of 5.88% (w/v) H2SO4 solution. Sol : 5.88% (w/v) solution represents that 100ml solution contains 5.88 gm of H2SO4

=

N 

wA 1000 5.88 1000    1.2N 98 100 E A Vml 2

 

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Relation between Moalrity and Normality :

N

wA 1000   E A Vml

wA 1000  MA Vml n  factor

wA 1000   n  factor M A Vml

N  Molarity  n  factor CAPS-64 Find the normalities of i) 0.1M HCl ii) 0.5M H2SO4 iii) 0.2M H3PO4 iv) 0.8M Ca(OH)2 v) 0.3M AlCl3 vi) 0.4M Al2(SO4)3 vii) 0.1M KMnO4 / H+ viii) 2M CH3COOH Sol : i) 0.1N HCl ii) 0.5 x 2 = 1N H2SO4 iii) 0.2 x 3 = 0.6N iv) 0.8 x 2 = 1.6N Ca(OH)2 v) 0.3 x 3 = 0.9N AlCl3 vi) 0.4 x 6 = 2.4N Al2(SO4)3 vii) 0.1 x 5 = 0.5N KMnO4 viii) 2x1 = 2N CH3COOH Law of dilution : is also applicable to normality i.e, before dilution and after dilution, no.of gm equivalents of solute remains constant. No.of gm equivalents =

weight of the solute gm. eq. wt. of solute

(or) No.of gm equivalents = Normality x Vlt. No.of milli g.equts = Normality x Vml After dilution  So, Before dilution N1 x V1 = N2 x V2 Example - 1 : Calculate no.of gm equivalents in a) 9.5g of MgCl2 b) 100ml of 0.1M H2SO4 Sol : a)

neq 

9.5  0.2 equts 95 2

b) neq  N Vli = 0.1  2 

100 = 0.02 1000

CAPS-65. Calculate no.of gm. equivalents of i) 100ml of 0.1M KMnO4 / H+ Sol : neq  ii)

200ml of 0.5M K2Cr2O7 / H+

Sol : neq  iii)

100  0.1 5  0.05 1000 200  0.5  6  0.6 1000

500ml of 0.2M H2SO4

Sol : neq 

500  0.2  2  0.2 1000 Sr. INTER

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1L of 0.3M Ca3(PO4)2

Sol : neq  1 0.3  6  1.8 v)

2L of 2M FeC2O4 / H+

Sol : neq  2  2  3  12 vi)

400ml of 0.8M oxalic acid

Sol : neq 

400  0.8  2  0.64 1000

Example : 10ml of 0.5M H2SO4 is diluted to 100ml. 25ml of this solution is taken and further diluted to 100ml. Find the final normality of the solution. Sol : i) 10  0.5  100  M 1  M 1  0.05 ii) 0.05  25  100  M 2  M 2  0.0125M  Final normality = 0.0125 x 2 = 0.025N Resultant Normality of mixture of strong acids (or) strong bases : If two strong acids (or) bases with normalities N1, N2 and volumes V1 , V2 are mixed with each other, then the total no.of gm equivalents of solute in final mixture should be the sum of gm equivalents of solute in all the solution mixed, if the solutes are non - reacting.

N mix 

N1V1  N 2V2 V1  V2

CAP-66. 100ml of 0.1M H2SO4 is mixed with 200ml of 0.2M HCl and diluted to 1lit. Then, find the normality of the resultant solution.

100  0.2  2  200  0.2  1 20  40  0.06 N = 1000 1000 Resultant Normality, when a strong acid is mixed with a strong base :  When an acid is added to a base, the resultant solution, may be acidic, basic (or) neutral depending on the no.of gm equivalents of acid and base added. Sol : Resultant normality =

Case - 1 : If neq Acid  neq Base Then, resultant solution is neutral, PH = 7   7 i.e,  H   OH   10 M at 25°C.

Case - 2 : If neq Acid  neq Base Then, resultant solution is acidic  PH < 7.  The excess gram equivalents of acid left over after neutralisation is given by :

neq Acid  neq Base = N A  VA  N A  VB  Normality of the resultant solution =

no.of gm equivalents of acid left over total vol.of solution in litres    H  mix 

N AVA  N BVB VA  VB

Case - 3 : If neq Acid  neq Base Then, resultant solution is basic  PH > 7. Sr. INTER

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CHEMISTRYv v v v v v v v v v v v v v v v v v v v v v v v STOICHIOMETRY  The excess gm equivalents of base left over after neutralisation in the solution is given by : N BVB  N AVA

 Normality of the resultant solution =   OH  mix 

no.of gm equivalents of baseleft over volume of solution in litres

N BVB  N AVA VA  VB

CAPS-67. Example - 1 : 20ml of 0.1M H2SO4 is added to 40ml of 0.5M NaOH solution. Then find the nature and concentration of the resultant solution. Sol : n m.eq H 2SO 4  20  0.1  2  4m.g.equts

n m.eq NaOH  40  0.05  1  2m.g.equts As, n eq acid  n eq base  Resultant solution is acidic

42 2    0.033N   H   60 60 Example - 2 : 100ml of 0.2M HCl is added to 200ml of 0.05MCa(OH)2. Then, find the nature of resultant solution. Sol : n m.eq HCl  100  0.2  20

n m.eq Ca  OH 2  200  0.05  2  20  As n eq Acid = n eq Base  Resultant solution is neutral. Example - 3 : 50ml of 2M H2SO4 is added to 100ml of 2M Ca(OH)2 and furtehr diluted to 1lit. Then, find the nature and concentration of the resultant solution. Sol : n m.eq H2SO4 = 50 x 2 x 2 = 200

n m.eq Ca(OH)2 = 100 x 2 x 2 = 400  As n eq Acid < n eq Base  Resultant solution is basic   OH  

400  200 200   0.2N 1000 1000

Molality (m) : It represents “The number of moles of solute present per kg of the solvent”.

no.of gm.moles of solute  Molality (m) = wt.of the solvent in kg

wA = M w  A B Kg

m

w A 1000  MA w B

where, wB = weight of the solvent in gms. Sr. INTER

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CAPS-68. Example -1 : Calculate the molaltiy of 10% (w/w) NaOH solution. Sol : 10% (w/w) NaOH  10gm of NaOH present in 100g of solution.  weight of solvent (wB) = 100 - 10 = 90g

m

10 1000   2.78m 40 90

Example - 2 : Calculate the weight of 6% (w/v) urea solution whose density is 1.2g/ml. Sol : 6% (w/v) urea solution  6 gm of urea in 100ml of solution  6 gm of urea in (100x1.2) g of solution  wB = 120 - 6 = 114.

 m

6 1000   0.8772m 60 114

Relation between Molarity & Molality : Example : The density of 1.5M urea solution is 1.05gm/ml. Calculate the molality of the solution. Sol : 1.5 M urea solution  1.5moles of urea in 1L of solution  1.5 moles of urea in (1000x1.05)g of solution  1.5x60g of urea in 1050g of solution  wB = 1050 - 90 = 960g of solvent

 m

90 1000   1.56m 60 960

NOTE : From the above conceptual work, the relation between molarity (M) and molality (m) can be put in the form of formula as :

m

1000  M 1000  d  M  M A

where

M = Molarity of solution MA = mol.wt of solute in gms. d = density of the solution in gm/ml

CAPS-69 1. Find the molality of 2% (w/v) Glucose solution whose density is 1.02 gm/ml. 2. Find the molality of 1.2M NaOH solution whose density is 1.5gm/ml. Molefraction (X) : “Molefraction of any component in a mixture represents its number of moles present in 1mole of the mixture”.  Let a solution be prepared by mixing n1 moles of solute and n2 moles of solvent. Then, the total no.of moles of solution becomes (n1 + n2).

n1  Hence, Molefraction of solute (X1) = n  n 1 2

n2  Molefraction of solvent (X2) = n  n 1 2

 X1  X 2  1

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CHEMISTRYv v v v v v v v v v v v v v v v v v v v v v v v STOICHIOMETRY  Sum of all the molefractions of all components present in a solution is unity. i.e if a solution is made up of ‘n’ no.of components then X1  X 2  X 3  ........X n  1 n

  Xi  1 i 1

NOTE : 1. All the units including volume terms will be temperature dependent, eg : molarity, normality, formality etc are temperature dependent. 2. Molefraction, % by mass, molality etc., as they does not involve volume terms, they are temperature independent. 3. Rise in temperature usually results in increase in volume, so molarity, normality etc decrease as temperature is increased. Example - 1 : A solution is made by mixing 64gm CH3OH and 46gm C2H5OH. Calculate the molefraction of methanol in the solution.

W 64  2 M 32

Sol : n CH3OH 

n C2 H5OH 

46 1 46

 X CH3OH 

n1 2   0.67 n1  n 2 2  1

Relation between molefraction and molality : Calculate the molefraction of solute in 1m aqueous urea solution. Sol : 1m solution  1mole urea in 1kg water

 X urea 

n urea 1   0.01768 1000 n urea  n H 2O 1  18

 In general, if the solvent is water, then the relation between molality and molefraction is given by

XA 

m m  55.55

If solvent is not water, then :

XA 

m 1000 m MB

MB = Molecular weight of solvent CAPS-70. 1. If 2gm of NaOH is dissolved in 180gm of water. Find the molefraction of NaOH. 2. Find the molefraction of 0.2m aqueous glucose solution. Volume strength : This method of expression of strength is used only for H2O2 solution. It represents : “The number of milliliters of oxygen gas liberated at STP by 1ml of H2O2 solution”. For example : i) 10 volume H2O2 means 1 vol of H2O2 solution on decomposition gives 10 volumes of O2 at STP. ii) 20 vol H2O2  1ml of H2O2 solution gives 20ml of O2 at STP by decomposition. iii) 20ml of 20 vol H2O2 : It gives 20x20 = 400ml of O2 at STP Sr. INTER

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Conversion of volume strength into % (w/v) : Example : Calculate %(w/v) strength of 10vol H2O2. Sol : 10 vol H2O2  1lit of H2O2 solution on decomposition gives 10lit of O2 at STP

2H 2 O 2   2H 2 O  O 2 2x34g ?

22.4 L at STP 10 L

10  2  34  30.36g 22.4

 1000ml H2O2 solution contains ------ 30.36g  100ml H2O2 solution contains ------ 3.036g = 3.036 % (w/v)  10 vol H2O2 20 vol H2O2 30vol H2O2 15vol H2O2

= 6.072 % (w/v) = 9.108 % (w/v) = 4.554 % (w/v)

100vol.H2O2  30.36% w / v H2O2 calledasperhydrol II. Conversion of volume strength into Molarity and Normality : Eg : Express 10vol. H2O2 as Molarity & Normality. Sol : 10 vol H2O2  3.036% (w/v)  3.036 gm H2O2 in 100ml of solution

M 

3.036 1000   0.893M 34 100

 Normality = Molarity x n - factor = 0.893 x 2 = 1.786 N  In general :

% concentration  17

 volume strength 56 Volume strength  Molarity 11.2 Volume strength  Normality  5.6

10 volume strength = 0.893 M = 1.786 N 20 volume strength = 1.786 M = 3.572 N 30 volume strength = 2.679 M = 5.358 N 40 volume strength = 3.572 M = 7.144 N 15 volume strength = 1.3395 M = 2.679 N CAPS-71. Example - 1 : What is the volume strength of 1N - H2O2 solution. Sol : 1N - H2O2 solution  1000ml solution contain 1g.eq H2O2  Produce 1g. eq. of O2  5600 ml of O2 at STP

 Volume strength =

5600  5.6vol 1000

 It is standard relation : 1N - H2O2 = 5.6 vol

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STOICHIOMETRY

Example - 2 : Calculate the volume strength of 10% (w/v) H2O2 solution. Sol : 10% (w/v) H2O2 solution  100ml solution contains 10gm H2O2

10 g.equt H 2 O 2 17

10  5600 ml O2 at STP 17

 Vol. strength =

10 5600   32.94vol 17 100 VOLUMETRIC ANALYSIS (OR) TITRATION

 It is the method of determination of strength of a solution with the help of an another solution of known strength.  It may also be defined as experimental method of determination of volume of a solution of known strength needed for a definite volume of an another solution of unknown strength. Types Titration : Acidimetry

i) Acid - Base Titration ii) Redox Titration

Alkalimetry

-

Permanganometry Dichrometry Iodometry Iodimetry etc.

iii) Precipitate titration iv) Complexometric titration v) Radio - titration vi) Conductometric titration vii) Potentiometric titration, etc.,  Out of which Acid - Base & Redox titrations are more important and discussed here : Terms used in volumetric analysis : Titre (or) Titrate or analyte : The substance, whose concentration is to be determined. Titrant : The substance dissolved in solution of known concentration. Standard solution : A solution whose concentration is known is called as standard solution. The titrants dissolved in standard solutions may be of two types. i) Primary standard titrants : Those substances whose solution can be prepared by weighing them directly Eg : Mohr’s salt, Hypo, CuSO4, AgNO3, K2Cr2O7, oxalic acid etc., ii) Secondary standard titrants : Those substances whose solution can not be prepared by weighing them directly as they react with substances surrounding them like atmospheric moisture CO2, O2 etc., Eg : KOH, NaOH, HCl, H2SO4, KMnO4 etc.  Their solution is standardised by titrating them with standard solutions. Preparation of standard solutions : A standard solution is prepared by dissolving an accurately weighed quantity of a highly pure material called a primary standard and diluting to an accurately known volume in a volumetric flask. Unknown solution : Solution of titrate whose concentration is unknown. End point : When the experiment of titration is stopped, it is called end point of titration. It is normally detected by sudden change in colour of the solution. Equivalence point : When complete reaction takes place between the solutions, it is called equivalent point of titration. Sr. INTER

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CHEMISTRYv v v v v v v v v v v v v v v v v v v v v v v v STOICHIOMETRY  Normally indicators are such selected that end point and equivalent point remain identical. Indicator : It is the substance used in titration in very small amount, which response the sudden change in colour of the solution.  In acid - base titration, the indicators used are either weak organic acids or weak organic bases. Some examples are : Acidic indicators : Phenolphthalein, litmus paper. Basic indicators : methyl orange, methyl red.  An indicator, whose PH range lie in the PH range of titration is perfect indicator for that titration. Selection of indicators :

Acid Strong Strong Weak

Base Strong Weak Strong

Indicator Any Methyl orange, methyl red etc Phenolphthalein etc

 In redox titrations, the sudden changes in colours due to reactants themselves and hence in most of the cases external indicators are not required. Principles of titration :  Titration means stoichiometry and hence it runs on moles as well as equivalent concept. But, for simplicity equivalent concept is preferred.  Principle : “Number of g.equivalents of all reactants reacted will be equal and the same number of g.equivalents of each product will form”.

 No.of g.equivalents = weight in gm  vol.of gas  N  VL gm.equt.wt

eq.vol

ACID - BASE TITRATIONS: The determination of bases by titration with a standard acid is called acidimetry and the determination of of concentration of acids by titration with a standard base is called alkalimetry.  For Acid - Base titrations, at equivalent point :

n eq Acid  n eq Base

weight of acid weight of base  Mol.wt.of acid  Mol.wt.of base

  N  VL Acid   N  VL Base REDOX TITRATIONS:

 For Redox titrations, at equivalent point : n eq oxidant  n eq reductant 

weight of oxidant weight of reductant  Mol.wt.of oxidant Mol.wt.of reductant

  N  VL oxidant   N  VL reductant

 Common oxidising agents which are used in redox titrations are : i) Acidified KMnO4 (Permanganometry) ii) Acidified K2Cr2O7 (dichrometry) iii) Iodine / Iodide (Iodimetry & Iodometry) etc., I) Titrations using acidified KMnO4 (Permanganometry) : In strongly acidic medium

MnO4  8H  5e   Mn 2  2H2O ; E0  1.51V Sr. INTER

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M 158   31.6 5 5

In weakly acidic or neutral medium :

MnO4  4H  3e   MnO2  2H2O;E0  1.69V E

M 158   52.6 3 3

In alkaline medium :

MnO 4  e    MnO 24 ; E 0  0.56V

E

M  158 1

NOTE : Among the common mineral acids H2SO4, HCl and HNO3 only H2SO4 is useful for providing acidic medium. HCl can not be used, since Cl- is oxidised to Cl2 by KMnO4. On the other hand, HNO3 itself acts as strong oxidising agent. Some, important titrations using KMnO4 are : a) KMnO4 vs oxalic acid :

2KMnO 4  5H 2 C 2 O 4  8H 2SO 4   K 2SO 4  2MnSO 4  10CO 2  8H 2 O Titrant Oxidant

Titrate Reductant

2MnO4  5C2O42 16H   2Mn 2  10CO2  8H2O

 Reaction is slow initially. But, after sometime, Mn2+ (Product), catalyse the reaction (Auto catalysis)  Medium : Acidic (by H2SO4)

 End point : Light pink colour of MnO 4 solution, KMnO4 acts as self indicator.. At equivalent point : i) 2mmole of KMnO4 reacts with 5mmole of oxalic acid

 m.eq.KMnO 4  meq of oxalic acid  N KMnO 4  VKMnO4  N oxalic acid  Voxalicacid 

1 1  M KMnO 4  VKMnO 4   M oxalic acid  Voxalic acid 2 5

 N  M  n  factor   w     M/ 5 KMnO4  E KMnO4  b)

 w     M/ 2 oxalicacid

M  31.6 ; E 5

oxalic acid ( H 2 C 2 O 4 .2 H 2 O )

126  63 2

2+

KMnO4 Vs Fe ions :

 Fe2+ is readily oxidised by MnO 4 to Fe3+  In the laboratory, for practical purposes ferrous ammonium sulphate (Mohr’s salt) is taken as a source of Fe2+ ions.

5Fe 2   MnO 4  8H    5Fe 3   Mn 2   8H 2 O Sr. INTER

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 4

 End point : MnO acts as self indicator and light pink colour of solution indicates end point.  Medium : Acidic (by H2SO4)  At equivalence point : i)

1 mole MnO 4 reacts with 5 mole Fe2+

ii)

N KMnO4  VKMnO4  N FAS  VFAS

iii)

5  M KMnO4  VKMnO4  1 M FAS  N FAS

iv)

 w  w      M / 5  KMnO4  M FAS

v)

158 392 E MnO  M   31.6 ; E FAS  5 4 5 1

NOTE : If titration is being carried out with solution of Ferrous oxalate, then :

Fe 2  C 2 O 24    Fe3  2CO 2  3e 

 n - factor = 3  E Fe C2 O 4   M 3 Estimation of Fe2+ and Fe3+ ions in solution :  In this estimation, the solution is titrated twice. First without reduction in which only Fe2+ reacts and Fe3+ remains unreacted.  In second step, same volume of solution is taken. Fe3+ ions are reduced to Fe2+ and then titrated. Here, i) In step - 1 : meq KMnO4 (let x) = meq of Fe2+ ii) In step - 2 : meq KMnO4 (let y) = meq of Fe2+ + meq of Fe3+  meq Fe3+ = (y - x) d. Analysis of sample containing oxalic acid and sodium oxalate :  Same volume of solution are titrated separately with NaOH and KMnO4 solution.  meq NaOH = meq oxalic acid and meq KMnO4 = meq of oxalic acid + meq of sodium oxalate. Titrations using K2Cr2O7 (dichrometry) :

 In volumetric analysis, Cr2 O 72  is also used commonly as oxidant yet weaker than MnO 4 , Ce 4 etc.,  In most of it’s applications it is reduced to Cr+3.

Cr2O72  14H  6e   2Cr3  7H2O ; E0  1.33V E Cr O2  2

7

M Cr O2 2

6

7

294  49 6

 These titrations are made in acidic solutions, but, not in neutral or alkaline medium. In alkaline medium, orange Cr2 O 72  changes to yellow CrO 24  .  It can be used as oxidising agent in present of HCl or Cl- also.  If same molarities of KMnO4 & K2Cr2O7 in acidic medium are used to titrate separately Fe2+ (aq), then

vol for MnO 4 6   1.2 vol for Cr2 O72 5 Thus, volume of MnO 4 required will be 1.2 times of volume of Cr2 O 72  . Sr. INTER

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3.

Iodometry & Iodimetry :  In Iodimetry I2 solution is used for titration, while in iodometry liberated I2 is titrated.  I2 acts as mild oxidising agent and is used for titrating several analytes. The titrations are based upon following half reaction :

 3I  ; E 0  0.536V I3  2e  Iodimetry : In these titrations, standard I2 solution is used to titrate easily oxidisable substances

 It includes the estimations of thiosulphates, sulphite, arsenite etc, 2S2 O32   I 2   S4 O 62  2I  SO32   I 2  H 2 O   SO 42   2H   2I  AsO 33  I 2  H 2 O   AsO 42   2H   2I  Iodometry : In iodometric titrations an oxidising agent is allowed to react with excess of KI (or I-) solution. The I2 liberated is titrated with hypo.  MnO4 / H , Cr2O27 / H , H2O2 / H , Cl2 , Br2 , O3 etc., can be used as oxidants to oxidise I  .  I2 thus produced is titrated with thiosulphate by using starch as indicator.

I 2  2S2 O32   2I   S4 O 62 n - factor for hypo = 1 n - factor for I2 = 2. NOTE : I2 has low solubility in water, but the complex I3 is highly soluble. So, I2 solution is prepared by dissolving I2 in concentrated solution of KI.

I 2  I    I3 I3 is therefore the actual species used in the titration. Different kind of problem profile on acid - base and redox titrations : Simple Problems CAPS-72. Example - 1 : How many ml of 0.2N - H2SO4 solution is needed to react completely with 20ml of 0.02N - NaOH solution ? Sol :

n meq

H2SO4 =

n meq

NaOH

0.2 x V = 20 x 0.02

 V = 2 ml Example - 2 : What volume of 0.1M - KMnO4 solution is needed for complete oxidation of 40ml of 0.2M H2C2O4 in presence of H2SO4. Sol : n meq KMnO4 = n H2C2O4 meq 5 x 0.1 x V = 40 x 0.2 x 2 V = 32 ml 7

M nO 4   Mn 2  n  factor  5  3

4

C2 O42   CO2  n  factor  2  Example - 3 : Calculate the weight of Na2CO3. 10H2O needed for complete neutralization of 100ml of 0.1M - H2SO4 solution. Sol : neq Na2CO3 . 10H2O = neq H2SO4 Sr. INTER

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CHEMISTRYv

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w 286

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STOICHIOMETRY

100  0.1 2 1000

 w = 2.86 g n - factor of Na2CO3 . 10H2O = 2 n - factor of H2SO4 = 2 Example - 4 : What volume of 0.4M - KMnO4 solution is needed for complete oxidation of 9.92gm hypo, Na2S2O3 . 5H2O in basic solution ? The un balanced reaction is : MnO 4  S2 O32  H 2 O  MnO 2  SO42  OH  Sol : neq KMnO4 = neq Na2S2O3 . 5H2O

V 9.92  248 1000 8

3  0.4 

 V  266.67ml 7

4

M nO4   M nO 2  n  factor  3 2

6

S2 O32   S O24  n  factor  8 Example - 5 : How many mg of magnesium should be added in 200ml of 1.02N - HCl solution to reduce its concentration to 1.0N ? (Assume no change in volume of solution on addition of magnesium) Sol : neqHCl taken = neqMg taken + neqHCl remained

200 w 200  1.0 1.02   24 1000 1000 2

 

 w = 0.048gm = 48mg Example - 6 : A certain amount of a mixture of H2C2O4 and NaHC2O4 requires equal volumes of 0.01M - KMnO4 and 0.2M - Ca(OH)2 solutions, in different experiments. Calculate the mole ratio of H2C2O4 and NaHC2O4 in the original mixture. Sol : Let ‘a’ mole of H2C2O4 and ‘b’ mole of NaHC2O4 are present in the original mixture with KMnO4. neqKMnO4 = neqH2C2O4 + neqNaHC2O4

0.1 5 

V  a 2  b2 1000

0.5V  2a  2b............. 1 1000

3

4

C 2 O 24    C O2 [x - factor = 2 ] with Ca(OH)2 : neqCa(OH)2 = neqH2C2O4 + neqNaHC2O4 (acid - base reactions)

0.2  2 

V  a  2  b 1 1000 Sr. INTER

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CHEMISTRYv 

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STOICHIOMETRY

0.4  V  2a  b...........  2  1000

On solving (1) and (2) ; a : b = 3 : 2 II. One substance reacts with more than one substance, Back titration, etc : CAPS-73. Example - 1 : 20ml of 0.2N - NaOH solution is mixed with 40ml of 0.3N - HCl solution. The excess acid requires 50ml of Na2CO3 solution for complete neutralization. Calculate the strength of Na2CO3 solution in gm/ lit. Sol :As HCl is reacting with NaOH as well as Na2CO3 solution.

neq HCl  neq NaOH  neq Na2CO3

40  0.3  20  0.2  50  N  N = 0.16 N = 0.16 143  22.88 gm / Lt Example - 2 : 0.03g of a metal is dissolved in 200ml of M/10 - H2SO4 solution. When the metal is dissolved completely, the excess acid required 50ml of

3M NaOH solution for complete neu4

tralization. Determine the equivalent weight of the metal. Sol : As H2SO4 solution is reacting with metal as well as NaOH solution.

neq H 2 SO4  neq metal  neq NaOH

200  0.1 2 0.03 50  0.75  1   1000 E 1000  E  12 Example - 3 : 1.0gm of an equimolar mixture of carbonates of two bivalent metals requires 15ml of 0.5N - HCl solution for complete neutralization. If one metal is calcium, what is the atomic weight of the other metal. Sol : Let the weight of CaCO3 present is x gm. Then the weight of other metal carbonate MCO3, is (1-x)gm. From question, moles of both carbonates are equal. Hence :

x 1 x  ........... 1 100 A  60 where A = at weight of metal ‘M’

neq CaCO3  neq MCO3  neq HCl

x 100

2

1  x   A  60 

15  0.5 1000

2

 From (1) and (2) : A = 1067. III. Dilution Based Problems : CAPS-74. Example -1 : 20ml of N - HCl solution is diluted to 100ml. 25ml of this diluted solution requires 40ml of NaOH solution for complete neutralization. Calculate the normality of NaOH solution ? Sol : From law of dilution : 20 x 1 = 100 x N2

 N2 = 0.2N ; neq HCl  neq NaOH Sr. INTER

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STOICHIOMETRY

25 x 0.2 = 4 0 x N  N = 0.125N Example -2 : 20ml of H2O2 solution is diluted to 50ml. 25ml of this diluted solution requires 20ml of 0.1N - KMnO4 solution for complete reaction. Calculate the volume strength of original H2O2 solution. Sol : nmeq H 2O2  nmeq KMnO4 25 x N = 20 x 0.1

 N H 2O2  0.008 N From law of dilution on H2O2 20 x N1 = 50 x 0.08  N1 = 0.2N = 0.2 x 5.6 = 1.12vol Example - 3 : 2.61gm sample of pyrolusite was boiled with 65ml of N-oxalic acid in presence of excess of dil. H2SO4. The liquid was then filtered and residue is washed. The filtrate and the washings were mixed and made upto 500ml. The solution requires 250ml of N/10 - KMnO4 for complete oxidation of excess oxalic acid. Calculate the percentage of pure MnO2 in the pyrolusite sample (Mn = 55). Sol : Here, oxalic acid is reacting with pure MnO2 as well as KMnO4. Hence,

neq oxalic acid  neq MnO2  neq KMnO4

250  1 65  1 w 10  w  1.74 gm   87 1000 1000 2  % of MnO2 inthe sample 

1.74 100  66.67% 2.61

Example - 4 : 200gm of a sample of KI is dissolved in 20ml of 1.2M - KMnO4 solution in acidic medium. After complete reaction, the solution is diluted to 150ml. 25ml of this diluted solution requires 40ml of 0.5M - K2Cr2O7 solution for complete reaction with excess KI. Calculate the % purity of KI sample (K = 39, I = 127). Sol: Here, KI will react with both KMnO4 & K2Cr2O7. Hence,

neq KI  neq KMnO4  neq K 2Cr2O7

w 166 1

 40  150   0.5  6 20 1.2  5  25    1000 1000

 w  139.44 gm 1

K I   I 20

n 

fa c to r  1 

7

M n O 4    M n 2 6

C r 2 O 7 2    C r 3

[The factor

n

 fa c to r  5 

n 

fa c to r  6 

40  150 will represent the total volume of K2Cr2O7 needed to react completely with all KI 25

left in 150ml solution] Sr. INTER

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CHEMISTRYv

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STOICHIOMETRY

139.44  100  69.72% 200

IV. Reactions in succession : CAPS-75. Example - 1 : 50ml of an aqueous solution of H2O2 was treated with an excess of KI solution in dil H2SO4. The liberated I2 required 20ml of 0.1N - Na2S2O3 solution for complete reaction. Calculate the concentration of H2O2 solution. Sol : neq H 2O2  neq I 2  neq Na2 S 2O3

 50  N  20  0.1  N  0.04 N Example - 2 :Hydroxylamine reduces iron (III) according to the equation, 4 Fe3  2 NH 2 OH   N 2 O  H 2O  4 Fe 2   4 H 

Iron (II) thus produced is estimated by titration with standard KMnO4 solution. The reaction, is M nO 4  5 Fe 2   8 H

  M n 2  5 F e 3  4 H 2O

A 10ml of hydroxylamine solution was diluted to one litre. 50ml of this diluted solution was boiled with an excess of Fe3+ solution. The resulting solution requires 12ml of 0.02M - KMnO4 solution for complete oxidation of Fe2+. Calculate the weight of NH2OH in one litre of original solution. Sol : Let the normality of original solution is xN. From, dilution formula : 10 x x = 1000 x N2

 N 2  0.01xN

neq NH 2OH  neq Fe2  neq KMnO4 

50  0.01x 12  0.02  5   x  2.4 1000 1000

 Weight of NH2OH per lt = N x eq.wt = 2.4  V.

33  39.6 gm 2

Double Indicator : When Na2CO3 is titrated against HCl (strong acid) the reaction takes place in two steps.

Step - I : Na2CO3  HCl   NaHCO3  NaCl Step - II : NaHCO3  HCl   NaCl  CO2  H2O

 Since, it is a titration between weak base and strong acid, methyl orange is perfect indicator. It will cause sudden change in colour at complete neutralization of Na2CO3, i.e, end of step II also. Hence, n-factor for Na2CO3 with methyl orange indicator will be 2.  But, if the titration is performed in presence of phenolphthalein, it will cause sudden change in colour at the completion of step 1 only. Hence, n-factor for Na2CO3 will become 1.  From the values of equilibrium constants, it may be clearly explained that step-II start only after completion of step-I. CAPS-76. Example - 1 : When a mixture of NaOH and Na2CO3 is treated with 0.2M - HCl solution, it required 40ml solution using phenolphthalein indicator. When the same amount of original mixture is titrated against same HCl solution using methyl orange indicator, it required 50ml acid solution. Calculate the % composition of original mixture by weight. Sr. INTER

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STOICHIOMETRY

Sol : Let the original mixture contains ‘a’ mole NaOH and b mole Na2CO3. When, mixture is titrated in presence of phenolphthalein,

neq HCl  eq NaOH  neq Na2CO3

40  0.2 1  a  1  b 1 1000

 a + b = 0.008 .............. (1) when, mixture is titrated in presence of methyl orange : neq NaOH  neq Na2CO3  neq HCl

 a 1  b  2 

50  0.2 1 1000

 a + 2b = 0.01 .............. (2) From (1) and (2), b = 0.002 ; a = 0.006 Hence, NaCl = 0.006mole = 0.006 x 40 = 0.240gm Na2CO3 = 0.002 mole = 0.002 x 106 = 0.212gm  % NaCl =

0.240  100  53.1% 0.452

 Na2CO3  46.9% Example - 2 : A certain amount of a mixture of NaOH and Na2CO3 is titrated with 0.5N - H2SO4 solution. When phenolphathalein is used as indicator, the volume of H2SO4 solution used for end point is 50ml. Then, methyl orange is added and titration is continued for the next end point, with the same acid solution. If the volume of acid solution used is 20ml, determine the masses of NaOH and Na2CO3 in the original mixture. Sol : Let the original mixture contains ‘a’ mole NaOH and ‘b’ mole Na2CO3 when, phenolphthalein is used

neq NaOH  neq Na2CO3  neq H 2 SO4

 a 1  b 1 

50  0.5 1000

 a  b  0.025........... 1 when, methyl orange is added, all NaOH reacted as well as all Na2CO3 is converted into NaHCO3.

 H2SO4 will be required only for NaHCO3 formed. Hence, neq NaHCO3  neq H 2 SO4

 b 1 

20  0.5 1000

 b = 0.010

and from (1) a = 0.015.  Mass of NaOH taken = 0.015 x 40 = 0.60gm and mass of Na2CO3 taken = 0.010 x 106 = 1.06gm VI. Hardness of Water :  The property of water to prevent lather formation with soap or detergent is called hardness of water. It causes due to presence of all metal salts in water, other than alkali metal salts. Types of hardness : 1. Temporary or Bicarbonate Hardness :  It is the hardness which may be removed by boiling the water sample. It is due to presence of bicarbonates Sr. INTER

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of hardness creating salts.  Ca  HCO3 2   CaCO3  CO2  H 2O

2.

Permanent Hardness :

 It is the hardness which may be removed only by chemical treatment. It is due to all hardness creating salts other than bicarbonates.  Different methods to remove hardness are limesoda process, zeolite process, ion - exchange method etc. Expression of Hardness : Hardness is expressed in ppm of CaCO3 equivalent to all hardness creating salts. CAPS-77. Example - 1 : 100ml of a sample of tap water requires 50ml of N/50 HCl solution for titration. Calculate the temporary hardness of tap water. Sol : HCl is required to react with bicarbonates present in water. Hence,

neq HCl  neq metal carbonates  neq CaCO3

50  1

50  w 1000 50

 w  0.05 gm  Temporary Hardness =

0.05  106  500 ppm of CaCO3 100

% of oleum and % of SO3 in oleum :  Oleum is a mixture of H2SO4 and SO3 obtained in contact process for the synthesis of H2SO4.  When oleum is diluted with water, SO3 present in the mixture is converted into H2SO4. In this way, the amount of H2SO4 obtained from the given amount of oleum is more than the amount of oleum taken due to addition of water to form H2SO4. If 100g sample of oleum forms 118g H2SO4, % oleum is said to be 118% and 18g of H2O reacts with SO3 present in oleum. With the help of above information, we can calculate the amount of SO3 present in 100g sample of given oleum. Eg : In 118% oleum 18g water reacts with SO3 according to the given reaction to form H2SO4. SO3 + H2O H2SO4 1mol 1mol 0 0 0 1mol 18g H2O, reacts with 80g (1 mole) of SO3 thus in the above sample of oleum, 100g oleum contains 80g SO3 or % of SO3 = 80. NOTE : Theoretically % oleum can’t be equal to or more than 122.5% as for this value whole 100g mass should be of SO3. Which can’t be considered as oleum. CAP-78. Calculate the % of oleum and % of SO3 in the same oleum whose 40g after dilution requires 1L of 0.977N - NaOH for neutralisation. Sol : Meq of NaOH required = 1000 x 0.977 = 977 meq. of H2SO4 formed from 40g oleum after dilution = 977 =

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= 47.878g

40g oleum forms - 47.873g H2SO4 100g oleum forms = = 119.63 H2SO4 Hence, the % of oleum is labelled as 119.68% Amount, of water added to 100g oleum = 19.68g as,. 18g of water reacts with 80g SO3. Determination of available Cl2 in Bleaching powder : Available Cl2 : with excess of acid, bleaching powder liberates Cl2 which is called available chlorine.

Cl2 is liberated with CO2 also Determination of available Cl2 in bleaching powder can be done by iodometric method. Iodometric method : The Cl2 liberated when bleaching powder is allowed to react with dil HCl or dil CH3COOH is known as available chlorine. For estimation of available chlorine, a weighed quantity of bleaching powder is treated with acetic acid and KI and the Cl2 liberated reacts with KI and liberates equal equivalents of I2. The amount of I2 liberated is estimated by treating with standard solution of sodium thiosulphate. The reactions are as follows : i) ii) iii) Let ‘w’g bleaching powder is dissolved in Vml water V1 ml of this solution is treated with excess of CH3COOH / KI and a few drops of KI are added in it. The I2 liberated is titrated with N-hypo solution. At the end point, blue colour of solution disappears. Let V2ml of hypo are used, then available Cl2 may be estimated as follows : Amount of bleaching powder in Vml solution meq of Cl2 in V1 ml solution = meq of I2 = meq of hypo = N x volume (in mL) = n x M x V2 In VmL solution, meq of Cl2 =

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In Vml, amount of Cl2 =

Cl2 obtained from wg bleaching powder =

Cl2 obtained from 100g bleaching powder =

% available Cl2 = Commercial samples of bleaching powder have 33 - 38% Cl2. CAP-79. 3.546g bleaching powder is dissolved in 100ml water. 25ml of this solution were treated with excess of CH3COOH and KI and then treated with 0.125N hypo solution. In this titration, 25ml hypo were used. Calculate the % of available Cl2 in this sample of bleaching powder. Amount of bleaching powder (w) = 3.546g Volume of solution (V) = 100mL Volume of solution taken for titration (V1) = 25ml Volume of hypo used (V2) = 25mL Normaltiy of hypo solution (N) = 0.125 % of available Cl2 = = 12.51% Some important redox titrations in volumetry: 1.

Estimation of

2.

Estimation of

Eq. mass of

Eq.mass of 3.

Estimation of

present in pyrolusite: ,

(or) Eq. Mass of 4.

. ,

Estimation of available chlorine in bleaching powder: ; Eq: mass of

5.

Estimation of

6.

Estimation of ozone:

7.

Estimation of

, Eq.mass of ; Eq.mass of Eq. mass of Sr. INTER

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Estimation of

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Estimation

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Eq.mass of Eq.mass of

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PART-9 (Significant Figures) There is always some degree of uncertainty in every scientific measurement except in counting. The uncertainty in measurement mainly depends upon two factors: (i) Skill and accuracy of the observer, (ii) Limitation of the measuring scale. To indicate the precision of a measurement, scientists use the term significant figures. The significant figures in a number are all the certain digits plus one doubtful digit. The number of significant figures gives the information that except the digit at extreme right, all other digits are preciese or reproducible. For example, mass of an object is 11.24g. This value indicates that actual mass of the object lies between 11.23g and 11.25g. Thus, one is sure of first three figures (1, 1 and 2) but the fourth figure is somewhat inexact. The total significant figures in this number are four. The following rules are observed in counting the number of significant figures in a given measured quantity: (i) All non-zero digits are significant. For example, 42.3 has three significant figures. 243.4 has four significant figures. 24.123 has five significant figures. (ii) A zero becomes significant figure if it appears between two non-zero digits. For example. 5.03 has three significant figures. 5.604 has four significant figures. 4.004 has four significant figures. (iii) Leading zeros or the zeros placed to the left of the number are never significant. For example, 0.543 has three significant figures. 0.045 has two significant figures. 0.006 has one significant figures. (iv) Trailing zeros or the zeros placed to the right of the number are significant. For example, 433.0 has four significant figures. 433.00 has five significant figures. 343.000 has six significant figures. (v) In exponential notation, the numerical portion gives the number of significant figures. For example. has three significant figures. has three significant figures. (vi) The non-significant figures in the measurement are rounded off. (a) If the figure following the last number to be retained is less than 5, all the unwanted figures are discarded and the last number is left unchanged, e.g., 5.6724 is 5.67 to three significant figures. (b) If the figure following the last number to be retained is great than 5, the last figure to be retained is increased by 1 unit and the unwanted figures are discarded, e.g., 8.6526 is 8.653 to four significant figures. (c) If the figure following the last number to be retained is 5, the last figure is increased by 1 only in case it happens to be odd. In case of even number the last figure remains unchanged. 2.3524 is 2.4 to two significant figures. 7.4511 is 7.4 to two significant figures. Sr. INTER

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