CHEMISTRY
v v v v v v v v v v v v v v v v v v v v v v v vSTOICHIOMETRY
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 : 2.
E
m C
)
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 CaCO33 2
CaCO CaOCO
ii) by heating NaHCO3
3 2 3 2 2
2NaHCO Na CO H OCO
iii) by burning in O2
2 2
CO CO
iv) by reaction of CaCO3 with HCl.
CaCO32HClCaCl2H O2 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 differentexperimental 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 B11.88 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 com-bined 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 :
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.
ii) 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.
H
S
O
2g
32g
16g
H
2S
H
2O
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
17.62 part N.
for H2O :
1 part H reacts with = 88.90 8.01
11.10 part O.
Thus, the ratio N : O :: 4.67 : 8.01
0.58. As the two ratio are same, thus law of reciprocal proportions are correct. 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) 2 3 2 2 3
1 3 2
g g g
N H NH
vol vol vol
ii) 2 2
1
2
1
1
1
2
g g gCO
O
CO
vol
vol
vol
iii) 2 2 2
1 1 2
g g g
H Cl HCl
vol vol vol
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 : 2C H2 25O2 4CO22H O2
2vol 5vol 4vol
40mL
5
2
x40mL4
2
x40mL 40mL 100mL 80mLSo, 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 num-bers and differ in chemical properties (Discovery of isobars)
Atoms of one element can be transformed into atoms of other element (Discovery of artificial transmuta-tion)
There are number of compounds which do not follow the law of constant proportions. Such, compounds are called non - stoichiometric compounds.Atoms and Molecules :
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.
They accepted the stable isotope of carbon
C12
with mass number of 12 as the standard.ATOMIC MASS UNIT (amu)
“The quantity, 112 mass of an atom of carbon -12
12C ” is known as the atomic mass unit and is abbreviated as amu.
The actual mass of one atom of carbon - 12 is 231.9924 10 g (or)
1.9924 10
26kg
1 amu = 1 1.9924 10 23 1.66 1024 1.66 1027 12 g kgATOMIC 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 112th part of the mass of one atom of
C-12”.
Atomic mass of an element 1 12
12
Mass of one atom of the element Mass of one atom of C
Atomic mass of an element
1
Mass of one atom of the element amu
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 amuACTUAL 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
24g
eg:- 1) The actual mass of H-atom = 24 241.008 1.66 10
1.6736 10
g
2) The actual mass of O-atom = 24 2316 1.66 10
2.656 10 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 37 25 35.5
100 100 amu
So, based on the average mass, atomic mass of chlorine is 35.5 amu
Average isotopic mass100 100
x y
a b
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. 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 ofC
13 1.1
Average atomic mass = 12 98.9 13 1.1
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
100
10.8 10
11
100
100
x
x
20%; 80% x y 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 112th of the mass of one atom of carbon 12.
Molecular mass =
Mass of one molecule of the substance 1
th mass of one atom of carbon 12
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 ofH SO
2 4 consists of 2 atoms of hydrogen, one atom of sulphur and four atoms of oxygen. Molecular mass of
H SO
2 4
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 24 2332 1.66 10
g
5.32 10
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 23 6.023 10 . This number is known as “Avogadro’s number”
N
A
, named in the honour of Amedeo Avogadro (1776-1856)Note : Mass of one atom of C-12 isotope 23
1.9924 10
g
Number of atoms present in 12g of C-12 23 12 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
O
2 is 32 amu
gram molecular mass ofO
2 = 32g. iii) Molecular mass of nitric acid(
HNO
3)
= 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 is32 1.66 10
29g
Gram molecular mass (or) one gram molecule of any substance contains 236.023 10 molecules. Calculating
a. Number of gram atoms: Number of gram atoms
Given weight of the element gram atomic mass
b. Number of atoms present in the given amount of the element
A
Given weight of the element N gram atomic mass
= no.of gm atoms
A
N
c. Mass of an element
Given weight of the substance gram atomic mass
e. Number of gm. molecules A
Given weight of the substance N gram atomic mass
f. Mass of a substance = No.of gm. molecules gm. molecular mass. CAPS-8
1. Calculate number of gram atoms in:
a. 48 gm of O2 b. 8 gm of H2
Solution:
a. one gm. atom of oxygen - 16 gms
? - 48 gms No.of gm atoms = 48 16=3 gm. atoms 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 23 23 6 1 6 10 1.5 10 24 NA 4
b. 16gm of oxygen contains - NA of atoms (1 gm atom)
4 gm of oxygen contains - ?
23 23
4 1
6 10 3 10
8NA 2 atoms
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
4. Calculate number of gm molecule in a. 90 gm of
C H O
6 12 6
180
gm
? - 90 gm 90 0.5 180 gm molecule. b. 71 gm 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 32NA of molecules 23 1 6 10 4 23 1.5 10 moleculesb. 44 gm of CO2 contains - NA of molecules (1 gm molecule)
88 gm of CO2 contains - 88 44NA of molecules 23 1 6 10 2 23 3 10 molecules
6. Calculate the mass of
a. 2.5 gm molecules of CH4 b. 1.2 gm molecules of C12H22O11 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.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 23 6.023 10 .
Number of atoms in 12g of C - 12
12 Mass of an atom of C 12 23 12 1.9924 10 23 6.023 10
This number is also known as Avagadro’s number named in honour of Amedeo Avogadro (1776-1856).
Thus, a mole contains 236.023 10 units. These units can be atoms, molecules, ions, electrons or anything
else. Examples:
i) 1 mole of H-atoms means 23
6.023 10 H-atoms.
ii) 1 mole of
CO
2 molecules means 6.023 10 23 CO2 molecules. iii) 1 mole of electrons means, 236.023 10 electrons. iv) 1 mole of ions means, 23
6.023 10 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
24g
. Mass of one mole Aluminium = 24 23
27 1.66 10
6.023 10
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 236.023 10 molecules of a substance is equal to its molecular mass in grams or gram - molecular mass or gram molecule.”
For Example : Molecular mass of
CO
2 is 44 amu Mass of one mole of CO2 44 1.66 1024 6.023 1023g 44g
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 :Number of moles (n) =
Mass of substance W in g
M g-atomic or g-molecular mass
M = atomic mass of element or formula mass of ionic solids or molar mass of compound.
Eg: ; 2 ; 4
58.5 18 16
NaCl H O CH
W W W
n n n
b) When number of entities are given:
Number of moles (n) = 23
A
number of entities present
6.023×10
i.e Avagadro constant, N
.Eg: 2 2 CO A number of CO molecules n = ; N Na A number of Na atoms n = N
c) For gaseous substances:
Number of moles (n) = Volume of gas at S.T.P in Lt
Eg: 2 Volume of O at S.T.P in Lt2 22.4 O n ; 4 Volume of 4at S.T.P in Lt 22.4 CH CH n
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 massEg: 1 mole of
H
2
2
g
1 mole ofO
2
32
g
1 mole of
SO
2
64
g
1 mole ofCO
2
44
g
1 mole ofCl
2
71
g
1 mole ofCH
4
16
g
1 mole ofH O
2
18
g
1 mole ofCaCO
3
100
g
1 mole ofC H O
6 12 6
180
g
1 mole ofC H O
12 22 11
342
g
1 mole of NaOH 40g
2g ofH
2, 32 g ofO
2, 64 g ofSO
2, 44 g ofCO
2,
71 g ofCl
2, 28 g ofN
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
0.5
mol
.
1mol
ofC H O
6 12 6 contains
6
N
A of ‘C’ - atoms
12
N
A of ‘H’ - atoms
6
N
A of ‘O’ -atoms.
Number of ‘C’ - atoms 23 24 6 6.023 10 3.6 10 Number of ‘H’ - atoms 23 24 12 6.023 10 7.2 10 Number of ‘O’ - atoms 23 24
6 6.023 10 3.6 10
2. Calculate number of electrons present in 4.5 g of
H O
2?
Sol. Number of moles 4.5 0.25 .
18 mol 18g of
H O
2 contains______10
N
A of electrons 4.5g ofH O
2 contains______
? 23 24 4.5 10 of electrons 18 =2.5 6 10 1.5 10 electrons. A N 3. Calculate the charge present on 1 mole of electrons. Sol. Charge on one electron = 19
Charge on one mole of electrons 19 23 1.609 10 6.023 10 96, 500 coloumbs =1 Faraday.
4. Calculate the number of ions present in 1.11 g of
CaCl
2 Sol : CaCl2 Ca22Cl1mol 1mol 1mol 111g NA 2NA
111 g of
CaCl
2 contains -3
N
A of ions1.11 g of
CaCl
2 contains -0.03
N
A of ions2 23 22
3 10 6 10 1.8 10
ions.
5. Calculate number of molecules present in a drop of water of volume 0.01mL? (density of
H O
2 is 1g/ mL)sol : Mass of 1 drop of
H O
20.001 1
0.01g 18 g of
H O
2 contains -N
A of molecules. 10 g
2 of 2H O
contains - ? 2 23 2010
6 10
3.3 10
18
molecules.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
Molecu-lar 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 100M
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 :2
H O
% 2 100 11.12% 18 H ; % 16 100 88.89% 18 O ii) In CaCO3 : 3 100 CaCO M 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 3.4 g CAPS - 11 :
Chlorophyll contains 2.68% of Mg by mass. Calculate the number of Mg atoms in 2g of chloro-phyll ?
Sol : 100g of chlorophyll contains ... 2.68g of Mg
2.68 24 mol of Mg 2g of chlorophyll contains .
2.68 2
24 100
mol of Mg = 3 2.23 10 Mole of Mg No. of Mg atoms = 3 23 2.23 10 6.023 10 = 21 1.345 10 atoms of Mg CAPS - 12 :Calculate the percentage composition of : MgSO4 . 7H2O Sol : MgSO4 . 7H2O =
24 96 7 18
= 246. i) % of Mg = 24 100 9.75% 246 ii) % of S = 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 Determination of chemical formula from percentage compsotion :
Types of chemical formulae
Emperical formula (E.F)
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 C2H4O2 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 % composition Mole ratio = % atomic mass Simple mole ratio Conversion into whole number E. F Write the symbols of elements present in the molecule 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 If simple mole ratio are fractional, thus multiply all with same number to
convert into whole no. or nearly whole no
Write the whole no.s obtained as subscript with the symbols of
elements
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 %
composition Mole ratio
Simple mole ratio Whole no. ratio E. F Fe 69.94 2 O 30.06 3 Fe2O3 69.94 1.25 55.85 1.25 1 2 1.25 30.06 1.88 16 1.88 1.5 2 1.25 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 M.F = E .F x n
n = 6
M. F = C6H122. 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 2. 2 BaI xH O
(10.407 )
g
2(9.520 ) BaI g 2 (0.887 ) xH O g 2 2Mole of H O
x
Mole of BaI
(or)0.887/18
2 9.520/ 391 x x
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.
% 2
2
100
18
weight of H O
H
weight of organic compound
% 2 1 2 1 0 0 1 4 w eig h t o f C O C w eig h t o f o rg a n ic co m p o u n d II. Estimation of Nitrogen : i) Duma’s method : % 2 28 100 22, 400 volume of N at STP N
weight of organic compound
ii) Kjeldhal’s method :
% 1 .4 N V N w e ig h t o f o r g a n ic c o m p o u n d
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.
iii) Estimation of Sulphur :
% 32 4 100
233
weight of BaSO S
weight of organic compound
iv) Estimation of halogens (Carius method)
% Cl =
35.5
100 143.5
w eight of A gC l
w eight of organic com pound
% 80 100 188 w eight of AgBr Br
w eight of organic compound
% 127 100 235 weight of AgI I
weight of organic compound
v) Estimation of Phosphorus : % 2 2 7
62
100
222
weight of Mg P O
P
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 com-pound gave 56mL N2 at STP. If the molecular weight of the compound is 59, then find the molecular formula of the compound.
Sol : % of 2 12 100 44 weight of CO C
weight of organic compound
= 12 0.528 100 61.07% 440.236 % of 2
2
100
18
weight of H O
H
weight of organic compound
= 2 0.324 100 15.25% 180.236 % of 228
100
22, 400
Volume of N at STP
N
weight of organic compound
28
56
100
23.73%
22, 400
0.295
Element Percentage Relative no. of atoms Simplest ratio C 61.07 5.09 3 H 15.25 15.25 9 N 23.73 1.695 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.
Sol : % . 49.3 146 6 100 . 100 12 Mol Mass C At mass % . 6.84 146 10 100 . 100 1 Mol Mass H At mass % . 43.86 146 4 100 . 100 16 Mol Mass O At mass 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 Relative no of atoms Simplest ratio Carbon 92.30 12 7.69 1 Hydrogen 7.70 1 7.70 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 unbal-anced.Eg :
H
2
O
2
H O
2
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 “Stoichio-metric coefficients”.
All the chemical equations can be treated as algebraic equations and coefficient 1 is not written. Eg : 12H2 12C l2 HC l(or) H2Cl22HCl
(or) 2H22Cl2 4HCl
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 chemi-cal 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 asaA bB
cC
dD
This equation gives information about i) Mole ratio :
a mole Ab mole Bc 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
dMD
g of DMA , 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 Cd vol D
Consider for example, the reaction represented by a balanced chemical equation
N2 g 3H2 g 2NH3 g Mole
ratio
1mol 3mol 2mol Molecule ratio Mass ratio 1xNA 1 molecule 28g 2xNA 3 molecules 6g 2xNA 2 molecules 34g Volume ratio
1vol 3vol 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 Analysisb) 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.
CAP-16. Calculate the weight of CO2 formed by complete combustion of 1.5 gm ethane. Sol :
C H
2 6
7
2
O
2
2
CO
2
3
H O
21 mol 2 mol
= 30 gm 2x44 gm
30gm C2H6 produces 2x44gm CO2 on complete combustion.
1.5gm C2H6 will produce 2 44 1.5 4.4
30 gm
CO2 2. 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
1.5 2 1 4.4 30 44 w w gm
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
4.4
30
44
14
7
w
w
gm
[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,
2 2 2 1 2 1 1 (2 36.5) (24 71) 73 95 M gO H C l M gC l H O m ol m ol m ol m ol g g g g 73g HCl produce 95g MgCl2 17g HCl produce ? 95 17 22.12 73 g
CAP-18. How many grams of oxygen are required to burn completely 570g of octane ? Sol : Balanced equation
8 18 2 2 2
2C H 25O 16CO 18H O
2mol 25mol
2x114g 25x32g
570
25 32 2000
2 114 g
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 : 4 3 1 1 53.5 1 g S g NH Cl NH HCl mol mol g mol 53.5g NH4Cl will give 3 1 26.25 53.5 mole NH = 0.5 mole From, PV = nRT1
V
0.5 0.0821 300
V = 12.315litCAP-20. Calculate the volume of H2 liberated at 27°C and 760mm pressure by treating 1.2g of Mg with excess of HCl. Sol :
2
2 21
1
24
22.4
Mg
HCl
MgCl
H
mol
mol
g
L at STP
24g Mg liberates 22.4 L of H2 1.2g Mg will liberate 22.4 1.2 1.12 24 L
Volume of H2 under given conditions can be calculated by applying : 1 1 2 2 1 2 PV PV T T P1 = 760mm P2 = 760mm T1 = 273K T2 = 27 + 273 = 300 K V1 = 1.12 L V2 = ? V2 = 760 1.12 300 273 760 =1.2308 LIII. Calculations based on volume - volume relationship These calculations are based on two laws :
i) Avagadro’s law ii) Gay - Lussac’s law For example :
2 2 3 1 22.4 3 2 ' 1 3 2 3 22.4 2 22.4 g g g L N H NH Avagadro s lawmol mol mol
L L
Under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes.
2 3 2 2 3
1 3 2
g g g
N H NH
vol vol vol
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 ?
Sol : 2 2 1 1 12 22.4 g g s C O CO mol mol g L 1000g ? 1000 22.4 12 lit 100L of air contains 21L of O2 ? 2 1000 22.4 12 L of O 1000 22.4 100 8888.5 12 21 L of air
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 ?
Sol : C H +SO3 8 2 3CO +4H O2 2 1vol 5 1 5 vol L L 1L of propane requires 5L of O2 20L of propane will require
2
5 20
100
L of O
at 760mm and 27°C.
This volume will be converted to STP conditions. Given conditions STP conditions P1=760mm P2=760mm V1=100lit V2 = ? T1 = 27 + 273 = 300K T2 = 273 1 1 2 2 2 1 2 760 100 273 91 300 760 PV P V V L T TDifferent 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.
Sol :
2
2 4 2 42
22
1
2 40
98
NaOH
H SO
Na SO
H O
mol
mol
gm
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 : 2KClO32KCl3O2 2mol 3mol 2x122.5gm 3x22.4L at STP 4.9 gm ? 4.9 3 22.4 1.344 2 122.3 L
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 reac-tant.
CAP-25. How many grams of NH3 will form by combination of 3gm H2 and 7gm N2 ? Sol : N23H2 2NH3
1mol 3mol 2mol 28gm 3x2=6gm 2x17=34gm
28gm N2 requires 6gm H2 for complete r x n 7 gm N2 requires < 3.0gm Hence, N2 is the limiting reagent.
Now, 28gm N2 will produce 2x17gm NH3
7 gm N2 will produce 8.5gm NH3
CAP-26. What volume of Cl2 gas will liberate at STP, when 2.61gm MnO2 is reacted with 2.92gm HCl ? Sol : MnO24HClMnCl2Cl22H O2
1mol 4mol 1mol 879gm 4x36.5g 22.4L
87gm MnO2 requires 4 36.5g HCl for complete reaction.
HCl is the L. R
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) 2KClO32KCl3O2 2mol 3mol 2x122.5g 3x32gm 49g 3 32 49 19.2 2 122.5 gm ii) C H2 43O2 2CO22H O2 1mol 3mol 28g 3x32g ? 1.92g 19.2 28 5.6 3 32 gm C2H4
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.
3 2 2 4
2
mol KClO
3
mol O
1
mol C H
2 x 122.5g 28gm 49g 28 49 5.6
2 122.5 g
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 :
4 2 2 2 4 4 2 4
2KMnO 5SO 2H OK SO 2MnSO 2H SO
2 2 2 3 2
4FeS 11O 2Fe O 8SO
Sol:
2
mole KMnO
.
4
5
mol SO
2
5
2
mol FeS
2
2x158g KMnO4 = 5 120 2 gm=300gm 31.6g KMnO4 2 5 60 31.6 30 2 158 gmFeS 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 CO 2
1mol 1mol 100g 44g ? 0.44g 100 0.44 1 44 gm
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 : 6NaOH 3Cl25NaClNaClO33H O2 6 mol 3mol 6 x 40 g 3 x 71 g 6 40 21.3 3 71 21.3g 24 g
The weight of NaOH sample required = 24 100 30 80 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 :100
Actual yield Percent yield
theoretical yield
CAP-31. When 3.9g Al(OH)3 is reacted with excess of HCl, 6.50gm AlCl3 is formed. Determine the percent-age yield of product ?
Sol : Al(OH)3 + 3HCl AAlCl3 + 3H2O
1mol 1mol 78gm 133.5g 3.9gm 3 133.5 3.9 6.675 78 gm AlCl
But, the amount formed is only 6.5 gm
Percentage yield = 6.5 100 97.38%
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 2.02 kg 2 6 30 2.02 0.6 2 50.5 kg C H
But, the yield of ethane is only 80% and hence, the amount of ethane produced is
80
0.6 0.48 100 kg
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 8 4 0.66 2.5 2 44 g NaHCO 3
Hence, % composition of the mixture is :
3 2 .52 1 00 63 % 4 N a H C O
100 63 37%
NaCl
CAP-34.2g of a mixture of CaCO3 and MgCO3 requires 2g of H2SO4 for complete reaction. Determine the per-centage composition of the original mixture.
Sol :
xg
(2
x g
)
3CaCO
MgCO
3}
2g
3 2 4 4 2 2 CaCO H SO CaSO CO H O 1mol 1mol 100g 98g xg 98 100x3 2 4 4 2 2 MgCO H SO MgSO CO H O 1mol 1mol 84g 98g
2
x g
98
2
84 x ghence, the total weight of H2SO4 required = 98 98(2 ) 2
100x84 x g x = 1.78
% CaCO3 = x/2100 = 89%
MgCO3 = 11%1%PART - 5
(EUDIOMETRY)
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 :
1. 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.
2 g 3 2 g 2 3 g
N H N H
1vol 3vol 2vol
Problem may be solved directly in terms of volume, in place of mole.
2. 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 + ...
3. The volume of solids or liquids is considered to be negligible in comparison to the volume of gas.
2 2 2
2H g O g 2H O l
2mole 1mole 2mole
2vol 1vol 0vol
4. Air is considered as a mixture of oxygen and nitrogen gases only. 5. 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?
Sol : 2 2 2 2 2 5 2 2 C H O CO H O 1vol. 5 2vol. 1vol C2H2 requires 5
40 ml C2H2 will require 5
2x40 = 100ml O2
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 : C H2 43O2 2CO22H O2
1vol 3vol 2vol 0vol 20 ml 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.
Sol : CO2(g )C(s) 2CO(g ) 1vol 2vol 600 ml 1200 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.
2 g s
2
gC O
C
C O
1vol 2vol
x ml 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 percent-age, by volume, of ozone, decomposed into oxygen.
Sol : Let only x ml of ozone decomposed
3 2
2O 3O
2vol 3vol
x ml 3/2x ml
Increase in volume = vol. of product
vol. of reactant Or, 10 32x x
x = 20
Hence, percentage decomposition of
3 20
O = ×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 +2O4 2 CO +2H O2 2
1vol 2vol 1vol 0vol 30ml 60ml 30ml 0
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, CO2,
SO2, SO3, etc
2 Alkalin e pyro gallo l so l. O2
3 Tur pe ntin e oil O3
4 Co n c. H2SO4, P2O5 etc M o istu re
5 H eated m e tal like Pd , Pt, etc H2
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 solu-tion, 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.
2 2 2 2 2
5
C H + O 2CO +H O 2
1vol. 5
2vol 2vol 0vol
x ml 5
2x ml 2x ml 0
2 2
2 C O + O 2CO
2vol 1vol 2vol
(40 - x)ml 1
2(40 - x) (40 - x)ml
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
10.410052x12
40x
2x
40x
5 1 10.4 100 2x 2 x = 16Method - 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
