http://www.goiit.com/posts/list/community-shelf-contribution-on-stoichiometry-12765.htm
Atomic weight of mass. A single atom or molecule is too small to weigh, even on the most sensitive balance made. For example, an atom of hydrogen has been found to weigh 1.67 x 10^¼¾g. Most atoms have a radius of 0.07 to 0.20 nanometer (nm), one nanometers is 10^9 m or 10A.
The actual weights of atoms or molecules, in grams, are cumbersome to use because of their extremely small magnitude. The relative weights of atoms are less cumbersome to use. By relative weight is meant the number of times heavier an atom is than another atom taken as a standard. This problem is complicated by the fact that most of the elements exist as isotopes. Chemists have chosen lighter isotopes of carbon, which was a value of exactly 12 atomic mass units (a.m.u.) designated as
Carbon-12:
When the relative weight (mass) of a specific isotope is referred to C-12, the value is called isotopic or nucleide weight (mass).
E.g. : Cl-35 (75% abundance) Cl-37 (25%)
= (35 x 75 + 37 x 25)/(100) = 35.5 a.m.u
The atomic weight of an element expressed in grams is called gram-atomic, abbreviated gram-atom.
Numerically, g-atoms= g/A = (weight of the element in grams) / (atomic weight of the element in grams)
It is known experimentally that one gram-atom of an element consists of 6.02 x 10^23 atoms, the Avogadro number (N)
Therefore, number of atoms
= g-atoms x N
= g/A x N
It is thus inferred that atomic weight is the mass of 6.02 x 10^23 atoms.
2. The Mole: The gram-formula weight of a substance is the formula weight expressed in grams, e.g. the gram-formula weight of NaCl is 58.5 g. For covalent compounds, that is called gram-molecular weight. For convenience the gram-formula weight, or gram-molecular weight, is abbreviated as mole.
Numerically, g-moles = (Weight of a substance in grams) / (Molecular Weight of the substance in grams) = g/M
Just as 1.0 g-atom of an element contains 6.02 x 10^23 atoms, so 1.0 mole of a covalent compound contains 6.02 x10^23 molecules of the compound.
Therefore,
Number of molecules = moles x N
= (g/m) x N.
Therefore, molecular weight of a compounds is the mass of N molecules (6.02 x 10^23 molecules).
Note 1: For metals and noble gases, atoms = molecule, and hence,
g-atom = g-mole
Note 2: A molecule has atoms and a more has g-atoms.
E.g. One mole of water contains:
2 g-atoms of hydrogen = 2 N atoms
1 g of oxygen = N atoms
Therefore, 1.0 mole of water, H2O have
(90/18) x 3 atoms
Note 3: Counting electrons
Let us find electrons in 4.4g of CO2 ( O = C = O) (M.W. = 44)
1 molecule has 8 + 6 + 8 = 22e-, and N molecules (i.e. 1 mole) has 22 x N-, and since we have 4.4/44 moles of CO2
O = C = O
2.6 2.4 2.6
No. of variety of electrons
= moles x no. of valency electrons per mole
= 4.4/44 x 16N
Note 5: 1 mole is N things you talk about e.g. 1 mole of mangoes is N (6.02 x 10^23 mangoes. Now think for a while and calculate how long should your dear mother live to make 1 mole of chapaties if she makes 20 chapaties per day. One way to wish long life to your mother is to say: "Mom, live long to make 1 mole (6.02 x 10¼½) of chapaties".
3. Gram-molecular Volume (G.M.V.): 22.4 litres of a gas at S.T.P. weigh 1g-mole, and hence contain N(6.02 x 10^23) molecules.
The G.M.V. is the volume occupied by 1 mole of gas or N molecules of a gas.
Numerically,
moles = Volume of gas in litres at s.t.p./22.4 litres.
= Volume of gas in c.c. (cm^3) at s.t.p./22400 c.c.
Therefore,
Number of molecules = moles x N
No. of molecules in 1 c.c. of gas
= 6.02 x 10^23/22400 = 2.7 x 10^19
and it is called Loschmidt's number.
4. One Faraday of electricity liberates one gram equivalent of substance. One Faraday is '1 mole of electrons, i.e.,
charge on Avogadro's number of electrons
1F = 96540 coulombs
= N (6.02 x 10^23) electrons
i.e. 1F = N x e (e = charge of single electron)
e = F/N = 96540 / 6.02 x 10^23
= 1.60 x 10^-19C.
Numerically, No. of Faradays
= 1 (amperes) x t (sec) / 96540
E.g. How many electrons pass in 0.001 sec. through a cross-section of a wire carrying a current of 0.1 milliamperes?
No. of electrons = No. of Faradays x N
= I x t / 96540 x N
= (0.10 x 10^-3) (0.001) / (96540 x 6.02 x 10^23)
= 6.243 x 10^11 electrons.
5. One Einstein (unit of energy used in photochemistry) is 1 mole of photons or quanta i.e., energy of Avogadro's number (N = 6.02 x 10^23) quanta.
Einstein = Nhv= Nhc/l
h = Planck's constant
= 6.63 x 10^-34 joules-sec.
c = velocity of light
= 3.0 x 10^8 m/sec.
6. Thus 1 mole = N atoms, N molecules, N ions, N electrons, N particles
Hence, N = Number of particles per mole
= (Number of particles) / (Number of moles)
Oxidation is increase in oxidation number and reduction is decrease in oxidation number. Oxidation number is an arbitrary number assigned to an element according to some rules and conventions.
Oxidation numbers are assigned as follows ;
(a) free element = 0
(b) oxygen in a compound = - 2 (- 1 for peroxides, + 2 in OF2)
(c) hydrogen in a compound = + 1 for metallic hydrides)
(d) group IA metals in a compound = + 1
(e) group IIA metals in a compound = + 2
(f) sum of the oxidation number of a neutral molecule = 0
(g) sum of the oxidation number of an ion = charge on the ion.
Assigning oxidation number is a skill that is learned by practice. Oxidation number per atom is called oxidation state. Oxidation number concept, though arbitrary, is very useful in balancing redox equations.
A clear relationship between oxides and oxyacides and other compounds of an element can be established by constructing a scale of oxidation states.
+ 7 ClO4- Cl 2O7
+ 6
+ 5 ClO3-
+ 4 ClO 2
+ 3
+ 2
+ 1 ClO- Cl2 O
0 Cl2
- 1 Cl-
The scale can then be used to convey the principle of disproportionation (e.g., of Cl2 to Cl- and ClO-).
Oxidation number equations are balanced as follows :
(a) assign oxidation numbers to the elements
(b) determine change in oxidation number per atom and per molecule
(c) make increase in oxidation number equal decrease in oxidation number by use of coefficients.
(d) balance the rest of equation by inspection.
The steps in balancing redox equation by the
(half reaction / equation) method are illustrated by the equation.
MnO4- + H+ + C2O4 ^-2 --> Mn^2+ + CO2 + H2O.
(a) separate the equation into half-reactions
(b) balance all atoms except oxygen and hydrogen
(c) balance oxygen by adding water ; balance hydrogen by adding H+
(d) balance charges by adding e-
(e) add balanced half-reactions to e-
(f) cancel terms and neutralise H¨ and OH- if necessary.
MnO4- + 8H+ + 5e- -->Mn^2- + 4H2O] x 2C2O4 ^2--> 2CO2 + 2e-] x 5
___________________________________
2MnO4 - +16H+ +4C2 O4^2- --> 2Mn^2+ +10CO2 + H2O
A useful practical exercise, illustratingsuccessive changes of oxidation state, is the reduction by Zn (or Zn amalgam) and dil. H2SO 4 of a solution of vanadate (e.g., sodium metablue, green and violet, according to the reduction scheme.
[VO3 ](yellow) --> [VO]^2+ (blue) --> V^3+ (green) --> V^2+ (violet)
Equivalent weight of a substance is that weight of the substance which will combine with or displace directly or indirectly 1.008 parts of hydrogen, 8 parts by weight of oxygen, 35.5 by parts of hydrogen, 8 parts by weight of oxygen, 35.5 by parts of chlorine and 108 parts by weight of silver.
1 g-equivalent of a substance means its equivalent weight in grams.
Substances combine in the ratio of their equivalents.
Equivalents (1) = equivalents (2)
(g1/E 1) = (g2/E2) or (g1/g2 ) = (E1/E2)
(wt. of Ist / wt. of 2nd) = (eq. wt. of Ist / eq. wt. of 2nd)
Equivalent weight of a substance is that weight of it which loses or gains N electrons (N = 6.024 x 10^23)
1 Faraday liberates 1 g-equivalent of a substance.
Average atomic mass called chemical atomic weight is computed from nuclide masses and their relative abundance (data obtained from mass spectrograph).
Cannizaro's Principle: 1 mole of a substance contains at least 1 g-atom of that element
Dulong and Petit's law : The product of atomic weight and specific heat of a solid element is 6.4 nearly. Therefore, approximate atomic
weight= 6.4 / Specific heat
A = E x V
where A = atomic weight
and E = equivalent weight
V = Valency of the element
Mitcherlich's Law of isomorphisms:
Isomorphous compounds have the same constitution, e.g.,
K2SO4.Al2 .(SO 4)3.24H2O Potash alum
K2SO4 .Cr2 (SO4)3 .24H2O Chrome alum
Equivalent weight of an oxidising agent (O.A) or a reducing agent (R.A)
= (its molecular weight) / (Total change in oxidation number)
An element can have 2 or more equivalent weights or valencies (transitional element for example) but has same atomic weight.
A = E1V1 = E2V2
E1V1 = E2 V2
(V2 / V1) = (E1 / E 2 )
Concentration Units:
The relative amount of substance in a solution or a mixture is known as its concentration. The concentration can be expressed in different units or measures.
1. Molarity (M): The molarity of a solution is the number of moles of solute present in 1 litre of solution. It is the most widely used unit employed for expressing concentration.
Numerically stated :
Molarity (M) = (Moles of solute) / (litres of solution)
= (milli moles of solute milli litres of solution) / (millilitres of solution)
where, moles = (grams of solute) / (M.W. of solute)
and, Milli-moles = (grams of solute x 1000 / M.W. of solute)
= (milligrams of solute / M.W. of solute)
Also,
molarity (M) x volume of solution in litres (V) = moles
and Molarity (M) x volume of solution in
millilitres (V) = milli-moles.
(a) A 0.1 M CaCl2 solution contains 0.1 mole of the salt per litre of solution, the concentration of Ca^2+ is 0.1 M and that of Cl- ion is 0.2 M (assuming complete ionisation).
(b) The terms active mass used in Law of Mass is Molarity or molar concentration when we deal with reversible reaction in solution.
2. Molality (m): The molality of a solution is number of moles of solute present in 1000 grams (1 kg) of the solvent.
Numerically expressed:
Molality (m) = (moles of solute / kg of solvent)
e.g., 0.1 urea solution in alcohol will mean 0.1 mole urea is dissolved in 1000g of alcohol. The unit is used in dealing with colligative properties of solutions.
3. Mole Fraction (X): In this system, the relative amount of a compound in a mixture is expressed as a fraction of the total amount of all the components, the amounts of each component being measured in moles.
To illustrate, suppose in a solution there are present n1 moles of A. and n2 moles of B. Total amount is n1 + n2 moles.
Then, mole fraction of A:
Xa = (moles of A)/(moles of A+moles of B)
= n2 / (n1 + n2)
Mole fraction of B:
Xb = (moles of A)/(moles of A+moles of B) = (n2 / n1 + n2 )
It is obvious that (mole-fraction of A) + (mole fraction of B) = Xa + Xb = 1.
The unit is convenient when a property of solution requires to be expressed in terms of a relative amounts of solute and solvent.
4. Normality (N): This unit is widely used in volumetric analysis and needs full elabortation.
The normality of a solution is the number of gram equivalent weights (or simply equivalents) of the solute present per litre of solution.
Numerically expressed :
N = (equivalents of solute)/ (litres of solution)
= (milli equivalents of solute) / (millilitres of solution)
where,
equivalents (eq.) = g / E and milli-equivalent (meq.) = g x 1000 / E
Here,
g = grams of substance
E = eq. wt. of substance.
It is obvious that :
N x V (volume of solution in litres)
= equivalents
= milli-equivalents (meq.)
Thus, there are two ways to obtain or calculate equivalents and milli-equivalents:
(i) eq. = g / E. Also eq. = N x V (litres)
(ii) meq. = g x 1000 / E Also,
meq. = N x V (mls)
Thus, the Law of Normality N1 V1 = N2 V2 actually means :
meq. (1) = meq. (2)
meq. (acid) = meq. (base)
meq. (Ox. agent) = meq. (Red. agent)
NV = g x 1000 / E
(g1 x 1000 / E1) = (g2 x 1000 / E2)
(g1 / E1) = (g2 / E2)
Substances combine or react in the ratio of their equivalents of milli-equivalents.
N = (equivalents (eq.) of solute) / (litres of solution)
= eq./litre = g / E litre
Therefore, N x E = g / litre
N x E = S (strength).
5. Relationship between Normality and Molarity :
one mole of H2SO 4 is 98 g
one equivalent of H2SO4 is 49 g.
98 g of H2SO4 dissolved in 1 litre of solution is 1 molar and 49 g of H2SO4 dissolved in 1 litre of solution is 1 Normal.
Hence 98 g of H2SO4 in 1 litre of solution is 1 molar but 2 Normal. We conclude by inspection that :
If Eq. wt. = M.W. / 2 as in the case of H2SO4
N = 2 x M (since 1 mole contains 2 equivalents)
If Eq. wt. = M.W. / 5 ( as in the case of KMnO4 in acid medium)
N = 5 x M (since 1 mole contains 5 equivalents)
If Eq. wt. = M.W. / 6 ( as in the case of K2Cr 2O7 in acid medium
N = 6 x M (since 1 mole contains 6 equivalents)
Molarity is bigger unit and smaller normality units are contained in it
If eq. wt. = M.W; N = M
Note1: N = Normality, M = Molarity
Note 2: Since molarity and normality involve volumes, they are affected by temperature.
Assignment 1
Concentration Units:
The relative amount of substance in a solution or a mixture is known as its concentration. The concentration can be expressed in different units or measures.
1. Molarity (M): The molarity of a solution is the number of moles of solute present in 1 litre of solution. It is the most widely used unit employed for expressing concentration.
Numerically stated :
Molarity (M) = (Moles of solute) / (litres of solution)
= (milli moles of solute milli litres of solution) / (millilitres of solution)
where, moles = (grams of solute) / (M.W. of solute)
and, Milli-moles = (grams of solute x 1000 / M.W. of solute)
= (milligrams of solute / M.W. of solute)
Also,
molarity (M) x volume of solution in litres (V) = moles
and Molarity (M) x volume of solution in
millilitres (V) = milli-moles.
(a) A 0.1 M CaCl2 solution contains 0.1 mole of the salt per litre of solution, the concentration of Ca^2+ is 0.1 M and that of Cl- ion is 0.2 M (assuming complete ionisation).
(b) The terms active mass used in Law of Mass is Molarity or molar concentration when we deal with reversible reaction in solution.
2. Molality (m): The molality of a solution is number of moles of solute present in 1000 grams (1 kg) of the solvent.
Numerically expressed:
Molality (m) = (moles of solute / kg of solvent)
e.g., 0.1 urea solution in alcohol will mean 0.1 mole urea is dissolved in 1000g of alcohol. The unit is used in dealing with colligative properties of solutions.
3. Mole Fraction (X): In this system, the relative amount of a compound in a mixture is expressed as a fraction of the total amount of all the components, the amounts of each component being measured in moles.
To illustrate, suppose in a solution there are present n1 moles of A. and n2 moles of B. Total amount is n1 + n2 moles.
Then, mole fraction of A:
Xa = (moles of A)/(moles of A+moles of B)
= n2 / (n1 + n2)
Mole fraction of B:
Xb = (moles of A)/(moles of A+moles of B) = (n2 / n1 + n2 )
It is obvious that (mole-fraction of A) + (mole fraction of B) = Xa + Xb = 1.
The unit is convenient when a property of solution requires to be expressed in terms of a relative amounts of solute and solvent.
4. Normality (N): This unit is widely used in volumetric analysis and needs full elabortation.
The normality of a solution is the number of gram equivalent weights (or simply equivalents) of the solute present per litre of solution.
Numerically expressed :
N = (equivalents of solute)/ (litres of solution)
= (milli equivalents of solute) / (millilitres of solution)
where,
equivalents (eq.) = g / E and milli-equivalent (meq.) = g x 1000 / E
Here,
g = grams of substance
E = eq. wt. of substance.
It is obvious that :
N x V (volume of solution in litres)
= equivalents
= milli-equivalents (meq.)
Thus, there are two ways to obtain or calculate equivalents and milli-equivalents:
(i) eq. = g / E. Also eq. = N x V (litres)
(ii) meq. = g x 1000 / E Also,
meq. = N x V (mls)
Thus, the Law of Normality N1 V1 = N2 V2 actually means :
meq. (1) = meq. (2)
meq. (acid) = meq. (base)
meq. (Ox. agent) = meq. (Red. agent)
NV = g x 1000 / E
(g1 x 1000 / E1) = (g2 x 1000 / E2)
(g1 / E1) = (g2 / E2)
Substances combine or react in the ratio of their equivalents of milli-equivalents.
N = (equivalents (eq.) of solute) / (litres of solution)
= eq./litre = g / E litre
Therefore, N x E = g / litre
N x E = S (strength).
5. Relationship between Normality and Molarity :
one mole of H2SO 4 is 98 g
one equivalent of H2SO4 is 49 g.
98 g of H2SO4 dissolved in 1 litre of solution is 1 molar and 49 g of H2SO4 dissolved in 1 litre of solution is 1 Normal.
Hence 98 g of H2SO4 in 1 litre of solution is 1 molar but 2 Normal. We conclude by inspection that :
If Eq. wt. = M.W. / 2 as in the case of H2SO4
N = 2 x M (since 1 mole contains 2 equivalents)
If Eq. wt. = M.W. / 5 ( as in the case of KMnO4 in acid medium)
N = 5 x M (since 1 mole contains 5 equivalents)
If Eq. wt. = M.W. / 6 ( as in the case of K2Cr 2O7 in acid medium
N = 6 x M (since 1 mole contains 6 equivalents)
Molarity is bigger unit and smaller normality units are contained in it
If eq. wt. = M.W; N = M
Note1: N = Normality, M = Molarity
Note 2: Since molarity and normality involve volumes, they are affected by temperature.
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
493
