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Atomic and Molecular Equivalent Masses


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Important definitions and theories of atomic mass, molecular mass and equivalent mass. This article also covers various other topics in chemistry.



ATOMIC, MOLECULAR AND EQUIVALENT MASSES



Important Definitions:

Avogadro’s hypothesis:

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

EQUIVALENT WEIGHT (On the basis of Hydrogen)
The weight of an element that combines with one part by weight of hydrogen.

Examples:
In H2O, 16 parts by weight of oxygen combine with 2 parts by weight of hydrogen. Therefore 8 parts by weight of oxygen will combine with 1 part by weight of hydrogen. Therefore equivalent weight of oxygen is 8.
In HCl, 35.5 parts by weight of chlorine combine with 1 part by weight of hydrogen. Therefore the equivalent weight of chlorine is 35.5.

MOLE CONCEPT
This is the unit that consists of 6.023 x 1023 particles.

Avogadro number (N):

It is the number of atoms present in exactly 12 grams of 6C12 isotope.
Its value is 6.023 × 1023.

Atomicity:
The number of atoms contained in one molecule of the element.

Vapour density:
Vapour Density is defined as the ratio of the mass of a certain volume of the gas or vapour to the mass of the same volume of hydrogen at the same temperature and pressure.

Atomic weight:
The relative atomic mass of an element is the mass of one atom of the element compared with the mass of one atom of hydrogen taken as one unit.

Gram atomic weight of an atom:
The atomic weight of an element expressed in grams is known as the gram atomic weight (or gram atom) of the element. For example, Gram atomic weight of carbon = 12 g Gram atomic weight of oxygen = 16 g.

Molecular mass:
The relative molecular mass of an element or a compound is the mass of one molecule of the element or compound compared with the mass of one atom of hydrogen taken as one unit.

Gram molecular weight:
The molecular weight of a substance expressed in grams is known as gram molecular weight of the substance. The gram molecular weight of oxygen is 32g and that of sulphuric acid is 98g.

Molar volume:
Volume occupied by one mole of any gas is called molar volume or gram molecular volume. It is 22.4L (or) 2.24 × 10-2 m3at S.T.P. It contains 6.023 × 1023 molecules.

Equivalent mass of an element:
Equivalent mass of an element is the number of parts by mass of that element which can displace or combine with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.46 parts by mass of chlorine or one equivalent mass of any other element. It is only a relative number and hence it does not have any units. When equivalent mass is expressed in gram, it is called gram equivalent mass.

Equivalent mass of an acid:
Equivalent mass of an acid is the number of parts by mass of the acids which contains 1.008 parts by mass of replaceable hydrogen.

Basicity:
Basicity of mineral acid is defined as the number of Replaceable hydrogen atoms present in one mole of the acid. Basicity of organic acid is defined as the number of carboxylic groups present in the acid

Equivalent weight of base:
Equivalent mass of the base is the number of parts by mass of the base required to neutralize one equivalent mass of an acid.

Acidity of a base:
Acidity of hydroxide base is defined as the number of replaceable hydroxyl ions present in one mole of the base.

Equivalent mass of salt:
Equivalent mass of a salt is the number of parts by mass of salt which reacts with one equivalent of mass of any other substance.

Equivalent weight of an oxidising agent:
Equivalent weight of oxidizing agent is the number of parts by mass of it, which contains 8 parts by mass of available oxygen. Available oxygen means, oxygen capable of being utilised for oxidation.

Equivalent mass of a reducing agent:
Equivalent weight of reducing agent is the number of parts by mass of it, which can be oxidized by 8 parts by mass of oxygen.

Normality of a solution:
Normal solution is a solution, which contains one gram equivalent mass of the substance dissolved in one litre of the solution.

Law of volumetric analysis:
When two solutions completely react with each other, the product of volume and normality of one solution will be equal to the product of volume and normality of the other solution.

Standard solution:
In a titration, concentration of either the solution in the burette or in the conical flask should be exactly known. The solution whose concentration is exactly known is called the standard solution. A standard solution can be prepared by dissolving a known mass of the substance in a known volume of the solution.

Avogadro’s hypothesis
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Vapour density
Vapour Density is defined as the ratio of the mass of a certain volume of the gas or vapour to the mass of the same volume of hydrogen at the same temperature and pressure.

Deduction of atomicity of Hydrogen:
In the reaction between hydrogen and chlorine, 1 vol. of hydrogen combines with 1 vol. of chlorine to form 2 vols. of hydrogen chloride.

H2 + Cl2 ? 2HCl
Applying Avogadro’s Law…

1 molecule of hydrogen + 1 molecule of chlorine ? 2 molecules of hydrogen chloride
Each molecule of hydrogen chloride must contain at least 1 atom of hydrogen.
Two molecules of hydrogen chloride, contain 2 atoms of hydrogen.
These 2 atoms of hydrogen must have come from 1 molecule of hydrogen.
Therefore, the molecule of hydrogen is diatomic and is written as H2.
Similarly atomicity of chlorine is 2.

Relation between Vapour Density & Relative Molecular Mass of a Gas:
Relative Molecular Mass:

It is defined as the ratio of the mass of 1 molecule of the gas or vapour to the mass of 1atom of hydrogen.
Mass of 1 molecule of the gas or vapour
Relative molecular mass of a gas = -------------------------------------------------
Mass of 1 atom of hydrogen

Relationship between equivalent weight of atomic weight:

Consider an element X of atomic weight A and valency n.
Suppose the element X combines with hydrogen to give the compound hydride, XHn.
X + nH ?XHn
1 atom of X combines with n atoms of hydrogen.
1 gram-atom of X combines with n gram-atoms of hydrogen.
Atomic weight of X = A
Atomic weight of hydrogen = 1.008
n × 1.008g of hydrogen combines with = A gram of X
A
1.008 grams of hydrogen combines with = ------------- × 1.008
n × 1.008

A
= ------
n

By definition, this gives the equivalent weight of the element X.

Atomic weight
Equivalent weight = -------------------
Valency

In the case of univalent elements, the atomic weight itself gives the equivalent weight. Elements having varying valency have different equivalent weight.
Equivalent weight of copper in cuprous oxide Cu2O is 63.5. (here, valency of Cu is one). Equivalent weight of copper in cupricoxide,
63.5
CuO is ------ = 31.75 (here, valency of Cu is 2).
2


Vapour Density:

It is defined as the ratio of the mass of a certain volume of the gas or vapour to the mass
of the same volume of hydrogen at the same temperature and pressure.

Mass of 1 volume of gas or vapour
Vapour density (V. D.) = -------------------------------------------
Mass of 1 volume of hydrogen
[Applying Avogadro’s Law]
Mass of 1 molecule of gas or vapour
Vapour density (V.D) = --------------------------------------------
Mass of 1 molecule of hydrogen
[Since hydrogen is diatomic]
Mass of 1 molecule of gas or vapour
V.D. = ---------------------------------------------
2 x Mass of 1 atom of hydrogen
[Multiplying both sides by 2, we get]
Mass of 1 molecule of gas or vapour
2 × V.D. = --------------------------------------------
Mass of 1 atom of hydrogen
2 × V.D. = Relative molecular mass of the gas or vapour

2 × Vapour Density = Molecular Weight.


Oxide formation method of determining equivalent mass of an element .

This method is used to find out the equivalent mass of those metals (Magnesium, Zinc) which can easily form their oxides.

2Mg + O2 ? 2MgO

Procedure:
? A known mass of a metal (m1) is heated in air or oxygen.
? The mass of metal oxide (m2), formed is found.
? In some cases (Cu, Pb, Sn etc.,) first the metal (m1) is converted into its salt.
? Then the metallic salt is converted into its oxide by heating the salts.
? Because the metals could not be directly converted into their oxides.
? The mass of metal oxide formed is found (m2).
? From the mass of metal oxide and the mass of metal, the mass of oxygen that has combined with the metal is found.
? The mass of oxygen that has combined with the metal = m2 – m1
? (m2 – m1) g of oxygen has combined with = m1g of metal
m1
8g of oxygen will combine with = --------- x 8
m2 – m1
By definition it refers to the equivalent mass of the metal.

Mass of metals
Equivalent mass of the metal = ------------------------------------------------------- x 8
Mass of oxygen that has combines with m1

Determination of equivalent mass of an element by hydrogen displacement method

This method is used to determine the equivalent mass of those metals (Na, Zn,Al, and Mg) which readily displace hydrogen from an acid or from a base or water.

Mg + 2 HCl ? MgCl2 + H2

Procedure:
? In this method a known mass of metal m (g) is added to acid or base or water.
? The volume of amount of hydrogen liberated (V1) is measured at room temperature and pressure.
? The pressure of moist hydrogen gas displaced (P) is measured.
? From the aqueous tension (p), the pressure of dry hydrogen gas displaced (P1) is measured.
? Using the values of standard pressure (P0) and standard temperature (T0) and room temperature (T1) the volume of hydrogen gas displaced is converted into volume of hydrogen gas displaced at STP by the gas equation

P1V1 P0V0
------ = -------
T1 T0

Calculation:
The mass of the metal displaced by V0 m3 of the dry hydrogen gas at STP = m.g
The mass of the metal displaced by 1.12 x 10-2 m-3
m
of the dry hydrogen gas at STP = -------------- = × 1.12 × 10–2m3
V0 m3

The mass of 1.12 x 10-2 m3 of at STP = 1.008g

(hence by definition it represents the equivalent mass of the metal)

Mass of the metal x 1.12 x 10-2 m
Equivalent mass of the metal = ----------------------------------------
V0m3

Determination of equivalent weight of an element by chloride displacement method

Principle: This method is used to find out the equivalent mass of those metals (Silver, Gold, Potassium, and Sodium) which can easily form their chlorides.

2Ag + Cl2 ? 2 AgCl

Procedure:
? A known mass of a metal (m1) is heated in the presence of chlorine.
? The mass of metal chloride (m2), formed is found.
? In case Silver, metal (m1) is converted into its salt by dissolving it in nitric acid.
? Then the salt formed is converted into its chloride by adding dilute HCl.
? The precipitated metal chloride is dried.
? The mass of metal chloride formed is found (m2).
? From these, the mass of chlorine that has combined with the metal is found.

Calculation:
The mass of chlorine that has combined with the metal = (m2 – m1)g
(m2 – m1) g of chlorine has combined with = m1g of metal
m1
35.465g of chlorine combined with = --------- x 35.46
m2 – m1
By definition it refers to the equivalent mass of the metal.
Mass of metal x 35.46
Equivalent mass of the metal = --------------------------------------------------------
Mass of chlorine that has combined with metal

Equivalent weight of an acid.

? The number of parts by mass of the acids which contains 1.008 parts by mass of replaceable hydrogen.
? Basicity of mineral acid is defined as the number of Replaceable hydrogen atoms present in one mole of the acid.
? Basicity of organic acid is defined as the number of carboxylic groups present in the acid.
Molecular mass of the Acid
? Equivalent mass of the Acid = ------------------------------------
Basicity of the Acid
? For monobasic acids the molecular mass and equivalent mass are the same.

Example:
Equivalent mass of hydrochloric acid.

The basicity of hydrochloric acid HCl is 1
The molecular mass of HCl = 1 + 35.46 = 36.46
Molecular mass of the Acid
Equivalent mass of the Acid = ------------------------------------
Basicity of the Acid
36.46
Equivalent mass of HCl = -------- = 36.46
1

Equivalent mass of Bases:
The number of parts by mass of the base required to neutralize one eq. mass of an acid.
Acidity of –OH is the number of replaceable -OH ions present in one mole of the base.
Molecular mass of the Base
Equivalent mass of the Base = ------------------------------------
Acidity of the Base
For monoacidic bases the molecular mass and equivalent mass are the same.

Example:
Equivalent mass of sodium hydroxide
Solution:
The acidity of sodium hydroxide NaOH is 1
The molecular mass of NaOH = (23 + 16 + 1) = 40

Molecular mass of the Base
Equivalent mass of the Base = ------------------------------------
Acidity of the Base
40
Equivalent mass of NaOH = --- = 40
1

Equivalent mass of salts:
The number of parts by mass of salt which reacts with one eq.of mass of any other substance.
Molecular mass of the salt
Equivalent mass of the salt = ---------------------------------------
Valency of the metal in the salt

Example:
The equivalent mass of cupric chloride.
Solution:
The Valency of copper in cupric chloride CuCl2 is 2
The molecular mass of CuCl2 = 63.5 + (2 × 35.46) = 134.42
Molecular mass of the salt
Equivalent mass of the salt = --------------------------------------
Valency of the metal in the salt
134.42
Equivalent mass of CuCl2 = -------- = 67.21
2

Note: In the case of salts like Na2SO4, the total valency of sodium is 2.


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