equivalent masses of elements
Equivalent masses can be determined by the following methods:
1. Hydrogen displacement method
2. Oxide method
3. Chloride method
Metal displacement method
This method is used to determine the equivalent mass of those metals
such as magnesium, zinc and aluminium which react with dilute acids and readily
Mg + 2HCl -> MgCl2 + H2
Zn + H2SO4 - > ZnSO4 + H2
2Al + 6HCl - > 2AlCl3 +
From the mass of the metal and the volume of hydrogen displaced, the
equivalent mass of the metal can be calculated.
This method is employed to determine the equivalent mass of those
elements which readily form their oxides eg. magnesium, copper etc. Oxide of an
element can be formed by direct method or by indirect method.
Magnesium forms its oxide directly on heating
2Mg + O2 ® 2 MgO
In the case of copper, its oxide is obtained in an indirect manner i.e.
copper is dissolved in concentrated nitric acid and the copper(II) nitrate
formed after evaporation is strongly heated to give copper (II) oxide.
Cu + 4HNO3 ® Cu(NO3)2 + 2H2O + 2 NO2
2Cu(NO3)2 ® 2CuO + 4 NO2 + O2
Mass of the element taken = w1 g
Mass of the oxide of the element = w2 g
Mass of oxygen = (w2-w1) g
- w1) g of oxygen has combined with w1 g of the metal.
8 g of oxygen
will combine with w1 / (w2 - w1) x 8
This value represents the equivalent mass of the metal.
The equivalent mass of those elements which
readily form their chlorides can be determined by chloride method. For example,
a known mass of pure silver is dissolved completely in dilute nitric acid. The
resulting silver nitrate solution is treated with pure hydrochloric acid when
silver chloride is precipitated. It is then filtered, dried and weighed. Thus
from the masses of the metal and its chloride, the equivalent mass of the metal
can be determined as follows :
Mass of the metal = w1 g
Mass of the metal chloride = w2 g
Mass of chlorine = (w2 - w2) g
(w2 - w1) g of chlorine has combined with w1
of the metal
35.46 g of chlorine will combine with 35.46 x w1 / (w2 - w1) g of the
This value gives the equivalent mass of the metal.
Uses of volumetric
If the volume of the acid is represented as V1, the normality
of the acid as N1, volume of base as V2 the normality of
the base as N2, then from the law of volumetric analysis it follows
V1 x N1 = V2 x N2
All volumetric estimations are based on the
above relationship. If any three quantities are known, the fourth one can readily
be calculated using the above expression.