All elements combine with each other according to the law of chemical combination, and the number of parts by which an element combines with 1 part by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine or one gram equivalent of any other substance is the value of the equivalent mass of the element. The formula to calculate the equivalent mass of an element is given by :

`text(Equivalent mass) = text(Atomic mass)/text(Valency)`

(a) Eq. wt. of metal = `text(Mass of metal)/text(Mass of hydrogen displaced) xx 1.008`

or ` = text(Mass of metal)/text(Mass of oxygen combined) xx 8.0`

or `= text(Mass of metal)/text(Mass of chlorine combined) xx 35.5`

e.g. In `H_2O , NH_3` and `CH_4` one mole hydrogen combines with `1/2` mole oxygen, `1/3` mole nitrogen and `1/4` mole carbon
respectively. Hence

Equivalent weight of oxygen `= 1/2 xx 16 = 8.0`
Equivalent weight of nitrogen `= 1/3 xx 14 = 4.67`
EquivaIent weight of carbon ` = 1/4 xx 12 = 3`

(b) Equivalent weight of acid `= text(Molecular weight of acid)/(text{Basicity (number of replaceable)}H^+)`
e.g. Equivalent weight of `H_2SO_4 = 98/2 = 49`

(c) Equivalent weight of base `= text(Molecular weight of base)/(text{Acidity (number of replaceable )} OH^-)`
e.g. Equivalent weight of `NaOH = 40/1 = 40`

Equivalent weight of `Ca(OH)_2 = 74/2 = 37`

(d) Eq. wt. of salt `= text(Molecular weight of salt)/text(Total positive valency of metal atoms)`
e.g. Equivalent weight of `Na_2CO_3 = 106/2 = 53`

(e) Equivalent weight of a substance that undergoes oxidation-reduction
`= text(Molecular weight)/text(Change in oxidation number)`

e.g. When `KMnO_4` reacts under acidic conditions, change in oxidation number (from `+7` to `+2`) is `5`, hence

Equivalent weight of `KMnO_4` in acidic medium `= 158/5 = 31.6`