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Equivalent weight - Definition, Examples, Practice problems & FAQs
Imagine going to a friend's house and their mother greets you with an assortment of snacks ranging from samosa to dhokla, from kachori to pakoda, etc. Some people might just gorge on sandwiches. Others might prefer eating samosa.
What would be the fallout of such behaviour? Each and every person would get their stomach full with any of the delicacies.
In a similar manner, in chemical terms, we use the term “equivalent weight”.
You might ask the purpose of such a term.
During chemical reactions, the tendency of an element to react with other elements can be monitored using this concept. So let's discuss this in detail.
Table of contents
- Equivalent weight
- Calculation of n-factor for acid-base reaction
- Calculation of n-factor for redox reaction
- Practice problem
- Frequently asked questions-FAQs
Equivalent weight
Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine, is known as the equivalent weight of that element.
Example: Find the equivalent weight of 
for its oxide formation.
Answer: 
From the above equation we can say,
16 grams of oxygen combined with 24 g of magnesium
1 gram of oxygen combine with 
g of magnesium
8 gram of oxygen combine with 
g of magnesium
Equivalent weight of magnesium = 12
For elements; Equivalent weight = 
Similarly
For acid; Equivalent weight = 
For base; Equivalent weight = 
For reducing agents; Equivalent weight = 
For oxidizing agents; Equivalent weight = 
In general, Equivalent weight = 
Calculation of 
Case 1: of acid = number of 
ions furnished per molecule of an acid
(number of replacable hydrogen present in an acid is called its basicity)
| Acid | Basicity |
 |
2 (dibasic acid) |
 |
3 (tribasic acid) |
|
|
1 (monobasic acid) |
 |
1 (monobasic acid) |
Note: 
Case 2: of base = number of 
ions furnisher per molecules of base
(number of replaceable hydroxide ion present in an acid is called its acidity)
| Base | Acidity |
 |
1 (mono-acidic base) |
 |
2 (di-acidic base) |
 |
3 (tri-acidic base) |
Calculation of n-factor for redox reactions
n-factor of reductants and oxidants are calculated on the basis of their change in oxidation state.
Case 1: When only one atom is undergoing a change in oxidation number either by reduction or oxidation.
Example: Oxidation: 
| Element | Initial O.S | Final O.S |
| C | +3 | +4 |
| O | -2 | -2 |
We can see, only oxidation state of carbon is changes from +3 to +4
(Note: per molecule 2 Carbon atomss are present in 
ion)
Example: Reduction: 
(in acidic medium)
| Element | Initial O.S | Final O.S |
 |
+6 | +3 |
| O | -2 | -2 |
We can see, only oxidation state of chromium is changes from +6 to +3
(Note: Per molecule, 2 chromium atom is present in 
ion
Case 2: If an atom from a compound oxidised or reduced but appears in two product in same oxidation state as a result of either only oxidation or only reduction then same rule is applicable
Example: Reduction: 
(in acidic medium)
| Element | Initial O.S | Final O.S |
 |
+7 | +2 & +2 |
| O | -2 | -2 |
We can see, only oxidation state of chromium is changes from +6 to +3
Case 3: If an atom from a compound oxidised or reduced and appears in two different product in two different oxidation state as a result of either only oxidation or only reduction.
Example: Reduction: 
| Element | Initial O.S | Final O.S |
|
+7 | +2 for  & +4 in  |
| O | -2 | -2 |
We can calculate n-factor only if we know the stoichiometric coefficients of , and 
in the simplest balanced chemical equation.
If a balanced equation is represented like this: 
where, 
Case 4: When one atom from a single molecule oxidised or reduced and appear in two different compounds, one is same as initial oxidation state and one in different oxidation state as a result of either only oxidation or only reduction.
Example:
| Element | Initial O.S | Final O.S |
 |
 |
-1 & 0 |
|
+6 | +3 |
Method 1: 
Apply the law of equivalence
Number of the equivalent of 
= No of the equivalent of 
=
Method 2: Balance out the chemical reaction in the simplest stochiometric coefficients
Out of 14 only 6 
have undergone a change in oxidation state i.e 
= 
Case 5: When two or more atom both are either oxidised or reduced as a result of either only oxidation or only reduction.
| Element | Initial O.S | Final O.S |
 |
 |
+3 |
 |
+3 | +4 |
= 
Case 6: when one atom from a compound is oxidised and another atom is reduced as a result of either only oxidation or only reduction.
| Element | Initial O.S | Final O.S |
|
 |
-1 |
 |
-2 | 0 |
= 
= 
You can consider n-factor by oxidation or reduction.
Case 7: Disproportionation reaction: Single-atom from a compound or molecule is reduced as well as oxidised
Type 1. When the number of electrons lost or gained are the same.
| Element | Initial O.S | Final O.S |
|
|
-2 in  & 0 in  |
Reduction: 
= 
Oxidation: 
= 
Type 2. When the number of electrons lost or gained are different.
| Element | Initial O.S | Final O.S |
|
 |
-1 in & +5 in  |
Out of 12 (from reactant) 10 acquire -1 charge () and 2 acquire +5 charge ()
Number of electron loss in oxidation = 
Number of electron loss in oxidation = 
Total change in electron for 6 
or 12 
So, change per molecule of 
For precipitation reaction
In a precipitation reaction, there is no change in the oxidation state of an atom. So, their n-factor can be calculated as charge on cation or anion
For 
Practice problems:
Q 1. Calculate the equivalent weight of 
for reaction 
; if molar mass of is M
a. 
b. 
c. 
d. 
Answer: (A)
of acid = number of ions furnished per molecule of acid
In this case; 
Equivalent weight of 
Q 2. Calculate the equivalent weight of 
for reaction 
; if the molar mass of is
a. 
b.
c. 
d. 
Answer: (A)
| Element | Initial O.S | Final O.S |
|
|
+3 |
|
+2 | +4 |
 |
-3 | +5 |
of = 
Equivalent weight of 
Q 3. Calculate n-factor of 
, for reaction 
a. 
b.
c.
d.
Answer: (A)
| Element | Initial O.S | Final O.S |
 |
in |
+2.5 in  |
of = 
Q 4. Calculate n-factor of nitrite ion for reaction 
a. 1
b. 2
c. 3
d. 4
Answer: (B)
| Element | Initial O.S | Final O.S |
|
 |
+5 |
of 
= 
Frequently asked questions-FAQs
Q 1. Is the n-factor of all compounds always the same?
Answer: No, the n-factor of many compounds depends on their reaction condition. The same compound can have a different n-factor.
E.g- 
Q 2. Is there any between equivalent weight and molecular weight?
Answer: molecular weight of any compound is the sum of the atomic mass of all atoms present in the compound and the equivalent weight is associated with the valence electron.
Q 3. What is the condition when equivalent weight and molar mass of compound are equal?
Answer: we know, 
. if the n-factor is one, then equivalent weight = molar mass of the compound.
Q 4. Why do we learn to calculate the equivalent weight of chemical species, not molar mass?
Answer: The molar mass of chemical species is always the same, but the equivalent weight of the same chemical species can be different in a different set of chemical reactions. In a few cases, molar mass and equivalent mass both are the same (in monobasic acid and monoacidic base).
Related Topics:
| Volume strength of H2O2 | Mole |
| Strength of oleum | ppm |
| Strength of solution | Density |
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