3. Equivalent weight or equivalent mass
• Equivalent weight of the given substance is
the number of parts by mass of the given
substance that displaces or combines with
1.008 parts H2 ; 8 parts oxygen and 35.5 parts
by mass of chlorine.
** We know that in any chemical reaction definite mass of one substance reacts with
particular mass of other substance. Ex – in water one part of H reacts with 16 part of O.
That’s mean-
“In a chemical reaction a definite mass of one substance is always equivalent to definite
mass of other substance this specific mass is called as its equivalent mass.”
5. Zn + H2SO4 ---- ZnSO4 + H2
65 98 2
2 parts by mass of hydrogen = 65 parts by mass of Zn
1 ------------”-------------------- = 65 /2 = 32.5
Similarly
2 parts by mass of hydrogen = 98 parts by mass of H2SO4
1 --------------- “ ----------------- = 98/2 = 49
* eq. weight is just a number it does not posses any unit
Equivalent weight can be worked out
experimentally also.
9. Equivalent weight of an element
Atomic mass of an element
Eq. Wt. of an element=
Valency
[Valency is the combining capacity of any substance]
Example:-
Eq. Wt of oxygen= 16/2= 8
Eq. wt of Mg= 24/2= 12
Eq wt of iodine = 127/1=127
Eq. wt of Cu in cupric salt = 63.6/2=31.8
Eq. wt of Cu in cuprous salt = 63.6/1=63.6
Variable
valency
effects eq
wt.
10. Find Eq wt of following
Al= 27
Fe= 56
Al
Fe2+
Fe3+
11. Equivalent weight of a normal salt
Formula weight of salt
Eq. Wt of a normal salt =
Total charge on cationic part
Example:-
Eq. Wt of NaCl = 23 + 35.5 /1 = 58.5
Eq. wt of Na2CO3 = (2 x 23) + 12 + (3 x 16 ) /2= 53
12. Equivalent weight of a Acid salt
Formula weight of salt
Eq. Wt of a acid salt =
number of replaceable H atoms
Example:-
Eq. wt of NaHCO3 = (1 x 23) + 1 + 12 + (16 x 3 ) /2 = 84
13. Equivalent weight of an Acid
Molecular weight of an acid
Eq. Wt of an acid =
Basicity of the acid
Example:-
Eq. wt of H2SO4 = (1 x 2) + 32 + (16 x 4 ) /2 = 98/2= 49
Eq. wt of HCl = 1 +35.5/1= 36.5/1 = 36.5
14. Equivalent weight of a base
Molecular weight of the base
Eq. Wt of a base =
Acidity of the base
Example:-
Eq. wt of Ca(OH)2 = 40 + 2(16 + 1 ) /2 = 74/2= 37
Eq. wt of NaOH = 23 + 16 + 1 = 40/ 1 = 40
15. Equivalent weight of a Oxidant (OA) or reductant (RA)
Formula mass
Eq. Wt of a OA / RA =
number of e- gained or lost by one molecule
126/2=63
16.
17. Gramequivalent weight
The mass of a substance in gm, numerically equal to its equivalent weight is called gm
equivalent weight or one gm equivalent of that substance.
mass of the substance in gm
Number of gm equivalents =
equivalent weight
Substance Mass taken Eq. wt. No. of gm. Eq.
H2SO4 4.9g 49 4.9/49=1
H2C2O4 10g 63 10/63=0.16
K2Cr2O7 2g 49 2/49=0.041
NaOH 0.4g 40 0.4/40=0.01
18. Lets understand this little more by connecting with our previous knowledge
2NaOH + H2SO4 --- Na2SO4 + 2H2O
• In case of mole we must have a balanced chemical equation to find of how
much of NaOH is reacting with How much H2SO4.
• Therefore came concept of eq wt and using that we can easily Find out that how
much of substance A is reacting with how much substance B.
19. Molecular wt and eq wt share similar relations
In case of Molecular wt
Mass
No of mole =
Mol. wt
In case of equivalent wt
Mass
No of gm equivalent =
eq. wt
Mass
No of gm eq.=
eq wt
Mass of sub Mass
No of gm eq.= or x X
Molecular wt / X Mol. Wt
No of gm eq = Mole x X [ X = valency/acidity/bascity/ charge on cation/ no of e- gained
or lost]
Not in reduced syllabus
20. 2NaOH + H2SO4 --- Na2SO4 + 2H2O
Lets find out no of gm eq when no of moles are given
2 mole x 1 1 mole x 2
No of gm eq = Mole x X [ X will have different values depending upon substance]
2 equivalent 2 equivalent
Therefore it is said that in a chemical reaction I equivalent of a substance
reacts with 1 eq of other substance.
21. Lets find out no of gm eq when no of moles are given
N2 + 3H2 --- 2NH3
No of gm eq = Mole x X [ X will have different values depending upon substance]
1 mole x 3 3 mole x 1
3 gm
equivalent
3gm
equivalent
22. Relationship between Gm Eq. Wt. (E),
Gm Atomic weight (A) and valency(V)
A
E =
V
A = E X V
Gm atomic mass = Gm eq. wt X valency
39. This Photo by Unknown Author is licensed under CC BY-NC-ND
Editor's Notes
As we know in any chemical reaction a definite mass of one substance reacts with definite mass of other substance for example in carbon mono oxide 12 gm of carbon reacts with 16 gm of oxygen in carbon di oxide 12 gm carbon reacts with 32 gm oxygen, in water 2 gm hydrogen reacts with 32 gm oxygen mean 1 gm reacts with 16 gm.
