Every element has fixed to combine mass for all other elements which are called equivalent mass or simply chemical equivalent.The equivalent masses of element are expressed by taking hydrogen, oxygen, and chlorine as reference element
equivalent mass of an element is defined as the number of parts by mass of an element which combines with or displaces directly or indirectly. 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine.
2Mg + O_{2 }→ 2MgO
\( 2\: × 24gm \:\:\:\:\:\: 32\: gm\)
32 parts by mass of oxygen combine with 2 x 24 parts by mass of Mg
8 parts by mass of oxygen combines with \( \frac { 2 × 24 × 8}{32}\) =12 P/M of Mg
∴ eq mass of Mg = 12
# Mg + H_{2}SO_{4 }→ MgSO_{4} + H_{2}
\(24gm \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:2×1.008 gm \)
2 x 1.008 gm Of H_{2} is displaced by 24 P/Mof Mg
ie. 1.008 P/M of H_{2 }is displaced by \( \frac {24}{2}\) P/M = 12 P/M of Mg
Hence, equivalent mass of Mg = 12
# Mg +Cl_{2} → MgCl_{2}
\(24gm \:\:\:\:\:\:\:\:\:\:2×35.5 gm \)
2 x 35.5 gm Of Cl_{2} combines with 24 P/Mof Mg
ie.35.5 gm Of Cl_{2} combines with \( \frac {24}{2}\) P/M = 12 P/M of Mg
Hence, equivalent mass of Mg = 12
It should be noted that one equivalent of any element always combines with one equivalent of other element. Therefore following relation always holds true for any reaction.
\( \frac {Equivalent\:Mass\:of\:A}{Equivalent\:Mass\:of\:B}\) = \( \frac {Mass\:of\:A}{Mass\:of\:B}\)
# Show that equivalent mass of carbon is 3.
C + O_{2} → CO_{2}
\(12\:gm\:\:\:32\:gm\)
32 P/M of oxygen combines with 12 P/M of carbon
8 P/M of oxygen combines with \( \frac {12 × 8}{32}\) = 3 P/ M of carbon
Hence, equivalent mass of Carbon = 3
# What do you mean by "equivalent mass of 'Mg' is 12 ?
ANS :
The equivalent mass of Mg is 12 means:
i) 12 P/m of magnesium displaces 1.008 P/m of hydrogen from dilute acid.
ii) 12 P/m of magnesium combines with 8 P/m of oxygen and 35.5 P/m of chlorine.
Gram equivalent mass
Equivalent mass of an element is expressed as the combining mass relative to the masses of another element i.r. it is simply a number. So, it has no unit expressed in gram, then it is called gram equivalent mass.
For example : (i) Equivalent mass of Mg = 12
(ii) Equivalent mass of H_{2}SO_{4 }= 49 i.e. 49 gm of H_{2}SO_{4 }= 1gm equivalent of H_{2}SO_{4}
Number of gram equivalnet = \( \frac {Mass\:in\:gram}{Gram\:equivalent\:mass}\)
# Calculate the number of gram equivalent present in 480 gram of Mg
Given mass (M) = 480 gm
Equivalent mass (E) = 12 gm
Hence, Number of gram equivalent = \( \frac{M}{E}\) = \( \frac {480}{12}\) = 40
Variable equivalent mass
When an element combines with oxygen or chlorine to give only one compound, the equivalent mass is fixed. If the element has variable valency, it combines with oxygen or chlorine to give more than one compound and element shows variable equivalent mass. For example Copper combines with oxygen to give tow oxides namely red oxide ( Cu_{2}O) and black oxide (CuO).
* In red oxide (Cu_{2}O)
16 P/M of oxygen combines with 2 x 63.5 P/M of Cu
8 P/M combines with \( \frac { 2 × 63.5 × 8} {16}\) = 63.5 P/M of Cu
Hence, equivalent mass of Cu in red oxide = 63.5
*In black oxide
16 P/M of oxygen combines with 63.5 P/M of Cu
8 P/M of oxygen combines with \( \frac { 63.5 × 8} {16}\) = 31.75 P/M of Cu
Hence, equivalent mass of Cu in black oxide = 31.75
Relation between atomic mass, equivalent mass, and valency
Let 'A', 'E' and 'n' be the atomic mass, equivalent mass and valency of an element respectively. The valency of an element is the number of Hydrogen atoms which combines with one atom of that element
→ n atoms of hydrogen combines with one atom of element
→ n x 1.008 P/M of hydrogen combines with 'A' P/M of element
→1.008 P/M of hydrogen combines with \( \frac {A}{n}\) P/M of element
which by definition is equivalent mass of element
∴ Equivalent mass (E) = \( \frac {Atomic\:Mass\:(A)}{Valency\:(n)}\)
Equivalent masses of substances
a) Equivalent mass of an element = \( \frac {Atomic\:Mass}{Valency}\)
Example: Equivalent mass of Mg = \( \frac {24}{2}\) = 12
Equivalent mass of carbon = \( \frac {12}{4}\) = 3
Equivalent mass of Aluminium = \( \frac {27}{3}\) = 9
b) Equivalent mass of acid = \( \frac {Molecular\:Mass}{Basicity}\)
Basicity of acid is the number of H^{+} ion displaced per molecule of acid.
- Equivalent mass of H_{2}SO_{4} = \( \frac {98}{2}\) = 49
- Equivalent mass of HCl = \( \frac {36.5}{1}\) = 36.5
- Equivalent mass of CH_{3}COOH = \( \frac {60}{1}\) = 60
- Equivalent mass of oxailic acid (anhydrous) i.e. (COOH)_{2} = \( \frac {90}{2}\) = 45
- Equivalent mass of oxailic acid (hydrated) i.e. (COOH)_{2}.2H_{2}O = \( \frac {126}{2}\) = 63
- Equivalent mass of H_{3}PO_{4} = \( \frac {98}{3}\) = 32.67
# Calculate equivalent mass of H_{2}SO_{4} in following two reactions
a ) NaOH + H_{2}SO_{4 }→NaHSO_{4} + H_{2}O
Here, basicity of H_{2}SO_{4} = 1
Hence, equivalent mass of H_{2}SO_{4} = \( \frac {98}{1}\) = 98
b) NaOH + H_{2}SO_{4 }→Na_{2}SO_{4} + H_{2}O
Here, basicity of H_{2}SO_{4} = 2
Hence, equivalent mass of H_{2}SO_{4} = \( \frac {98}{2}\) = 49
c) Equivalent mass of base
\( \frac {Molecular\:Mass}{Acidity}\)
Acidity of acid is the number ofOH^{-} ion displaced per molecule of base.
- Equivalent mass of NaOH = \( \frac {40}{1}\) = 40
- Equivalent mass of KOH = \( \frac {56}{1}\) = 56
- Equivalent mass of Ca(OH)_{2} = \( \frac {74}{2}\) = 37
- Equivalent mass of NH_{4}OH = \( \frac {35}{1}\) = 35
* For basic oxide , Equivalent mass = \( \frac {Molecular\:Mass}{2 × Number\: of \: oxygen\: atoms\: per\:molecule}\) ]
- Equivalent mass of CaO = \( \frac {56}{2}\) = 28
- Equivalent mass of Al_{2}O_{3} = \(\frac {102}{6}\) = 17
d) Equivalent mass of salts
Equivalent mass of salts = \( \frac {Molecular\:Mass}{Total\:number\:of\:positive\:or\:negative\:charge\:per\:molecule}\)
- Equivalent mass of Na_{2}CO_{3} = \( \frac {106}{2}\) = 53
- Equivalent mass of NaCl = \( \frac {58.5}{1}\) = 58.5
- Equivalent mass of KCl = \( \frac{74.5}{1}\) = 74.5
- Equivalent mass of CaCO_{3} = \( \frac{100}{2}\) = 50
- Equivalent mass of Na_{2}CO_{3}.10H_{2}O = \( \frac {286}{2}\) = 143
e) Equivalent mass of radicals
Equivalent mass of radicals = \( \frac {Mass\:of\:radical}{Charge}\)
- Equivalent mass of SO_{4}^{- - }= \( \frac {96}{2}\) = 48
Equivalent mass of CO_{3}^{- - } = \( \frac {60}{2}\) = 30
The equivalent mass of comopund can also be calculated by the sumation of mass of cation and anion.
-Equivalent mass of H_{2}SO_{4} = Equivalent mass of H^{+} + Equivalent mass of SO_{4}^{- - } = 1 + 48 = 49
Equivalent mass of Na_{2}CO_{3} = Equivalent mass of Na^{+} + Equivalent mass of CO_{3}^{- - }= 13 + 30 = 43
f) Equivalent mass of oxidising or reducing agent
Equivalent mass of oxidising or reducing agent = \( \frac {Molecular\:Mass}{Change\:in\:Oxidation\:Number\:(Or\:Number\:of\:electrons\:involved\:per\:molecule}\)
Example: KMnO_{4} (Oxidising agent)
KMnO_{4} shows different equivalent mass in different medium
i) Acidic medium
Molecular mass of KMnO_{4} = 158
Change in oxidation number = 5
Hence, equivalent mass = \( \frac {158}{5}\) = 31.6
ii) Alkaline medium
Molecular mass of KMnO_{4} = 158
Change in oxidation number = 1
Hence, equivalent mass = \( \frac {158}{1}\) = 158
c) Neutral medium
Molecular mass of KMnO_{4} = 158
Change in oxidation number = 3
Hence, equivalent mass = \( \frac {158}{3}\) = 52.66
# Equivalent mass of K_{2}Cr_{2}O_{7} (Oxidising agent)
Molecular mass = 284
Change in oxidation number = 2 x 3 = 6
Equivalent mass = \( \frac {284}{6}\) = 49
# Equivalent mass of oxalic acid (Reducing agent)
Total change in oxidation number = 1 x 2 = 2
Molecular mass (anhydrous) = 90 i.e. Eq. mass = \( \frac {90}{2}\) = 45
Molecular mass (hydrated) = 126 i.e. Eq. mass = \(\frac{126}{2}\) = 63
n
A. By Hydrogen Displacement Method
Metals above hydrogen in electrochemical series can displace hydrogen from dilute mineral acid (non-oxidising acids) since they are more electropositive than hydrogen. Thus,equivalent mass of such metals like (Zn, Fe, Al, Pb, Sn, Hg) can be determined by hydrogen displacement method. In this method accurately weighed the quantity of metal is treated with the excess of non-oxidising acids like dil. HCl or dil. H_{2}SO_{4}.
Mg + H_{2}SO_{4} → MgSO_{4} + H_{2} ↑
Zn + 2HCl → ZnCl_{2} + H_{2}↑
The volume of hydrogen gas displaced during the reaction is collected over water in eudiometer tube. The volume of H_{2} gas evolved after completion of the reaction is noted at lab temperature and pressure. Then this volume is converted into NTP by using combined gas equation. The volume of H_{2} gas displaced at NTP is then multiplied by 0.000089 gm to get the mass of hydrogen gas displaced (since the mass of 1 ml of H_{2} at NTP= 0.000089 gm). Then the equivalent mass of metal is given by:
Equivalent mass = \( \frac {Mass\:of\:metal\:taken}{Mass\:of\:hydrogen\:displaced} × 1.008 \)
Calculation:
At lab condition,
Mass of metal taken = x gm (known)
Volume of H_{2} gas displaced (V_{1}) = ? ml ( Obtained from eudiometer reading in lab)
Lab temperature = t_{1}^{0}C = (t_{1} + 273) K
Aqueous tension = f mmHg
Atmospheric pressure = P mmHg (Pressure of moist gas)
Pressure exerted by dry gas (P_{1}) = ( P - f) mmHg
The volume of hydrogen gas displaced in lab condition should be converted into colume at NTP.
At NTP,
The volume of H_{2} gas displaced (V_{2}) = ?
Temperature at NTP = 273K
Pressure at NTP = 760 mmHg
Using combined gas equation ,
\( \frac {(P - f) V_1}{T_1}\) = \( \frac {P_2 V_2}{T_2}\)
or, V_{2} = \( \frac {(P - f) V_1T_2}{T_1 P_2}\)
or, V_{2} = \( \frac {(P - f) }{T_1} × \frac{273}{760}\)
Now,
Mass of H_{2} gas displaced at NTP = V_{2} x 0.000089 gm
So, V_{2} x 0.000089 gm of H_{2} is displaced by x gm of metal
1.008 gm of H_{2} is displaced by \( \frac { x × 1.008}{V_2 × 0.000089}\) gm of metal
Hence, equivalent mass = \( \frac { x × 1.008}{V_2 × 0.000089}\)
Limitation
i) This method is not applicable to determine the equivalent mass of less active metals which lies below hydrogen in electrochemical series like Cu, Ag, Au since they cannot displace hydrogen from non-oxidising acids.
B. By Oxide formation method
This method is applicable when metal can form oxide directly or indirectly. If a metal cannot displace hydrogen from nonoxidising acids but can form oxide then oxide formation method to determine its equivalent mass. In this method, a fixed mass of metal is converted to oxide directly or indirectly .
(i) Some metals like Mn, Al, Mg can be converted into their oxides directly when heated with the excess of oxygen.
(ii) Some metals like Zn, Cu, Pb do not react easily with oxygen. Such metals are converted into their oxides indirectly. For this, a fixed mass of metal is completely dissolved in conc. HNO_{3} to get metal nitrate, metal nitrate thus obtained is then heated strongly to get metal oxide.
Cu + 4HNO_{3} → Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O
2Cu(NO_{3})_{2} 2CuO + 4NO_{2} + O_{2}
Then, the mass of oxygen which combines with metal is calculated and the equivalent mass of metal is given by the formula:
Equivalent mass = \( \frac {Mass\: of \: metal \:taken} {Mass\:of\:oxygen} × 8\)
Calculation
A fixed mass of metal is taken in a crucible and is converted into its oxide directly or indirectly. Then, the crucible is cooled and weighed until the constant mass is observed. The constant mass is then noted and the mass of oxygen which combines with metal is calculated.
Let, mass of empty crucible = M_{1 }gm
Mass of (crucible + metal ) = M_{2 }gm
∴ Mass of metal = (M_{2} - M_{1}) gm
Mass of (crucible + metal oxide) = M_{3} gm
Mass of metal oxide = ( M_{3} - M_{1}) gm
Mass of oygen = ( M_{3} - M_{1}) - (M_{2} - M_{1}) = (M_{3} - M_{2}) gm
Now, (M_{3} - M_{2}) parts by mass of oxygen combines with (M_{2} - M_{1}) P/M of metal
8 P/M of oxygen combines with \( \frac {M_2 - M_1}{M_3 - M_2} × 8\) P/M of metal
Hence, equivalent mass = \( \frac {M_2 - M_1}{M_3 - M_2} × 8\)
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