Determination of Equivalent Mass

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  • Equivalent mass of a substance is the number of parts by mass of the substance which combines with or displaces or contains 1.008 parts by mass of hydrogen, 8 part by mass of oxygen or 35.5 part by mass of chlorine. If the equivalent mass is expressed in grams then it is called gram equivalent mass (GEM).
  • Illustration: 1.008 parts by weight of Hydrogen combines with 35.5 parts by weight of chlorine to give 36.5 parts by weight of HCI. Thus the equivalent mass of chlorine is 35.5.
  • Equivalent mass has no unit because it is a pure ratio. Some important equivalent masses are H = 1, O = 8, Cl = 35.5

Determination of Equivalent Mass

Method – I (Hydrogen Displacement Method):

Principle:

  • The known mass of a metal to react with dilute acids and volume of hydrogen produced in the reaction is measured and the equivalent mass of an element is calculated using formula

Equivalent mass 01

  • This method is useful for the metals which react, or dissolves in mineral acids and liberates hydrogen gas. e.g. Mg, Zn, Al, Ca, Zn, Sn etc.

Procedure:

  • Clean and weigh accurately a piece of metal (Mg / Zn / Al) whose equivalent mass is to be found and place it in a conical flask containing distilled water.
  • The mouth of the conical flask is fitted with a cork through which side tube (gas carrying tube) and a thistle tube are inserted as shown in the diagram. Using side tube the conical flask is connected to graduated (calibrated) test tube called Eudiometer. Eudiometer tube is completely filled with water and is inverted on the side tube as shown in the diagram. The bottom part of thistle tube is dipped in the water in the flask. Mineral acid like HCl is added to the flask through the thistle funnel.
  • The reaction between the metal and acid takes place and hydrogen gas is liberated which is collected in eudiometer by downward displacement of water.
  • When the evolution of gas has stopped the mouth of the eudiometer tube is closed with a thumb and carried to a tray of water where the level of water inside and outside the tube is equalised and the volume of hydrogen gas is read at atmospheric pressure.

Observations:

  • The weight of metal piece = W g
  • The volume of hydrogen collected = V dm³
  • Atmospheric pressure  = P mm of Hg
  • Aqueous tension = f mm of Hg
  • Room temperature = t °C
  • Absolute temperature = T K

Calculations:

Now the volume of hydrogen at S.T.P. from the above data using following formula is calculated.

Equivalent mass 03



Where Volume of hydrogen at S.T.P.  Vo dm³,

Pressure at S.T.P. Po mm of Hg = 760 mm = 1.013 x 105 N/m2

The absolute temperature at S.T.P.  To K = 273 K

Equivalent mass 04



Notes:

  • Since Hydrogen is collected over water and it is moist. Hence, Pressure of hydrogen = (P – f) mm of Hg is used.
  • This method does not give accurate results.

Example – 1:

  • 0.205 g of a metal on treatment with a dil. acid gave 106.6 ml of hydrogen collected over water at 755 mm pressure and 17 °C. Calculate the equivalent mass of the metal if aqueous tension at 17 °C is 14.4 mm.
  • Solution:
  • Given: W = 0.205 g, V = 106.6 ml, P = 755 mm of Hg, f = 14.4 mm of Hg, (P – f) = 755 – 14.4 = 740.6 mm of Hg, t = 17 °C, T = 17 + 273 = 290 K, Po = 760 mm of Hg, To = 273 K.

Equivalent mass 05

Ans: the equivalent mass of the metal is 23.5

Example – 2:

  • 0.33 g of a metal on dissolving in dilute sulphuric acid, liberates 113 ml of dry hydrogen at NTP. Determine the equivalent mass of the metal.
  • Solution:
  • Given: W = 0.33 g, Vo = 113 ml = 0.113 dm³

Equivalent mass 06

Ans: the equivalent mass of the metal is 32.71

Example – 3:

  • 0.05 g of a metal on treatment with a dilute acid gave 51 ml of hydrogen collected over water at 751 mm pressure and 27O C. Calculate the equivalent mass of the metal.
  • Solution: 
  • Given: W = 0.05 g, V = 51 ml, P = 751 mm of Hg, f = 0 (not mentioned in problem),  P – f = P, t = 27 °C, T = 27 + 273 = 300 K, Po = 760 mm of Hg, To = 273 K

Equivalent mass 07  Ans: the equivalent mass of the metal is 12.21



Example – 4:

  • 0.0396 g of metal was completely dissolved in dilute hydrochloric acid and hydrogen evolved is mixed with dry oxygen. Then the mixture is sparked. 13.75 ml of dry oxygen at 27 °C and 680 mm pressure is required for complete combustion of oxygen formed. Find the equivalent mass of the metal.
  • Solution: 
  • Given: W = 0.0396 g, P = 680 mm of Hg, f = 0 (not mentione), t = 27O C, T = 27 + 273 = 300 K, Po = 760 mm of Hg, To = 273 K

Combustion of hydrogen is represented as

2 H2(g) + O2(g)   → 2H2O(g))

2 vol          1 vol           2 vol

Thus  1vol of oxygen combines with 2 vol of hydrogen.

Using Gay-Lussac’s law of combining volume we can say that



13.75 ml of oxygen combines with 2 x 13.75 = 27.5 ml of hydrogen.

∴  V =  27.5 ml

Equivalent mass 08

Ans: the equivalent mass of the metal is 19.80

Method – II (Oxide Formation Method):

Procedure:

  • A known mass of an element is reacted with oxygen. Mass of the oxide formed is measured. The mass of oxygen in the oxide is calculated using formula

Mass of oxygen = Mass of the oxide – Mass of the element



The equivalent mass is calculated using the formula

Example – 5:

  • 0.4 g of metal, when heated in air, gave 0.72 g of the metal oxide. Find the equivalent mass of the metal.
  • Solution:
  • Given: Mass of metal = 0.4 g, Mass of oxide = 0.72 g

Mass of oxygen = 0.72  – 0.4  = 0.32 g



Ans: The equivalent mass of the metal is 10.

Example -7:

  • If the mass of the copper taken is 0.324 g and mass of the product on heating its nitrate is 0.406 g. calculate atomic mass of copper.
  • Solution:
  • Given: Mass of metal = 0.324 g, Mass of oxide = 0.406 g,

Mass of oxygen = 0.406 – 0.324 = 0.082 g

Ans: Hence equivalent mass of the metal is 31.61.

Example – 8:

  • 1.08 g of metal oxide on heating decomposes to give pure metal and 56.0 ml of oxygen at NTP. What is the equivalent mass of metal?
  • Solution:
  • Given: Mass of metal oxide = 1.08 g, Volume of oxygen at NTP = 56.0 ml = 0.056 dm3.

One mole of any gas occupies 22.4 dm3 by volume at NTP.



 

Mass of metal = Mass of oxide – Mass of oxygen = 1.08 – 0.08 = 1 g

Ans: The equivalent mass of the metal is 100.



Example – 9:

  • 0.139 g of a metal, when dissolved in dilute hydrochloric acid, evolved 29.5 ml of hydrogen when collected over water at 13 °C and 741 mm pressure. What would be the mass of the oxygen present in 100 g of the oxide of the metal if the aqueous tension at 13 °C is 11.2 mm?
  • Solution:
  • Given: W = 0.139 g, V = 29.5 ml, P = 741 mm of Hg, f = 11.2 mm of Hg, T = 13 °C = 13 + 273 = 286 K, PO = 760 mm of Hg, TO = 273 K.

Let ‘x’ g be the mass of oxygen in the oxide

Mass of oxide = 100 g, Mass of oxygen = x g

∴ Mass of metal = (100 – x) g

∴  57.6 x = 800 – 8x



∴  65.6 x = 800

∴  x = 800/ 65.6 = 12.20 g

Ans: 100 g of metal contains 12.20 g of oxygen

Method – III (Reduction Method):

Procedure:

  • A known mass of a metal oxide is reduced to metal. Mass of the metal obtained is measured. The mass of oxygen in the oxide is calculated using formula

Mass of oxygen = Mass of oxide – Mass of element

The equivalent mass is calculated using the formula

Example – 10:

  • 1.44 g of the metal oxide on reduction gave 0.8 g of metal. Find the equivalent mass of the metal.
  • Solution:
  • Given: Mass of metal = 0.8 g, Mass of oxide = 1.44 g, Mass of oxygen = 1.44  – 0..8 = 0.64 g

Ans: The equivalent mass of the metal is 10.

Example – 11:

  • On heating 0.8567 g of copper oxide in a current of hydrogen, the resulted in the formation of 0.6842 g of copper. Find the atomic mass of copper.
  • Solution:
  • Given: Mass of copper = 0.6842 g, Mass of oxide = 0.8567 g

Mass of oxygen = 0.8567 – 0.6842  = 0.1725 g

Ans: The equivalent mass of the metal is 31.63.

Method – IV (Equivalent Mass by Chloride Formation Method):

Procedure:

  • A known mass of a metal is reacted with chlorine. Mass of the chloride obtained is measured. The mass of chlorine in the chloride is calculated using formula

Mass of chlorine = Mass of chloride – Mass of the element

The equivalent mass is calculated using the formula

 

Example – 12:

  • 2.00 g of metal yielded 2.656 g of its chloride. Find the equivalent mass of the metal.
  • Solution:
  • Given: Mass of metal = 2.00 g, Mass of chloride = 2.656 g

Mass of chlorine = 2.656 – 2.00 = 0.656 g

Ans: The equivalent mass of the metal is 108.2.

Example – 13:

  • The chloride of a metal contained 52.85 % of metal. What is the equivalent mass of the metal?
  • Solution:
  • Given: % of metal = 52.85,

% of Chlorine = 100 – 52.85 = 47.15, Consider 100 g of chloride

Mass of metal = 52.85 g, Mass of chloride = 47.15 g

Ans: The equivalent mass of the metal is 39.79.

Method – V (Equivalent Mass by Double Decomposition Method):

Procedure:

  • In this method, a known mass of a compound (say AB) is treated with a known mass of another compound say (CD). By the exchange of radicals, the new compound is formed. The mass of new compound formed is found.

AB    +   CD  →  AD   +   CB

Then atomic mass is calculated using following formula.

Example – 14:

  • 0.106 g of sodium carbonate was treated with an excess of calcium carbonate and mass of calcium carbonate was found to be 0.1 g. Find the equivalent mass of calcium carbonate if that of sodium carbonate is 53.
  • Solution:
  • Given: Mass of Na2CO3 =  0.106 g. Mass of CaCO3 = 0.1 g, Eq. Mass of  Na2CO3 = 53,
  • To Find: Eq. Mass of  CaCO3 = ?

Ans: The equivalent mass of CaCO3 is 50.

Example – 15:

  • 0.194 g of chloride of a certain metal, when dissolved in water and treated with an excess of silver nitrate yield 0.50 g of silver chloride. Calculate the equivalent mass of the metal. (Ag = 108, Cl = 35.5)
  • Solution:
  • Given: Mass of metal chloride =  0.194 g, Mass of silver chloride = 0.50 g,
  • To Find: Eq. Mass of  Metal = ?

Let the equivalent mass of metal be E.

∴  0.194 × (108 + 35.5) = 0.50 ( E + 35.5)

∴  0.194 × 143.5 =0.50 E + 17.75

∴  27.839 – 17.75 = 0.50 E

∴  10.089 = 0.50 E

∴  E = 10.089 / 0.50 = 20.17

Ans: The equivalent mass of metal is 20.17.

Example – 16:

  • 1.520 g of the hydroxide of a metal gave 0.995 g of its oxide. Calculate the equivalent mass of the metal.
  • Solution:
  • Given: Mass of hydroxide =  1.520 g, Mass of oxide = 0.995 g

Let the equivalent mass of metal be E.

∴  1.520 × (E + 8) = 0.995 ( E + 17)

∴  1.520 E +  12.16  = 0.995 E + 16.915

∴  1.520 E – 0.995 E   =  16.915 – 12.16

∴  0.525 E = 4.755

∴  E = 4.755 / 0.525  = 9.06

Ans: The equivalent mass of metal is 9.06.

Example – 17:

  • 1.0 g of an acid when completely acted upon by magnesium gave 1.301 g of anhydrous magnesium salt. Find the equivalent mass of the acid. Mg = 24, H = 1.
  • Solution:

Atomic mass of Mg = 24

Equivalent mass of Mg = Atomic mass /valency =  24 / 2 = 12

Mass of acid =  1.0 g

Mass of magnesium salt = 1.301 g

Let equivalent mass of acid be E.

∴  1 × (E + 12) = 1.301 ( E + 1)

∴  E + 12  = 1.301 E + 1.301

∴  12 – 1.301  = 1.301 E – E

∴  0.301 E = 10.699

∴  E = 10.699 / 0.301  = 35.54

Equivalent mass of acid = 1 + 35.54 = 36.54

Ans: Hence equivalent mass of acid is 36.54.

Example – 18:

  • Chloride of a metal ‘M’ contains 47.23% of the metal. 1.00 g of this metal displaced from a compound 0.88 g of another metal N. Find equivalent masses of M and N respectively.
  • Solution:

In chloride % of metal = 47.23, hence % of chlorine = 100 – 47.23 = 52.77

Let us consider 100 g of chloride

Mass of metal = 47.23 g, Mass of chlorine = 52.77 g

Hence equivalent mass of the metal M is 31.77.

Given that, 1.00 g of metal M displaced from a compound 0.88 g of another metal N.

Hence equivalent mass of N = 31.77 x 0.88 = 27.96

Ans: The equivalent mass of M is 31.77 and that of N is 27.96..

Example – 19:

  • 4.215 g of metallic carbonate was heated in a hard glass tube and CO2 evolved was found to measure 1336 ml at 27 °C and 700 mm of pressure. What is the equivalent mass of the metal? (IITJEE 1976).
  • Solution:
  • Given: V = 1336 ml, P = 700 mm of Hg, T = 27 °C = 27 + 273 = 300 K, PO = 760 mm of Hg, TO = 273 K

Molecular mass of CO2 = 12 + 16 x 2 = 44 g

1 mole of CO2 at STP occupies 22.4 dm3 by volume.

44 g of CO2 at STP occupies 22.4 dm3 by volume.

Mass of metal carbonate =  4.215 g

Mass of CO2 evolved = 2.2 g

Mass of metal oxide  = 4.215 – 2.2 = 2.015 g

Let the equivalent mass of metal be E.

∴  4.215 × (E + 8) = 2.015 ( E + 30)

∴ 4.215 E + 33.72  = 2.015 E + 60.45

∴ 4.215 E –  2.015 E =  60.45 – 33.72

∴  2.2 E = 26.73

∴  E = 26.73 / 2.2  = 12.15

Ans: The equivalent mass of metal is 12.15.

Example – 20:

  • For dissolution of 1.08 g of metal 0.49 g of sulphuric acid was required. If the specific heat of metal is 0.06. Find the atomic mass of the metal.
  • Solution:

Mass of metal =  1.08 g

Mass of sulphuric acid = 0.49 g

Let the equivalent mass of metal be E.

Hence equivalent mass of metal is 108.

Specific heat = 0.06

Actual atomic mass = Equivalent mass x valency = 108 x 1 = 108

Ans: The atomic mass of the metal is 108.

Method – VI (Metal Displacement Method):

Procedure:

  • In this method known mass of metal is added to the solution of a salt of the other (Placed lower in electrochemical series).
  • Metal A + Salt of metal B → Salt of metal A + Metal B
  • The precipitate formed is washed dried and carefully weighed.
  • Equivalent mass is calculated using following formula.

Example – 21:

  • 1.296 g of silver metal was displaced when 0.382 g of copper was added to the solution of silver sulphate. If the equivalent mass of silver metal is 108. Find that of copper.
  • Solutions:
  • Given: Mass of silver = 1.296 g, Mass of copper = 0.382 g, Eq. mass of silver = 108
  • To Find: Eq. mass of copper =?

Ans: The equivalent mass of copper is 31.83

Example – 22:

  • 1.8 g of iron displaces 2.04 g copper from copper sulphate solution. If copper has an equivalent mass of 31.7. Find that of iron.
  • Solution: Mass of iron = 1.8 g, Mass of copper = 2.04 g. Eq. mass of copper = 31.7
  • To Find: Eq. mass of iron =?

Ans: The equivalent mass of iron is 27.97.

Example – 23:

  • 2.47 g of CuO obtained by oxidising 1.986 g of copper by nitric acid. 0.335 g of copper was precipitated by 0.346 g of zinc from CuSO4. Find the equivalent mass of copper and zinc.
  • Solutions: Mass of CuO = 2.47 g. Mass of copper = 1.986 g.
  • To Find: the equivalent mass of copper and zinc =?

Mass of oxygen = 2.47 – 1.986 = 0.484 g

Eq. mass of copper = 32.83

Ans: The equivalent mass of copper is 32.83 and that of zinc zinc is 33.91

Method – VII (Faraday’s Electrolysis Method):

Use of Faraday’s first law of electrolysis:

Statement :

  • The weight of any substance deposited or liberated or dissolved at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

Steps Involved:

  • A known quantity of electricity (Q = i t) is passed through electrolyte solution and mass (w) of the substance deposited or liberated during electrolysis is measured.
  • Using the following relation value of electrochemical equivalent (z) is calculated.

w = z i t

  • Then equivalent mass is calculated by the formula

Equivalent mass = 96500 x z

Example – 24:

  • On passing a current of 0.5 A through a solution of a salt of a metal for 32 minutes 0.3158 g of the metal was deposited. What is the equivalent mass of the metal?
  • Solution:
  • Given: i = 0.5 A, t = 32 min = 32 x 60 s , W =  0.3158

By Faraday’s first law of electrolysis,

W = z i t

Now, Equivqlent mass = 96500 x z

Equivalent mass = 96500 x 3.29 x 10-4 = 31.74

Ans: The equivalent mass of metal is 31.74

Example – 25:

  • On passing a current of 0.6 A through a solution of a salt of copper for 20 minutes 0.24 g of the copper was deposited. What is the equivalent mass of copper?
  • Solution:
  • Given: i = 0.6 A, t = 20 min = 20 x 60s, W =  0.24 g

By Faraday’s first law of electrolysis,

W = z i t

Now, Equivalent mass = 96500 x z

Equivalent mass = 96500 x 3.33 x 1-4 = 31.84

Ans: The equivalent mass of copper is 31.84

Use of Faraday’s second law of electrolysis:

Statement :

  • When the same quantity of electricity is passed through different electrolytes (generally connected in series), the weights of different substances deposited or liberated or dissolved at the respective electrodes are directly proportional to their chemical equivalents (equivalent weights).

Steps Involved:

  • The same quantity of electricity is passed through the solution of different electrolytes, the masses of different substances liberated or evolved as a result of electrolysis are noted.
  • Then equivalent mass is calculated by the formula.

Where W1 = mass of the first substance deposited

W2 = mass of the second substance deposited

E1 = Equivalent mass of the first substance

E2 = Equivalent mass of the second substance

Example – 26:

  • An electric current is passed through two cells containing CuSO4 and AgNO3 solutions respectively connected in series. The masses of copper and silver deposited are 0.424 g and 1.44 g respectively. Find the equivalent mass of copper if that of silver is 108.
  • Solution:
  • Given: WCu = 0.424 g,  WAg = 1.44 g, EAg = 108
  • To Find: ECu =?

By Faraday’s second law of electrolysis,

Ans: The equivalent mass of copper is 31.8

Example – 27:

  • The same quantity of electricity that liberated 2.158 g of silver was passed through a solution of a gold salt and 1.314 g of gold was deposited. The equivalent mass of silver is 107.9. Calculate the equivalent mass of gold. also find oxidation state and valency of gold.
  • Solution:
  • Given: WAg = 2.158 g, WAu = 1.314 g, EAg = 107.9
  • To Find: EAu =?

By Faraday’s second law of electrolysis,

Ans: The equivalent mass of gold is 65.7

Ans: The equivalent mass of gold is 65.7 and its oxidation state is+3. Valency is 3.

Equivalent-Mass of Acids:

  • One gram equivalent mass of an acid is that mass of it which contains one gram equivalent mass of replaceable hydrogen atoms.
  • Thus the equivalent mass of an acid depends on the replaceable hydrogen atoms it contains per mole. The number of replaceable hydrogen atoms present in a molecule of an acid is called the basicity of the acid.

  • Illustration – 1:

Molecular mass of HCl = 1 + 35.5 = 36.5

  • Illustration – 2:

Molecular mass of H2SO4 = 2  + 32 + 64 = 98

Equivalent-Mass of Base:

  • One gram equivalent mass of a base is that mass of it which contains one gram equivalent mass of the hydroxyl radical.
  • Thus the equivalent mass of a base depends on the number of hydroxyl radicals it contains per mole. The number of hydroxyl radical present in a molecule of a base is called the acidity of the base.

  • Illustration – 1:

Molecular mass of NaOH = 23 + 16 + 1 = 40

  • Illustration – 2:

Molecular mass of Ca(OH)2= 40 + (16+1) x 2 = 74

Equivalent-Mass of Salts:

  • Equivalent mass of a simple salt is that mass of it which contains one gram equivalent of the metal or a radical
  • Illustration – 1:

Molecular mass of KCl = 39 + 35.5 = 74.5

In this case, KCl contains  1 gram equivalent of K and 1 gram equivalent of Cl. Hene equivalent mass of KCl is 74.5 / 1 = 74.5.

  • Illustration – 2:

Molecular mass of AlCl3 = 27 + 35.5 x 3  = 133.5

In this case AlCl2 contains  1 gram equivalent of Al and 3 gram equivalent of Cl. Hene equivalent mass of KCl is 133.5 / 3 = 44.5.

Equivalent mass of a salt is also that mass of it, which will combine with one gram equivalent of another substance.

  • Illustration: To find equivalent mass of Na2CO3

Na2CO3  reacts with HCl as

Na2CO3   + 2HCl  → 2 NaCl  + CO2 + H2O

Molecular mass of Na2CO3 = 23 x 2  + 12 x 1  + 16 x 3 = 106

one gram equivalent of Na2CO3 reacts with 2 gram equivalent of HCl. Hence equivalent mass of Na2CO3 is 106 / 2 = 53.

Equivalent Mass of Oxidising and Reducing Agents:

Note:

  • Metals with variable valency show variable equivalent masses depending upon their valency in the compound. For e.g. in oxides FeO, Fe2O3 and Fe3O4 the equivalent masses of Fe are 28.18.6 and 21 respectively.

Gram Equivalent:

  • The equivalent mass expressed in grams is called gram equivalent mass (GEM)

Milliequivalent:

  • A milliequivalent is one-thousandth of an equivalent mass of any substance is the equivalent mass expressed in milligrams. It is the unit which is used to express the concentration of electrolytes in tissue fluids of animals and plants.
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