In this article, we shall study the determination of the equivalent mass of acid, base, and salt.

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 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.

Illustration – 3:

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

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.

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Equivalent Masses of Acids, Bases, and Salts

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In the last few articles, we have studied the hydrogen displacement method, oxide formation method, reduction method, chloride formation method, and double displacement method to determine the equivalent mass of metal. In this article, we shall study use of Faraday’s laws of electrolysis to determine the equivalent mass of a substance.

The 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

Method – VII (Faraday’s First Law of Electrolysis):

Statement :

The mass 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

Numerical Problems on faraday’s First Law of Electrolysis

Example – 01:

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?

Given: i = 0.5 A, t = 32 min = 32 x 60 s , W =  0.3158

Solution:

By Faraday’s first law of electrolysis,

W = z i t

z = W/ (i t) = 0.3158/(0.5 x 32 x 60)

z = 3.29 x 10^-4 g/C

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 – 02:

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?

Given: i = 0.6 A, t = 20 min = 20 x 60s, W =  0.24 g

Solution:

By Faraday’s first law of electrolysis,

W = z i t

z = W/ (i t) = 0.24/(0.6 x 20 x 60)

z = 3.33 x 10^-4 g/C

Now, Equivalent mass = 96500 x z

Equivalent mass = 96500 x 3.33 x 10^-4 = 31.84

Ans: The equivalent mass of copper is 31.84

Method – VIII (Faraday’s Second Law of Electrolysis):

Statement:

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

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

Numerical Problems on faraday’s Second Law of Electrolysis

Example – 01:

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.

Given: W_Cu = 0.424 g,  W_Ag = 1.44 g, E_Ag = 108

To Find: E_Cu =?

Solution:

By Faraday’s second law of electrolysis,

Ans: The equivalent mass of copper is 31.8

Example – 02:

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.

Given: W_Ag = 2.158 g, W_Au = 1.314 g, E_Ag = 107.9

To Find: E_Au =?

Solution:

By Faraday’s second law of electrolysis,

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

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Use of Laws of Electrolysis

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In the last two articles, we have studied the hydrogen displacement method, oxide formation method, reduction method, and chloride formation method to determine the equivalent mass of metal. In this article, we shall study double displacement method and metal displacement method.

The 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

Method – V (Double Displacement 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 the new compound formed is found.

AB    +   CD  →  AD   +   CB

Then equivalent mass is calculated using following formula.

Numerical Problems on Double Displacement Method

Example – 01:

0.106 g of sodium carbonate was treated with an excess of calcium carbonate and the 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.

Given: Mass of Na₂CO₃ =  0.106 g. Mass of CaCO₃ = 0.1 g, Eq. Mass of  Na₂CO₃ = 53,

To Find: Eq. Mass of  CaCO₃ = ?

Solution:

Ans: The equivalent mass of CaCO3 is 50.

Example – 02:

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)

Given: Mass of metal chloride =  0.194 g, Mass of silver chloride = 0.50 g,

To Find: Eq. Mass of  Metal = ?

Solution:

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 – 03:

1.520 g of the hydroxide of metal gave 0.995 g of its oxide. Calculate the equivalent mass of the metal.

Given: Mass of hydroxide =  1.520 g, Mass of oxide = 0.995 g

Solution:

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 – 04:

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 – 05:

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 – 06:

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).

Given: V = 1336 ml, P = 700 mm of Hg, T = 27 °C = 27 + 273 = 300 K, P_O = 760 mm of Hg, T_O = 273 K

Solution:

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 – 07:

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.

Numerical Problems on Metal Displacement Method

Example – 01:

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.

Given: Mass of silver = 1.296 g, Mass of copper = 0.382 g, Eq. mass of silver = 108

To Find: Eq. mass of copper =?

Solutions:

Ans: The equivalent mass of copper is 31.83

Example – 02:

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.

Given: Mass of iron = 1.8 g, Mass of copper = 2.04 g. Eq. mass of copper = 31.7

To Find: Eq. mass of iron =?

Solution: 

Ans: The equivalent mass of iron is 27.97.

Example – 03:

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.

Given: Mass of CuO = 2.47 g. Mass of copper = 1.986 g.

To Find: the equivalent mass of copper and zinc =?

Solution: 

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

Previous Topic: Equivalent Mass by Oxide Formation, Chloride Formation, and Reduction Method

Next Topic: Equivalent Mass Using Faraday’s Laws of Electrolysis

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Double Displacement Method

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In the last article, we have studied the hydrogen displacement method to find the equivalent mass of an element. In this article, we shall study three more methods oxide formation method, reduction method, and chloride formation method.

The 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.

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

Numerical Problems on Oxide Formation Method:

Example – 01:

0.4 g of metal, when heated in air, gave 0.72 g of the metal oxide. Find the equivalent mass of the metal.

Given: Mass of metal = 0.4 g, Mass of oxide = 0.72 g

Solution:

Mass of oxygen = 0.72  – 0.4  = 0.32 g

Ans: The equivalent mass of the metal is 10.

Example – 02:

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 the chemical equivalent of copper.

Given: Mass of metal = 0.324 g, Mass of oxide = 0.406 g,

Solution:

Mass of oxygen = 0.406 – 0.324 = 0.082 g

Ans: Hence chemical equivalent of the metal is 31.61.

Example – 03:

1.08 g of metal oxide on heating decomposes to give pure metal and 56.0 ml of oxygen at NTP. What is the chemical equivalent of metal?

Given: Mass of metal oxide = 1.08 g, Volume of oxygen at NTP = 56.0 ml = 0.056 dm³.

Solution:

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 chemical equivalent of the metal is 100.

Example – 04:

0.139 g of 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?

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, P_O = 760 mm of Hg, T_O = 273 K.

Solution:

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

Numerical Problems on Reduction Method:

Example – 01:

1.44 g of the metal oxide on reduction gave 0.8 g of metal. Find the equivalent mass of the metal.

Given: Mass of metal = 0.8 g, Mass of oxide = 1.44 g, Mass of oxygen = 1.44  – 0..8 = 0.64 g

Solution:

Ans: The equivalent mass of the metal is 10.

Example – 02:

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.

Given: Mass of copper = 0.6842 g, Mass of oxide = 0.8567 g

Solution:

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 – 01:

2.00 g of metal yielded 2.656 g of its chloride. Find the equivalent mass of the metal.

Given: Mass of metal = 2.00 g, Mass of chloride = 2.656 g

Solution:

Mass of chlorine = 2.656 – 2.00 = 0.656 g

Ans: The equivalent mass of the metal is 108.2.

Example – 02:

The chloride of a metal contained 52.85 % of metal. What is the equivalent mass of the metal?

Given: % of metal = 52.85,

Solution:

% 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.

In the next article, we shall study double displacement method and metal displacement method to determine equivalent mass of a substance.

Previous Topic: Equivalent Mass by Hydrogen Displacement Method

Next Topic: Equivalent Mass by Double Displacement Method

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Oxide Formation, Reduction, and Chloride Formation Method

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In the last few articles, we have studied the concept of atomic mass and methods to determine it. In this article, we shall study the hydrogen displacement method to determine the equivalent mass of a metal.

The 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

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

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.

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

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

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

Notes:

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

Numerical Problems

Example – 01:

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.

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, P_o = 760 mm of Hg, T_o = 273 K.

Solution:

Ans: the equivalent mass of the metal is 23.5

Example – 02:

0.33 g of metal on dissolving in dilute sulphuric acid, liberates 113 ml of dry hydrogen at NTP. Determine the equivalent mass of the metal.

Given: W = 0.33 g, V_o = 113 ml = 0.113 dm³

Solution:

Ans: the equivalent mass of the metal is 32.71

Example – 03:

0.05 g of metal on treatment with a dilute acid gave 51 ml of hydrogen collected over water at 751 mm pressure and 27 °C. Calculate the equivalent mass of the metal.

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, P_o = 760 mm of Hg, T_o = 273 K

Solution: 

Ans: the equivalent mass of the metal is 12.21

Example – 04:

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.

Given: W = 0.0396 g, P = 680 mm of Hg, f = 0 (not mentione), t = 27O C, T = 27 + 273 = 300 K, P_o = 760 mm of Hg, T_o = 273 K

Solution: 

Combustion of hydrogen is represented as

2 H_2(g) + O_2(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

Ans: the equivalent mass of the metal is 19.80

In the next article, we shall study oxide formation method, reduction method, and chloride formation method to determine equivalent mass.

Previous Topic: Atomic Mass by Dulong Petit’s Law

Next Topic: Equivalent Mass by Oxide formation, Chloride Formation, and Reduction Method

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Hydrogen Displacement Method

The post Equivalent Mass By Hydrogen Displacement Method appeared first on The Fact Factor.

In the last article, we have studied the concept of atomic mass and to calculate the average atomic mass. In this article, we shall study the Cannizzaro’s method for determination of atomic mass. This method was proposed by Cannizzaro, Stanislao (1826-1910), an Italian chemist.

Principle of Cannizzaro’s Method:

An atom is the smallest part of an element that can be present in a molecule of a compound. Hence the smallest weight of an element contained in the molecular mass of its compounds shall be the atomic mass of that element.

Steps Involved in Cannizzaro’s Method:

1. Collect as many compounds of the element as possible.
2. Determine molecular mass of each compound.
3. Determine percentage composition of these compounds.
4. Calculate the relative mass of that particular element in the molecular mass of each compound from the molecular mass of the compound and its percentage composition.
5. The highest common factor (HCF) of the values obtained gives atomic mass.

Problems Based on Cannizzaro’s Method:

Example – 01:

For series of volatile compounds following data is obtained. Using it calculate atomic mass of carbon.

Solution:

By Cannizzaro’s method HCF of the numbers in last column is 12. Hence the atomic mass of carbon is 12 g

Example – 02:

The pecentage of carbon in its four compounds is 92.2; 62.0; 40.0 and 15.8 respectively. The vapour densities of these compounds are 39; 29; 30 and 38 respectively. Deduce atomic mass of the carbon.

Solution:

HCF of the numbers in last column is 12. Hence the atomic mass of carbon is 12 g

Example – 03:

The vapour densities of five compounds of a certain element are 23, 26, 22, 8.5 and 24 respectively. The percentage of the same element in these compounds are 91.3, 53.8, 63.7; 82.4; and 97.7 respectively. Find atomic mass of the element.

Solution:

By Cannizzaro’s method HCF of the numbers in last column is 14. Hence the atomic mass of the element is 14 g

Example – 04:

Vapour densities of three substances referred to hydrogen as unity were 45, 70 and 25 and percent mass of certain element contained in each were 22.22, 42.86 and 40 respectively. Find the probable atomic mass of the element.

Solution:

By Cannizzaro’s method HCF of the numbers in last column is 20. Hence the atomic mass of the element is 20 g

Example – 05:

Vapour densities of seven compounds of phosphorous phosphoric oxide, phosphorous oxide, phosphorous trichloride, phosphorous pentafluoride, phosphorous oxychloride, phosphorous pentasulphide and tetra phosphorous trisulphide were 150, 110, 70, 63, 77, 111, 114 and percent mass of phosphorous contained in each were 43.7, 56.4, 22.5, 24.8, 20.2, 27.9 and 56.4 respectively. Find the probable and exact atomic mass of phosphorous.

Solution:

By Cannizzaro’s method the approximate HCF of the numbers in last column is 31.1. Hence the probable atomic mass of phosphorous is 31.1 g

To find exact atomic mass we can consider any one compound in the list. Let us consider first compound phosphoric acid. % of phosphorous = 43.7% of oxygen = 100 – 43.7 = 56.3, Mass of phosphorous = 43.7 g Mass of oxygen = 56.3 g

Equivalent mass of an element = (Mass of an element in compound / Mass of oxygen in the compound) × 8

∴ Equivalent mass of an element = (43.7 g / 56.3 g) × 8 g = 6.21 g

Now, Valency = Approx atomic mass / Equivalent Mass = 31.1 / 6.21 = 5 (nearest whole number)

Corrected atomic mass = Equivalent mass x valency = 6.21 g x 5 = 31.05 g

Thus the exact atomic mass of phosphorous is 31.05.

Example – 06:

A metal forms three volatile chlorides containing 23.6, 38.2 and 48.3 per cent of chlorine respectively. The vapour densities of chlorides are 74.6, 92.9 and 110.6 respectively. The specific heat of the metal is 0.055. Find the exact atomic mass of the metal and formulae of its chlorides.

Solution:

The approximate HCF (The least mass) is 114. Hence probable atomic mass of metal is 114.

To find exact atomic mass we can consider any one compound in the list. Let us consider first chloride (100 g).

% of metal = 76.4

% of chlorine = 23.6

Mass of metal = 76.4 g

Mass of chlorine = 23.6 g

Equivalent mass of metal = Mass of metal in chloride x 35.5 / Mass of chlorine in metal chloride

Equivalent mass of metal = 76.4 x 35.5 / 23.6 = 114.9

Valency = Approximate atomiic mass / Equivalent mass = 114 / 114.9 = 1 (Nearest whole number)

Actual atomic mass = Equivalent mass x valency = 114.9 x 1 = 114.9

To find molecular formulae of the chlorides:

 Let x be the valency of the metal, hence its molecular formula is MClx.

First Chloride:

Molecular mass of the first chloride  = 114.9 + 35.5x = 149.2

∴    35.5x = 149.2 – 114.9

∴    35.5x = 34.3

x = 1 (Nearest whole number)

Hence the formula of the first chloride is MCl.

Second Chloride:

Molecular mass of the second chloride = 114.9 + 35.5x = 185.8

∴    35.5x = 185.8 – 114.9

∴     35.5x = 70.9

∴       x = 2

Hence the formula of the second chloride is MCl₂.

Third Chloride:

Molecular mass of the third chloride = 114.9 + 35.5x = 221.2

∴      35.5x = 221.2 – 114.9

∴      35.5x = 106.3

∴       x = 3

Hence the formula of the third chloride is MCl₃.

The exact atomic mass of the metal is 114.9

And formulae of chlorides are MCl, MCl₂, MCl₃ respectively.

In the next article, we shall study to determine atomic mass using the law of isomprphism.

Previous Topic: The Concept of Atomic Mass

Next Topic: Atomic Mass Using the Law of Isomorphism

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The smallest particle of an element which can take part in a chemical reaction is called an atom. According to Dalton’s atomic theory atom is the smallest particle of an element which is indivisible. In modern research, it is proved that the atom is divisible into its constituent particles like electrons, protons, and neutrons. In this article, we shall understand the concept of atomic mass and gram atomic mass (GAM).

Atomic Mass:

Water contains 11.19 % of hydrogen and 88.89% of oxygen. Thus hydrogen and oxygen combine with each other in a ratio 1 : 8 by mass. Besides in water, there are 2 atoms of hydrogen and 1 atom of oxygen. From these two observations, it follows that the mass of oxygen atom is 16 times that of the hydrogen atom. In this hydrogen based system mass of hydrogen is taken as unity and masses of other element were determined relative to the mass of an atom of hydrogen.

A later atom of oxygen was chosen as reference atom because by taking its mass as 16 units, the relative atomic masses of other elements were very close to whole numbers. Oxygen has 3 isotopes. Hence the standard of oxygen was considered as inappropriate. Hence instead of taking the average atomic mass of the mixture of oxygen, stable isotope of carbon (C-12) was taken as standard. Hence in 1961 International Union of Chemists selected the most stable isotope of carbon (C – 12) as a standard atom to compare masses of various elements.

The relative atomic mass of an element is a mass of one atom of the element compared with the mass of an atom of ⁶C₁₂ isotope taken as 12000 units.

Average Atomic Mass:

Isotopes are the atoms of the same element having the same atomic number containing the same number of protons and electrons but different numbers of neutrons hence they possess different mass numbers.

The observed atomic mass of the atom of the element is the average atomic mass of the element taking into consideration the natural abundance of the element. For example, atomic masses of chlorine’s two isotopes are 36 u and 37 u. u stands for unified mass. They are found in the ratio 3: 4 in nature. Hence average atomic mass of chlorine is

The gram atomic mass of an element is atomic mass expressed in grams (GAM). e.g. The gram atomic mass of chlorine is 35.5 g

Thus one gram hydrogen atom means 1.008 g of hydrogen. one gram atom of carbon means 12 g of carbon. 2 gram atom of chlorine means 2 × 35.5 = 71 g of chlorine.

Number of Atoms in Gram Atom:

By Avogadro’s law, one gram atom of an element contains 6.023 × 10²³ atoms. Thus 1 gram atom (i.e. 1.008 g) of hydrogen contains 6.023 × 10²³ atoms. Thus the mass of each atom of hydrogen is

Mass in gram = number of gram atom ×  gram atomic mass

Definition of Valency:

The valency of an element is the number of electrons an atom of the element can accept or donate in the formation of molecule of a compound. OR the number of a hydrogen atom, which can combine with or displaced by one atom of an element is called the valency of the element.

Relation Between Atomic Mass, equivalent Mass, and Valency:

Let A, E, and v be the atomic mass, equivalent mass, and valency of element X. Then the formula of its compound with hydrogen is XH_v.

Thus v parts by mass of hydrogen will combine with A parts by mass of X.

i.e. 1 part by mass of hydrogen will combine with A/v parts by mass of X.

By definition A/v is equivalent mass of the element.

Thus E = A/v

At. Mass (A) = Equ. Mass (E) x Valency (v)

Calculation of Average Atomic Mass by Relative Abundance Method:

Example – 01:

Naturally, occurring lead is found to contain four isotopes 1.40 % ⁸²Pb₂₀₄ isotope with isotopic mass 203.973,  24.10 % ⁸²Pb₂₀₆ isotope with isotopic mass 205.974, 22.10 % ⁸²Pb₂₀₇ isotope with isotopic mass 206.976 and 52.40 % ⁸²Pb₂₀₈ isotope with isotopic mass 207.977. Calculate the average atomic mass of lead.

Solution:

Hence average atomic mass of lead is 207.2 u.

Example – 02:

Naturally, occurring neon is found to contain three isotopes 90.92 % ¹⁰Ne₂₀ isotope with isotopic mass 9.9924, 8.82 % ¹⁰Ne₂₂ isotope with isotopic mass 21.9914, 0.26 % ¹⁰Ne₂₁ isotope with isotopic mass 20.9940 Calculate the average atomic mass of neon.

Solution:

Hence average atomic mass of neon is 20.17 u.

Example – 03:

Naturally, occurring lithium is found to contain two isotopes 8.24 % ³Li₆ isotope with isotopic mass 6.0151 and 91.76 % ³Li₇ isotope with isotopic mass 7.0160 Calculate the average atomic mass of lithium.

Solution:

Hence average atomic mass of lithium is 6.934 u.

Example – 04:

Naturally, occurring silicon is found to contain three isotopes 92.23 % ¹⁴Si₂₈ , 4.67 % ¹⁴Si₂₉, 3.10 % ¹⁴Si₃₀ Calculate the average atomic mass of silicon.

Solution:

Hence average atomic mass of silicon is 28.1 u

Example – 05:

In naturally occurring neon the fractional abundance of various isotopes is as follows 0.9051 of ¹⁰Ne₂₀, 0.0027 of ¹⁰Ne₂₁, 0.0922 of ¹⁰Ne₂₂. Calculate the average atomic mass of neon.

Solution:

Average atomic mass =0.9051 × 20  + 0.0027 ×  21  + 0.0922 × 22

= 18.102 + 0.057 + 2.028 = 20.187 u

Hence average atomic mass of neon is 20.187 u.

Example – 06:

Nitrogen occurs in nature in the form of two isotopes with atomic mass 14 and 15 respectively. If the average atomic mass of nitrogen is 14.0067. What is the percentage abundance of the two isotopes?

Solution:

Let % abundance of the isotope with atomic mass 14 be ‘x’. Hence that of the isotope with atomic mass 15 will be (100 -x).

% abundance of isotope of nitrogen with atomic mass 14 is 99.33 and  with atomic mass 15 is 0.67

Example – 07:

Boron occurs in nature in the form of two isotopes with atomic mass 10 and 11 respectively. If the average atomic mass of boron is 10.80 u. What is the percentage abundance of the two isotopes?

Solution:

Let % abundance of the isotope with atomic mass 10 be ‘x’. Hence that of isotope with atomic mass 11 will be (100 -x).

% abundance of isotope of boron with atomic mass 10 is 20 and 

% abundance of the isotope of boron with atomic mass 11 is 80.

Example – 08:

Chlorine has two stable isotopes Cl – 35 and Cl -37, with atomic masses 34.968 u and 36.956 u respectively. If the average atomic mass is 35.452 u, calculate the % abundance of isotopes.

Solution:

Let % abundance of the isotope with atomic mass 35 be ‘x’. Hence that of the isotope with atomic mass 37 will be (100 -x).

% abundance of isotope of chlorine with atomic mass 35 is 75.65 and 

% abundance of isotope of chlorine with atomic mass 37 is 24.35

In the next article, we shall study Cannizzaro’s method to determine atomic mass.

Previous Chapter: Laws of Chemical Combinations

Next Topic: Atomic Mass by Cannizzaro’s Method

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