In previous articles, we have studied Cannizzaro’s method and law of isomorphism method to determine atomic mass. In this article, we shall study to calculate atomic mass by Dulong Petit’s law.

Specific Heat:

The amount of heat required to raise the temperature of one mole of an element from 287.5 K to 288.5 K is called specific heat of the element. Its S.I. unit is J/mol/ K.

The product of atomic mass and specific heat of the element is called atomic heat of the element.

Dulong-Petit’s Law:

The product of specific heat and the atomic mass of an element in the solid-state is approximately equal to 6.4. OR   Atomic heat of a solid element is nearly equal to 6.4.

Limitations of Dulong-Petit’s Law:

• This law is applicable to elements which are in solid state.
• This law Is applicable to the heavier element.  It is not applicable to lighter elements having high melting points.
• This law gives approximate atomic mass.

Steps Involved:

• Find the equivalent mass of the element by any method mentioned in topic equivalent mass.
• Find approximate atomic mass using relation, Approx. atomic mass × Specific heat = 6.4
• Find the valency of the element using relation, Approx. atomic mass = equivalent mass × valency
• Find the nearest whole number for the calculated valency and use this whole number as valency of that element.
• Use following formula to calculate the corrected atomic mass of the element, Corrected atomic mass = Equivalent mass × valency

Numerical Methods:

Example – 01:

The specific heat of a metal A is found to be 0.03; its equivalent mass is 69.66. Calculate the valency and exact atomic mass of an element.

Solution:

By Dulong-Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.03 = 213.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 213.33 / 69.66 = 3.06

∴ Valency = 3 (Corrected to nearest whole number)

Now, corrected atomic mass = Equivalent mass × valency  =  69.66 × 3 = 208.98 u

Ans: Hence the valency is 3 and its atomic mass is 208.98 u.

Example – 02:

A solid element of equivalent mass 9 has specific heat 1 J/g/K. calculate its atomic mass.

Solution:

Specific heat = 1 J/g/K = 1 / 4.188 = 0.2388

By Dulong-Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.2388 = 26.80

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 26.80 / 9 = 2.98

∴ Valency = 3 (Corrected to nearest whole number)

Now, corrected atomic mass = Equivalent mass × valency  =  9 × 3 = 27 u

Ans: Hence the valency is 3 and its atomic mass is 27 u.

Example – 03:

Equivalent mass of barium is 68.68 and its valency is 2. Find its atomic mass.

Solution:

Equivalent mass of barium = 68.68, valency of barium = 2

Atomic mass = Equivalent mass x valency = 68.68 x 2 = 137.6

Thus atomic mass of barium is 137.6.

Example – 04:

The specific heat of a metal was found to be 0.03 and its equivalent mass is 103.5. Find the atomic mass of the metal.

Solution:

By Dulong-Petit’s Law,  Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.03 = 213.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 213.33 / 103.5 = 2.05

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency =  103.5 × 2 = 207 u

Ans: atomic mass is 207 u.

Example – 05:

The specific heat of metal M that forms sulphide MS is 0.032 cal g^-1 deg^-1. What is the equivalent mass of the metal?

Solution:

By Dulong-Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.032 = 200

Now metal sulphide has formula MS

Hence valency of metal is 2

Now, approx. atomic mass = equivalent mass × valency

∴ Equivalent mass = approx. atomic mass / valency  = 200 / 2 = 100

Ans: equivalent mass is 100 u.

Example – 06:

The chloride of a metal was found to contain 26.2 % of chlorine. The specific heat of the metal is 0.033. Calculate the atomic mass and equivalent mass of the metal. Also write the molecular formula of metal chloride.

Solution:

Consider 100 g of metal chloride

% of chlorine = 26.2

% of metal = 100 – 26.2 = 73.8

Mass of chlorine = 26.2 g, Mass of metal = 73.8 g

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.033 = 194

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 194 / 100 = 1.94

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency  =  100 × 2 = 200 u

Valency of metal (M) is 2 and that of chlorine is 1. Hence the formula of metal chloride is MCl₂.

Ans: the atomic mass of the metal is 200 and its equivalent mass is 100. The formula of metal chloride is MCl₂.

Example – 07:

1 g of metal having specific heat 0.060205 combines with oxygen to form 1.08 g of oxide. Find atomic mass and valency of the metal. Also, write the molecular formula of the metal oxide.

Solution:

Mass of metal = 1 g, Mass of oxide = 1.08 g

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

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.060205 = 103.1

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 103.1 / 100 = 1.03

∴  Valency = 1 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency =  100 × 1 = 200 u

Valency of metal (M) is 1 and that of oxygen is 2. Hence the formula of metal chloride is M₂O.

Ans: the atomic mass of the metal is 100 and its valency is 1, the formula of metal chloride is M₂O.

Example – 08:

A metallic oxide contains 47.06 % of oxygen. The specific heat of metal is 0.25. Calculate the atomic mass of the metal. Write molecular formula of metal oxide.

Solution:

Consider 100 g of metal oxide

% of oxygen = 47.06

% of metal = 100 – 47.06 = 52.94

Mass of oxygen = 47.06 g, Mass of metal = 52.94 g

Equivalent mass = (52.94 × 8) / 47.06 = 9

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.25 = 25.6

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 25.6 / 9  = 2.8

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency  =  9 × 3 = 27 u

Valency of metal (M) is 3 and that of oxygen is 2. Hence the formula of metal chloride is M₂O₃.

Ans: the atomic mass of the metal is 27 and the formula of metal oxide is M₂O₃.

Example – 09:

0.54 g of a metal combines with 0.48 g of oxygen to form its oxide. Its specific heat is 0.22 cal per deg. What is the atomic mass of the metal?

Solution:

Mass of oxygen = 0.48 g, Mass of metal = 0.54 g

Equivalent mass = (0.54 × 8) / 0.48 = 9

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.22 = 29.09

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 29.09 / 9 = 3.2

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 9 × 3 = 27 u

Ans: the atomic mass of the metal is 27.

Example – 10:

0.45 g of metal displaced 560 ml of hydrogen at STP from acid. Specific heat of metal is 0.214. Calculate the equivalent mass and atomic mass of the metal.

Solution:

Mass of metal = 0.45 g

Volume of hydrogen = 560 ml = 0.560 dm³ at STP

Mass of 0.560 dm³ of hydrogen at STP 

= (Molecular mass of gas x volume of gas in dm³)  / 22.4

Mass of hydrogen displaced = (2 x 0.560) / 22.4 = 0.05 g

Equivalent mass of metal = 0.45 / 0.05 = 9

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.214 = 29.9

Now, approx. atomic mass = equivalent mass × valency

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 29.9 / 9 = 3.3

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 9 × 3 = 27 u

Ans: the atomic mass of the metal is 27 and its equivalent mass is 9.

Example – 11:

0.2160 g of metal, when treated with an excess of dilute sulphuric acid gave 80.4 cc of moist hydrogen measured at 20 °C and 770 mm of pressure. The specific heat of the metal is 0.0955. Calculate the valency and exact atomic mass of the metal. The aqueous tension at 20 °C is 17.5 mm.

Solution:

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.0955 = 67.02

Now, approx. atomic mass = equivalent mass × valency

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 67.02 / 32.36 = 2.07

∴  Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency 

= 32.36 × 2 = 64.72 u

Ans: the atomic mass of the metal is 64.72 and its valency is 2.

Example – 12:

0.45 g of metal gave 176.6 ml of hydrogen at 23 °C and 743 mm pressure when treated with dilute sulphuric acid. Calculate the equivalent mass of the metal. Aqueous tension at 23 °C is 21 mm. If the specific heat of the metal is 0.091, calculate the exact atomic mass of the metal.

Solution:

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.091 = 70.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 70.33 / 32.32 = 2.1

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 32.32 × 2 = 64.64 u

Ans: the atomic mass of the metal is 64.64 and its valency is 2.

Example – 13:

A metal M forms a volatile chloride, containing 80% of chlorine. The vapour density of the chloride is 66.75. Calculate the exact atomic mass of the element.

Solution:

Molecular mass = 2 x Vapour Density = 2 x 66.75 = 133.5

% of chlorine = 80

% of the element = 100 – 80 = 20

Mass of element = 20 g, Mass of chlorine = 80 g

Equivalent mass of metal = (20 x 35.5) / 80 = 8.875

Let valency of the metal be ‘x’, hence the formula of the chloride is MCl_x.

Atomic mass of metal = Equivalent mass x valency = 8.875 x

Molecular mass of chloride = Atomic mass of metal + Atomic mass of chlorine × x

Molecular mass of chloride = 8.875 x + 35.5 × x = 133.5

∴   8.875 x + 35.5 × x = 133.5

∴   44.375 x = 133.5

∴   x = 133.5 /44.375 = 3.01

∴ Valency = 3 (Corrected to nearest whole number)

Actual atomic mass = Equivalent mass x valency = 8.875 x 3 = 26.625

Ans: The exact atomic mass of the element is 26.635

Example – 14:

The specific heat of a metal A is found to be 0.03. 10 g of metal on evaporation of nitric acid gave 18.9 g of pure dry nitrate. Calculate the equivalent mass and exact atomic mass of the metal.

Solution:

Mass of metal = 10 g

By double decomposition method

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.03 = 213.33

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 213.33 / 69.66  = 3.06

∴ Valency = 3 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency  =  69.66 × 3 = 208.98 u

Ans: the atomic mass of the metal is 208.98 and its equivalent mass is 69.66.

Example – 15:

1 g of metallic bromide dissolved in water gave with the excess of silver nitrate, 1.88 g of silver bromide. Calculate the accurate atomic mass of the element, if its specific heat is 0.15 cal (atomic mass of Ag is 108 and that of bromine is 80).

Solution:

Molecular mass of silver bromide = 108 + 80 = 188 g

Mass of bromine in 1.88 g of silver bromide = (80/180) x 1.8 = 0.8 g

Mass of metallic bromide = 1g, Mass of bromine  = 0.8 g

Mass of metal = 1 – 0.8 = 0.2 g

By double decomposition method

By Dulong Petit’s Law, Atomic mass × Specific heat = 6.4 (Approx.)

∴ Approx. atomic mass = 6.4 / Sp. heat = 6.4 /0.15 = 42.67

Now, approx. atomic mass = equivalent mass × valency

∴ Valency = approx. atomic mass / equivalent mass = 42.67 / 20 = 2.1

∴ Valency = 2 (Corrected to nearest whole number)

Now, Corrected atomic mass = Equivalent mass × valency = 20 × 2 =40 u

Ans: the atomic mass of the metal is 40.

In next article, we shall study the concept of equivalent mass and hydrogen displacement method to determine it.

Previous Topic: Atomic Mass Using Law of Isomorphism

Next Topic: Equivalent Mass by Hydrogen Displacement Method

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Atomic Mass by Dulong Petit’s Law

The post Atomic Mass by Dulong Petit’s Law appeared first on The Fact Factor.

In the last article, we have studied determination of atomic mass by Cannizzaro’s method. In this article, we shall study the law of isomorphism and its use to find the atomic mass of a substance. This law was given by a German chemist Eilhard Mitscherlich.

Isomorphism:

The phenomenon of two or more substances displaying similarity or identity of crystalline form is called isomorphism. Such substances are called isomorphs or isomorphous to each other.

Characteristics of Isomorphous Substances:

The crystals of isomorphous substances have the same shape.

• If crystals of one substance are suspended in a saturated solution of another, the former continuous to grow as latter is deposited all over it. Thus they form overgrowth on each other.
• They can form a mixed crystal with each other.

Examples:

• Ferrous sulphate (FeSO₄.7H₂O) and magnesium sulphate (MgSO₄.7H₂O).
• Potassium perchlorate (KClO₄)and potassium permangnate (KMnO₄).
• Potassium chromate (K₂CrO₄)and potassium sulphate (K₂SO₄).
• Ammonium alum ((NH₄)₂SO₄.Al₂(SO₄)₃.24H₂O) and potash alum (K₂SO₄.Al₂(SO₄)₃.24H₂O)

Law of Isomorphism:

This law was given by Mitscherlich in 1819. It states that “Substance having similar internal structure exhibit identity of crystalline form”.

Thus we can conclude that

• isomorphous substances should have similar chemical formulae.
• the elements forming isomorphous substances must have the same valency
• In isomorphous compounds, the ratio between masses of two elements which combine with the same combined mass of all other elements is the same as the ratio between their atomic masses. Mathematically.

To Find Atomic Mass Using Law of Isomorphism: (Steps Involved):

• Find percentage composition of a compound containing an element whose atomic mass is to be found.
• Write the formula of the compound using given the formula of the isomorphous substance.
• Calculate molecular mass of the compound.
• Assume atomic mass of the element as ‘x’.
• Write percentage formula for the element.
• Find the value of x. Which gives the atomic mass of the element.

Nunerical Problems:

Example – 01:

The sulphate of metal contains 20.9 % of the metal and is isomorphous with ZnSO₄.7H₂O. What is the probable atomic mass of the metal?

Solution:

The sulphate is isomorphous with  ZnSO₄.7H₂O.

Hence by the law of isomorphism, its chemical formula should be  MSO₄.7H₂O.

Let atomic mass of metal M be ‘x’

∴  The molecular mass of metal sulphate = (x + 32 + 16 × 4) + 7(1 × 2 + 16)

∴ The molecular mass of metal sulphate = (x + 32 + 64) + 7(18) = x + 96 + 126 = x + 222

∴  100x = 20.9x + 20.9 × 222

∴ 100x – 20.9x = 4639.80

∴ 79.1x = 4639.80

∴  x = 4639.80 / 79.1 = 58.65

Thus the probable atomic mass of the metal is 58.65.

Example – 02:

The oxides of two elements A and B are isomorphous. The metal A whose atomic mass is 52, forms a chloride whose vapour density is 79. The oxide B contains 47.1 % of oxygen. Calculate the atomic mass of B.

Solution:

Let valency of the element be ‘x’. Hence its chloride formula is ACl_x.

The molecular mass of chloride = 2 × its vapour density = 2 × 79 = 158 g

The molecular mass of chloride ACl_x. = 52 + 35.5  x = 158

∴   35.5  x = 158 – 52 = 106

∴  x  = 106 / 35.5 = 3 (Nearest whole number)

Thus valency of element A is 3.

The oxides of two elements A and B are isomorphous.

Hence the valency of A and B should be the same. Hence valency of element B is also 3.

The oxide of B contains 47.1 % of oxygen.

i.e. it contains 100 – 47. 1= 52.9 % of element B.

Mass of oxygen = 47.1 g Mass of element = 52.9 g

Equivalent mass of B = (52.9 x 8)/47.1 = 8.99

Atomic mass of B = Equivalent mass x valency = 8.99 x 3 = 26.97

Thus the atomic mass of the element B is 26.97.

Example – 03:

A metal has 22.64% of metal in its sulphate. The metallic sulphate is isomorphous with MgSO₄.7H₂O. Calculate the atomic mass of the metal.

Solution:

The sulphate is isomorphous with MgSO₄. 7H₂O..

Hence by the law of isomorphism, its chemical formula should be MSO₄. 7H₂O..

Let atomic mass of metal M be ‘x’

The molecular mass of metal sulphate = (x + 32 + 16 × 4) + 7(1 × 2 + 16)

The molecular mass of metal sulphate = (x + 32 + 64) + 7(18) = x + 96 + 126 = x + 222

∴ 100x = 22.64x + 22.64 × 222

∴ 100x – 22.64x = 5026.08

∴ 77.36x = 5026.08

∴ x = 5026.08/ 77.36 = 64.97 = 65

Thus the probable atomic mass of the metal is 65.

Example – 04:

Magnesium sulphate contains 9.75% of magnesium and 39.02% of sulphate whereas zinc sulphate contains 22.6% of zinc and 35.5% sulphate. If the atomic mass of zinc is 65, find that of magnesium, if both the sulphates are isomorphous.

Solution:

Let formula of zinc sulphate be ZnSO₄.xH₂O.

% of zinc = 22.6, % of sulphate = 35.5, % of water = 100 -(22.6 + 35.5) = 100 – 58.1 = 41.9

The molecular mass of zinc sulphate = (65+ 32 + 16 × 4) + x(1 × 2 + 16)

The molecular mass of zinc sulphate =(65 +32 + 64) + 18x = 161 + 18x

∴ 1800x = 41.9(18x +  161)

∴ 1800x = 754.2x +  6745.9

∴ 1800x – 754.2x = 6745.9

∴ 1045.8x = 6745.9

∴ x = 6745.9 / 1045.8 = 6.5 = 7

The sulphate is isomorphous with  magnesium sulphate.

Hence formula of magnesium sulphate is MgSO₄.7H₂O.

Let atomic mass of magnesium be be ‘x’

The molecular mass of magnesium sulphate = (x + 32 + 16 × 4) + 7(1 × 2 + 16)

The molecular mass of magnesium sulphate = (x + 32 + 64) + 7(18) = x + 96 + 126 = x + 222

∴ 100x = 9.75x + 9.75 × 222

∴ 100x –  9.75x = 2164.5

∴ 90.25x = 2164.5

∴ x = 2164.5/90.25 = 23.98 = 24

Thus the atomic mass of the metal is 24.

Example – 05:

1g of the chloride of a metal when treated with the excess of silver nitrate produced 0.965 g of dry silver chloride. Calculate the atomic mass of the metal, given that it forms a sulphate which is isomorphous with BaSO₄.

Solution:

By double decomposition method

∴ 0.965 E + 34.26 = 143.5

∴ 0.965 E  = 109.24

∴  E  = 113.2

Now the required sulphate is isomorphous with BaSO₄, hence the formula for the sulphate is MSO₄. Hence the valency of metal is 2.

Atomic mass = equivalent mass x valency = 113.2 × 2 = 226.4

Thus atomic mass of the metal is 226.4

Example – 06:

Potassium selenate is isomorphous with potassium sulphate and contains 35.75% of selenium. Find the atomic mass of selenium (Se).

Solution:

Potassium selenate  is isomorphous with potassium sulphate K₂SO₄,

hence the formula for the Potassium selenate is K₂SeO₄.

Let atomic mass of metal selenium be be ‘x’

The molecular mass of metal sulphate =39 × 2 + x + 16 × 4  =78 + x + 64 = x + 142

% of selenium in sulphate

∴ 100x = 35.75x + 35.75 × 142

∴ 100x –  35.75x = 5076.5

∴  64.25x =5076.5 ∴

x  = 79.01

Thus the atomic mass of the seleniumis 79.01.

Example – 07:

Potassium permanganate is isomorphous with potassium perchlorate KClO₄ and contains 34.81 % of manganese. Find the atomic mass of manganese.

Solution:

Potassium permanganate  is isomorphous with potassium perchlorate KClO₄,

hence the formula for the Potassium permanganate is KMnO₄.

Let atomic mass of metal selenium be be ‘x’

The molecular mass of ptassium permanganate =39  + x + 16 x 4  =39 + x + 64 = x + 103

% of selenium in sulphate

∴ 100x = 34.815x + 34.81 × 103

∴ 100x –  34.815x = 3585.43

∴ 65.19x = 3585.43

∴ x =  54.99 = 55

Thus the atomic mass of the manganese is 55.

Example – 08:

Chrome alum is isomorphous with potash alum, K₂SO₄, Al₂(SO₄)₃. 24 H₂O and is found to contain 10.42% of chromium. Find the atomic mass of chromium.

Solution:

Chrome alum is isomorphous with potash alum, K₂SO₄, Al₂(SO₄)₃. 24 H₂O,

hence the formula for the chrome alumn is K₂SO₄, Cr2(SO₄)₃. 24 H₂O.

Let atomic mass of chromium be be ‘x’

The molecular mass of chrome alum is

= 39 × 2 + 32 + 16 × 4 + 2 x  + (32 + 16 × 4 ) x 3  + 24 × (1 ×2 + 16)

molecular mass of chrome alum = 78 + 32 + 64 + 2x + (32 + 64) x3 + 24x (2 + 16)

molecular mass of chrome alum = 78 + 32 + 64 + 2x + 96 x3 + 24x 18

molecular mass of chrome alum = 78 + 32 + 64 + 2x + 96 x3 + 24x 18

molecular mass of chrome alum = 78 + 32 + 64 + 2x + 288 + 432

molecular mass of chrome alum =  2x + 894

% of chromium

∴  200x = 10.42 (2x + 894)

∴  200x = 20.84 x + 9315.48

∴  200x – 20.84 x = 9315.48

∴  179.16x =9315.48

∴  x =  51.99 =52

Thus the atomic mass of the chromium is 52.

In the next article, we shall stdy determination of atomic mass by Dulong Petit’s law.

Previous Topic: Atomic Mass by Cannizzaro’s Method

Next Topic: Atomic Mass by Dulong Petit’s Law

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Law of Isomorphism Method

The post Atomic Mass Using Law of Isomorphism 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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