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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The ratio of the radius of cations (r⁺) to the radius of the anion (r^–) is known as the radius ratio of the ionic solid.

The significance of radius ratio:

• It is useful in predicting the structure of ionic solids.
• The structure of an ionic compound depends upon stoichiometry and the size of ions.
• In crystals, cations tend to get surrounded by the largest possible number of anions around it.
• Greater the radius ratio, greater is the coordination number of cations and anions.
• If cations are extremely small and anions are extremely large, then the radius ratio is very small. In such case packing of anions is very close to each other and due to repulsion between anions, the system becomes unstable. Hence the structure changes to some suitable stable arrangement.
• The radius ratio at which anions just touch each other, as well as central cation, is called the critical radius ratio.

Effect of Radius Ratio on Coordination Number:

A cation would fit exactly into the octahedral void and would have a coordination number of six if the radius ratio were exactly 0.414. Similarly, a cation would fit exactly into the tetrahedral void and would have a coordination number of four, if the radius ratio were exactly 0.225.

Let us consider a case in which a cation is fitting exactly into the octahedral void of close pack anions and have the coordination number of six, in this case, the radius ratio is exactly 0.414.

When the radius ratio is greater than this, then the anions move apart to accommodate larger cation. This situation is relatively unstable. If the radius ratio is further increased the anions will move farther and farther apart till to reach a stage at which more anions can be accommodated. Now, the bigger cation moves to bigger void i.e. octahedral void whose coordination number is 8. This happens when the radius ratio exceeds 0.732.

In case of radius ratio becomes less than 0.414, the six anions will not be able to touch the smaller cation. To touch the cation, the anions starts overlapping with each other, which is an unstable situation. Hence smaller cation moves to smaller void i.e. tetrahedral void and coordination number decreases from 6 to 4.

Note: Although a large number of ionic substances obey this rule, there are many exceptions to it.

Radius Ratio of Interstitial Voids:

The vacant space left in the closest pack arrangement of constituent particles is called an interstitial void or interstitial site.

Tetrahedral Voids:

Locating Tetrahedral Voids:

Let us consider a unit cell of ccp or fcc lattice [Fig. (a)]. The unit cell is divided into eight small cubes. Each small cube has atoms at alternate corners as shown. Each small cube has 4 atoms. When joined to each other, they make a regular tetrahedron as shown in the figure. Thus, there is one tetrahedral void in each small cube and eight tetrahedral voids in total in the unit cell. Each of the eight small cubes has one void in one unit cell of ccp structure. The ccp structure has 4 atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.

Radius Ratio of Tetrahedral Void:

A tetrahedral site in a cube having a tetrahedral void of radius ‘r’ is as shown at the centre of the cube. Let ‘R’ be the radius of the constituent particle of the unit cell. Let ‘a’ be each side of the cube.

Consider face diagonal AB (Right-angled triangle ABC). We have,

AB² = a² + a²

∴  AB² = 2 a²

∴  AB² = √2 a

Now the two-sphere touch each other along face diagonal AB

∴  2R = √2 a

∴  R = √2 a/2

∴  R =  a/√2  ……. (1)

Consider body diagonal AD (Right-angled triangle ABD). We have,

AD² = AB² + BD²

∴  AD² = (√2 a)² + a²

∴  AD² = 2 a² + a² = 3 a²

∴  AD = √3 a

Now the two spheres at the diagonals touch the tetrahedral void sphere along body diagonal AD

∴  2 R + 2r  = √3 a

∴  R + r  = (√3/2)a  ……..  (2)

Dividing equation (2) by (1)

Thus the radius ratio for the tetrahedral void is 0.225

Octahedral Voids:

Locating Octahedral Voids:

Let us consider a unit cell of ccp or fcc lattice [Fig. (a)]. The body centre of the cube, C is not occupied but it is surrounded by six atoms on face centres. If these face centres are joined, an octahedron is generated. Thus, this unit cell has one octahedral void at the body centre of the cube.

Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. [Fig. (b)]. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centre) and two belonging to two adjacent unit cells.

Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 th of each void belongs to a particular unit cell.

For cubic close-packed structure:
Number of octahedral void at the body-centre of the cube = 1
12 octahedral voids located at each edge and shared between four unit cells

Thus number of octahedral valve at 12 edges =12 x 1/4 = 3
Thus, the total number of octahedral voids in unit cell = 1 + 3 = 4

In the ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to this number of atoms.

Radius Ratio of Octahedral Void:

A octahedral site in a cube having octahedral void of radius r is as shown at the centre of the cube. Let R be the radius of the constituent particle of the unit cell. Let a be each side of the cube. In this case we can see that a = 2R

Consider Right angled triangle ABC. We have,

AB² = AC² + CB²

∴  (2R + 2r)² = a² + a²

∴  (2R + 2r)² = 2 a²

∴  2R + 2r = √2 a

But from figure a = 2R

∴  2R + 2r = √2 x 2R

∴   R +  r = √2  R

∴   r = √2  R – R

∴   r = (√2  – 1)R

∴   r = (1.414  – 1)R

∴   r = 0.414R

Thus radius ratio for the octahedral void is 0.414

Science > Chemistry > Solid State > Radius Ratio and its Significance

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In this article, we shall study two types of voids formed during hexagonal closed packing namely, a) Tetrahedral voids and b) octahedral voids.

Let us consider a closed pack hexagonal packing in the first layer as shown. There are empty spaces between the particles (sphere) are called voids.

All the voids are equivalent in the first layer they have marked alternately as ‘a’ and ‘b’. The spheres of the second layer can be either placed on voids which are marked ‘a’ or ‘b’ but it is impossible to place spheres on both types of voids simultaneously. When spheres of the new layer are placed on voids marked ‘a’ then voids marked as ‘b; remain unoccupied.

Thus there is no void above ‘a’ in the second layer but there is void above ‘b’ even in the second layer. The void between the first layer and second layer at ‘a’ is tetrahedral void, while the void between the first layer and the second layer at ‘b’ is octahedral void.

T = Tetrahedral void and O = Octahedral void

Tetrahedral Voids:

Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called tetrahedral voids because a tetrahedron is formed when the centres of these four spheres are joined. These voids have been marked as ‘T’ in the figure.

Octahedral Voids:

At other places, the triangular voids in the second layer are above the triangular voids in the first layer, and the triangular shapes of these do not overlap then the octahedral void is formed. One of them has the apex of the triangle pointing upwards and the other downwards. These voids have been marked as ‘O’ in the figure.

Number of Voids:

The number of these two types of voids depends upon the number of close-packed spheres.

Let the number of close-packed spheres be N, then, the number of octahedral voids generated = N and the number of tetrahedral voids generated = 2N

Characteristics of Tetrahedral Voids:

• The vacant space or void is surrounded by four atomic spheres. Hence co-ordination number of the tetrahedral void is 4.
• The atom in the tetrahedral void is in contact with four atoms placed at four corners of a tetrahedron.
• This void is formed when a triangular void made coplanar atoms (first layer) is in contact with the fourth atom above or below it (second layer).
• The volume of the void is much smaller than that of the spherical particle.
• If R is the radius of the constituent spherical particle, then the radius of the tetrahedral void is 0.225 R.
• If the number of close-packed spheres is N, then the number of tetrahedral voids is 2N.

Characteristics of Octahedral Voids:

• The vacant space or void is surrounded by six atomic spheres. Hence, the coordination number of the tetrahedral void is 6.
• The atom in the octahedral void is in contact with six atoms placed at six corners of an octahedron.
• This void is formed when two sets of equilateral triangles pointing in the opposite direction with six spheres.
• The volume of the void is small.
• If R is the radius of the constituent spherical particle, then the radius of the octahedral void is 0.414 R.
• If the number of close-packed spheres is N, then the number of octahedral voids is N.Characteristics of Voids

Packing Voids in Ionic Solids

We have already seen that when particles are close-packed resulting in either cubic closed packing (ccp) or hexagonal close packing (hcp) structure, two types of voids are generated. The number of octahedral voids present in a lattice is equal to the number of close-packed particles, the number of tetrahedral voids generated is twice this number.

Ionic solids are formed from cations and anions. The charge is balanced by an appropriate number of both types of ions so that net charge on solid is zero (neutral).

In ionic solids, the bigger ions (usually, anions) form the close-packed structure and the smaller ions (usually, cations) occupy the voids. If the smaller ion is small enough then tetrahedral voids are occupied, if bigger, then octahedral voids are occupied.

Not all octahedral or tetrahedral voids are occupied. In a given compound, the fraction of octahedral or tetrahedral voids that are occupied, depends upon the chemical formula of the compound, as can be seen from the following examples.

Example 1:

A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make cubic closed packing (ccp) and those of the element X (as cations) occupy all the octahedral voids. Obtain the formula of the compound.

Solution:

Given that the cubic closed packing (ccp) lattice is formed by the element Y.

The number of octahedral voids generated would be equal to the number of atoms of Y present in it.

Since all the octahedral voids are occupied by the atoms of X, their number would also be equal to that of the element Y. Thus, the atoms of elements X and Y are present in equal numbers or 1:1 ratio.

Therefore, the formula of the compound is XY.

Example 2:

Atoms of element B form hexagonal close packing (hcp) lattice and those of element A occupy 2/3rd of tetrahedral voids. Obtain the formula of the compound formed by the elements A and B.

Solution:

The number of tetrahedral voids formed is equal to twice the number of atoms of element B

Only 2/3rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3):1 or 4:3

Hence the formula of the compound is A₄B₃.

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In this article, we shall study the different types of close packing of the constituent particles in solids.

The Principle of Packing in Solids:

In a crystal particle (atoms, molecules or ions) occupy the lattice points in the crystal lattice. These particles may be of various shapes and hence mode of packing of these particles will change according to their shape.

Constituent particles are considered rigid incompressible spheres of equal size. The constituent particles try to get closely packed together so that the maximum closely packed structure is attained. The maximum closely packed structure is such that the minimum empty space (void) is left. The closer the constituent particles lie, the greater is the stability of the packed system. By this arrangement, a maximum possible density of the crystal and stability of the crystal is achieved.

Close Packing in One Dimension:

There is only one way of arranging spheres in a one-dimensional close-packed structure, that is to arrange them in a row and touching each other.

In this arrangement, each sphere is in contact with two of its neighbours. The number of nearest neighbours of a particle is called its coordination number. Thus, in a one-dimensional close-packed arrangement, the coordination number is 2.

Close Packing in Two Dimensions:

A two-dimensional close-packed structure can be generated by placing the rows of close-packed spheres side by side horizontally. This can be done in two different ways.

a) Square Close Packing:

The second row may be placed in contact with the first one such that the spheres of the second row are exactly above those of the first row. The spheres of the two rows are aligned horizontally as well as vertically. If we call the first row as ‘A’ type row, the second row being exactly the same as the first one, is also of ‘A’ type. Similarly, we may place more rows to obtain AAA type of arrangement.

In this arrangement, each sphere is in contact with four of its neighbours. Thus, the two-dimensional coordination number is 4. Also, if the centres of these 4 immediate neighbouring spheres are joined, a square is formed. Hence this packing is called square close packing in two dimensions.

b) Hexagonal Close Packing:

In this arrangement, the second row may be placed above the first one in a staggered manner such that its spheres fit in the depressions of the first row.

If the arrangement of spheres in the first row is called ‘A’ type, the one in the second row is different and may be called ‘B’ type. When the third row is placed adjacent to the second in a staggered manner, its spheres are aligned with those of the first row. Hence this row is also of ‘A’ type. The spheres of the similarly placed fourth row will be aligned with those of the second row (‘B’ type). Hence this arrangement is of ABAB type.

In this arrangement, there is less free space and this packing is more efficient than the square close packing. Each sphere is in contact with six of its neighbours and the two-dimensional coordination number is 6. The centres of these six spheres are at the corners of a regular hexagon hence this packing is called two-dimensional hexagonal close packing.

It can be seen in the figure that in these rows there are some voids (empty spaces). These are triangular in shape. The triangular voids are of two different types. In one row, the apexes of the triangles are pointing upwards and in the next layer downwards.

Close Packing in Three Dimensions:

A three-dimensional close-packed structure can be generated by stacking (placing) layers of close-packed spheres in two dimensions one over the other. This can be done in the following ways.

Three-dimensional close packing from a two-dimensional square layer:

In such an arrangement, the second layer is placed over the first layer such that the spheres of the upper layer are exactly above those of the first layer. In this arrangement spheres of both the layers are perfectly aligned horizontally as well as vertically as shown. Similarly, we may place more layers one above the other.

If the arrangement of spheres in the first layer is called ‘A’ type, all the layers have the same arrangement. Thus this lattice has AAA…. type pattern. The lattice thus generated is the simple cubic lattice, and its unit cell is the primitive cubic unit cell.

The coordination number for such arrangement is 6. The volume occupied by particles is 52% i.e. there is 48% void space. Of all metals in the periodic table, only polonium crystalizes in this crystal form.

Three-dimensional close packing from two-dimensional hexagonal close-packed layers:

This close-packed structure can be generated by placing layers one over the other.

All the voids are equivalent in the first layer they have marked alternately as ‘a’ and ‘b’. The spheres of the second layer can be either placed on voids which are marked ‘a’ or ‘b’ but it is impossible to place spheres on both types of voids simultaneously. When spheres of a new layer are placed on voids marked ‘a’ then voids marked as ‘b; remain unoccupied.

Let us take a two-dimensional hexagonal close-packed layer ‘A’ and place a similar layer above it such that the spheres of the second layer are placed in the depressions of the first layer. Since the spheres of the two layers are aligned differently, let us call the second layer as B.

T = Tetrahedral void and O = Octahedral void

Placing the third layer over the second layer can be done in two ways.

It can be observed from the figure that not all the triangular voids of the first layer are covered by the spheres of the second layer.

This gives rise to two different types of holes or voids. Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called tetrahedral voids because a tetrahedron is formed when the centres of these four spheres are joined. At other places, the triangular voids in the second layer are above the triangular voids in the first layer, and the triangular shapes of these do not overlap. One of them has the apex of the triangle pointing upwards and the other downwards. These voids have been marked as ‘O’ in the figure above.

Covering Tetrahedral Voids (Hexagonal Closed Packing):

Tetrahedral voids of the second layer may be covered by the spheres of the third layer. In this case, the spheres of the third layer are exactly aligned with those of the first layer. Thus, the pattern of spheres is repeated in alternate layers. This pattern is often written as ABAB ……. pattern. This structure is called a hexagonal close-packed (hcp) structure.

This sort of arrangement of atoms is found in many metals like magnesium and zinc.

B) Covering Octahedral Voids (Cubic Closed Packing):

The third layer may be placed above the second layer in a manner such that its spheres cover the octahedral voids. When placed in this manner, the spheres of the third layer are not aligned with those of either the first or the second layer. This arrangement is called ‘C’ type.

Only when the fourth layer is placed, its spheres are aligned with those of the first layer as shown. This pattern of layers is often written as ABCABC ……….. This structure is called cubic close-packed (ccp) or face-centred cubic (fcc) structure. Metals such as copper and silver crystallise in this structure.

Note:

Both these types of close packing are highly efficient and 74% space in the crystal is filled. In either of them, each sphere is in contact with twelve spheres. Thus, the coordination number is 12 in either of these two structures.

Characteristics of Hexagonal Close Packing:

• In this three dimensional arrangement of the unit cell, spheres of the third layer are placed on triangular-shaped tetrahedral voids of the second layer.
• The spheres of the third layer lie exactly above the spheres of the first layer.
• The arrangements of the first layer and third layer are identical.
• The arrangement of hexagonal close packing is represented as ABAB type.
• Packing efficiency is 74%

Characteristics of Cubic Close Packing:

• In this three dimensional arrangement of the unit cell, spheres of the third layer are placed in the positions of tetrahedral voids having apices upward.
• In this arrangement, the spheres of the third layer do not lie exactly above the spheres of the first layer. Actually, the spheres of the fourth layer lie exactly above the spheres of the first layer.
• The arrangements of the first layer and third layer are different. The arrangements of the first layer and fourth layer are identical.
• The arrangement of cubic close packing is represented as ABCABC type.
• Packing efficiency is 74%

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In this article, we shall study to solve problems to calculate the atomic radius, the distance between atoms in the unit cell and to decide the type of crystal structure.

Example – 01:

A naturally occurring gold crystallizes in face centred cubic structure and has a density of 19.3 g cm^-3. Calculate the atomic radius of gold. (Au = 197 g mol^-1)

Given: Density of gold  = 19,3 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1.Atomic mass of gold = M = 197 g mol^-1.Type of crystal structure = fcc

To Find: Atomic radius of gold =?

Solution:

The number of atoms in the unit cell of a face centred cubic structure is n = 4

Ans: The atomic radius of gold is 144 pm

Example – 02:

Sodium metal crystallizes in the bcc structure with an edge length of unit cell 4.29 x 10^-8 cm. Calculate the atomic radius of sodium metal.

Given: Edge length = a = 4.29 x 10^-8 cm, Type of crystal structure = bcc

To Find: Atomic radius of sodium =?

Solution:

Ans: The atomic radius of sodium is 186 pm

Example – 03:

Niobium crystallizes in bcc structure and has a density of 8.55 g cm^-3. Calculate its atomic radius, if its atomic mass is 93.

Given: Density of niobium = 8.55 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1.Atomic mass of niobium = M = 93 g mol^-1.Type of crystal structure = bcc

To Find: Atomic radius of niobium =?

Solution:

The number of atoms in the unit cell of body centred cubic structure is n = 2

Ans: Atomic radius of niobium is 143.1 pm

Example – 04:

Sodium metal crystallizes in the body-centred cubic unit cell. If the distance between the nearest Na atom is 368 pm, calculate the edge length of the unit cell.

Given: the distance between nearest Na atom is 368 pm,

To Find: Edge length of unit cell =?

Solution:

The nearest atoms in bcc crystal structure are along the body diagonal of the cube and they are in contact with each other.

2r = 368 pm

∴ r = 184 pm

Ans: The edge length of the cell is 424.9 pm

Example – 05:

Copper crystallizes in fcc structure. If two neighbouring Cu atoms are at a distance of 234 pm, find a) edge length, b) volume of a unit cell. Also, find the distance between the next neighbouring atoms.

Given: the distance between nearest Cu atom is 234 pm,

Solution:

The nearest atoms in fcc crystal structure are along the face diagonal of the cube and they are in contact with each other.

2r = 234 pm

∴ r = 117 pm

a = 331 x 10^-10 m = 3.31 x 10^-8 m

Volume of unit cell = a³ = (3.31 x 10^-8 m)³

Volume of unit cell = (3.31) ³ x 10^-24 m³ = 36.24 x 10^-24 m³

The next neighbouring atom is along the edge of the cell. Thus the distance between next neighbouring atom is ‘a’ i.e. 331 pm

Ans: The edge length = 331 pm, the volume of unit cell = 36.24 x 10^-24 m³, the distance between next neighbouring atom is 331 pm

Example – 06:

A metal occurs as body centred cube and has a density of 7.856 g cm^-3. Calculate the atomic radius of metal. Atomic mass of metal = 58 g/mol

Given: Density of metal = 7.856 g cm^-3, Avogadro’s number N = 6.022 x 10²³ mol^-1.Atomic mass of metal = M = 58 g mol^-1.Type of crystal structure = bcc

To Find: Atomic radius of metal =?

Solution:

Ans: The atomic radius is 125.7 pm

Example – 07:

An element germanium crystallizes in the bcc type crystal structure with an edge of unit cell 288 pm and the density of the element is 7.2 g cm^-3. Calculate the number atoms present in 52 g of the crystalline element. Also, calculate the atomic mass of the element.

Given: Density of germanium = 7.2 g cm^-3, edge length = 288 pm = 288 x 10^-12 m = 288 x 10^-10 cm = 2.88 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Given mass of germanium = 52 g, Type of crystal structure = bcc

To Find: the number atoms present in 52 g of the crystal =? atomic mass of the element = M = ?

Solution:

The number of atoms in the unit cell of bcc structure is n = 2

For bcc there are 2 atoms in a unit cell

Number of atoms in given mass = 2 x 3.023 x 10²⁹ = 6.046 x 10²⁹

Example – 08:

An atom crystallises in fcc crystal lattice and has a density of 10 g cm^-3 with unit cell edge length of 100 pm. Calculate the number atoms present in 1 g of the crystal.

Given: Density = 10 g cm^-3, edge length = 100 pm = 100 x 10^-12 m = 100 x 10^-10 cm = 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Given mass of crystal = 1 g, Type of crystal structure = fcc

To Find: the number atoms present in 1 g of the crystal

Solution:

The number of atoms in the unit cell of fcc structure is n = 4

For fcc there are 4 atoms in a unit cell

Number of atoms in given mass = Number of atoms in unit cell x Number of unit cells

Number of atoms in given mass = 4 x 10²³

Ans: Number of atoms in given mass = 4 x 10²³

Example – 09:

Aluminium having atomic mass 27 g mol^-1, crystallizes in face centred cubic structure. Find the number of aluminium atom in 10 g of it. Also, find the number of unit cells in the given quantity.

Given: Atomic mass = 27 g mol^-1, Avogadro’s number N = 6.022 x 10²³ mol^-1. Given mass = 10 g, Type of crystal structure = fcc

To Find: the number of aluminium atom in 10 g =? Number of unit cells =?

Solution:

Mass of each atom = Atomic mass / Avogadro’s number

Mass of each atom = 27 g mol^-1/ 6.022 x 10²³ mol^-1 = 4.483 x 10^-23 g

The number of atoms in unit cell of fcc structure is n = 4

Hence mass of unit cell = 4 x 4.483 x 10^-23 g = 1.793 x 10^-22 g

Number of unit cells = Given mass/ mass of unit cell = 10 g/ 1.793 x 10^-22 g = 5.58 x 10²²

Number of aluminium atom = 4 x number of unit cell

Number of aluminium atom = 4 x 5.58 x 10²² = 2.23 x 10²³

Ans: Number of aluminium atoms = 2.23 x 10²³ and number of unit cells = 5.58 x 10²²

Example – 10:

A unit cell of iron crystal has edge length 288 pm and density 7.86 g cm^-3. Find the number of atoms per unit cell and type of crystal lattice. Given the molar mass of iron 56 g/mol.

Given: Density = 7.86 g cm^-3, edge length = 288 pm = 288 x 10^-12 m = 288 x 10^-10 cm = 2.88 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Molecular mass of iron = M = 56 g mol^-1,

To Find: the number of atoms per unit cell and type of crystal lattice

Solution:

Number of atoms per unit cell = 2,

Now body centred cubic structure has 2 atoms in the unit cell. Hence the crystal structure must be bcc

Ans: The crystal structure is bcc

Example – 11:

The edge length of a unit cell of a cubic crystal is 4.3 A°, having atomic mass 89 g/mol. If the density of crystal is 9.02 g cm^-3, find the number of atoms in a unit cell.

Given: Density = 9.02 g cm^-3, edge length = 4.3 A0 = 4.3 x 10^-10 m = 4.3 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Molecular mass = M = 89 g mol^-1,

To Find: the number of atoms in a unit cell.

Solution:

Ans: Number of atoms per unit cell = 5

Example – 12:

The density of silver having atomic mass 107.8 g /mol is 10.7 g cm^-3. If the edge length of a cubic unit cell is 405 pm, find the number of silver atoms in a unit cell and predict its type.

Given: Density = 10.7 g cm^-3, edge length = 405 pm = 405 x 10^-12 m = 405 x 10^-10 cm = 4.05 x 10^-8 cm, Avogadro’s number N = 6.022 x 10²³ mol^-1. Molecular mass = M = 107.8 g mol^-1,

To Find: the number of silver atoms in a unit cell and predict its type.

Solution:

The number of atoms per unit cell = 4,

Now face centred cubic structure has 4 atoms in the unit cell. Hence the crystal structure must be fcc

Ans: The crystal structure is fcc

Example – 13:

A crystalline solid is made up of two elements ‘A’ and ‘B’. Atoms of A are present at the corners and atoms of B are present at face centres. One atom of A is missing from the corner. Find the simplest formula of the solid.

Solution:

Atoms of A are present at 7 corners

Hence the number of atoms of A in the lattice = 7 x 1/8 = 7/8

Atoms of B are present at 6 face centres

Hence the number of atoms of B in the lattice = 6 x 1/2 = 3

Thus the ratio of atoms of A to atoms of B in crystal = 7/8 : 3 = 7 : 24

Hence the simplest formula of the solid is A₇B₂₄

Example – 14:

A solid has a structure in which ‘W’ atoms are located at the corners of the cube, ‘O’ atoms are at the centre of the edges and Na atoms at the centre of the cube. Find the formula of the compound.

Solution:

Atoms of W are present at 8 corners

Hence the number of atoms of W in the lattice = 8 x 1/8 = 1

Atoms of O are present at 12 edge centres

Hence the number of atoms of O in the lattice = 12 x 1/4 = 3

Atoms of Na are present at the centre

Hence the number of atoms of Na in the lattice = 1

Thus the ratio of atoms of W to atoms of O to atoms of Na in crystal = 1:3:1

Hence the simplest formula of the solid is WO₃Na i.e. NaWO₃

Science > Chemistry > Solid State > Numerical Problems on Type of Crystal Structure

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