Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Law of Isomorphism Method
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 AClx.
The molecular mass of chloride = 2 × its vapour density = 2 × 79 = 158 g
The molecular mass of chloride AClx. = 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.
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