In the previous article, we have studied the law of definite proportions. In this article, we shall study the law of multiple proportions. The law of multiple proportions was given by British scientist John Dalton in 1803.

Statement:

When two elements combine to form more than one compound, then the different weights of one element combining with a fixed weight of the other element are in simple numerical ratio with each other.

Explanation & illustration :

This law is applicable to pairs of elements which can form more than one compound.

Illustration 1:  

Carbon and oxygen combine together to give two compounds carbon dioxide (CO₂) and carbon monoxide (CO)

Thus the ratio of different weights of oxygen (32 and 16) combining with a fixed weight of carbon (12) is 32 : 16 i.e. 2 :1, which is simple whole number ratio.

Illustration 2:

Hydrogen and oxygen combine together to give two compounds of water (H₂O) and hydrogen peroxide (H₂O₂)

Thus the ratio of different weights of oxygen (16 and 32) combining with a fixed weight of hydrogen (2) is 16 : 32 i.e. 1 :2, which is simple whole number ratio.

Illustration 3:

Nitrogen combines with oxygen to form the various oxides.

Thus the ratio of different weights of oxygen (8, 16, 24, 32, 40) combining with fixed weight of nitrogen (14) is  8 :16 : 24 : 32 : 40 i.e. 1:2:3:4:5, which is simple whole number ratio.

Limitations of the Law of Multiple Proportions:

The existence of isotopes of hydrogen like H¹ or H² causes discrepancies similar to that observed in the law of constant proportions.  Hence the same isotope or mixture of isotope should be used throughout the preparation of a series of compounds.

Explanation of the Law of Multiple Proportions on the Basis of Dalton’s Atomic Theory:

According to Dalton’s atomic theory, compounds are formed by the combination of atoms of different elements in the ratio of simple whole numbers.

Atoms of elements have a fixed weight. Hence, it follows that when elements combine to form more than one compound, the different weights of one which combines with a fixed weight of the other must be in the ratio of simple whole numbers. This explains the law of multiple proportions

Numerical Problems:

Example – 01:

Hydrogen and oxygen are known to form two compounds. The hydrogen content in one of them is 5.93 % while in other it is 11.2 %. Show that the data illustrate the law of multiple proportions.

Solution:

Compound – 1:

Let us consider 100 g of compound – 1

Mass of hydrogen = 5.93 g

Mass of oxygen = 100 g – 5.93 g = 94.06 g

Thus, 5.93 g of hydrogen combines with 94.07 g of oxygen.

∴ 1 g of hydrogen combines with

94.07/5.93 = 15.86 g of oxygen.   …………………….. (1)

Compound – 2:

Let us consider 100 g of compound – 2

Mass of hydrogen = 11.2  g

Mass of oxygen = 100 g – 11.2 g = 88.8 g

Thus, 11.2 g of hydrogen combines with 88.8 g of oxygen.

∴ 1 g of hydrogen combines with

88.8/11.2 = 7.92 g of oxygen.   …………………….. (2)

From statements (1) and (2), the ratio of different masses of oxygen combining with fixed mass of hydrogen (1 g) is  15.86 : 7.92 i.e. 2 : 1, which is simple whole number ratio. Thus, the data illustrate the law of multiple proportions.

Example – 02:

Carbon and oxygen are known to form two compounds. The carbon content in one of them is 42.9 % while in other it is 27.3 %. Show that the data illustrate the law of multiple proportions.

Solution:

Compound – 1:

Let us consider 100 g of compound – 1

Mass of carbon = 42.9 g

Mass of oxygen = 100 g – 42.9 g = 57.1 g

Thus, 42.9 g of carbon combines with 57.1 g of oxygen.

∴ 1 g of carbon combines with

57.1/42.9 = 1.33 g of oxygen.   …………………….. (1)

Compound – 2:

Let us consider 100 g of compound – 2

Mass of carbon = 27.3 g

Mass of oxygen = 100 g – 27.3 g = 72.7 g

Thus, 27.3 g of carbon combines with 72.7 g of oxygen.

∴ 1 g of carbon combines with

72.7/27.3 = 2.66 g of oxygen.   …………………….. (2)

From statements (1) and (2), the ratio of different masses of oxygen combining with fixed mass of carbon (1 g) is 1.33 : 2.66 i.e. 1 : 2, which is simple whole number ratio. Thus, the data illustrate the law of multiple proportions.

Example – 03:

A metal forms two oxides. The higher oxide contains 80% of metal. 0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidized. Show that the data illustrate the law of multiple proportions.

Solution:

Higher oxide:

Let us consider 100 g of higher oxide

Mass of metal = 80 g

Mass of oxygen = 100 g – 80 g = 20 g

Thus,80 g of metal combines with 20 g of oxygen.

∴ 1 g of metal combines with

20/80 = 0.25 g of oxygen.   …………………….. (1)

Lower oxide:

Mass of lower oxide = 0.72 g

Mass of higher oxide = 0.8 g

The higher oxide contains 80% of metal.

Mass of metal in higher oxide = 80/100 x 0.8 = 0.64 g

Mass of metal in lower oxide = 0.64 g

Mass of oxygen on lower oxide = 0.72 – 0.64 = 0.08g

Thus 0.64 g of metal combines with 0.08 g of oxygen.

∴ 1 g of carbon combines with 0.125 g of oxygen.

Thus the ratio of different masses of oxygen combining with fixed mass of hydrogen (1 g) is 0.25 : 0.125 i.e. 2 : 1, which is simple whole number ratio. Thus, the data illustrate the law of multiple proportions.

Example – 04:

Two oxides of metal contain 27.6 % and 30 % of oxygen respectively. If the formula of the first oxide is M₃O₄ what is the formula of the second oxide?

Solution:

Let ‘x’ be the atomic mass of the metal.

First oxide: % of oxygen = 27.6, % of metal = 100 – 27.6 = 72.4

The formula of the first oxide is M₃O₄. Thus the atomic ratio of M to O in oxide is 3 : 4.

Thus the atomic mass of the metal is 56.

Second oxide: % of metal = 70 % of oxygen = 100 – 70 = 30

Thus, the formula of the second oxide is M₂O₃.

In the next article, we shall study the law of reciprocal proportions.

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