LAW OF MILTIPLE PROPORTION:

When two elements combine to form two or more than two different compounds then the different masses of one element B which combine with fixed mass of the other element bear a simple ratio to one another.

For example: Carbon forms two oxides in oxygen

The ratio of masses of oxygen in CO and CO₂ for fixed mass of carbon (12)
is 16: 32 = 1: 2.

ILLUSTRATIVE EXAMPLE (9): The law of multiple proportions is illustrated by the pair of compounds:

(A) Sodium chloride and sodium bromide     

(B) Water and heavy water

(C) Sulphur dioxide and sulphur trioxide       

(D) Magnesium hydroxide and magnesium oxide

SOLUTION:  In SO₂ 32 gram of suphur react with 32 gram of oxygen. Similarly for SO₃ fixed mass of sulphur (32 gram) react with 48 gram of oxygen. The ratio of oxygen’s mass = 32:48 = 2:3 which support law of multiple proportions. Hence (C) is correct answer.

ILLUSTRATIVE EXAMPLE (10): Carbon is found to form two oxides, which contain 42.9% and 27.3% of carbon respectively. Show that these figures illustrate the law of multiple proportions.

SOLUTION:  

Step 1: To calculate the percentage composition of carbon and oxygen in each of the two oxides

Step 2:  To calculate the weights of carbon which combine with a fixed weight i.e., one part by weight of oxygen in each of the two oxides.

In the first oxide, 57.1 parts by weight of oxygen combine with carbon = 42.9 parts.

1 part by weight of oxygen will combine with carbon = 42.9/57.1=0.751

In the second oxide 72.7 parts by weight of oxygen combine with carbon = 27.3 parts.

1 part by weight of oxygen will combine with carbon =27.3/72=0.376

Step 3:  To compare the weights of carbon which combine with the same weight of oxygen in both the oxides-The ratio of the weights of carbon that combine with the same weight of oxygen (1 part) is 0.751: 0.376 or 2:1

Since this is a simple whole number ratio, so the above data illustrate the law of multiple proportions.

TRY YOURSELF:

Exercise (1): Metal M and chlorine combine in different proportions to form two compounds A and B. The mass ratio M: Cl is 0.895: 1 in A and 1.791: 1 in B. What law of chemical combination is illustrated?

Exercise (2): By means of the following analytical results show that law of multiple proportions is true:

Exercise (3): 1 g of a metal, having no variable valency, produces 1.67 g of its oxide when heated in air. Its carbonate contains 28.57% of the metal. How much oxide will be obtained by heating 1 g of carbonate?

Exercise (4): Phosphorous and chlorine form two compounds. The first contains 22.54% by mass of phosphorous and the second 14.88% of phosphorous. Show that these data are consistent with law of multiple proportions.

TRY YOURSELF: SOLUTION:

(1) Mass of metal which combine with 1 part of chlorine are in the ratio of 1:2, which is a simple ratio. Hence, law of multiple proportions is illustrated.

(2) The masses of mercury which combine with 1 part of chlorine are in the ratio of 2:1 which is simple ratio. Hence, law of multiple proportions is illustrated.     

(3) 0.477 g

(4) Prove yourself ….

Limitation of law of multiple proportions: Non Stoichiometric compound do not follow this law for example Fe_0.93O₁.