Law of Definite Proportions

Science > Chemistry > Laws of Chemical Combinations > Law of Definite Proportions

In the last article, we have studied the law of conservation of mass. In this article, we shall study the law of definite proportions. The law of definite proportions was given by French chemist Joseph Proust in 1799. The law of definite proportions is also known as the law of constant or fixed proportion.

Statement:

A pure chemical compound, irrespective of its source or the method of preparation, always contains the same elements combined in a fixed or constant ratio by weight.

Explanation: 

If the compound AB is formed by two different methods and in the first method, if m gm of A combines with n gm of B and In the second method, if x gm. of A combine with y gm of B, then according to this law,

m/n = x/y

Illustrations:

We get water from different sources like rain, streams, lakes, springs, wells, tanks, rivers, sea, chemical reactions etc. But in each case, the water contains the same elements hydrogen and oxygen combining in the ratio 2:16 i.e. 1: 8 by weight. Thus irrespective of the source and method of preparation water contains the same elements combining in the same (constant) proportion.

We get carbon dioxide from different sources like atmosphere, photosynthesis reaction, springs, chemical reactions etc. But in each case, the carbon dioxide contains the same elements carbon and oxygen combining in the ratio 12:32 i.e. 3: 8 by weight. Thus irrespective of the source and method of preparation carbon dioxide contains the same elements combining in the same (constant) proportion.

Limitations:

Isotopes are the atoms of the same element which have the same atomic number but different atomic weights.  The law is not applicable to those elements which are found in different isotopic forms. For e.g. Chlorine has two isotopes having weights 35 and 37. In hydrogen chloride (HCl) containing the chlorine atom with atomic weight 35, the ratio of the weight of hydrogen to that of chlorine is 2:35. In hydrogen chloride (HCl) containing the chlorine atom with atomic weight 37, the ratio of the weight of hydrogen to that of chlorine is 2:37. These two ratios are not the same.

The converse of this law is not true. There are some compounds in which the same elements combine together in the same proportion, give different compounds.For e.g. Dimethyl ether (CH₃ – O – CH₃) and Ethyl alcohol (C₂H₅OH) have the same molecular formula C₂H₆O, and hence they contain the same elements carbon, hydrogen, and oxygen combining in the proportion 24 : 6 :16 i.e. 12 : 3: 8 by weight. But these two compounds are entirely different having different properties. Such compounds are called isomers.

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

According to Dalton’s atomic theory, a compound is formed by the combination of atoms of different elements in simple numerical proportions. Atoms of an element are identical in all respects, thus the weight of atoms are also fixed. Therefore, a chemical compound always contains the same elements combining in the same constant proportion by weight. This explains the law of definite proportion.

Numerical Problems:

Example – 01:

1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the mass of copper that remained was 1.098 g. In another experiment, 1.179 g of copper was dissolved in the nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The mass of cupric oxide formed was 1.476 g. Show that the results illustrate the law of definite proportions.

Solution:

Experiment -1:

Mass of cupric oxide = 1.375 g

Mass of copper = 1.098 g

Mass of oxygen = 1.375 g – 1.098 g = 0.277 g

Ratio of mass of copper to that of oxygen is

1.098 g / 0.277 g = 3.96  ………….  (1)

Experiment -2:

Mass of cupric oxide = 1.476 g

Mass of copper = 1.179 g

Mass of oxygen = 1.476 g  – 1.179 g = 0.297 g

Ratio of mass of copper to that of oxygen is

1.179 g / 0.297 g = 3.97  ………….  (2)

Within experimental limits, the results of experiment 1 and 2 show that the ratios of mass of copper to the mass of oxygen in both the samples of cupric oxide are the same. Hence, the results illustrate the law of definite proportions.

Example – 02:

2.8 g of calcium oxide prepared by heating limestone was found to contain 0.8 g of oxygen. When 1 .0 g of oxygen is treated with calcium, 3.5 g of calcium oxide was obtained. Show that the result illustrates the law of definite proportions.

Solution:

Experiment -1:

Mass of calcium oxide = 2.8 g

Mass of oxygen = 0.8 g

Mass of calcium = 2.8 g – 0.8 g = 2.0 g

Ratio of mass of calcium to that of oxygen is

2.0 g / 0.8 g =  2.5  ………….  (1)

Experiment -2:

Mass of calcium oxide = 3.5 g

Mass of oxygen = 1.0 g

Mass of calcium = 3.5 g – 1.0 g = 2.5 g

Ratio of mass of calcium to that of oxygen is

2.5 g / 1.0 g = 2.5  ………….  (2)

The results of experiment 1 and 2 show that the ratios of mass of calcium to the mass of oxygen in both the samples of calcium oxide are the same. Hence, the results illustrate the law of definite proportions.

Example – 03:

A sample of pure magnesium carbonate was found to contain 28.5 % of magnesium, 14.29 % of carbon, and 57.14 % of oxygen. Applying the law of constant proportion, find the mass of magnesium, carbon, and oxygen in 15.0 g of another sample of magnesium carbonate.

Solution:

In MgCO3 % of Mg = 28.5, % of C = 14.29, % of O = 57.14

Mass of second sample of MgCO₃ = 15.0 g

Applying the law of constant proportion

Mass of Mg in the second sample =  = 15.0 g x  28.5 / 100 =  4.28 g

Mass of C in the second sample = 15.0 g x 14.29 / 100 =  2.14 g Mass of C in

Mass of O in the second sample =  15.0 g – (4.28 g + 2.14 g) =  8.58 g

Thus, 15.0 g of MgCO3 contains 4.28 g of magnesium, 2.14 g of carbon, and 8.58 g of oxygen.

Example – 04:

A sample of pure calcium carbonate was found to contain 40 % of calcium, 12 % of carbon, and 48 % of oxygen. Applying the law of constant proportion, find the mass of calcium, carbon, and oxygen in 2.44 g of another sample of calcium carbonate.

Solution:

% of Ca in CaCO3 = 40, % of C in CaCO3 = 12, % of O in CaCO3 = 48

Mass of second sample of CaCO3 = 2.44 g

Applying the law of constant proportion

Mass of Ca in second sample = 2.44 g x 40/100 =  0.98 g

Mass of C in second sample = 2,44 g x 12/100 =  0.29 g

Mass of C in second sample = 2.44 g – (0.98 g + 0.29 g) =  1.17 g

Thus, 2.44 g of CaCO3 contains 0.98 g of calcium, 0.29 g of carbon, and 1.17 g of oxygen.

Example – 05:

Mass of copper oxide obtained by treating 2.16 g of metallic copper with nitric acid and on subsequent ignition was 2.70 g. In another experiment, 1.15 g of copper oxide on reduction yielded 0.92 g of copper. Show that the results are in accordance with the law of fixed proportions.

Solution:

Experiment -1:

Mass of copper oxide = 2.70 g

Mass of copper = 2.16 g

Mass of oxygen = 2.70 g – 2.16 g  = 0.54 g

Ratio of mass of copper to that of oxygen is

2.16 / 0.54 = 4  ………….  (1)

Experiment -2:

Mass of copper oxide = 1.15 g

Mass of copper = 0.92 g

Mass of oxygen = 1.15 g – 0.92 g = 0.23 g

Ratio of mass of copper to that of oxygen is

0.92 / 0.23 = 4  ………….  (2)

The results of experiment 1 and 2 show that the ratios of the mass of copper to the mass of oxygen in both the samples of copper oxide are the same. Hence, the results are in accordance with the law of fixed proportions.

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