Law of Constant or Definite Proportion


 
 
Concept Explanation
 

Law of Constant or Definite Proportion

Law of Constant Proportions or Law of Definite Proportions: French chemist, Joseph Proust analysed the chemical composition (types and percentage of elements present) of a large number of compounds and came to the conclusion that the proportion of each element in a compound is constant (or fixed), irrespective of where the compound came from or who prepared it. On this basis he proposed the law of constant proportions, according to which “In a chemical substance (or compound), the elements are always present in definite proportions (or ratios) by mass.”

Example: In a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always 1: 8, whatever the source of water. Thus, if 9 g of water is decomposed, 1 g of hydrogen and 8 g of oxygen are always obtained. Similarly, carbon dioxide (CO_2) always contains carbon and oxygen in the ratio of 3: 8. If a sample of  (CO_2)contains 36 g of carbon then it is compulsory that the sample has 96 g Oxygen.

Example: Copper oxide was prepared by two different methods. In one case, 1.75 g of the metal gave 2.19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide. Show that the given data illustrate the law of constant proportions.

Case l : Mass of copper = 1.75 g          And mass of copper oxide = 2.19 g.

So, mass of oxygen  = Mass of copper oxide — Mass of copper = 2.19 — 1.75 = 0.44g

Now, in first sample of copper oxide compound. Mass of copper : Mass of oxygen = 1.75 : 0.44 = 3.98 : 1 = 4: 1

Case 2:  Mass of copper = 1.14g         And mass of copper oxide = 1.43 g.

So, mass of oxygen = Mass of copper oxide — Mass of copper = 1.43 — 1.14 = 0.29g

Now, in second sample of copper oxide compound. Mass of copper : Mass of oxygen

=1.14:0.29 = 3.93:1 =4 : 1

From the above calculations we can see that the ratio (or proportion) of copper and oxygen elements in the two samples of copper oxide compound is the same, i.e. 4: 1. So, the given data verify the law of constant proportions.

ELEMENT SYMBOL ATOMIC NUMBER ATOMIC MASS
HYDROGEN H 1 1
HELIUM He 2 4
CARBON C 6 12
NITROGEN N 7 14
OXYGEN O 8 16
FLUORINE F 9 19
SODIUM Na 11 23
MAGNESIUM Mg 12 24
SULPHUR S 16 32
CHLORINE Cl 17 35

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

In a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always 1: 8, whatever the source of water.

If 18 g of water is decomposed, then how much grams of hydrogen we get ?

Right Option : B
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Explanation
Question : 2

SO_2 gas was prepared by

(1) burning sulphur in oxygen,

(2) reaching sodium sulphite with dilute H_2SO_4 and

(3) heating copper with conc. H_2SO_4

And it was found that in each case sulphur and oxygen combined in the ratio of 1:1. The data illustrates the law of:

Right Option : C
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Explanation
Question : 3

The percentage by weight of O_2 in CaSO_4 ( O = 16, S = 32, Ca = 40) IS

Right Option : C
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Explanation
 
 


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