Law of Constant or Definite Proportion
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 () always contains carbon and oxygen in the ratio of 3: 8. If a sample of
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 |
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 27 g of water is decomposed, then how much grams of hydrogen we get ? | |||
| Right Option : C | |||
| View Explanation | |||
The percentage by weight of | |||
| Right Option : C | |||
| View Explanation | |||
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 oxygen we get ? | |||
| Right Option : B | |||
| View Explanation | |||
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