Physics |
Chemistry |
Biology |
Mathematics |

Science > Chemistry > Basic Concepts of Chemistry > You are Here |

- John Dalton FRS (6 September 1766 – 27 July 1844) was an English chemist, meteorologist and physicist.
- He is best known for his pioneering work in the development of atomic theory, and his research into colour blindness (Daltonism).
- His atomic theory is considered as the foundation of modern atomic theory.

### 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 (H
_{2}O) and hydrogen peroxide (H_{2}O_{2})

- 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 H
^{1}or H^{2}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

### Applications of Law of Multiple Proportions:

#### Example – 1:

- 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 – 2:

- 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 – 3:

- 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 – 4:

- 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

Element | % | Atomic Mass | Atomic Ratio |
---|---|---|---|

M | 72.4 | x | 72.4/x |

O | 27.6 | 16 | 27.6/16 |

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

Element | % | Atomic Ratio | Simplest Ratio | Whole Number Ratio |
---|---|---|---|---|

M | 70 | 70/56 = 1.25 | 1.25/1.25 = 1 | 2 |

O | 30 | 30/16=1.875 | 1.875/1.25 = 1.5 | 3 |

Thus, the formula of the second oxide is M_{2}O_{3}.

Science > Chemistry > Basic Concepts of Chemistry > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |

Thanks it was very helpful and satisfying i understood the concept clearly. Thanks again for such a great help.