 ## Law of Multiple Proportions

The law of multiple proportions states that if two elements A and B combine together to form more than one compound, then, the several masses of A, which separately combine with a fixed mass of B, are in a simple ratio.

This law recognizes the fact that two elements may combine to form more than one product. For example, carbon and oxygen can combine to form carbon(II)oxide and carbon(IV)oxide; nitrogen combines with oxygen to form three possible oxides - nitrogen(I)oxide, nitrogen(II)oxide and nitrogen(IV)oxide.

If we consider one of the elements to be of fixed mass in the different products, then, the other element will be of varied mass which can be seen to be in a simple ratio.

Example: (1). In CO and CO2 If we consider carbon to be of fixed mass in the two products, then we have

 CO CO2 12g of carbon + 16g of oxygen 12g of carbon + 32 g of oxygen Carbon is of fixed mass: 16g of oxygen 32g of oxygen Ratio:       1 2

The different masses of oxygen that separately combine with the fixed mass (12g) of carbon in the two products (CO and CO2) are in a simple ratio of 1:2. However, if we consider the mass of oxygen fixed in the two products, then the different masses of carbon, which separately combine with the fixed mass of oxygen, can also be expressed in a simple ratio as shown below:

 CO CO2 12g of carbon + 16g of oxygen 12g of carbon + 32 g of oxygen To make oxygen of equal mass, multiply the masses of the elements here by 2 or divide the masses of elements in the other column by 2.   :. 24g of carbon + 32g of oxygen 12g of carbon + 32g of oxygen Oxygen is of fixed mass: 24g of carbon 12g of carbon Ratio:       2 1

The ratio of the masses of carbon in the two products, CO and CO2 that separately combine with the fixed mass of oxygen is therefore 2:1 respectively.

2. In the oxides of nitrogen: N2O, NO and NO2. If we consider the mass of nitrogen to be fixed, then:

 N2O NO NO2 28g of N + 16g of OX 14g of N + 16g of OX 14g of N + 32g of OX Divide the mass of elements here by 2 to bring the mass of nitrogen to be same in all products   :. 14g of N + 8g of OX 14g of N + 16g of OX 14g of N + 32g of OX Mass of nitrogen is fixed: 8g of OX 16g of OX 32g of OX Ratio:       1 2 4

Therefore, the ratio of the different masses of oxygen that will combine with a fixed mass of nitrogen to form the products N2O, NO and NO2 is 1:2:4 respectively.

Considering the mass of oxygen fixed:

 N2O NO NO2 28g of N + 16g of OX 14g of N + 16g of OX 14g of N + 32g of OX :. 28g of N + 16g of OX 14g of N + 16g of OX Divide the masses of elements here by 2 to bring the mass of oxygen fixed   7g of N + 16g of OX Mass of oxygen is fixed: 28g of N 14g of N 7g of N Ratio:       4 2 1

Therefore, the ratio of the different masses of nitrogen which combine with the fixed mass of oxygen to form the products N2O, NO and NO2 is 4:2:1 respectively.

Note: to confirm whether a statement obeys the law of definite proportion or multiple proportions, what you should do is to keep the mass of one of the components of the compound fixed at 1g, and deduce the masses of the other component(s) in combination with it in the different compounds given.

If the masses of the other component(s) are the same in all the compounds, it means the composition of the different compounds is fixed, i.e., the law of definite proportion is satisfied. But if the masses are different, it means that the law of multiple proportions is satisfied, and the different masses can be expressed in a simple ratio.

Example: 1. An element X forms two oxides containing 77.47 and 69.62 per cent of X respectively. (a) what law is satisfied? (b) if the first oxide has the formula XO, what is the formula of the second oxide?

Solution: (a) In the first oxide, O is 22.53 g and X is 77.47g. 1g of oxygen combines with

77.47/22.53 = 3.4g of X

In the second oxide, O is 30.38g and X is 69.62g. 1g of oxygen combines with

69.62/30.38 = 2.3g of X

The masses of X in combination with a fixed mass of oxygen in both compounds are different, therefore, the law of multiple proportions is satisfied. Notice that percentage compositions can be expressed as composition in mass.

(b) Expressing the different masses of X in a simple ratio:

 X of 1st oxide X of 2nd oxide 3.4 2.3 3.4/2.3 2.3/2.3 1.5 or 3/2 1 or 3 2 or 1 2/3

If the 1st oxide is XO, the 2nd is X2/3O

X2/3O is same as X2O3

2. In two separate experiments, 0.125g and 0.11g of oxygen combine with a metal X to give V and W respectively. An analysis showed that V and W contain 0.5g and 0.44g of X respectively. What law is represented by the data above?

Solution: In V, 0.5g of X combine with 0.125g of oxygen. 1g of X will combine with

0.125/0.5 = 0.250g of oxygen

In W, 0.44 g of X combine with 0.11g of oxygen. 1g of X combine with

0.11/0.44 = 0.250g of oxygen

The masses of oxygen in the two compounds, V and W, which separately combine with a fixed mass of X are the same - the law of definite or constant proportion is satisfied. Notice that you will obtain similar result if you make the mass of oxygen fixed (at 1g) and deduce the masses of X in the two compounds.

 Like This Post? Please Share!

Copyright , All Rights Reserved Free Chemistry Online | About Us | Usage of Content | Total Disclosures | Privacy Policy