Law of multiple proportion- source : pinterest.com

## Law of multiple proportion-

This law was given by John Dalton. According to this law,

”  When two elements combine to form two or more than two compounds, the weights of one of the element which combine with a fixed weight  of other,bear a simple whole number ratio.”

OR

” When two elements A and B combine to form more than one chemical compounds then different weights of A, which combine with a fixed weight of B, are in a proportion of simple whole numbers.”

### Experimental Verification-

Nitrogen and oxygen combine to form five oxides such as nitrous oxide (N2O) ,nitric oxide (NO),nitrogen trioxide (N2O3),nitrogen tetraoxide (N2O4) and nitrogen pentaoxide (N2O5).

### N2O5         28                 80                  14                                         40

Ratio of oxygen in different compounds which combine with same weight of nitrogen

## Problem 1 :

Tin combines with oxygen to form two compounds having the following composition ;

% of  Sn         % of Oxygen

Compound A            78.77               21.23

Compound B            88.12              11.88

Show that the above data follows the law of multiple proportion.

## Solution –

In compound A ,

21.23 part of oxygen combines with 78.77 part Sn

1 part of oxygen combines with = 78.77 x 1 /21.23

= 3.7 part tin

In compound B ,

11.88 part of oxygen combines with 88.12 part Sn

1 part of oxygen combines with = 88.12 x 1 /11.88

= 7.4 part tin

Ratio of Sn with one part of oxygen = 3.7  :  7.4

=  1    :   2

## Problem 2 :

Carbon combines with hydrogen to form compounds having the following composition ;

% of  C         % of H

Compound A            75.0               25.0

Compound B            85.7             14.3

Compound C            92.3             7.7

Show that the above data illustrate the law of multiple proportion.

## Solution –

In compound A ,

75.0 part of Carbon  combines with 25.0  part Hydrogen

1 part of Carbon combines with = 25.0 x 1 / 75.0

= 0.33 part hydrogen

In compound B ,

85.7 part of Carbon  combines with  14.3  part Hydrogen

1 part of Carbon combines with = 14.3 x 1 / 85.7

= 0.166 part hydrogen

In compound C ,

92.3 part of Carbon  combines with  7.7  part Hydrogen

1 part of  Carbon combines with = 7.7 x 1 /  92.3

= 0.08 part hydrogen

Ratio of Hyrogen  with one part of Carbon = 0.33 :  0.166 : 0.08

=  4     :   2  :    1

So data follows law of multiple proportion.