Question

CuS contains 33.3% of sulphur, CuO contains 20.1% of oxygen and SO3 contains 40% of sulphur. show that the data illustrates the law of reciprocal proportion.​

Answer 1

Solution :-

As per the given data ,

$\bigstar \:\mathtt{\underline{CuS (Copper \:sulphide )}}$

Amount of Sulphur = 33.3 g

Amount of copper present = 100 - 33.3 = 66.7 g

If 66.7 g of copper contains 33.3 g of sulphur

Then , 1 g of copper will contain = 33.3/66.7 = 1/2 g of sulphur

Similarly ,

$\bigstar \:\mathtt{\underline{CuO (Copper \: oxide)}}$

Amount of oxygen = 20.1  g

Amount of copper present = 100 - 20.1  = 79.9 g

If 79.9 g of copper contains 20.1  g of oxygen  

Then , 1 g of copper will contain = 20.1 /79.9=  1 / 4 g of oxygen

The m/m ratio of S :O = 4 /2 = 2

$\bigstar \:\mathtt{\underline{SO_3 (Sulphur \:trioxide)}}$

Amount of sulphur= 40 g

Amount of oxygen present = 100 - 40  = 60 g

Hence ,

S:O = 40 / 60 = 2 /3

According to law of reciprocal proportions ,

"The weights of the two or more elements which separately react with same weight of third element are also the weights of these elements which react with each other or in simple multiple of them "

Hence , the following data illustrates law of reciprocal proportions