Hydrogen sulphide `(H_(2)S)` contains `94.11 %` sulphur. Sulphur dioxide `(SO_(2))` contains `50 %` oxygen and water `(H_(2)O)` contains `11.11 %` hydrogen. Shpw that the results are in agreement with the law of reciprocal proportions.

Explanation

Let us fix `1g` of sulphur (S) as the fixed weight
In hydrogen sulphide `(H_(2)S)`
Weight of sulphur `= 94.11 g`
Weight of hydrogen `=100 - 94.11 =5.89 g`
`94.11g` of sulphur have combined have compound with hydrogen `= 5.89 g`
`1.0 g` sulphur has combined with hydrogen `= (5.89)/(94.11)g`
In sulphur dioxide `(SO_(2))`
`50.0 g` of sulphur have combined with oxygen `= 50.0 g`
`1.0 g` of sulphur has combined with oxygen `= 1.0 g`
Thus, the ratios by weight of hydrogen and oxygen combining with a fixed weight of sulphur in the two compounds is
or `(5.89)/(94.11):1or5.89:94:11` (This ratio is not the same)
In water `(H_(2)O)`
Weight of hydrogen `= 11.11 g`
Weight of oxygen `= 100-11.11 = 88.89 g`
`11.11 : 88.89`
Let us compare the ratios (i) and (ii). There are related to each other as :
`(5.89)/(94.11):(11.11)/(88.89)or0.0625:0.1336or1:2`
Since these ratios are the simple whole number multiples of each other, the Law of Reciprocal Proportion is illustrated.