Copper sulphide `(Cus)` contains `66.5 % Cu` copper axide `(CuO)` contains `79.9% Cu` and sulphur trioxide `(SO_(3))` containts `40% S`. Shows that these compounds obey the law of recipreocal proportions.
Strategy: Elements `S` and `O` chemically combine seperately with element `Cu` to from `Cus` adn `CuO`. Elements `S` and `O` also chemicaly combine with each other to form `SO_(3)`. Find the ratio of masses of `S` to `O` combining with a given mass of `Cu` and also find the ratio of masses of `S` to `O` in`SO_(3)`.

Explanation

In CuS, if Cu is `66.6%` then Sulphur is
In `SO_(3)`, oxygen : sulphur mass ratio is `60 : 40`
Now in `CuS, 33.4` parts of sulphur combines with
`66.6` parts of Cu
So, 40 parts of sulphur will combine with `(66.6xx40)/(33.4)=79.8` parts.
Now, the ratio of masses of Cu and Oxygen which combines with same mass of sulphur separately is `79.8 : 60`
`Cu : O` ratio by mass in CuO is `79.9 : 20.1`
So, Ratio -1 : Ratio - 2 `=(79.8)/(60)xx(20.1)/(79.9)=1:3`
This ratio is a simple whole number ratio, so the law of reciprocal proportions is proven.