Question

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)`.

Answer 1

In `CuS(100G)`,
`m_(Cu) = 66.5g, m_(S) = (100 g) - (66.5 g) = 33.5 g`
`:.` Ratio of masses `(m_(Cu) : m_(O)) = (66.5 g) : (33.5 g)`
`= (66.5)/(66.5) , (33.5)/(66.5) = 1 : 0.503`
In `CuO (100 g)`,
`m_(Cu) = 79.99 g, m_(O) = (100 g) - (79.9 g) = 20.1 g`
Ratio of masses `(m_(Cu) : m_(S)) = 79.9 g: 20.1 g`
`= (79.9)/(79.9): (20.1)/(79.1) = 1 = 0.251`
Thus, the ratio of masses of `S` to `O` combining with a given mass of Cu (1 part) is
`0.503: 0251 = (0.503)/(0.251) :(0.251)/(0.251) = 3:1`
In `SO_(3) (100 g)`,
`m_(S) = 40g, m_(O) = (100g) - (40 g) = 60 g`
Ratio of masses `(m_(S) : m_(O)) = (40 g) : (60g) = 2:3`