`1.59 g` of first sample fo cupric oxide `(CuO)` on comple reduction by hydrogen `(H_(2))` gas gave `1.27 g` of pure copper `(Cu)` metal. Secound pure sample of curpic oxide weighing `3.18 g` yieled `2.54 g` of pure copper metal on complete refuction by hydrogen gas. Show that the law of definite proportions is valid.
Strategy: Find the ratio by mass of copper to oxygen in both the samples.

Explanation

Pure sample `-I`
`m_(CuO) = 1.59 g`
`m_(Cu) = 1.27 g`
`:. M_(O) = (1.59 g) - (1.27 g) = 0.32 g`
Now, `m_(cu) : m_(O) = 1.27g : 0.312 g = ((1.27)/(0.32)): ((0.32)/(0.32)) = 3.9 : 1`
Pure sample - `II`
`m_(CuO) = 3.18 g`
`m_(Cu) = 2.54 g`
`:. m_(O) = (3.18 g) - (2.54 g) = 0.64 g`
Now, `m_(Cu) : m_(O) = (2.54 g): (0.64 g) = ((2.54)/(0.64)) : ((0.64)/(0.64))`
As both the pure sample contain the samples contan the same ratio by mass of copper to oxygen, the law of definite proptions is valid.