Question

1.375 g of CuO was reduced by hydrogen and 1.098 g of Cu was obtained. In another experiment, 1.178 g of Cu was dissolved in nitric acid and the resulting copper nitrate was converted into CuO by ignition. The weight of CuO formed was 1.476 g. Show that these results prove the law of constant composition.

Answer 1

Hint:Recall the law of constant composition (or the definite proportion), and apply its knowledge to prove that the formation of copper oxide (CuO) has the same proportion (i.e., ratio) of copper and oxygen in both the given conditions.

Complete answer:
First, let’s understand what is the law of constant composition, according to the law of constant composition sometimes called Proust’s law states that a given chemical compound always contains its component elements in a fixed ratio and this does not depend on the source or method of preparation of that chemical compound.
Let’s calculate the ratio of copper and oxygen present in the first case,
The weight of copper in 1.375 g of CuO is 1.098 g,
$ \Rightarrow $ the weight of oxygen = 1.375 – 1.098 = 0.277 g
$\therefore $ the ratio of their weight,
$ \Rightarrow $ weight of copper : the weight of oxygen
$ \Rightarrow $ 1.098 : 0.277
$ \Rightarrow $ 3.9 : 1 $ \simeq $ 4 : 1
Now, let’s calculate the ratio for the second case,
The weight of copper that was added was 1.178g and it formed 1.476 g of CuO,
$ \Rightarrow $ the weight of oxygen = 1.476 – 1.178 = 0.298
$\therefore $ the ratio of their weight,
$ \Rightarrow $ the weight of copper : the weight of oxygen
$ \Rightarrow $ 1.78 : 0.304
$ \Rightarrow $ 3.9 : 1 $ \simeq $4 :1
In both cases, the ratio is the same i.e., 4 :1

Therefore, the law of constant composition is well defined in these conditions of formation of copper oxide (CuO).

Note:
While calculating the weight of oxygen in the second case remember to put the value carefully .i.e., first write the weight of CuO and then the weight of Cu not the other way around, otherwise you will get a negative value of the weight which is not possible.