Question

Copper oxide was prepared by 2 different methods. In case 1, 1.75 g of metal gives 2.19 grams of oxide. In case 2, 1.14 grams of metal gives 1.43 grams of oxide. How we can illustrate this according to the law of constant proportion

Answer 1

In case 1, mass of copper = 1.75 g
Mass of copper oxide = 2.19 g

% of copper in the oxide =Mass of copper/Mass of copper oxide×100 = $\frac{1.75}{2.19}\times 100=79.9$%

∴% of oxygen =100−79.9=20.1%

In case 2, the mass of copper = 1.14 g
Mass of copper oxide = 1.43 g
% of copper in the oxide =$\frac{1.14}{1.43}\times 100=79.7$%
% of oxygen = 100 - 79.7 = 20.3 %.

Thus, copper oxide prepared by any of the given methods contains copper and oxygen in the same proportion by mass (with the experimental error). Hence, it proves the law of constant proportions.