Question

1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the mass of copper that remained was 1.098 g . In another experiment 1.179 g copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide . The mass of cupric oxide formed was 1.476 g . Show that these results illustrate the law of constant composition.

Answer 1

first experiment :

Copper oxide = 1.375 g

Copper left = 1.098 g

Oxygen present = 1.375 - 1.098 = 0.277 g

hence % of oxygen in CuO = 0.277 X 100 /1.375 = 20.14 %

Second experiment :

copper taken = 1.179 g

Copper oxide formed = 1.476 g

Oxygen present = 1.476 - 1.179 = 0.297 g

hence % of oxygen of CuO = 0.297 X 100 / 1.476 = 20.12 %

Percentage of oxygen is same in both the above cases , so the law of constant proportion is illustrated...