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

1.375 g of cupric oxide was reduced by heating in a current of hydroge

1.	1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098 g In another experiment, 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g. Show that these result illustrate the law of constant composition.
Answer» Solution : In the first EXPERIMENT: Mass of CuO taken = 1.375 g Mass of Cu obtained = 1.098 g `therefore` Mass of oxygen that combined with `Cu = 1.375 - 1.098 = 0.277g` Thus, the percentage of oxygen in the given sample of `CuO = 0.277/(1.375)XX 100 = 20.14 %` In the serial experiment: Since, 1.178 g of Cu yielded 1.467 g of CuO, the mass of oxygen added would be: `=1.476 - 1.178 = 0.298 g` `therefore` Percentage of oxygen in this SIMPLE `=(0.298 xx 100)/1.476 = 20.19%` Since, the percentage of oxygen is almost the same (within the limit of experimental errors) in the two cases, the data are in accordance to the LAW of constant proportion and prove it.