Copper oxide was prepared by the following methods: (a) In one case, 1.75 g of the metal were dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.(b) In the second case, 1.14 g of metal dissolved in nitric acid were precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43g of copper oxide.(c) In the third case, 1.45 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide.Find the percentage.

Copper oxide was prepared by the following methods: (a) In one case,

1.	Copper oxide was prepared by the following methods: (a) In one case, 1.75 g of the metal were dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.(b) In the second case, 1.14 g of metal dissolved in nitric acid were precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43g of copper oxide.(c) In the third case, 1.45 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide.Find the percentage.
Answer» In the FIRST experiment2.19 G of CuO contained 1.75 g of Cu100 g of CuO ------?\frac { 1.75 }{ 2.19 } \QUAD \TIMES \quad 100\quad =\quad 79.91\quad g 2.191.75 ×100=79.91gb) In the second experiment1.43 g of CuO contained 1.14 g of Cu100 g of CuO ------?\frac { 1.14 }{ 1.43 } \quad \times \quad 100\quad =\quad 79.72\quad g 1.431.14 ×100=79.72gc) In the third experiment1.83 g of CuO contained 1.45 g of Cu100 g of CuO ------?\frac { 1.45 }{ 1.83 } \quad \times \quad 100\quad =\quad 79.23\quad g 1.831.45 ×100=79.23gFrom the above three experiments, nearly same amount of COPPER oxide.Hence, the above data illustrate the Law of constant composition.