Question

In an experiment, 4.90 g of copper oxide was obtained from 3.92 g of copper. In another experiment, 4.55 g of copper oxide gave, on reduction, 3.64 g of copper. Show with the help of calculations that these figures verify the law of constant proportions.

Answer 1

The law of constant proportion states that in a compound, the element is always present in definite proportions by mass.

$2\mathrm{Cu}+{\mathrm{O}}_2\to 2\mathrm{CuO}$

4.90 g of copper oxide was obtained from 3.92 g of copper. Therefore, the percentage of copper in 4.90 g of copper oxide will be,

$$\begin{gathered} =\frac{\mathrm{mass}\mathrm{of}\mathrm{copper}}{\mathrm{mass}\mathrm{of}\mathrm{copper}\mathrm{oxide}}\times 100 \\ =\frac{3.92}{4.90}\times 100 \\ =80\% \end{gathered}$$

Now, 3.64 g of copper is obtained from 4.55 g of copper oxide. So, the percentage of copper in this sample of copper oxide will be,

$$\begin{gathered} =\frac{\mathrm{mass}\mathrm{of}\mathrm{copper}}{\mathrm{mass}\mathrm{of}\mathrm{copper}\mathrm{oxide}}\times 100 \\ =\frac{3.64}{4.55}\times 100 \\ =80\% \end{gathered}$$

Thus, we can see that the percentage of copper in both samples of copper oxide is the same. This means that copper and oxygen always combine in a fixed ratio by mass to form copper oxide. Thus, the law of constant proportions is satisfied here.