Question

In a experiment, 4.90g of copper oxide was obtained from 3.92g of copper. In another experiment, 4.55g copper oxide gave on reduction 3.64g of copper. Show that these figures verify the law of constant proportion.

Answer 1

The law of constant proportion is also known as the law of definite proportion. This law states that in a compound, the elements are always present in definite proportions by mass. The reactio between copper and oxygen leading to the formation of copper oxide is written as follows $2Cu+O_2\to 2CuO$
As per the given data, 4.90g of copper oxide was obtained from 3.92g of copper. Therefore, the percentage of copper in 4.90g of CuO will be
$=\left(\frac{massofcopper}{massofCuO}\right)\times 100$
$=\left(\frac{3.92}{4.9}\right)\times 100$
$=80\%$
As per the data of the second experiment, 3.64g of copper is obtained from 4.55g of CuO. So, percentage of copper in this sample of copper oxide will be
$=\left(\frac{massofcopper}{massofCuO}\right)\times 100$
$=\left(\frac{3.64}{4.55}\right)\times 100$
$=80\%$
Thus, we see that percentage of copper in both samples of CuO is the same. This means that copper and oxygen always combine in a fixed ratio by mass to form CuO. Thus the law of constant proportions is satisfied here.