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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.

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Solution

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 reaction between copper and oxygen leading to the formation of copper oxide is written as follows

2Cu + O2 → 2CuO

As per the given data, 4.90 g of copper oxide was obtained from 3.92 g of copper. Therefore, the percentage of copper in 4.90g of CuO will be

= (mass of copper / mass of CuO) X 100

= (3.92 / 4.9) X 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

= (mass of copper / mass of CuO) X 100

= (3.64 / 4.55) X 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.


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