In an experiment 4.90g of copper oxide was obtained from 3.92g of copper . In another experiment 4.55g of copper oxide gave on reduction 3.64g of copper . Show with the help of calculations that these figures verify the law of constant propotions.
In an experiment 4.90g of copper oxide was obtained from 3.92g of copper . In another experiment 4.55g of copper oxide gave on reduction 3.64g of copper . Show with the help of calculations that these figures verify the law of constant propotions.
The law of constant proportion or law of definite proportion states that in a compound, the elements are always present in definite proportions by mass.In the given question at first Cu is reacting with O2 to form CuO.
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 %
In 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. So the given data verify the law of constant proportion or definite proportion.
The law of constant proportion or law of definite proportion states that in a compound, the elements are always present in definite proportions by mass.In the given question at first Cu is reacting with O2 to form CuO.
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 %
In 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. So the given data verify the law of constant proportion or definite proportion.
