Question

In an experiment , 1.288 g of copper oxide was obtained from 1.03 g of copper. In another experiment, 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. Show that these figures verify the law of constant proportions.​

Answer 1

Answer:

In order to solve this problem we have to calculate the ratio of copper (Cu) and oxygen (O₂) in the two samples of copper oxide (CuO) compound. Now :

(a) In the first experiment :

Mass of Copper (Cu) = 1.03 g

Mass of Copper oxide (CuO) = 1.288 g

So, Mass of oxygen (O₂) = Mass of copper oxide (CuO) - Mass of copper (Cu)

⇒ 1.288 - 1.03

⇒ 0.258 g

Now, in the first sample of copper oxide compound :

» Mass of Copper (Cu) : Mass of Oxygen (O₂)

⇒ 1.03 : 0.258

⇒ 1.03/0.258 : 1

⇒ 3.99 : 1

⇒ 4 : 1

(b) In the second experiment :

Mass of Copper (Cu) = 2.938 g

Mass of Copper Oxide (CuO) = 3.672 g

So, Mass of oxygen (O₂) = Mass of Copper Oxide (CuO) - Mass of Copper (Cu)

⇒ 3.672 - 2.938

⇒ 0.734 g

Now, in the second sample of copper oxide compound :

» Mass of copper (Cu) : Mass of Oxygen (O₂)

⇒ 2.938 : 0.734

⇒ 2.938/0.734 : 1

⇒ 4 : 1

From the above calculations we can see that the ratio of Copper and oxygen elements in the two samples of copper oxide compound is the same 4 : 1. So, the given figures verify the law of constant proportions.