Question

Copper and oxygen combine to form two oxides, the red cuprous (Cu2O) oxide and the black cupric oxide (CuO). Show that these compounds follow law of multiple proportion.

Answer 1

2Cu+O2=2CuO
4Cu+ O2=2Cu2O
We know that mass of O=16u & mass of Cu=63.5u.
In Cu2O& CuO if we fix the mass of Oxygen as 16 then the ratio of mass of Cu in both the oxides will be
2.63.5:63.5
127:63.5 which is 2:1
This is a simple whole number ratio hence these compounds follow law of multiple proportions

Answer 2

Answer:

According to law of multiple proportions, when two elements combine with each other to form two or more than two compounds, then the ratio of masses of one element is simple whole number when it combines with  fixed ratio of other.

For the given question, reactions taking place are:

            4Cu + O₂ → 2Cu₂O

Mass: 4×63.5   32

         = 254

            2Cu + O₂ → 2CuO

Mass: 2×63.5   32

          = 127

Now on fixing the mass of oxygen, the ratio of mass of copper in both oxides will be:

$=\frac{254}{127} =\frac{2}{1} = 2:1$

This is a simple whole number ratio.

Thus this proves that the compounds red cuprous oxide(Cu₂O) and black cupric oxide (CuO) follows law of multiple proportions.

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