Question

Copper and oxygen combine to form two oxides, which contain 88.7% and 79.8% of copper respectively. Show that these data illustrate Law of multiple proportion​

Answer 1

Answer:

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:

in first component

mass of copper=88.7%g

mass of oxygen= 100-88.7=11.3g

mass of oxygen combines with copper 88.7%g=11.3%g

mass of oxygen combines with one gram of copper =11.3/88.7=0.127

in second component

mass of copper= 79.8%g

mass of oxygen= 100-79.8=20.2g

mass of oxygen combines with copper 79.8%g=20.2g

mass of oxygen combines with one gram copper = 20.2/79.8=0.253g

ratio of mass of oxygen combined with fixed mass of copper=0.127/0.253=0.127:0.253=1:2

mass of oxygen combines with the one gram of copper in the two oxide in the ratio of 1:2. So the reaction obey slow of multiple proportion