copper forms two oxides. On heating 1 g of each in hydrogen 0.888 g and 0.798 g of the metal was obtained. Show that the results are in agreement with the law of multiple proportions.

For the first oxide, 1 g gives 0.888 g of copper.

Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.

For the second oxide, 1 g gives 0.798 g of copper.

Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.

So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.

Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.

let the two oxides of Cu be named A and B

oxide A

mass of Cu in 1g of oxide A = 0.888g

mass of O in 1g of oxide A =1-0.888=0.112g

oxide B

mass of Cu in 1g of   oxide B = 0.799g

mass of O in g of oxide B = 1-0.799=0.201g

In oxide A , 1g Cu is combined with oxygen = 0.112 x 1/0.888 = 0.126g

In oxide B , 1g Cu is combined with oxygen = 0.201 x 1/0.799 = 0.252g

the different weight of oxygen combined with a fixed weight (1g) of Cu are in ratio 0.126 : 0.252  or 1: 2