# 2 oxides of a certain metal was separately heated in a current of hydrogen until constant weights were obtained. The water produced in each case was carefully collected and weighed. 2 grams of each oxide gave respectively 0.2517 grams and 0.4526 grams of water. Show that these results establish the law of multiple proportion.

The law of multiple proportion in all the given cases can be verified by combining the weight of oxygen with the weight of the metal in two different oxides. Therefore, following are the steps that needs to be followed:
Step 1: To calculate the weight of oxygen in each oxide
Weight of the each oxide = 2g
Weight of water produced in case 1 = 0.2517g
Weight of water produced in case 2 = 0.4526g
18g of H2O is equivalent to 16g of oxygen
Which means that 18g of water contains 16g of oxygen
Therefore, 0.2517g of water contains (16)/(18)*0.2517 = 0.2237g
And, 0.4526g of water contains (16)/(18)*0.4526 = 0.4023g

Step 2: To calculate the weight of oxygen when combined with 1g of metal in each oxide
Case 1
Weight of metal oxide = 2g
Weight of oxygen =0.2237g
Weight of metal = 2-0.2237 =1.7763g
Weight of oxygen combining with 1.7763g of metal = 0.2237g
Weight of oxygen combining with 1g of metal = (0.2237)/(1.7763) = 0.1259g
Case 2
Weight of metal oxide = 2g
Weight of oxygen =0.4023g
Weight of metal = 2-0.4023 =1.5977g
Weight of oxygen combining with 1g of metal = (0.4023)/(1.5977) = 0.2518g

Case 3:
To compare the weights of oxygen
The ratio of 0.1259:0.2582 is same as 1:2

This ratio explains the law of multiple proportions.