verification of law of constant proportion in easy words
Answers
Verification of law of constant proportion
Many compounds can be made by different methods. For example, two samples of the compound copper oxide, CuO, were obtained, one by decomposition of copper carbonate, CuCO3, and another by decomposition of copper nitrate, Cu(NO3)2
. From each of these samples,
a mass of 8g of copper oxide was taken and each was treated independently with hydrogen gas. Both gave 6.4 g copper and 1.8 g water. Let us see how does this is a verify of the law of constant proportion.
The reaction of copper oxide with hydrogen yielded two known substances, namely, the compound water and the element copper. It is known that, in the compound water, H2O, the elements H and O are in the proportion 1:8 by mass. This means that in 9g water there are 8g of the element oxygen. Therefore, 1.8g water contains (8x1.8/9 = 1.6)g oxygen. This oxygen has come from 8g copper oxide. It means that 8g of both the samples of copper oxide contained 6.4g copper and 1.6g oxygen; and the proportion by mass of copper and oxygen in it is 6.4:1.6, that is, 4:1. Thus, the experiment showed that the proportion by mass of the constituent elements in different samples of a compound is constant.
Now let us see what the expected proportion by mass of the constituent elements of copper oxide would be from its known molecular formula CuO. To find out this, we need to use the known atomic masses of the elements. The atomic masses of Cu and O are 63.5 and 16 respectively. This means that the proportion by mass of the constituent elements Cu and O in the compound CuO is 63.5 : 16 which is 3.968:1, or approximately 4:1.
The experimental value of proportion by mass of the constituent elements matched with the expected proportion calculated from the molecular formula. Thus, the law of constant proportion is verified.