Question
Question: 5.06 g of pure CuO on complete reduction by heating in a current of hydrogen gave 4.04 g of metallic...
5.06 g of pure CuO on complete reduction by heating in a current of hydrogen gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was carefully dried and ignited 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions.
Solution
. The Law of constant proportions states that the compounds that are made up of the same type of elements which are present in constant ratio by mass.
If both the compounds have a fixed ratio of elements, then the compound follows a law of constant proportions.
Complete step by step answer:
First, let us see what the law of constant proportions is.
The Law of constant proportions states that the compounds that are made up of the same type of elements which are present in constant ratio by mass.
For example - The CuO given to us in question will consist of a fixed ratio of elements by mass. The Cu and O will be constant by mass in both the CuO compounds.
If it is true then the above given compound follows that law of constant proportions otherwise not.
Let us see the first compound.
Mass of CuO (w1) = 5.06 g
Mass of Cu (w2) = 4.04 g
Mass of O = Mass of CuO - Mass of Cu
Mass of O = 5.06 - 4.04
Mass of O = 1.02
If we see the ratio of mass of both elements as -
Mass of OMass of Cu=1.024.04
Mass of OMass of Cu= 3.9
Now, let us see the another compound
Mass of CuO = 1.63 g
Mass of Cu = 1.3 g
Mass of O = Mass of CuO - Mass of Cu
Mass of O = 1.63 - 1.3
Mass of O = 0.33
If we see the ratio of mass of both elements as -
Mass of OMass of Cu=0.331.3
Mass of OMass of Cu= 3.9
As the ratio is constant in both cases; So, it follows the law of constant proportions.
Note: It must be noted that law of constant proportions is also called Proust’s law or law of definite proportions. There are some compounds that do not follow these laws. Some of the natural polymers can vary in their compositions.