Question

On analysis it was found that the black oxide of copper and red oxide of copper contain 80% and 89% of copper respectively. Show that this data is in accordance with law of multiple proportions.

Answer 1

Hint: An oxide is a chemical compound which contains at least one oxygen atom with any another element. Oxide is an anion of oxygen represented as ${{O}^{2-}}$ In this state the oxygen is present in -2 oxidation state and acts as an anion which carries negative charge.

Complete answer:
Black oxide of copper = $CuO$
Red oxide of copper = $C{{u}_{2}}O$
% copper in $CuO$= 80%
% copper in $C{{u}_{2}}O$= 20%
Give % of copper in $C{{u}_{2}}O$ = 89%
So % of oxygen will be 11%
20 g of oxygen contains 80 g of metal
So 8 g of oxygen combine with $\dfrac{80\times 8}{20}=32g$
11 g of oxygen combines with 89 g of metal
So 8 g oxygen combines with $\dfrac{89\times 8}{11}=64.72g$
The weight of metal are = 32:64.72 = 1:2
Thus for the fixed weight of oxygen the amount of metal combined in 1:2.
The law of multiple proportions is obeyed.

Note:
Law of multiple proportions are also known by the name Dalton’s law by the name of scientist John Dalton which first expressed it. This states that if two elements form more than one compound present between them then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.