Question

Explain law of multiple proportion and law of reciprocal proportion in a chemical reaction with suitable examples.

★ Neither copy from internet and paste over here nor spam.

All the very best !

Answer 1

Answer:

⇒ Law of Multiple Proportion

→Whenever the same two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers.

Let us take an example of Carbon dioxide

• In every sample of carbon dioxide, there is 32.0 g of oxygen present for every 12.0 g of carbon. By dividing 32.0 by 12.0, this simplifies to a mass ratio of oxygen to carbon of 2.66 to 1.  
• There is another compound that forms from the combination of carbon and oxygen called carbon monoxide.
• Every sample of carbon monoxide contains 16.0 g of oxygen for every 12.0 g of carbon.
• This is a mass ratio of oxygen to carbon of 1.33 to 1. In the carbon dioxide, there is exactly twice as much oxygen present as there is in the carbon monoxide.

⇒Law of reciprocal proportion

→The law of reciprocal proportions says that if we know the proportion of elements in compound AB and the proportion of elements in compounds BC, we can determine the proportion of elements in compound AC

Let us taken an example of Methane and water

• The molecule weight of carbon is 12 g/mol, and the molecule weight of hydrogen is 1 g/mol. Since we have 4 atoms of hydrogen for every atom of carbon, the proportion is 12:4, which can be simplified down to 3:1.
• Now, let's look at water, H2O. The proportion of elements is 16:2, or 8:1 (oxygen has a molecular weight of 16)
• So, methane and water both contain a hydrogen and one other element. According to this law if we combine carbon and oxygen (the other element in both compounds) it should be in a ratio of 3:8, or a simple multiple of that ratio.
• We get 3:8 because, in methane, carbon is a ratio of 3, and in water, oxygen is a ratio of 8. Let's see if this is true, when carbon and oxygen combine they form carbon dioxide, CO2, which has a proportion of 12:32. This is equal to 3:8, exactly as we predicted!

Answer 2

$\textbf{\large Law of Multiple Proportions (by John Dalton)}$

   $\textsf{When one element combines with another element to form two or more}\\\textsf{different compounds. The mass of one element, which combines with a}\\\textsf{constant mass of other, bears a simple ratio.}$

       $\textsf{\textbf{Example - }}\\\textsf{\textbf{\hspace{5.345em} N \hspace{1.73em} O}}\\\textsf{\hspace{.5em} N_2O \hspace{2.6em} 20 \hspace{1.76em} 16}\\\textsf{\hspace{.5em} N_2O_2 \hspace{2.16em} 20 \hspace{1.76em} 32}\\\textsf{\hspace{.5em} N_2O_3 \hspace{2.16em} 20 \hspace{1.76em} 48}\\\textsf{\hspace{.5em} N_2O_4 \hspace{2.16em} 20 \hspace{1.76em} 64}\\\textsf{\hspace{.5em} N_2O_5 \hspace{2.16em} 20 \hspace{1.76em} 80}$

          $\textsf{For constant mass of N_2 ,}\\\textsf{\: \: mass of O_2 \: = 16 : 32 : 48 : 64 : 80}\\\textsf{\hspace{6.44em} = 1 : 2 : 3 : 4 : 5}$

$\textbf{\large Law of Reciprocal Proportions (by Jeremias Richter)}$

   $\textsf{The ratio of the weights of two elements A \& B which combine separately}\\\textsf{with a fixed weight of third element C is either same or in a simple mutiple}\\\textsf{of the ratio of weights in whcih A \& B combine directly with the other.}$

        $\textsf{A \rightarrow \boxed{C}_{\: (Constant)} \leftarrow B \: \: and,}\\\\\textsf{\hspace{3.5em} A \leftrightarrow B}\\\\\left( \dfrac{A}{B}\right)_{Direct} : \left( \dfrac{A}{B}\right)_{Separately} = 1 : 1 \textsf{\:\: (Simple ratio)}$

       

       $\textsf{\textbf{Example - }}$

           $\textsf{C \xrightarrow{CH_4} \boxed{H}_{\: (Constant)} \xleftarrow{2 \: * H_2O} O \: \: and,}\\\\ \textsf{\hspace{4em} C \xrightarrow{CO_2} O }\\\\\left( \dfrac{3}{8}\right)_D : \left( \dfrac{3}{8}\right)_S = 1 : 1$