Question

Define the law of multiple propositions. Explain with one example.

Answer 1

Law of Multiple Proportions Definition

When two elements combine to form two or more compounds, the ratio in which one of the elements combines with a fixed mass of the other is always simple.

"If an element combines with another element to produce two or more than two compounds, the weight of one which combines with the constant weight of the other, bears a simple ratio to one another."

For example Hydrogen and Oxygen combine to form water H2O and hydrogen peroxide H2O2. Two atoms of hydrogen combine with one atom of oxygen in the case of water, while two atoms of hydrogen combine with two atoms of oxygen in the case of hydrogen peroxide. The ratio of oxygen atoms combining with a fixed number of hydrogen atoms in these two compounds is 1:2.

In terms of masses also, every 2.016 g of hydrogen ( g.at. mass of hydrogen is 1.008g) combines with 15.999 g of oxygen to form water and 31.998g of oxygen to form hydrogen peroxide. Hence the ratio is 15.999 : 31.998 or 1:2 which is a simple ratio.

Law of Multiple Proportions Examples

Below you could see examples

Solved Examples

Question 1: What is the ratio of copper and oxygen in cuprous oxide?
Solution:

Copper forms two oxides namely Cuprous oxide (Cu2O) and Cupric oxide (CuO). 

Atomic mass of copper is 63. 

In terms of oxygen 1 atom of oxygen combines with 2 atoms or 2 X 63 parts by mass of Copper in cuprous oxide and 1 atom of oxygen combines with 1 atom or 1X 63 parts by mass of copper in cupric oxide. 

Thus the ratio is 2:1 which is a simple ratio.




Question 2: A metal forms two oxides with first oxide having 1 and second having 3 oxygen atoms and their masses are 74g and 164 g respectively. If the g.at.mass of oxygen is 16g, what is the atomic mass of the metal. prove that the two compounds follow the law of multiple proportions. 
Solution:

First Oxide contains 16 g of oxygen and 74-16 g = 58 g of metal. 

Since 1 atom of oxygen has to combine with minimum of 1 atom of metal to form its stable oxide, the atomic mass of metal should be 58.

Second Oxide has 3 X 16 = 48 g of oxygen and 164-48 = 116 g of metal.(since the formula shows two atoms of metal we can directly take 2 X 58=116 g) and has a formula M2O3

To prove the law of multiple proportions, in case of oxide M2O3, 16 g of oxygen combine with 116/3 g of metal = 38.66 g of metal.

Thus the metal ratios will be 38.66 : 58 or 1 : 1.5 (aprox) or 2: 3 which is a simple ratio.




Question 3: Hydrogen and oxygen forms two compounds. The hydrogen content in these compounds is 42.9% and 27.3%. Prove these compounds agree with the law of multiple proportions. 
Solution:

In the first compound, the mass of hydrogen is 42.9 g. and the mass of oxygen will be 100 - 42.9 = 57.1g.

Therefore the mass of hydrogen that combines with 1 g of oxygen = 42.9 / 57.1 = 0.75g

In the second compound the mass of hydrogen is 27.3g and the mass of oxygen will be 100 - 27.3 = 72.7g

Therefore the mass of hydrogen that combines with 1 g of oxygen = 27.3 / 72.7 = 0.375g

Thus the ratio of hydrogen combining with a fixed mass of oxygen is 0.75 : 0.375 or 2:1 which proves the law of multiple proportions

Answer 2

$\huge\mathfrak\red{Answer :}$

$\huge\textbf{Law of Multiple Proportion :}$

When fixed mass of an element react with different amount of another same element then it bears a simple whole number ratio..
======================================

$\huge\textbf{Example :}$

$C$ + $\dfrac{1}{2}O_{2}$ ---> $CO$

Here,

$C$ = 12 gm

$\dfrac{1}{2}O_{2}$ = 16 gm

* $C$ + $O_{2}$ ---> $CO_{2}$

$C$ = 12 gm

$O_{2}$ = 32 gm

→ Ratio of $\dfrac{1}{2}O_{2}$ : $O_{2}$

= 16 : 32

= 1 : 2