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Law of Multiple Proportions
The law of multiple proportions states that if two elements A and B combine together to form more than one compound, then, the several masses of A, which separately combine with a fixed mass of B, are in a simple ratio.
This law recognizes the fact that two elements may combine to form more than one product. For example, carbon and oxygen can combine to form carbon (II) oxide and carbon (IV) oxide; nitrogen combines with oxygen to form three possible oxides - nitrogen (I) oxide, nitrogen (II) oxide and nitrogen (IV) oxide.
If we consider one of the elements to be of fixed mass in the different products, then, the other element will be of varied mass which can be seen to be in a simple ratio.
Example:
(1). In CO and CO2
If we consider carbon to be of fixed mass in the two products, then we have
|
CO |
CO2 |
|
12g of
carbon + 16g of oxygen |
12g of
carbon + 32 g of oxygen |
|
Carbon is of
fixed mass:
16g of oxygen |
32g of
oxygen |
|
Ratio:
1 |
2 |
The different masses of oxygen that separately combine with the fixed mass (12g) of carbon in the two products (CO and CO2) are in a simple ratio of 1:2.
However, if we consider the mass of oxygen fixed in the two products, then the different masses of carbon, which separately combine with the fixed mass of oxygen, can also be expressed in a simple ratio as shown below:
|
CO |
CO2 |
|
12g of
carbon + 16g of oxygen |
12g of
carbon + 32 g of oxygen |
|
To make
oxygen of equal mass, multiply the masses of the elements here by 2
or divide the masses of elements in the other column by 2.
:. 24g of carbon + 32g of oxygen |
12g of
carbon + 32g of
oxygen |
|
Oxygen is of
fixed mass: 24g of carbon |
12g of
carbon |
|
Ratio:
2 |
1 |
The ratio of the masses of carbon in the two products, CO and CO2
that separately combine with the fixed mass of oxygen is therefore 2:1 respectively.
2. In the oxides of nitrogen: N2O, NO and NO2. If we consider the mass of nitrogen to be fixed, then:
|
N2O |
NO |
NO2 |
|
28g of N +
16g of OX |
14g of N +
16g of OX |
14g of N +
32g of OX |
|
Divide the
mass of elements here by 2 to bring the mass of nitrogen to be same
in all products
:. 14g of N + 8g of OX |
14g of N +
16g of OX |
14g of N +
32g of OX |
|
Mass of
nitrogen is fixed: 8g of OX |
16g of OX |
32g of OX |
|
Ratio:
1 |
2 |
4 |
Therefore, the ratio of the different masses of oxygen that will combine with a fixed mass of nitrogen to form the products N2O, NO and NO2 is 1:2:4 respectively.
Considering the mass of oxygen fixed:
|
N2O |
NO |
NO2 |
|
28g of N +
16g of OX |
14g of N +
16g of OX |
14g of N +
32g of OX |
|
:. 28g of N + 16g of OX
|
14g of N +
16g of OX |
Divide the
masses of elements here by 2 to bring the mass of oxygen fixed
7g of N +
16g of OX |
|
Mass of
oxygen is fixed: 28g of N |
14g of N |
7g of N |
|
Ratio:
4 |
2 |
1 |
Therefore, the ratio of the different masses of nitrogen which combine with the fixed mass of oxygen to form the products N2O, NO and NO2 is 4:2:1 respectively.
Note: to confirm whether a statement obeys the law of definite proportion or multiple proportions, what you should do is to keep the mass of one of the components of the compound fixed at 1g, and deduce the masses of the other component (s) in combination with it in the different compounds given.
If the masses of the other component (s) are the
same in all the compounds, it means the composition of the different compounds
is fixed, i.e., the law of definite proportion is satisfied. But if the masses
are different, it means that the law of multiple proportions is satisfied, and
the different masses can be expressed in a simple ratio.
Example:
1. An element X forms two oxides containing 77.47 and 69.62 per cent of X respectively. (a) what law is satisfied? (b) if the first oxide has the formula XO, what is the formula of the second oxide?
Solution:
(a) In the first oxide, O is 22.53 g and X is 77.47g.
1g of oxygen combines with
77.47/22.53
= 3.4g of X
In the second oxide, O is 30.38g and X is 69.62g.
1g of oxygen combines with
69.62/30.38
= 2.3g of X
The masses of X in combination with a fixed mass of oxygen in both compounds are different, therefore, the law of multiple proportions is satisfied.
Notice that percentage compositions can be expressed as composition in mass.
(b) Expressing the different masses of X in a
simple ratio:
|
X of 1st
oxide |
X of 2nd
oxide |
|
3.4 |
2.3 |
|
3.4/2.3
|
2.3/2.3 |
|
1.5 or 3/2 |
1 |
|
or 3 |
2 |
|
or 1 |
2/3 |
If the 1st oxide is XO, the 2nd is X2/3O
X2/3O is same as X2O3
2. In two separate experiments, 0.125g and 0.11g of oxygen combine with a metal X to give V and W respectively. An analysis showed that V and W contain 0.5g and 0.44g of X respectively. What law is represented by the data above?
Solution:
In V, 0.5g of X combine with 0.125g of oxygen.
1g of X will combine with
0.125/0.5
= 0.250g of oxygen
In W, 0.44 g of X combine with 0.11g of oxygen.
1g of X combine with
0.11/0.44
= 0.250g of oxygen
The masses of oxygen in the two compounds, V and W, which separately combine with a fixed mass of X are the same - the law of definite or constant proportion is satisfied. Notice that you will obtain similar result if you make the mass of oxygen fixed (at 1g) and deduce the masses of X in the two compounds.
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