LAW OF MULTIPLE PROPORTIONS: EXPLANATION, APPLICATIONS - CHEMISTRY - 2024

The law of multiple proportions is one of the principles of stoichiometry and was first formulated in 1803 by the chemist and mathematician John Dalton, to offer an explanation of the way in which chemical elements combine to form compounds..

In this law it is expressed that if two elements combine to generate more than one chemical compound, the proportion of the masses of element number two when integrating with an invariable mass of element number one will be in small integer relationships.

John dalton

In this way, it can be said that from the law of definite proportions formulated by Proust, the law of conservation of mass proposed by Lavoisier and the law of definite proportions, the idea of ​​atomic theory was arrived at (a milestone in the history of chemistry), as well as the formulation of formulas for chemical compounds.

Explanation

Joining two elements in different proportions always results in unique compounds with different characteristics.

This does not mean that the elements can be associated in any relationship, since their electronic configuration must always be taken into account to determine which links and structures can be formed.

For example, for the elements carbon (C) and oxygen (O), only two combinations are possible:

- CO, where the ratio of carbon to oxygen is 1: 1.

- CO ₂, where the ratio of oxygen to carbon is 2: 1.

Applications

The law of multiple proportions has been shown to apply more precisely in simple compounds. Similarly, it is extremely useful when it comes to determining the ratio required to combine two compounds and form one or more through a chemical reaction.

However, this law presents errors of great magnitude when applied to compounds that do not present a stoichiometric relationship between their elements.

Likewise, it shows great flaws when it comes to using polymers and similar substances due to the complexity of their structures.

Solved exercises

First exercise

The mass percentage of hydrogen in a water molecule is 11.1%, while in hydrogen peroxide it is 5.9%. What is the ratio of hydrogen in each case?

Solution

In the water molecule, the hydrogen ratio is equal to O / H = 8/1. In the peroxide molecule it is a O / H = 16/1

This is explained because the relationship between both elements is closely linked to their mass, so in the case of water there would be a ratio of 16: 2 for each molecule, or what is equal to 8: 1, as illustrated. That is, 16 g of oxygen (one atom) for every 2 g of hydrogen (2 atoms).

Second exercise

The nitrogen atom forms five compounds with oxygen that are stable under standard atmospheric conditions (25 ° C, 1 atm). These oxides have the following formulas: N ₂ O, NO, N ₂ O ₃, N ₂ O ₄ and N ₂ O ₅. How can this phenomenon be explained?

Solution

By means of the law of multiple proportions we have that oxygen binds to nitrogen with an invariable mass proportion of this (28 g):

- In N ₂ O the ratio of oxygen (16 g) to nitrogen is approximately 1.

- In NO, the ratio of oxygen (32 g) to nitrogen is approximately 2.

- In N ₂ O ₃ the ratio of oxygen (48 g) to nitrogen is approximately 3.

- In N ₂ O ₄ the ratio of oxygen (64 g) to nitrogen is approximately 4.

- In N ₂ O ₅ the ratio of oxygen (80 g) to nitrogen is approximately 5.

Third exercise

You have a couple of metal oxides of which one contains 27.6% and the other has 30.0% by mass of oxygen. If the structural formula of oxide number one was determined to be M ₃ O ₄. What would be the formula for oxide number two?

Solution

In oxide number one, the presence of oxygen is 27.6 parts out of 100. Therefore, the amount of metal is represented by the total amount minus the amount of oxygen: 100-27.4 = 72, 4 %.

On the other hand, in oxide number two, the amount of oxygen is equal to 30%; that is, 30 parts per 100. Thus, the amount of metal in this would be: 100-30 = 70%.

It is observed that the formula of oxide number one is M ₃ O ₄; this implies that 72.4% of metal is equal to three atoms of the metal, while 27.6% of oxygen is equal to four atoms of oxygen.

Therefore, 70% of the metal (M) = (3 / 72.4) x 70 atoms of M = 2.9 atoms of M. Similarly, 30% of oxygen = (4 / 72.4) x 30 O atoms = 4.4 M atoms.

Finally, the ratio or ratio of metal to oxygen in oxide number two is M: O = 2.9: 4.4; that is, it is equal to 1: 1.5 or, which is equal, 2: 3. So the formula for the second oxide would be M ₂ O ₃.

References

1. Wikipedia. (2017). Wikipedia. Recovered from en.wikipedia.org
2. Leicester, HM, Klickstein, HS (1952) A Source Book in Chemistry, 1400-1900. Recovered from books.google.co.ve
3. Mascetta, JA (2003). Chemistry the Easy Way. Recovered from books.google.co.ve
4. Hein, M., Arena, S. (2010). Foundations of College Chemistry, Alternate. Recovered from books.google.co.ve
5. Khanna, SK, Verma, NK, Kapila, B. (2006). Excel with Objective Questions in Chemistry. Recovered from books.google.co.ve