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## Definition of Law of Reciprocal Proportions

The law of reciprocal proportions states that, If two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they do so are either the same as or a simple

multiple of the ratio of the masses in which they combine.

## Other Definitions

- The law was proposed by Jeremias Richter which he used for neutralization of ratios of metals with acids.

It was stated that when two substances A and B are chemically reactive with two other substances C and D. Then the ratio of the quantities of C and D has the same amount of A as the ratio of the quantities of C and D has with B.

- The law of reciprocal proportions (also called the law of equivalent proportions or the law of permanent ratios ) is one of the basic laws of stoichiometry. It relates the proportions in which elements combine across some different elements. Jeremias Richter first formulated it in 1791.
- The acceptance of the law allowed tables of element equivalent weights to be drawn up. Chemists widely used these equivalent weights in the 19th century. The other laws of stoichiometry are the law of definite proportions and the law of multiple proportions. Law of definite proportions refers to the fixed composition of any compound formed between element A and element B. And the law of multiple proportions describes the stoichiometric relationship between two or more different compounds formed between element A and element B.
- The law states that if two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they combine are either the same or are in simple multiple ratios of the masses in which they combine with each other.

## Explanation of the Law with Example

The above statement might be difficult to understand, but it can be easily explained with help of an illustration. Consider reactions: A reacts with C to form D and B reacts with C to form E. The reactions are shown below:

A + C – > D

B + C – > E

Let m_{A}, m_{B} and m_{C} be the mass consumed of A, B, and C in the above reactions. The mass of C, m_{C} is maintained constant in both the above reactions. r_{AB} is the ratio of mass of A and B.

r_{AB} = m_{A} / m_{B}

Consider one more reaction between A and B. A and B react to form AB

A + B – > AB

Let m’_{A} and m’_{B} be the mass consumed of A and B in the above reaction. The ratio of these two masses be r’_{AB}

r’_{AB} = m’_{A} / m’_{B}

As per the law, r_{AB} is the same or multiple of r’_{AB}.

r_{AB} = n x r’_{AB}

r_{AB} / r’_{AB} = n

## History of Law of Reciprocal Proportions

The law of reciprocal proportions was proposed in essence by Richter, following his determination of neutralization ratios of metals with acids.

In the early 19th century, it was investigated by Berzelius, who formulated it as follows: When two substances, A and B, have an affinity for two others, C and D, the ratio of the quantities C and D, which saturate the same amount of A are the same as that between the quantities C and D which saturate the same amount of B.

Later, Jean Stas showed that within experimental error, the stoichiometric laws were correct.

## Limitations of Law of Reciprocal Proportions

Here are some limitations of this law

- The presence of isotopes causes inconsistency in calculations. This can lead to a deviation in the final value of ratios.
- The law fails when a non-stoichiometric compound is formed in the chemical reaction.
- Hence the same isotope or mixture of isotope should be used throughout the preparation of a series of compounds. Since there are few elements which will combine with the third element and also combine with each other.

## Examples of Law of Reciprocal Proportions

Here are some examples of law of reciprocal proportions

### Example-1

Consider 3 grams of C reacting with 1 g of H to form methane. Also, 8 g of O reacting with 1 g of H to form water.

Here, the mass ratio of carbon and oxygen is 3:8

Similarly, 12g of C reacts with 32g of O to form CO_{2}.

The mass ratio of carbon and oxygen is 12:32 = 3:8

The mass ratio in which C and O combine with each other is the same as the mass ratio in which they separately combine with a fixed mass of H.

### Example-2

Oxygen and sulfur react with copper to give copper oxide and copper sulfide, respectively. Sulfur and oxygen also react with each other to form SO_{2}.

Therefore,in CuS:Cu:S = 63.5:32, in CuO:Cu:O = 63.5:16, S:O = 32:1 S:O = 2:1

Now in SO_{2} :S:O = 32:32 S:O = 1:1Thus the ratio between the two ratios is the following:2:1.

### Example-3

Let’s take a look at methane, CH4. Let’s figure out the proportion of elements.

The molecule weight of carbon is 12g/mol, and the molecule weight of hydrogen is 1g/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, H_{2}O. 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 (t he 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, CO_{2}, which has a proportion of 12:32. This is equal to 3:8, exactly as we predicted.

### Example-4

As an example, 1 gram of sodium (Na = A) is observed to combine with either 1.54 grams of chlorine (Cl = B) or 5.52 grams of iodine (I = C). (These ratios correspond to the modern formulas NaCl and NaI). The ratio of these two weights is 5.52/1.54 = 3.58.

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