Law of Reciprocal Proportions by Jeremias Richter

07th Jun 2019 @ 4 min read

The law of reciprocal proportions is also known as the law of equivalent proportions or the law of permanent ratios. It along with the law of definite and multiple proportions is one of the fundamental laws of stoichiometry. The law was proposed by German chemist Jeremias Richter in 1791. The is similar to the law of multiple proportions.

Statement

The law states when two elements separately combine with the fixed amount of a third element, the ratio of masses of two elements is the same or multiple of the ratio of masses in which they combine with one another.

Explanation

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.

$\text{A}+\text{C} \longarrow \text{D}$

$\text{B}+\text{C} \longrightarrow \text{E}$

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.

$r_\text{AB} = \frac{m_\text{A}}{m_\text{B}}$

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

$\text{A}+\text{B} \longrightarrow \text{AB}$

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.

$r'_\text{AB} = \frac{m'_\text{A}}{m'_\text{B}}$

As per the law, r_AB is the same or multiple of r'_AB.

$r_\text{AB} = n\times r'_\text{AB}$

$\frac{r_\text{AB}}{r'_\text{AB}} = n$

where n is a multiplier.

Examples

Example 1

Oxygen and sulphur can react with copper to form cupric oxide (CuO) and cupric sulphide (CuS) respectively. They also can react with each other to form sulphur dioxide (SO₂). The reactions are as follows:

$\underset{63.5\,\text{g}}{\text{Cu}}+\underset{32\,\text{g}}{\text{S}} \longrightarrow \underset{95.5\,\text{g}}{\text{CuS}}$

$\underset{63.5\,\text{g}}{\text{Cu}}+\underset{16\,\text{g}}{\text{O}} \longrightarrow \underset{79.5\,\text{g}}{\text{CuO}}$

$\underset{32\,\text{g}}{\text{S}_{}}+\underset{32\,\text{g}}{\text{O}_2} \longrightarrow \underset{64\,\text{g}}{\text{SO}_2}$

As we see in the above reactions, the amount of copper in the first two reactions is fixed. So, the ratio of the mass of S to O is $\frac{32}{16}=\frac{2}{1}$ .

$\therefore r_\text{SO}=2$

For the third reaction, the ratio of the mass of $\frac{32}{32}=\frac{1}{1}$

$\therefore r’_\text{SO}=1$

Thus, the first ratio (r_SO) is twice the second ratio (r'_SO). So, we conclude the law of reciprocal proportion is valid.

Example 2

12 g of carbon reacts with 4 g hydrogen to 16 g methane. Also, 12 g of carbon combines with 32 g oxygen to form carbon dioxide. We also know that water is formed from oxygen and hydrogen. These sentences can be expressed in the below reactions.

$\underset{12\,\text{g}}{\text{C}_{}}+\underset{4\,\text{g}}{2\text{H}_2} \longrightarrow \underset{16\,\text{g}}{\text{CH}_4}$

$\underset{12\,\text{g}}{\text{C}_{}}+\underset{32\,\text{g}}{\text{O}_2} \longrightarrow \underset{44\,\text{g}}{\text{CO}_2}$

$\underset{4\,\text{g}}{2\text{H}_2}+\underset{32\,\text{g}}{2\text{O}_2} \longrightarrow \underset{36\,\text{g}}{2\text{H}_2\text{O}}$

Taking the ratio of the amount of hydrogen to oxygen in first two reactions, we get $r_\text{HO}=\frac{4}{32}=\frac{1}{8}$ .

For the third reaction, $r’_\text{HO}=\frac{4}{32}=\frac{1}{8}$ .

As we can see, in this example, both the ratios are equal, which confirms the law of reciprocal proportions.

Example 3

Halogens are electronegative elements; they react with metal by gaining an electron from them. Consider below two reactions.

$\underset{46\,\text{g}}{2\text{Na}_{}}+\underset{71\,\text{g}}{\text{Cl}_2} \longrightarrow \underset{117\,\text{g}}{2\text{NaCl}_{}}$

$\underset{46\,\text{g}}{2\text{Na}_{}}+\underset{253.8\,\text{g}}{\text{I}_2} \longrightarrow \underset{299.8\,\text{g}}{2\text{NaCl}_{}}$

In the above reactions, 46 g of sodium reacts with 71 g chlorine and 253.8 g iodine to form their respective halides. The ratio of the mass of chorine to iodine, $r_\text{ClI}=\frac{71}{253.8}$ .

We will not convert this fraction to decimals.

Now, it is also known that halogens can react between themselves to form an interhalogen compound. Iodine trichloride is one of such compounds.

$\underset{253.8\,\text{g}}{\text{I}_{2}}+\underset{213\,\text{g}}{3\text{Cl}_2} \longrightarrow \underset{466.8\,\text{g}}{2\text{ICl}_3}$

The ratio of the mass of chlorine to iodine in the above reaction, $r'_\text{ClI}=\frac{213}{253.8}=3\times \frac{71}{253.8}$ .

The ratio of the two ratios r_ClI : r’_ClI = 1:3. Thus, r_ClI is one-third of r’_ClI, or r’_ClI is thrice of r_ClI. Finally, we can say the example obeys the law of reciprocal proportions.

Example 4

Phosphorus can react with hydrogen as well as chlorine as shown in the reactions below. Also, hydrogen and chlorine can combine to form hydrogen chloride.

$\underset{31\,\text{g}}{\text{P}_{}}+\underset{3\,\text{g}}{1.5\text{H}_2} \longrightarrow \underset{43\,\text{g}}{\text{PH}_3}$

$\underset{31\,\text{g}}{\text{P}_{}}+\underset{106.5\,\text{g}}{1.5\text{Cl}_2} \longrightarrow \underset{137.5\,\text{g}}{\text{PCl}_3}$

$\underset{1\,\text{g}}{\tfrac{1}{2}\text{H}_{2}}+\underset{106.5\,\text{g}}{\tfrac{}{}1.5\text{Cl}_2} \longrightarrow \underset{137.5\,\text{g}}{\tfrac{}{}\text{PCl}_3}$

The mass of phosphorus is fixed in the above reaction to 31 g. The ratio of chlorine to hydrogen is $r_\text{ClH}=\frac{106.5}{3}=35.5$ and $r'_\text{ClH}=\frac{35.5}{1}=35.5$ . Thus, both ratios are the same.

Limitations of 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.

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Laws Of Chemical Combinations

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