Equivalent Mass
The idea of equivalent mass to compare chemically different elements! [1]
Atoms combine with each other to form chemical compounds, such that the elements are always present in definite proportions by mass and this property can be used to make chemically different molecules with different mass, equal.
But how can we equate such elements? It's like balancing two different compounds which are chemically dissimilar, which is similar to adding \(2\) oranges and \(3\) apples!
But in the world of chemistry nothing is impossible. If you don't get it, here's something which might help.
Consider the following compounds:
\(\ce{HCl}:\) \(1\) gram of hydrogen reacts with \(35.5\) grams of chlorine.
\(\ce{AgCl}:\) \(108\) grams of silver reacts with \(35.5\) grams of chlorine.
From the above compounds, we can deduce that
\[\text{1g of }\ce{H2}\equiv \text{108g of }\ce{Ag},\]
just as we do in modular arithmetic! We also see that the above congruence just indicates that the combining capacities (or valency) of each of those atoms can be related to each other. A careful observation reveals that the valency of hydrogen equals the valency of silver. Let us now try to define equivalent mass:
All elements combine with each other according to the laws of chemical combination, and the number of parts by which an element combines with \(1\) part by mass of hydrogen, or \(8\) parts by mass of oxygen, or \(35.5\) parts by mass of chlorine, or one gram equivalent of any other element, is the value of the equivalent mass of the element.
Calculating Equivalent Mass
Equivalent mass, as said earlier, depends upon the valency of the element. The formula to calculate the equivalent mass of an element is given by
\[\text{Equivalent Mass}=\dfrac{\text{Atomic Mass}}{\text{Valency}}.\]
Sometimes if we don't know the valency of the atom, we cannot find the equivalent mass. But then, we have different solutions to get rid of this problem. We have many methods to determine the equivalent mass of an element, and we will discuss in detail three most popular methods:
1. Hydrogen Displacement Method
We know that metals react with acids to produce hydrogen. Hydrogen displacement method just uses this fact to calculate the equivalent mass. Here some mass of an element reacts with an acid to produce some amount of hydrogen gas under standard temperature and pressure (STP).
Next, we formulate the mass of the metal required to displace \(11200\text{ cm}^3\) of hydrogen, which is nothing but \(1\text{ g}\) of hydrogen as we learned in mole concept. Thus we arrive at this formula:
\[\text{Equivalent Mass}=\dfrac{\text{Mass of the Metal}\times 11200}{\text{Volume of Hydrogen Liberated (STP)}}.\]
\(3\) grams of metal reacts with \(\ce{dil_.HCl}\) and liberates \(2800\text{ cm}^3\) of hydrogen gas at STP. Calculate the equivalent mass of the metal.
Plugging in the given values in the previous equation, we get
\[\begin{align} \text{Equivalent Mass}&=\dfrac{3\times 11200}{2800}=12.\ _\square \end{align}\]
2. Oxide Method
We have studied that metals react with oxygen and produce metal oxides, and the oxide method evaluates the mass of the metal oxide generated when some mass of a metal readily reacts with oxygen. Then, as we need a standard ratio, we determine the mass of the metal that combines with \(8\text{ g}\) of oxygen, which in turn is the equivalent mass of the metal:
\[\text{Equivalent Mass}=\dfrac{\text{Mass of the Metal}\times 8}{\text{Mass of Oxygen in the Oxide}}.\]
The mass of a metal in its oxide is \(75\%\) of the mass of the whole compound. Calculate its equivalent mass.
Let us assume that the mass of the metal oxide is \(x.\) Then the mass of the metal would be \(\frac{75x}{100}\) and the mass of oxygen in it would be \(\frac{25x}{100}.\) Thus, the equivalent mass can be found by using the oxide method:
\[\begin{align} \text{Equivalent Mass}&=\dfrac{\frac{75x}{100}\times 8}{\frac{25x}{100}}\\ &=\dfrac{75\times 8}{25}= 24.\ _\square \end{align}\]
3. Chloride Method
This is another alternative method used when the above two methods aren't compatible to calculate the equivalent mass. According to this method, when some mass of an element reacts with chlorine to form the respective chloride, we calculate the ratio at which a metal combines with \(35.5\text{ g}\) of chlorine, which gives us the equivalent mass:
\[\text{Equivalent Mass}=\dfrac{\text{Mass of the Metal}\times 35.5}{\text{Mass of Chlorine in the Chloride}}.\]
\(54\text{ g}\) of silver reacts with \(\ce{HNO3}\) to form silver nitrate; later a displacement reaction takes place when \(\ce{NaCl}\) is added to \(\ce{AgNO3}\) to form \(71.75\text{ g}\) of \(\ce{AgCl}\). Find the equivalent mass of silver.
The mass of \(\ce{AgCl}\) formed is \(71.75\text{ g},\) and thus the amount of chlorine inside it is \(71.75-54=17.75\) grams. Thus, plugging these values in the chloride equation, we get
\[\begin{align} \text{Equivalent Mass}&=\dfrac{54\times 35.5}{17.75}=54\times 2=108.\ _\square \end{align}\]
Finding Equivalent Masses of Compounds
So far we have deduced the methods to find the equivalent mass of elements. Now, let us take this a step further and find out the methods to formulate the equivalent mass of compounds.
1. Equivalent Mass of Acids
An acid as we know is a compound which contains one or more hydrogen atoms that can be replaced by a metal. And the number of hydrogen atoms which can be replaced by a metal in an acid gives us the basicity of the compound. Let's have a look at some examples:
Find the basicity of the following acids: \(\ce{H2SO4}, \ce{HCl}, \ce{CH3COOH}, \ce{HNO3}\).
We have
\[\begin{align} 1.& \ \ce{H2SO4} &&\longrightarrow \text{Has two replaceable hydrogen atoms} &&\rightarrow \text{Basicity = 2}\\ 2.& \ \ce{HCl} &&\longrightarrow \text{Has one replaceable hydrogen atom} &&\rightarrow \text{Basicity = 1}\\ 3.& \ \ce{CH3COOH} &&\longrightarrow \text{Has one replaceable hydrogen atom} &&\rightarrow \text{Basicity = 1}\\ 4.& \ \ce{HNO3} &&\longrightarrow \text{Has one replaceable hydrogen atom} &&\rightarrow \text{Basicity = 1}. \end{align}\]
Observe that in the \(3^{\text{rd}}\) example, though we see there are \(4\) hydrogen atoms, only one is replaceable, as the other three are held together with some forces. \(_\square\)
But what's the use of the basicity? How is it related to the equivalent mass of an acid? We will look into it now. Note that the equivalent mass was related to the parts of hydrogen which is replaceable; similarly, the equivalent mass of an acid can be related to the number of hydrogen atoms that can be replaced, or simply to the basicity of the acid. The relationship is as follows:
\[\text{Equivalent Mass of an Acid}=\dfrac{\text{Molecular Mass of Acid}}{\text{Basicity}}.\]
Find the equivalent mass of boric acid \(\ce{H3BO3}\).
As we have seen earlier, we need to find the number of replaceable hydrogen atoms. We know boric acid dissociates as follows:
\[\ce{H3BO3}+\ce{H2O}\rightarrow \ce{H4BO4-}+\ce{H+}.\]
Thus it has only one replaceable hydrogen atom, which means its basicity is \(1\). And we also know the molecular mass of boric acid is \(62\). Hence the equivalent mass is
\[\text{Equivalent Mass of }\ce{H3BO3}=\dfrac{62}{1}=62.\ _\square\]
2. Equivalent Mass of Bases
Let us now investigate the equivalent mass of the opposite compounds of acids, bases. As we discussed about basicity, here we will discuss something similar, (you guessed it right) acidity. While this is not to be confused with what happens in your stomach, it should also not be taken as the number of replaceable hydrogen atoms in a base.
Acidity of an atom is a bit different; it is the number of molecules of a monobasic acid (or an acid with basicity 1) required to completely neutralize one molecule of the base. And accordingly the equivalent mass of a base is given by the formula
\[\text{Equivalent Mass of an Base}=\dfrac{\text{Molecular Mass of Base}}{\text{Acidity}}.\]
Find the equivalent mass of sodium hydroxide \(\ce{NaOH}\).
Let us see how \(\ce{HCl}\) which is a monobasic acid reacts with \(\ce{NaOH}\):
\[\ce{NaOH}+\ce{HCl}\rightarrow \ce{NaCl + H2O}.\]
Thus the number of molecules of \(\ce{HCl}\) required to completely neutralize one molecule of \(\ce{NaOH}\) is \(1\), and thus the acidity of sodium hydroxide is \(1\). The mass as we know is \(40\), and thus we can compute the equivalent mass now:
\[\text{Equivalent Mass of NaOH}=\dfrac{40}{1}=40.\ _\square\]
3. Equivalent Mass of Salts
Finding the equivalent mass of salts is quite interesting; it's more like the reaction between the acids and bases. As such it is defined and can be calculated as follows: the equivalent mass of a salt is defined as the mass of the salt formed when one equivalent of an acid is completely neutralized by one equivalent of a base.
For example, the reaction between \(\ce{HCl}\) and \(\ce{NaOH}\) gives rise to \(\ce{NaCl}\) in the following manner:
\[\ce{HCl + NaOH -> NaCl + H2O}.\]
See that the equivalent mass of \(\ce{HCl}=36.5\text{ g}\) (as it is monobasic), and the equivalent mass of \(\ce{NaOH}=40\text{ g}\) \((\)as we calculated earlier, and the mass of \(\ce{NaCl}\) formed is \(58.5\text{ g}).\) Thus the equivalent mass of the salt is \(58.5\text{ g}\) as one equivalent each of an acid and a base have reacted to form the salt.
If \(x\) grams of a metal gives \(y\) grams of its oxide, the equivalent mass of the metal (in grams) is \(\text{__________}.\)
The formula to formulate the equivalent mass of salts is pretty complicated and the justification of it is beyond the scope of this wiki. But here is the general formula:
\[\text{Equivalent Mass of a Salt}=\dfrac{\text{Molecular Mass of the Salt}}{\frac{\text{Net positive charge of the cation}}{\text{Net negative charge on the anion}}}.\]
See Also
References
- Riley, C. The Search For Balance. Retrieved from http://www.ccriley.com/lifestyle/the-search-for-balance/