# 0.4 Gram equivalent concept  (Page 3/4)

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$\text{Gram equivalent (geq)}=\frac{g}{E}$

This formula is widely used to express grams of substance in terms of gram equivalent and vice-versa.

## Relation between moles and gram equivalents (geq)

Gram equivalents is given by :

$\text{geq}=\frac{g}{E}$

Substituting for equivalent weight, we have :

$\text{geq}=\frac{g}{E}=\frac{xg}{{M}_{O}}$

Moles is given by :

$n=\frac{g}{{M}_{O}}$

Combining expressions, we have :

$\text{geq}=xn$

$\text{gram equivalent}=\text{valence factor}X\text{moles}$

## Gram equivalent concept

Consider the example of formation of water :

$2{H}_{2}+{O}_{2}\to 2{H}_{2}O$

Here, 2 moles of hydrogen combines with 1 mole of oxygen to form 2 moles of water molecule. In terms of mass, 4 gm of hydrogen combines with 32 gm of oxygen to form 36 gm of water molecule. The relevant proportions involved with this equation are :

$\text{Coefficients of balanced equation = 2:1:2}$

$\text{Molecules/moles = 2:1:2}$

$\text{Mass = 4:32:36 = 1:8:9}$

On the other hand, equivalent weights of hydrogen, oxygen and water are 1, 8 and 9. Clearly, proportion of mass in which chemical entities participate is exactly the proportion of equivalent weights! Note that mole concept depends on the coefficient of balanced chemical equation. On the other hand, the equivalent weight concept is independent of coefficient of balanced chemical equation. If we know that hydrogen and oxygen combines to form water molecule, then we can say straightway that entities are in the proportion of equivalent weights - without any reference to coefficients in the balanced chemical equation.

$\text{equivalent weight of hydrogen}\equiv \text{equivalent weight of oxygen}\equiv \text{equivalent weight of water}$

Note that there is no mention of coefficients of balanced chemical equation in the equivalent weight relation. We should, however, understand that two techniques of analyzing chemical reactions are essentially equivalent. We need to consider coefficients involved in balanced chemical equation for applying mole concept. On the other hand, coefficients are not considered when using equivalent weight concept, but we need to know the corresponding valence factor of each entity. It is important to realize that above relation is not a relation connected by "equal to (=)" sign. Rather, they are connected by equivalent sign (≡). As such, we still need to apply unitary method to analyze the relation.

Gram equivalent concept is a step head in this context. The gram equivalents of participating entities are same. For the case of formation of water, the proportion of mass of hydrogen, oxygen and water is 1 gm: 8 gm : 9 gm. Now, we know that the gram equivalents of entities are obtained by dividing mass by equivalent weight. Hence, gram equivalents of three entities are 1/1 = 1, 8/8=1 and 9/9 = 1. Thus, gram equivalents of participating entities are same. If gram equivalents of hydrogen is 2, then gram equivalents of oxygen and water will also be 2. As such :

$\text{x gm equivalents of hydrogen}=\text{x gm equivalents of oxygen}=\text{x gm equivalents of water}$

Significant aspect of this relation is that it is connected with equal to (=) sign and as such relieves us from applying unitary method altogether.

Problem : 100 gm of a mixture nitrates of two metals A and B are heated to constant weight of 50 gm, containing corresponding oxides of the metals. The equivalent weights of A and B are 103 and 31 respectively. What is the percentage composition of A and B in the mixture.

Solution : Here, we make use of the fact that :

Equivalent weight of metal nitrate = $\text{Equivalent weight of metal nitrate}=\text{Equivalent weight of metal}+\text{Equivalent weight of nitrate - radical}$

Therefore,

$\text{Equivalent weight of nitrate of A}=103+\frac{\left(14+3X16\right)}{1}=103+62=165$

$\text{Equivalent weight of oxide of A}=103+\frac{16}{2}=103+8=111$

$\text{Equivalent weight of nitrate of B}=31+62=93$

$\text{Equivalent weight of oxide of B}=31+8=39$

Let the mass of A in the mixture be x gm. Then mass of B is 100-x gm. Applying concept of equivalent weight concept to chemical reaction,

165 gm of nitrate of A yields 111 gm of A’s oxide. Therefore, x gm of A’s nitrate yields :

$⇒\text{mass of A’s oxide}=\frac{111x}{165}=0.67x$

Similarly,

$⇒\text{mass of B’s oxide}=\frac{39\left(100-x\right)}{93}=0.42X100-0.42x=42-0.42x$

According to question,

$⇒0.67x+.42X100-0.42x=50$

$⇒0.25x=50-42=8$

$⇒x=32\phantom{\rule{1em}{0ex}}gm$

$⇒\text{Mass of A}=x=32\phantom{\rule{1em}{0ex}}gm$

$⇒\text{Mass of B}=100-x=68\phantom{\rule{1em}{0ex}}gm$

Thus, mixture contains 32 % of A and 68 % of B.

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