Equivalent Concept & n-Factor

Mole–mole analysis is powerful, but it requires you to balance the chemical equation first. There's a parallel framework — the equivalent concept — that lets you skip balancing entirely. It is especially useful for acid–base neutralisations and redox reactions.

The core idea: in any chemical reaction, equal equivalents of reactants combine to give equal equivalents of products.

For a reaction (balanced or not):

$aA + bB \longrightarrow cC + dD$

the relation is simply:

$\frac{w_A}{E_A} = \frac{w_B}{E_B} = \frac{w_C}{E_C} = \frac{w_D}{E_D}$

where $w$ is the given (or formed) mass and $E$ is the equivalent weight of each substance.

Unlike mole–mole, this doesn't depend on stoichiometric coefficients at all. The trade-off: you must know each substance's equivalent weight — which is itself a function of how the substance behaves in the reaction.

Equivalent weight is the mass of an element or compound that reacts with, displaces, or supplies a fixed reference quantity:

• 1.008 g of $\ce{H2}$, or
• 8 g of $\ce{O2}$, or
• 35.5 g of $\ce{Cl2}$

Unlike molecular weight (which is fixed for a molecule), equivalent weight changes depending on the reaction the substance is in.

Example 1. $\ce{2Mg + O2 -> 2MgO}$. From the equation, 48 g of Mg reacts with 32 g of O₂. Scaling down: 12 g of Mg reacts with 8 g of O₂. By the definition above:

$\boxed{\text{Eq. wt of Mg} = 12 \text{ g}}$

Example 2. $\ce{Zn + H2SO4 -> ZnSO4 + H2}$. 65.5 g of Zn displaces 2 g of H₂. Per the definition, the mass of Zn that displaces 1.008 g of H₂ is:

$\text{Eq. wt of Zn} = \frac{65.5}{2} \approx 32.75 \text{ g}$

The general formula that ties everything together:

$\text{Equivalent weight} = \frac{\text{Atomic / Molecular weight}}{n\text{-factor}}$

So the problem reduces to: what's the $n$-factor for this substance, in this reaction?

The $n$-factor depends on what category of compound you're dealing with and how it reacts:

A classic trap — the $n$-factor of an acid depends on how it actually reacts, not just its formula. Consider $\ce{H2SO4}$:

• $\ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O}$ → both $\ce{H+}$ replaced → $n = 2$
• $\ce{NaOH + H2SO4 -> NaHSO4 + H2O}$ → only one $\ce{H+}$ replaced → $n = 1$

Same acid, two different $n$-factors. Always check the equation, not just the formula.

Common substances — memorise the pattern:

The phosphorus acids are a classic JEE trap because not every hydrogen in their formula is acidic. Only hydrogens bonded to oxygen are acidic. Hydrogens bonded directly to phosphorus are not acidic and don't dissociate.

The rule of thumb: count only the O–H groups, never the P–H or C–H bonds. The formula $\ce{H3PO3}$ looks like it should have basicity 3, but the actual structure has one H bonded directly to the phosphorus — so only the other two can dissociate.

Number of equivalents = how many "reaction units" you have in a sample. The formula has three equivalent forms — pick whichever matches what you're given:

$\text{No. of equivalents} = \frac{w}{E} = \frac{w \times n}{M} = (\text{moles}) \times (n\text{-factor})$

where $w$ is the given mass, $E$ is the equivalent weight, $M$ is the molecular weight, and $n$ is the $n$-factor.

Quick example — for 18 g of Al ($M = 27$, $n$-factor = 3):

$\text{No. of equivalents} = \frac{18}{27} \times 3 = \frac{2}{3} \times 3 = 2 \text{ equivalents}$