1
$\begingroup$

I realize that n-factor is an outdated concept however I am interested in its formal definition when it was introduced.

We have many definitions now on how to calculate n-factor depending on the type of reaction and compounds involved.

  1. For acids, n-factor is defined as the number of H+ ions replaced by 1 mole of acid in a reaction.
  2. For bases, n-factor is defined as the number of OH– ions replaced by 1 mole of base in a reaction.
  3. The n-factor for such salts is defined as the total moles of catioinic/anionic charge replaced in 1 mole of the salt
  4. The n-factor of such salts is defined as the number of moles of electrons exchanged (lost or gained) by one mole of the salt.

However, none of these definitions explain what n-factor actually is. Is there a formal definition for n-factor?

$\endgroup$
4
  • 1
    $\begingroup$ n-factor is nothing but saying that if n moles of a substance react in the simplified chemical equation than n moles of that substance is called an equivalent and n-facror=1/n. It's an number used to convert between moles and equivalents. $\endgroup$ – Nisarg Bhavsar May 28 at 5:26
  • 4
    $\begingroup$ I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books. The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions. $\endgroup$ – M. Farooq May 29 at 4:19
  • 1
    $\begingroup$ You still find the term normalité for normality when describing a titration (known e.g., as analyse volumétrique). Example Uni Rennes, or other French speaking countries like CEAEQ's method to determine acidity in waste waters MA315AlcAc10 by 2016, etc. But the advantages of molarity are recognized (e.g., virtual UniSciel). $\endgroup$ – Buttonwood May 29 at 11:03
  • 1
    $\begingroup$ The "n-factor" is like "ICE tables", a strange concept that is used in schools in some regions in the world but that no chemist uses in real life and most chemists have never heard of. $\endgroup$ – Loong May 29 at 16:47
3
$\begingroup$

Instead of asking what n-factor really was, a better question would have been what an equivalent was (But that would have been a duplicate)

According to the above question, the formal definition of an equivalent is as follows:

An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction.

In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

  1. react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
  2. react with or supply one mole of electrons in a redox reaction.

Therefore given any chemical reaction, one equivalent of each reactant reacts to form one equivalent of each product, irrespective of the stoichiometric coefficients

$$\ce{aA + bB -> cC + dD}$$

In all such scenarios, the n-factor of the compound is defined as,

$$\textit{n}\text{-factor} = \frac{M_\mathrm{molar}}{M_\mathrm{equiv}}$$

However, why are there so many different definitions for calculating the same thing. This is because an equivalent doesn't have one fixed definition, rather depending upon the situation, the definition of an equivalent changes and hence the n-factor definition changes as well.

Therefore there is no formal definition per se for the n-factor, simply because it was never formally introduced. As M.Farooq pointed out in the comments,

I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books. The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions.

$\endgroup$
-4
$\begingroup$

Yes,the definitions are correct. n-factor is required when finding the equivalent weight of the substance.

In case of acids equivalent wt.=molecular weight of acid/basicity(no. of replaceable hydrogen atoms) here,the basicity is the n-factor

important observation:replaceable atoms of hydrogen are those which are attached to oxygen atoms. And in the same way, in case of bases equivalent wt.=molecular weight of base/acidity(no. of replaceable hydroxyl ions

in case of bases equivalent wt.=molecular weight of salt/no. of cations *valency of the cation

And so on, The denominator in each case is the n factor .The numerator (molecular wt.)being the same factor in the equivalent weights of different types of substances,the denominator plays an important role of changing the equivalent wt. according to the substance.

Equivalent weight of a substance is that amount of the substance that reacts with or can displace 1.008g or 1.008 parts of hydrogen or 35.5 g or 35.5 parts of chlorine or 8 g or 8 parts of oxygen.

important observation:the equivalent wt. of hydrogen is 1.008g or parts ,of oxygen =8g or parts,of chlorine =35.5g or parts

And equivalent wt. in turn is required to find the normality of a substance.

normality=Amount of solute in 1000ml or 1L of solvent/equivalent wt. of solute

$\endgroup$
2
  • 2
    $\begingroup$ Since the English used is already incredibly hard to comprehend, it is almost impossible to understand what is being tried to convey here. There is no way to tell the word-formulas apart from sentences. $\endgroup$ – Martin - マーチン May 29 at 10:05
  • $\begingroup$ The answer should mention that "equivalent weight" is an obsolete concept that (since more than 150 years) is no longer used in chemistry. $\endgroup$ – Loong May 29 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.