# Mole concept

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Stoichiometry is a branch of chemistry which deals with quantitative relationships of the reactants and products in a chemical reaction. The quantity of substance involved in measurement has different states (solid, liquid and gas) and forms (mixture and compound).

The stoichiometry calculations are basically arithmetic calculations, but confounded by scores of measuring terminologies. These terms are required to represent atomic or molecular level of association involved in chemical reactions. Besides chemical substance reacts or combines in definite proportion which we account by valence factor. Corresponding to valence factor, there are measurements which measures the proportion in which one element or group of elements reacts or combines with another element or group of elements.

$xA+yB\to {A}_{x}{B}_{y}$

A good part of stoichiometry is about mastering measuring terms. In this module, we seek to have a clear idea of mole and associated concept which is central to stoichiometry calculation.

## Moles

Moles measure quantity of substance. There is a subtle ambiguity about treating “moles” – whether as “mass” or “number”. We take the position that it measures “amount of substance”, which can be either be expressed in terms of mass or in terms of numbers. The two approaches are equivalent and need not be a source of ambiguity any further. We only need to interpret the meaning as appropriate in a particular context.

One mole is defined as ${N}_{o}$ entities of the substance, where ${N}_{o}$ is equal to Avogadro’s number. ${N}_{o}$ is generally taken as $6.022X{10}^{23}$ . If need be, then we may use still closer approximation of this number.

## Relation between mole and molecular weight

The mole and molecular weight are designed to be related. Argument goes like this :

1: Molecular weight is arithmetic sum of atomic weights of constituent atoms.

2: Atomic weight is expressed in “atomic mass unit (a.m.u)”.

3: The atomic mass unit is the mass of ${\frac{1}{12}}^{\mathrm{th}}$ of C-12 atom. This is average weight of constituent 6 protons and 6 neutrons forming nucleus of C-12 atom. The a.m.u, therefore, is close to the mass of either a single proton or single neutron as a constituent of C-12. The a.m.u is numerically equal to :

$1a.m.u=1.66X{10}^{-27}\phantom{\rule{1em}{0ex}}kg=1.66X{10}^{-24}\phantom{\rule{1em}{0ex}}gm$

4: On the scale of a.m.u, the atomic weight of a carbon atom is exactly “12” and that of hydrogen is “1.008” or about “1”.

5: The number of C-12 atoms present in the quantity expressed in grams but numerically equal to atomic weight (as measured in a.m.u.) is given as :

${N}_{o}=\frac{\text{grams numerically equal to atomic weight}}{\text{Mass of one C-12 atom in gram}}=\frac{12}{12X1.66X{10}^{-24}}$

The important point about this ratio is that “12” as atomic weight in the numerator and “12” as the number of a.m.u units in a single atom of C-12 cancels out, leaving us with a constant ratio :

$⇒{N}_{o}=\frac{1}{1.66X{10}^{-24}}=6.022X{10}^{23}$

Clearly, this ratio is same for all elements. It means that the numbers of atoms in the grams numerically equal to atomic weight is ${N}_{o}$ for all elements. Putting it equivalently, the mass of a collection ${N}_{o}$ atoms of an element is equal to grams numerically equal to atomic weight.

how do they get the third part x = (32)5/4
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