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For a general reaction of the type :
$$xA+yB\to {A}_{x}{B}_{y}$$
We say that “x moles of A reacts with y moles of B”.
This is a very powerful deduction popularly known as “mole concept”. It connects the measurement of mass neatly with the coefficients which are whole numbers. Further as moles are connected to mass in grams, measurement of quantities can be easily calculated in practical mass units like grams.
Problem : Find the mass of KCl and oxygen liberated when 24.5 grams of $KCl{O}_{3}$ is heated. Assume 100 % decomposition on heating. Given, ${M}_{K}=39$ .
Solution : On heating $KCl{O}_{3}$ decomposes as :
$$2KCl{O}_{3}\to 2KCl+3{O}_{2}$$
In order to apply mole concept, we need to find the moles of $KCl{O}_{3}$ in the sample.
$$\text{moles of}KCl{O}_{3}=\frac{24.5}{\text{molecular wt of}KCl{O}_{3}}$$
Here, molecular weight of $KCl{O}_{3}$ is :
$$\text{molecular wt of}KCl{O}_{3}=39+35.5+3X16=122.5$$
Putting this value in the equation of moles :
$$\Rightarrow \text{moles of}KCl{O}_{3}=\frac{24.5}{\text{molecular wt of}KCl{O}_{3}}=\frac{24.5}{122.5}=0.2$$
Applying mole concept to the equation, we have :
$$2\phantom{\rule{1em}{0ex}}\text{mole of}KCl{O}_{3}\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}2\phantom{\rule{1em}{0ex}}\text{moles of}KCl\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}3\phantom{\rule{1em}{0ex}}\text{moles of oxygen}$$
$$1\phantom{\rule{1em}{0ex}}\text{mole of}KCl{O}_{3}\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}1\phantom{\rule{1em}{0ex}}\text{moles of}KCl\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}\frac{3}{2}\phantom{\rule{1em}{0ex}}\text{moles of oxygen}$$
$$0.2\phantom{\rule{1em}{0ex}}\text{mole of}KCl{O}_{3}\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}0.2\phantom{\rule{1em}{0ex}}\text{moles of}KCl\phantom{\rule{1em}{0ex}}\stackrel{\frown}{=}\phantom{\rule{1em}{0ex}}\frac{0.6}{2}\phantom{\rule{1em}{0ex}}\text{moles of oxygen}$$
At this point, we need to convert the mass measurement from mole to grams,
$${m}_{\text{KCl}}={n}_{\text{KCl}}X{M}_{\text{KCl}}=0.2X\left(39+35.5\right)=0.2X74.5\phantom{\rule{1em}{0ex}}\text{gm}=14.9\phantom{\rule{1em}{0ex}}\text{gm}$$
$${m}_{{O}_{2}}={n}_{{O}_{2}}X{M}_{{O}_{2}}=\frac{0.6}{2}X32=0.6X16\phantom{\rule{1em}{0ex}}\text{gm}=9.6\phantom{\rule{1em}{0ex}}\text{gm}$$
The concept of mole is applicable to identical entities (atoms, molecules, ions). Thus, its direct application is restricted to pure substances – irrespective of its state (solid, liquid and gas). For all practical purposes, we treat mole as an alternative expression of mass. They are connected to each other via molecular weight for a given pure substance. However, measurement of gas is generally reported in terms of volume in stoichiometric calculation. We would, therefore, need to convert the volume of the gas to mass/mole.
If the gas in question is ideal gas, then we can easily connect volume to mole directly. Ideal gas presents a special situation. The collisions are perfectly elastic and there is no intermolecular force between molecules. The volume occupied by gas molecules is negligible with respect to the volume of ideal gas. It means that the size of molecule has no consequence as far as the volume of ideal has is concerned. These negligibly small molecules, however, exchange momentum and kinetic energy through perfectly elastic collisions. Clearly, volume depends solely on the numbers of molecules present – not on the mass or size of molecules. This understanding leads to Avogadro’s hypothesis :
All ideal gases occupy same volume at a given temperature and pressure. One mole of ideal gas occupies 22.4 litres i.e. 22,400 ml at standard temperature (273 K) and pressure (1 atmosphere) condition.
The gas volume for 1 mole is known as molar volume. The Avogadro’s hypothesis provides an important relation between volume and moles of gas present. This relationship does not hold for real gas. But we are served well in our calculation by approximating real gas as ideal gas. The real gas like hydrogen, oxygen etc. may still be close to the ideal gas molar volume at normal temperature and pressure conditions.
Problem : Air contains 22.4 % oxygen. How many moles of oxygen atoms are there in 1 litre of air at standard condition ?
Solution : The volume of oxygen is 0.224 litre. We know that 22.4 litres contains 1 mole of oxygen molecules. Hence, 0.21 litres contains “n” moles given by :
$$n=\frac{0.224}{22.4}=0.01\phantom{\rule{1em}{0ex}}\text{moles of oxygen molecules}$$
We know that one molecule of oxygen contains 2 atoms. Therefore, numbers of moles of oxygen atoms is 0.02 moles.
Problem : Calculate the numbers of molecules in 300 cc of oxygen at 27° C and 740 mm pressure.
Solution : We first need to convert the given volume at given temperature and pressure to that at standard temperature and pressure condition. Applying gas law,
$$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}$$
$$\Rightarrow {V}_{2}=\frac{{P}_{1}{V}_{1}{T}_{2}}{{P}_{2}{T}_{1}}=\frac{740X300X273}{760X300}=265.82\phantom{\rule{1em}{0ex}}cc$$
Number of molecules of oxygen is :
$$\Rightarrow N=\frac{265.82X6.022X{10}^{23}}{22400}=7.146X{10}^{21}$$
Stoichiometric analysis is about analyzing mass of different chemical species or concentrations of solutions involved in chemical reactions. We employ varieties of different concepts and hypotheses. These principles are primarily based on molecular association of various chemical species in chemical reactions. Most important among these are mole, equivalent weight and gram equivalent weight concepts. Each of these concepts relates amount of different species in chemical reaction in alternative ways. It means that these concepts have contextual superiority over others (suitability in a particular context), but are essentially equivalent.
Stoichiometric analysis is broadly classified as gravimetric or volumetric analysis depending on whether calculations are mass based (gravimetric) or concentration of solution based (volumetric). There is no strict division between two approaches as we encounter reactions which require considerations of both aspects i.e. mass of species and concentration of solution.
In general, gravimetric analysis covers displacement reactions, action of acid on metals, calculations based on balanced equations, decomposition reaction etc. On the other hand, volumetric analysis covers wide ambit of titration including neutralization and redox reactions. Many of the chemical reactions, however, need combination of two analytical approaches. For example, a sequence of chemical process may involve decomposition (gravimetric analysis) and titration (volumetric analysis).
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