Category : JEE Main & Advanced
Chemical stoichiometry
Stoichiometry (pronounced “stoy-key om-e-tree”) is the calculation of the quantities of reactants and products involved in a chemical reaction. That means quantitative calculations of chemical composition and reaction are referred to as stoichiometry.
Basically, this topic involves two types of calculations.
(a) Simple calculations (gravimetric analysis) and
(b) More complex calculations involving concentration and volume of solutions (volumetric analysis).
There is no borderline, which can distinguish the set of laws applicable to gravimetric and volumetric analysis. All the laws used in one are equally applicable to the other i.e., mole as well as equivalent concept. But in actual practise, the problems on gravimetric involves simpler reactions, thus mole concept is convenient to apply while volumetric reactions being complex and unknown (unknown simple means that it is not known to you, as it’s not possible for you to remember all possible reactions), equivalent concept is easier to apply as it does not require the knowledge of balanced equation.
(1) Gravimetric analysis : In gravimetric analysis we relate the weights of two substances or a weight of a substance with a volume of a gas or volumes of two or more gases.
Problems Involving Mass-Mass Relationship
Proceed for solving such problems according to the following instructions,
(i) Write down the balanced equation to represent the chemical change.
(ii) Write the number of moles below the formula of the reactants and products. Also write the relative weights of the reactants and products (calculated from the respective molecular formula), below the respective formula.
(iii) Apply the unitary method to calculate the unknown factor (s).
Problems Involving Mass-Volume Relationship
For solving problems involving mass-volume relationship, proceed according to the following instructions,
(i) Write down the relevant balanced chemical equations (s).
(ii) Write the weights of various solid reactants and products.
(iii) Gases are usually expressed in terms of volumes. In case the volume of the gas is measured at room temperature and pressure (or under conditions other than N.T.P.), convert it into N.T.P. by applying gas equation.
(iv) Volume of a gas at any temperature and pressure can be converted into its weight and vice-versa with the help of the relation, by \[PV=\frac{g}{M}\times RT\] where \[g\] is weight of gas, \[M\] is mole. wt. of gas, \[R\] is gas constant.
Calculate the unknown factor by unitary method.
Problems Based on Volume-Volume Relationship
Such problems can be solved according to chemical equation as,
(i) Write down the relevant balanced chemical equation.
(ii) Write down the volume of reactants and products below the formula to each reactant and product with the help of the fact that \[1gm\] molecule of every gaseous substance occupies 22.4 litres at N.T.P.
(iii) In case volume of the gas is measured under particular (or room) temperature, convert it to volume at NTP by using ideal gas equation.
Take the help of Avogadro’s hypothesis “Equal volume of different gases under the similar conditions of temperature and pressure contain the same number of molecules”.
(2) Volumetric analysis : It is a method which involves quantitative determination of the amount of any substance present in a solution through volume measurements. For the analysis a standard solution is required. (A solution which contains a known weight of the solute present in known volume of the solution is known as standard solution.)
To determine the strength of unknown solution with the help of known (standard) solution is known as titration. Different types of titrations are possible which are summerised as follows,
(i) Redox titrations : To determine the strength of oxidising agents or reducing agents by titration with the help of standard solution of reducing agents or oxidising agents.
Examples:
\[\begin{align}
& \underline{\begin{align}
& \underline{\begin{align}
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{K}_{2}}C{{r}_{2}}{{O}_{7}}+4{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+4{{H}_{2}}O+3[O] \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[2FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+O\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{H}_{2}}O]\times 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{align}} \\
& \,\,6FeS{{O}_{4}}+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+7{{H}_{2}}S{{O}_{4}}\to 3Fe{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}7{{H}_{2}}O \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+3{{H}_{2}}O+5[O] \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[2FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+O\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{H}_{2}}O]\times 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{align}} \\
& 10FeS{{O}_{4}}+2KMn{{O}_{4}}+8{{H}_{2}}S{{O}_{4}}\to 5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O \\
\end{align}\]
Similarly with \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
\[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+5{{H}_{2}}{{C}_{2}}{{O}_{4}}\to \]
\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+10C{{O}_{2}}\]etc.
(ii) Acid-base titrations : To determine the strength of acid or base with the help of standard solution of base or acid.
Example: \[NaOH+HCl\to NaCl+{{H}_{2}}O\]
and \[NaOH+C{{H}_{3}}COOH\to C{{H}_{3}}COONa+{{H}_{2}}O\] etc.
(iii) Iodiometric titrations : This is a simple titration involving free iodine. This involves the titration of iodine solution with known sodium thiosulphate solution whose normality is \[N\]. Let the volume of sodium thiosulphate is \[V\,ml\].
\[{{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}\]
\[n=2\], \[n=1\]
Equivalents of \[{{I}_{2}}=\]Equivalent of \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]
\[\therefore \] Equivalents of \[{{I}_{2}}=N\times V\times {{10}^{-3}}\]
Moles of \[{{I}_{2}}=\frac{N\times V\times {{10}^{-3}}}{2}\]
Mass of free \[{{I}_{2}}\] in the solution \[=\left[ \frac{N\times V\times {{10}^{-3}}}{2}\times 254 \right]\,g\].
(iv) Iodometric titrations : This is an indirect method of estimation of iodine. An oxidising agent is made to react with excess of solid \[KI\]. The oxidising agent oxidises \[{{I}^{-}}\] to \[{{I}_{2}}\]. This iodine is then made to react with \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] solution.
Oxidising Agent
\[(A)+KI\to {{I}_{2}}\xrightarrow{2N{{a}_{2}}{{S}_{2}}{{O}_{3}}}2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}\]
Let the normality of \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] solution is \[N\] and the volume of thiosulphate consumed to \[V\,ml\].
Equivalent of \[A=\]Equivalent of \[{{I}_{2}}=\]Equivalents of \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\]
Equivalents of \[{{I}_{2}}\] liberated from \[KI=N\times V\times {{10}^{-3}}\]
Moles of \[{{I}_{2}}\] liberated from \[KI=\frac{N\times V\times {{10}^{-3}}}{2}\]
Mass of \[{{I}_{2}}\] liberated from \[KI=\left[ \frac{N\times V\times {{10}^{-3}}}{2}\times 254 \right]g\].
(v) Precipitation titrations : To determine the anions like \[C{{N}^{-}},\ AsO_{3}^{3-},\ PO_{4}^{3-},\ {{X}^{-}}\] etc, by precipitating with \[AgN{{O}_{3}}\] provides examples of precipitation titrations.
\[NaCl+AgN{{O}_{3}}\to AgCl\downarrow +NaN{{O}_{3}}\] \[KSCN+AgN{{O}_{3}}\to AgSCN\downarrow +KN{{O}_{3}}\]
End point and equivalence point : The point at which titration is stopped is known as end point, while the point at which the acid and base (or oxidising and reducing agents) have been added in equivalent quantities is known as equivalence point. Since the purpose of the indicator is to stop the titration close to the point at which the reacting substances were added in equivalent quantities, it is important that the equivalent point and the end point be as close as possible.
Normal solution : A solution containing one gram equivalent weight of the solute dissolved per litre is called a normal solution; e.g. when 40 g of NaOH are present in one litre of NaOH solution, the solution is known as normal (N) solution of NaOH. Similarly, a solution containing a fraction of gram equivalent weight of the solute dissolved per litre is known as subnormal solution. For example, a solution of NaOH containing 20 g (1/2 of g eq. wt.) of NaOH dissolved per litre is a sub-normal solution. It is written as N/2 or 0.5 N solution.
Formula used in solving numerical problems on volumetric analysis
(1) Strength of solution = Amount of substance in g \[litr{{e}^{-1}}\]
(2) Strength of solution = Amount of substance in g moles \[litr{{e}^{-1}}\]
(3) Strength of solution = Normality ´ Eq. wt. of the solute
= molarity \[\times \] Mol. wt. of solute
(4) \[\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume in litre}}\]
(5) \[\text{Number of moles}=\frac{\text{Wt}\text{. in }gm}{\text{Mol}\text{. wt}\text{.}}=M\times {{V}_{(in\,l)}}\]
\[=\frac{\text{Volume in litres}}{22.4}\] at NTP (only for gases)
(6) Number of millimoles \[=\frac{\text{Wt}\text{. in }gm\text{ }\times \text{1000}}{\text{mol}\text{. wt}\text{.}}\]
\[=\text{Molarity}\times \text{Volume in }ml.\]
(7) Number of equivalents
\[=\frac{\text{Wt}\text{. in }gm}{\text{Eq}\text{. wt}\text{.}}=x\times \text{No}\text{. of moles}\times \text{Normality}\times \text{Volume in litre}\]
(8) Number of milliequivalents (meq.)
\[=\frac{\text{Wt}\text{. in }gm\times \text{1000}}{\text{Eq}\text{. wt}\text{.}}=\text{normality}\times \text{Volume in }ml.\]
(9) Normality\[=x\times \text{No}\text{. of millimoles}\]
\[=x\times \text{Molarity}=\frac{\text{Strength in }gm\,litr{{e}^{-1}}}{\text{Eq}\text{. wt}\text{.}}\]
where \[x=\frac{\text{Mol}\text{. wt}\text{.}}{\text{Eq}\text{. wt}\text{.}}\], x = valency or change in oxi. Number.
(10) Normality formula, \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]
(11) % by weight \[=\frac{\text{Wt}\text{. of solvent}}{\text{Wt}\text{. of solution}}\times 100\]
(12) % by volume \[=\frac{\text{Wt}\text{. of solvent}}{\text{Vol}\text{. of solution}}\times 100\]
(13) % by strength\[=\frac{\text{Vol}\text{. of solvent}}{\text{Vol}\text{. of solution}}\times 100\]
(14) Specific gravity
\[=\frac{\text{Wt}\text{. of solution}}{\text{Vol}\text{. of solution}}=\text{Wt}\text{. of 1 }ml.\text{ of solution}\]
(15) Formality \[=\frac{\text{Wt}\text{. of ionic solute}}{\text{Formula Wt}\text{. of solute}\times {{V}_{in\,l}}}\]
(16) Mol. Wt. = V.D ´ 2 (For gases only)
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