Chemical stoichiometry
Category : NEET
Chemical Stoichiometry
Calculation based on chemical equations is known as chemical stoichiometry. Stoichiometry can be broadly classified into two groups: (1) Gravimetric analysis (Stoichiometry-I), (2) Volumetric analysis (Stoichiometry-II)
(1) Gravimetric analysis (Stoichiometry-I) : With the help of chemical equation, we can calculate the weights of various substances reacting and weight of substances formed. For example,
\[MgC{{O}_{3}}\xrightarrow{{}}MgO+C{{O}_{2}}\uparrow \]
This equation implies :
(i) 1 mol of \[MgC{{O}_{3}}\] gives 1 mol of \[MgO\] and 1 mol of \[C{{O}_{2}}\].
(ii) 84 g of \[MgC{{O}_{3}}\] (Mol. wt. of \[MgC{{O}_{3}}\]) gives 40 g of \[MgO\] and 44 g of \[C{{O}_{2}}\].
Hence, chemical equation provide us information regarding :
(i) Molar ratio of reactants and products.
(ii) Mass ratio between reactants and products.
(iii) Volume ratio between gaseous reactant and products.
The calculation based upon chemical equation (Stoichiometry–I) are based upon three types namely :
(a) Mass-mass relationship (b) Mass-volume relationship (c) Volume-volume relationship
(2) Volumetric analysis (Stoichiometry-II) : 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 : To determine the reducing agents with the help of standard iodine solution is known as iodiometry.
For example: \[\underset{\text{Reducing agent}}{\mathop{A{{s}_{2}}{{O}_{3}}}}\,+2{{I}_{2}}+2{{H}_{2}}O\to A{{s}_{2}}{{O}_{3}}+4HI\]
\[N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\]
(iv) Iodometric titrations : To determine the oxidising agent indirectly by titration of liberated \[{{I}_{2}}\] with the help of standard hypo solution is known as iodometric titrations.
Examples: Oxidising agents such as \[KMn{{O}_{4}},\,{{K}_{2}}C{{r}_{2}}{{O}_{7}},\,CuS{{O}_{4}}\], ferric salts, etc. are reduced quantitatively when treated with large excess of KI in acidic or neutral medium and liberate equivalent amount of \[{{I}_{2}}\].
\[2CuS{{O}_{4}}+4KI\to C{{u}_{2}}{{I}_{2}}+2{{K}_{2}}S{{O}_{4}}+{{I}_{2}}\]
\[K{{r}_{2}}C{{r}_{2}}{{O}_{7}}+7{{H}_{2}}S{{O}_{4}}+6KI\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+4{{K}_{2}}S{{O}_{4}}+7{{H}_{2}}O+3{{I}_{2}}\]
This \[{{I}_{2}}\] is estimated with hypo
\[{{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\]
(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.
\[NaCl+AgN{{O}_{3}}\to AgCl\downarrow +NaN{{O}_{3}}\]; \[KSCN+AgN{{O}_{3}}\to AsSCN\downarrow +KN{{O}_{3}}\]
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
(2) Strength of solution = Amount of substance in g moles \[litr{{e}^{-1}}\]
(3) Strength of solution = Normality ´ Eq. wt. of the 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)ent
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