The word ‘stoichiometry’ is derived from two Greek words — stoicheion (meaning, element) and metron (meaning, measure). Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants required or the products produced in a chemical reaction, let us study what information is available from the balanced chemical equation of a given reaction. Let us consider the combustion of methane. A balanced equation for this reaction is as given below:
\(
\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})
\)
Here, methane and dioxygen are called reactants and carbon dioxide and water are called products. Note that all the reactants and the products are gases in the above reaction and this has been indicated by the letter \((g)\) in the brackets next to its formula. Similarly, in case of solids and liquids, (s) and (l) are written respectively.
The coefficients 2 for \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are called stoichiometric coefficients. Similarly the coefficient for \(\mathrm{CH}_{4}\) and \(\mathrm{CO}_{2}\) is one in each case. They represent the number of molecules (and moles as well) taking part in the reaction or formed in the reaction.
Thus, according to the above chemical reaction,
Limiting Reagent
Many a time, reactions are carried out with the amounts of reactants that are different than the amounts as required by a balanced chemical reaction. In such situations, one reactant is in more amount than the amount required by balanced chemical reaction. The reactant which is present in the least amount gets consumed after some time and after that further reaction does not take place whatever
be the amount of the other reactant. Hence, the reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent.
Reactions in Solutions
A majority of reactions in the laboratories are carried out in solutions. Therefore, it is important to understand as how the amount of substance is expressed when it is present in the solution. The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways.
Let us now study each one of them in detail.
Balancing a chemical equation
According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides
\(4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \quad\) (a) balanced equation
\(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}\) (s) \(\quad\) (b) balanced equation
\(\mathrm{P}_{4}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s}) \quad\) (c) unbalanced equation
Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms on each side of the equations. However, equation (c) is not balanced. In this equation, phosphorus atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the equation.
\(\mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s}) \quad\) balanced equation
Now, let us take the combustion of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\). This equation can be balanced in steps.
Step 1: Write down the correct formulas of reactants and products. Here, propane and oxygen are reactants, and carbon dioxide and water are products.
\(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) unbalanced equation
Step 2: Balance the number of \(C\) atoms: Since 3 carbon atoms are in the reactant, therefore, three \(\mathrm{CO}_{2}\) molecules are required on the right side.
\(
\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O} \text { (l) }
\)
Step 3: Balance the number of \(H\) atoms: on the left there are 8 hydrogen atoms in the reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will be required for eight hydrogen atoms on the right side.
\(
\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O} \text { (l) }
\)
Step 4: Balance the number of \(O\) atoms: There are 10 oxygen atoms on the right side \((3 \times 2=6\) in \(\mathrm{CO}_{2}\) and \(4 \times 1=4\) in water). Therefore, five \(\mathrm{O}_{2}\) molecules are needed to supply the required 10 \(\mathrm{CO}_{2}\) and \(4 \times 1=4\) in water). Therefore, five \(\mathrm{O}_{2}\) molecules are needed to supply the required 10 oxygen atoms.
\(
\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O} \text { (1) }
\)
Step 5: Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. All equations that have correct formulas for all reactants and products can be balanced. Always remember that subscripts in formulas of reactants and products cannot be changed to balance an equation.
Example: 1.1
Calculate the amount of water \((\mathrm{g})\) produced by the combustion of \(16 \mathrm{~g}\) of methane.
Answer:
The balanced equation for the combustion of methane is :
\(
\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})
\)
(i) \(16 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) corresponds to one mole.
(ii) From the above equation, 1 mol of \(\mathrm{CH}_{4}(\mathrm{~g})\) gives 2 mol of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).
\(2 \mathrm{~mol}\) of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)=2 \times(2+16)\)
\(
=2 \times 18=36 \mathrm{~g}
\)
\(1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{O}=18 \mathrm{~g} \mathrm{~H}_{2} \mathrm{O} \Rightarrow \frac{18 \mathrm{~g} \mathrm{~H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{O}}=1\)
Hence, \(2 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{O} \times \frac{18 \mathrm{~g} \mathrm{~H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{O}}\)
\(
=2 \times 18 \mathrm{~g} \mathrm{~H}_{2} \mathrm{O}=36 \mathrm{~g} \mathrm{~H}_{2} \mathrm{O}
\)
Example: 1.2
How many moles of methane are required to produce \(22 \mathrm{~g} \mathrm{~CO}_{2}(\mathrm{~g})\) after combustion?
Answer:
According to the chemical equation, \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(44 \mathrm{~g} \mathrm{CO}_{2}(\mathrm{~g})\) is obtained from \(16 \mathrm{~g} \mathrm{~CH}_{4}(\mathrm{~g})\). \(\left[\therefore 1 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\right.\) is obtained from \(1 \mathrm{~mol}\) of \(\left.\mathrm{CH}_{4}(\mathrm{~g})\right]\)
Number of moles of \(\mathrm{CO}_{2}(\mathrm{~g})\)
\(
=22 \mathrm{~g} \mathrm{~CO}_{2}(\mathrm{~g}) \times \frac{1 \mathrm{molCO}_{2}(\mathrm{~g})}{44 \mathrm{gCO}_{2}(\mathrm{~g})}
\)
\(=0.5 \mathrm{~mol} \mathrm{~CO}_{2}(\mathrm{~g})\)
Hence, \(0.5 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) would be obtained from \(0.5 \mathrm{~mol} \mathrm{CH}_{4}(\mathrm{~g})\) or \(0.5\) \(\mathrm{mol}\) of \(\mathrm{CH}_{4}(\mathrm{~g})\) would be required to produce \(22 \mathrm{~g} \mathrm{CO}_{2}\) (g).
Example: 1.3
\(50.0 \mathrm{~kg}\) of \(\mathrm{N}_{2}(\mathrm{~g})\) and \(10.0 \mathrm{~kg}\) of \(\mathrm{H}_{2}(\mathrm{~g})\) are mixed to produce \(\mathrm{NH}_{3}\) (g). Calculate the amount of \(\mathrm{NH}_{3}(\mathrm{~g})\) formed. Identify the limiting reagent in the production of \(\mathrm{NH}_{3}\) in this
Answer: A balanced equation for the above reaction is written as follows :
\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\)
Calculation of moles:
Number of moles of \(\mathrm{N}_{2}\)
\(=50.0 \mathrm{~kg} \mathrm{~N}_{2} \times \frac{1000 g \mathrm{~N}_{2}}{1 \mathrm{~kg} \mathrm{~N}_{2}} \times \frac{1 \mathrm{~mol} \mathrm{~N}_{2}}{28.0 \mathrm{~g} \mathrm{~N}_{2}}\)
\(=17.86 \times 10^{2} \mathrm{~mol}\)
Number of moles of \(\mathrm{H}_{2}\)
\(
\begin{aligned}
&=10.00 \mathrm{~kg} \mathrm{H}_{2} \times \frac{1000 \mathrm{~g} \mathrm{H}_{2}}{1 \mathrm{kgH}} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2}}{2.016 \mathrm{gH}_{2}} \\
&=4.96 \times 10^{3} \mathrm{~mol} \\
&\text { According to the above equation, } 1 \mathrm{~mol} \\
&\mathrm{~N}_{2}(\mathrm{~g}) \text { requires } 3 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g}), \text { for the reaction. } \\
&\mathrm{Hence}_{\text {of }} \text { for } 17.86 \times 10^{2} \mathrm{~mol} \text { of } \mathrm{N}_{2} \text {, the moles } \\
&\text { of } \mathrm{H}_{2}(\mathrm{~g}) \text { required would be } \\
&17.86 \times 10^{2} \mathrm{~mol}_{2} \times \frac{3 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})}{1 \mathrm{~mol} \mathrm{~N}_{2}(\mathrm{~g})} \\
&=5.36 \times 10^{3} \mathrm{~mol} \mathrm{H}_{2}
\end{aligned}
\)
But we have only \(4.96 \times 10^{3} \mathrm{~mol} \mathrm{H}_{2}\). Hence, dihydrogen is the limiting reagent in this case. So, \(\mathrm{NH}_{3}(\mathrm{~g})\) would be formed only from that amount of available dihydrogen i.e., \(4.96 \times 10^{3} \mathrm{~mol}\)
Since \(3 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})\) gives \(2 \mathrm{~mol} \mathrm{NH}_{3}(\mathrm{~g})\)
\(4.96 \times 10^{3} \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g}) \times \frac{2 \mathrm{~mol} \mathrm{NH}_{3}(\mathrm{~g})}{3 \mathrm{molH}_{2}(\mathrm{~g})}\)
\(=3.30 \times 10^{3} \mathrm{~mol} \mathrm{NH}_{3}(\mathrm{~g})\)
\(3.30 \times 10^{3} \mathrm{~mol} \mathrm{NH}_{3}(\mathrm{~g})\) is obtained.
If they are to be converted to grams, it is done as follows :
\(1 \mathrm{~mol} \mathrm{NH}_{3}(\mathrm{~g})=17.0 \mathrm{~g} \mathrm{NH}_{3}(\mathrm{~g})\)
\(=3.30 \times 10^{3} \times 17 \mathrm{~g} \mathrm{NH}_{3}(\mathrm{~g})\)
\(=56.1 \times 10^{3} \mathrm{~g} \mathrm{NH}_{3}\)
\(=56.1 \mathrm{~kg} \mathrm{NH}_{3}\)
Mass percent or Weight Percent (w/W%)
It is obtained by using the following relation:
Mass per cent \(=\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100\)
Example: 1.4
A solution is prepared by adding \(2 \mathrm{~g}\) of a substance A to \(18 \mathrm{~g}\) of water. Calculate the mass percent of the solute.
Answer:
Mass per cent of \(A=\frac{\text { Mass of } A}{\text { Mass of solution }} \times 100\)
\(
\begin{aligned}
&=\frac{2 \mathrm{~g}}{2 \mathrm{~g} \text { of } A+18 \text { g of water }} \times 100 \\
&=\frac{2 \mathrm{~g}}{20 \mathrm{~g}} \times 100 \\
&=10 \%
\end{aligned}
\)
Mole Fraction
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘ \(\mathrm{B}\) ‘ and their number of moles are \(n_{\mathrm{A}}\) and \(n_{\mathrm{B}}\), respectively, then the mole fractions of \(A\) and \(B\) are given as:
Mole fraction of A
\(
\begin{aligned}
&x_{\mathrm{A}}=\frac{\text { No. of moles of } A}{\text { No. of moles of solutions }} \\
&=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}
\end{aligned}
\)
Mole fraction of B
\(
\begin{aligned}
&x_{\mathrm{B}}=\frac{\text { No. of moles of } \mathrm{B}}{\text { No. of moles of solutions }} \\
&=\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}
\end{aligned}
\)
Molarity
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution. Thus,
Molarity \((M)=\frac{\text { No. of moles of solute }}{\text { Volume of solution in litres }}\)
Suppose, we have \(1 \mathrm{M}\) solution of a substance, say \(\mathrm{NaOH}\), and we want to prepare a \(0.2 \mathrm{M}\) solution from it.
\(1 \mathrm{M} \mathrm{NaOH}\) means \(1 \mathrm{~mol}\) of \(\mathrm{NaOH}\) present in 1 litre of the solution. For \(0.2 \mathrm{M}\) solution, we require \(0.2\) moles of \(\mathrm{NaOH}\) dissolved in 1 litre solution.
Hence, for making \(0.2 \mathrm{M}\) solution from \(1 \mathrm{M}\) solution, we have to take that volume of \(1 \mathrm{M} \mathrm{NaOH}\) solution, which contains \(0.2 \mathrm{~mol}\) of \(\mathrm{NaOH}\) and dilute the solution with water to 1 litre.
Now, how much volume of concentrated (1M) NaOH solution be taken, which contains \(0.2\) moles of \(\mathrm{NaOH}\) can be calculated as follows:
If \(1 \mathrm{~mol}\) is present in \(1 \mathrm{~L}\) or \(1000 \mathrm{~mL}\) solution
then, \(0.2 \mathrm{~mol}\) is present in
\(
\begin{aligned}
&\frac{1000 \mathrm{~mL}}{1 \mathrm{~mol}} \times 0.2 \mathrm{~mol} \text { solution } \\
&=200 \mathrm{~mL} \text { solution }
\end{aligned}
\)
Thus, \(200 \mathrm{~mL}\) of \(1 \mathrm{M} \mathrm{NaOH}\) are taken and enough water is added to dilute it to make it 1 litre.
In fact for such calculations, a general formula, \(M_{1} \times V_{1}=M_{2} \times V_{2}\) where \(M\) and \(V\) are molarity and volume, respectively, can be used. In this case, \(M_{1}\) is equal to \(0.2 \mathrm{M} ; V_{1}=1000 \mathrm{~mL}\) and, \(\mathrm{M}_{2}=1.0 \mathrm{M} ; \mathrm{V}_{2}\) is to be calculated.
Substituting the values in the formula:
\(
\begin{aligned}
&0.2 \mathrm{M} \times 1000 \mathrm{~mL}=1.0 \mathrm{M} \times V_{2} \\
&\therefore V_{2}=\frac{0.2 \mathrm{M} \times 1000 \mathrm{~mL}}{1.0 \mathrm{M}}=200 \mathrm{~L}
\end{aligned}
\)
Note that the number of moles of solute \((\mathrm{NaOH})\) was \(0.2\) in \(200 \mathrm{~mL}\) and it has remained the same, 1.e., \(0.2\) even after dilution ( in 1000 \(\mathrm{mL})\) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to \(\mathrm{NaOH}\). But keep in mind the concentration.
Example: 1.5
Calculate the molarity of \(\mathrm{NaOH}\) in the solution prepared by dissolving its \(4 \mathrm{~g}\) in enough water to form \(250 \mathrm{~mL}\) of the solution.
Answer:
Since molarity (M)
\(
\begin{aligned}
&=\frac{\text { No. of moles of solute }}{\text { Volume of solution in litres }} \\
&=\frac{\text { Mass of } \mathrm{NaOH} / \text { Molar mass of } \mathrm{NaOH}}{0.250 \mathrm{~L}} \\
&=\frac{4 \mathrm{~g} / 40 \mathrm{~g}}{0.250 \mathrm{~L}}=\frac{0.1 \mathrm{~mol}}{0.250 \mathrm{~L}} \\
&=0.4 \mathrm{~mol}{\mathrm{L}^{-1}} \\
&=0.4 \mathrm{M}
\end{aligned}
\)
Note that the molarity of a solution depends upon temperature because the volume of a solution is temperature-dependent.
Molality
It is defined as the number of moles of solute present in \(1 \mathrm{~kg}\) of solvent. It is denoted by \(\mathrm{m}\). Thus, Molality \((\mathrm{m})=\frac{\text { No. of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)
Example: 1.6
The density of \(3 \mathrm{M}\) solution of \(\mathrm{NaCl}\) is \(1.25 \mathrm{~g} \mathrm{~mL}^{-1}\). Calculate the molality of the solution.
Answer:
\(\mathrm{M}=3 \mathrm{~mol} \mathrm{~L}^{-1}\)
Mass of \(\mathrm{NaCl}\)
in \(1 \mathrm{~L}\) solution \(=3 \times 58.5=175.5 \mathrm{~g}\)
Mass of
\(1 \mathrm{~L}\) solution \(=1000 \times 1.25=1250 \mathrm{~g}\)
(since density \(=1.25 \mathrm{~g} \mathrm{~mL}^{-1}\) )
Mass of water in solution \(=1250-75.5\)
\(=1074.5 \mathrm{~g}\)
Molality \(=\frac{\text { No. of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)
\(
=\frac{3 \mathrm{~mol}}{1.0745 \mathrm{~kg}}=2.79 \mathrm{~m}
\)
Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that the molality of a solution does not change with temperature since mass remains unaffected with temperature.
Normality
The number of gram equivalents of the solute dissolved per litre of the solution. It is denoted by ‘ \(N\) ‘ :
\(
\text { Normality }=\frac{\text { Number of gram equivalents of solute }}{\text { Volume of solution (lit.) }}
\)
\(\because\) Gram equivalents of solute
\(
=\frac{\text { Weight of solute }(\mathrm{gm})}{\text { Equivalent weight of solute }}
\)
ppm. (Parts per million)
The parts of the component per million parts \(\left(10^6\right)\) of the solution.
\(
\mathrm{ppm}=\frac{w}{w+W} \times 10^6
\)
where, \(w=\) weight of solute, \(W=\) weight of solvent
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