Stoichiometry and Stoichiometric Calculations: When we add too much salt or sugar to lemonade, the solution either becomes too salty or too sweet and disrupts the sweet and sour taste of the lemonade. Hence it is essential to know the actual amount of salt or sugar to be dissolved. Similarly, it is essential to know the masses of the reactants and the products in chemical reactions. The branch of chemistry that deals with the calculation of masses of the reactants and the products involved in a chemical reaction is known as stoichiometry. The word ‘stoichiometry’ is derived from two Greek words: stoicheion, element, and metron, measure. Let’s explore this concept.

What is Stoichiometry?

Stoichiometry deals with the calculation of masses and volumes of reactants and products of a chemical reaction. It uses balanced chemical equations to calculate the amounts of reactants and products. 

A balanced chemical equation is an equation in which both sides of the equation has an equal number of atoms. 

For example, the balanced chemical equation for the combustion of methane is-

\({\rm{C}}{{\rm{H}}_{\rm{4}}}\left( {\rm{g}} \right){\rm{ + 2}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right)\)

Here, the compounds to the left of the arrow, methane and dioxygen, are the reactants, and the compounds to the right, carbon dioxide and water, are the products. The letter \(\left( {\rm{g}} \right)\) indicates that all the reactants and the products are in the gaseous phase. For solids and liquids, \(\left( {\rm{s}} \right)\) and \(\left( {\rm{L}} \right)\) are written, respectively. The number \(2\) for \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are called stoichiometric coefficients, which is one for \(\mathrm{CH}_{4}\) and \(\mathrm{CO}_{2}\) Stoichiometric coefficients are the numbers that indicate the number of molecules (and moles as well) taking part in the reaction or are formed in the reaction. Thus, in the above chemical reaction

\(1\) mole of \({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g)}}\) reacts with \(2\) moles of \(\mathrm{O}_{2}(\mathrm{~g})\) to give \(1\) mole of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(2\) moles of \({{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\) 

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Or, \(1\) molecule of \(\mathrm{CH}_{4}(\mathrm{~g})\) reacts with \(2\) molecules of \(\mathrm{O}_{2}(\mathrm{~g})\) to give \(1\) molecule of \(\mathrm{CO}_{2}(\mathrm{~g})\), and \(2\) molecules of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) 

Or, \(22.4\) L of \(\mathrm{CH}_{4}(\mathrm{~g})\) reacts with \((22.4 \times 2) 44.8\) L of \(\mathrm{O}_{2}(\mathrm{~g})\) to give \(22.4\) L of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \((22.4 \times 2) 44.8 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Or, \(16 \mathrm{~g}\) of \(\mathrm{CH}_{4}(\mathrm{~g})\) reacts with \((2 \times 32) 64 \mathrm{~g}\) of \(\mathrm{O}_{2}(\mathrm{~g})\) to give \(44 \mathrm{~g}\) of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \((2 \times 18) 36\) g of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) All of the above data can be interconverted as:

mass \(\leftrightharpoons\) moles \(\leftrightharpoons\) no.of molecules

\(\frac{{{\rm{ Mass }}}}{{{\rm{ Volume }}}}{\rm{ = Density}}\)

Applying Conversion Factors to Stoichiometry

Almost all stoichiometric problems can be solved in just four simple steps. These are-

1. Balancing the chemical equation.
2. Converting units of a given substance to moles.
3. Calculating the moles of a substance using mole ratio.
4. Converting moles to desired units.

Let us consider-

\({\rm{Fe}} + {{\rm{O}}_2} \to {\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}\)

Step 1. Balancing the Chemical Equation

The Law of conservation of mass governs the balancing of a chemical equation according to which the number of atoms on the reactant and product side should always be equal. 

The balancing of the equation is done by placing numerical coefficients before the various molecules and atoms to ensure that the number of atoms on the left side of the arrow corresponds exactly to the number of elements on the right. In the unbalanced equation, there are-

Left-hand side	Right-hand side
\(\mathrm{Fe}-1\)	\({\rm{Fe – 2}}\)
\({\rm{O – 2}}\)	\({\rm{O – 3}}\)

Balancing the above equation by trial and error method, we get-

\(4{\rm{Fe}} + 3{{\rm{O}}_2} \to 2{\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3}\)

In the balanced equation, there are-

Left-hand side	Right-hand side
\({\rm{Fe – 4}}\)	\({\rm{Fe – 2 \times 2 = 4}}\)
\(0-3 \times 2=6\)	\(0-2 \times 3=6\)

Step 2. Converting Units of a Given Substance to Moles

Converting Grams to Moles

The gram formula mass of a compound (or element) is calculated by adding the atomic weights of every atom in the compound. Atomic weights on the periodic table are expressed in terms of amu (atomic mass units), but by convenience, amu corresponds to the gram formula mass. For example, a mole of a \(12\) amu carbon atom will weigh \(12\) grams.

\({\rm{Number \,of \,Moles = }}\frac{{{\rm{ Given\, mass }}}}{{{\rm{ Molar\, Mass }}}}\)

Conversion between Volume of a Gas and Moles at STP

At STP, a mole of gas will always occupy a volume of \(22.4 \mathrm{~L}\)

\({\rm{Number \,of\, moles = }}\frac{{{\rm{ Given \,volume\, at \,STP (in\, litres) }}}}{{{\rm{22}}{\rm{.4L}}}}\)

Conversion between Volume of a Gas and Moles

The Ideal Gas equation is \(\mathrm{PV}=\mathrm{nRT}\)

Where \({\rm{n}}\) represents the number of moles.

\({\rm{P}}\) representing the pressure in atm, 

\({\rm{V}}\) representing volume in litres, 

\({\rm{T}}\) representing temperature in Kelvins, and 

\({\rm{R}}\) the gas constant.

Arranging the above equation for n, we get:

\({\rm{n = }}\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\)

Conversion between Individual Particles and Moles

Avogadro’s number provides the conversion factor from the number of particles to moles. In every mole of a substance, there are \(6.022 \times 10^{23}\) formula units of particles, whether it is a compound, molecule, atom, or ion. 

\({\rm{Number\, of\, Moles = }}\frac{{{\rm{ Given\,number\, of\, species }}}}{{{\rm{6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}}}\)

Conversion between Solutions and Moles

Molarity is defined as the number of moles of solute dissolved in litres of the solvent.

\({\rm{Number\, of\, Moles = Molarity \times Volume\, of\, solution }}\left( {{\rm{in litres}}} \right)\)

Molality is defined as the number of moles of solute dissolved in kilograms of the solvent.

\({\rm{Number\, of\, Moles = Molality \times Mass\, of\, the\, solvent\, in\, kilograms}}\)

Step 3. Converting Moles Back to the Units Requested

Solved Examples

Given the following equation at STP:

\({{\rm{N}}_{\rm{2}}}{\rm{(g) + }}{{\rm{H}}_{\rm{2}}}{\rm{(g)}} \to {\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{(g)}}\)

Determine what volume of \({{\rm{H}}_{\rm{2}}}{\rm{(g)}}\) is needed to produce \(224 \mathrm{~L}\) of \(\mathrm{NH}_{3}(\mathrm{~g})\)

Solution:
Step 1: Balancing the equation.

\({{\rm{N}}_{\rm{2}}}{\rm{(g) + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2N}}{{\rm{H}}_{\rm{3}}}{\rm{(g)}}\)

Step 2: Convert the given quantity to moles. 

\(22.4 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) is occupied by \(1\) mole of \(\mathrm{NH}_{3}\)

\(224 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) is occupied by \(\frac{1 \times 224}{22.4}\) moles \(=10\) moles of \(\mathrm{NH}_{3}\)

Step 3:

\(2\) moles of \(\mathrm{NH}_{3}\) is produced by \(3\) moles of \({{\rm{H}}_{\rm{2}}}\)

\(10\) moles of \(\mathrm{NH}_{3}\) is produced by \(\frac{3 \times 10}{2}\) moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = 15}}\) moles of \({{\rm{H}}_{\rm{2}}}\)

Step 4: Conversion to Desired Units

\(1\) mole of \({{\rm{H}}_{\rm{2}}}\) occupies \({\rm{22}}{\rm{.4}}\,{\rm{L}}\)

\(15\) moles of \({{\rm{H}}_{\rm{2}}}\) occupies \(15 \times 22.4 \mathrm{~L}=336 \mathrm{~L}\)

Summary

Stoichiometric calculations are crucial to estimating the mass of the reactant and products in a chemical equation. These calculations make use of the balanced chemical equation. In stoichiometry, all the given units need to be converted to moles and then to the desired unit. At STP, a mole of gas will always occupy a volume of \({\rm{22}}{\rm{.4}}\,{\rm{L}}\) and an Avogadro number of particles. Avogadro’s number provides the conversion factor from the number of particles to moles. This page explains the concept of stoichiometric estimation and the steps followed in carrying out the calculation. It also explains the conversion of various units to moles and from moles to the desired unit.

Frequently Asked Questions

Q.1. What is the first step in all stoichiometric calculations?
Ans: The first step in any stoichiometric calculation is to ensure that the chemical reaction under question is always balanced.

Q.2. Why is stoichiometry important?
Ans: Stoichiometry helps us to predict the masses of a reactant participating in a chemical reaction, the masses of the products formed, and how much of the reactant is left over.

Q.3. What is the mole ratio?
Ans: Mole ratio is the ratio of coefficients of reactants and products of a balanced chemical equation. It is a conversion factor that relates the amounts in moles of any two substances in a chemical reaction. 

Q.4. What is the stoichiometric coefficient?
Ans: The stoichiometric coefficient is a numerical value beside atoms, ions and molecules in a chemical reaction. These numbers balance the number of each element on both the reactant and product sides of the equation. The stoichiometric coefficients can be fractions also, but whole numbers are often preferred for easy handling.

Q.5. What is one mole?
Ans: A mole is defined as the amount of material containing \(6.022 \times 10^{23}\) particles. The particles can be atoms, ions or molecules.

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