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:
CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (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 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 O2 and H2O are called stoichiometric coefficients. Similarly the coefficient for CH4 and CO2 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,
- One mole of CH4 (g) reacts with two moles of O2 (g) to give one mole of CO2 (g) and two moles of H2O (g)
- One molecule of CH4 (g) reacts with 2 molecules of O2 (g) to give one molecule of CO2 (g) and 2 molecules of H2O (g)
- 7 L of CH4 (g) reacts with 45.4 L of O2 (g) to give 22.7 L of CO2 (g) and 45.4 L of H2O (g)
- 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of H2O (g).
From these relationships, the given data can be interconverted as follows:
Mass ßàmoles ßànumber of molecules
Mass / Volume = Density
1.10.1 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 amountgets consumed after sometime 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. In performing stoichiometric calculations, this aspect is also to be kept in mind.
1.10.2 Reactions in Solutions
A majority of reactions in the laboratories are carried out in solutions. Therefore, it isimportant 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.
- Mass per cent or weight per cent (w/w %)
- Mole fraction
Let us now study each one of them in detail.
1.Mass per cent
It is obtained by using the following relation:
Mass per cent = (Mass of the solute / Mass of the Solution) X 100
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 ‘B’ and their number of moles are nA and nB , respectively, then the mole fractions of A and B are given as:
Mole fraction of A = No of moles of A / No of moles of solution = nA / (nA + nB)
Mole fraction of B = No of moles of B / No of moles of solution = nB / (nA + nB)
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) = no of moles of solute / volume of solution in liters
Suppose, we have 1 M solution of a substance, say NaOH, and we want to prepare a 0.2 M solution from it.1 M NaOH means 1 mol of NaOH present in 1 litre of the solution. For 0.2 M solution, we require 0.2 moles of NaOH dissolved in 1 litre solution. Hence, for making 0.2M solution from 1M solution, we have to take that volume of 1M NaOH solution, which contains 0.2 mol of 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 NaOH can be calculated as follows: If 1 mol is present in 1L or 1000 mL solution
then, 0.2 mol is present in
(1000 mL / 1 mol) X 0.2 mol solution
= 200 mL solution
Thus, 200 mL of 1M NaOH are taken and enough water is added to dilute it to make it 1 litre.
Note that the number of moles of solute (NaOH) was 0.2 in 200 mL and it has remained the same, i.e., 0.2 even after dilution ( in 1000 mL) as we have changed just the amount of solvent (i.e., water) and have not done anything with respect to NaOH. But keep in mind the concentration.
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.
Thus, Molality (m) = No of moles of solute / mass of solvent in Kg