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Stoichiometry

Table of Content

The Mole

Chemical Equations and Stoichiometry

The Limiting Reagent

The Yield of Product (Percentage Yield)

Chemical stoichiometry deals with the determination of quantities of reactants or products of a chemical reaction. The word “stoichiometry” is derived from greek work “stoichion” means element and “metry” means measure. Stoichiometry is divided into two subsections.

Gravimetric analysis and

Volumetric analysis.

The problems on gravimetric and volumetric analysis can be solved using two well known concepts i.e. mole concept and equivalent concept. But generally, the problems on gravimetric analysis are solved using mole concept since mole concept is easier to apply in such cases while problems on volumetric analysis are solved making use of equivalent concept since it does not require the use of balanced chemical reaction.
However, there is no hard and fast rule that these guidelines should be followed strictly. We can make use of equivalent concept of gravimetric problems and mole concept for volumetric problems.

Will explore the topic stoichiometry under the fiollowing sub topics

n Factor Calculation

Law of Equivalents

Volumetric Analysis

Gravimetric Analysis

Let us first understand the mole concept and its use in chemical reactions and then we will learn about equivalent concept.

The Mole

The concept of amount of substance is confined to the chemical measurement.

The amount of substance of a system is proportional to the number of elementary entitled (which may be atoms or molecules or ions or specified group of such particles) of that substance present in the system.

Let us take elements Ag, Mg and Hg with masses equal to their atomic masses in grams, and then to our surprise, each element contains equal number of atoms.

This is not only limited to atoms but also applicable to molecules. For example, if we molecules like CO2, NO2 and SO2 with masses equal to their molecular masses in grams, then they would also contain equal number of molecules. This specified number of atoms or molecules is referred as a “mole”.

Thus a system containing a specified number (6.023 × 1023) of elementary entitles is said to contain 1 mole of the entities. Thus 1 mole of an iron sample mean that the sample contain 6.023 × 1023 atoms of iron. Similarly, 1 mole of NaCI crystal contains 6.023 × 1023 ion pairs (Na+ CI–).

This specific number 6.023 × 1023 elementary entities is called Avogadro number or constant (NAV). The SI unit for amount of substance is the mole. The mole is defined as the amount of substance in exactly 12 g of carbon-12. One mole of any substance contains the same number of elementary entitles as there are carbon atoms in exactly 12 g of carbon-12.

The mass of specific number (6.023 × 1023) of elementary entities is equal to atomic mass for atoms and molecular mass for molecules.

Let M g/mole be the molecular mass of a species. Thus M g be the mass of 1 mole (equal to the mass of 6.023 × 1023 molecules) of the species. Then, x g of the species contain  mole. Hence

Number of moles of a species = $\frac{\textup{Weight\ (grams)}}{\textup{Atomic\ or\ molecular\ mass\ (g/mole)}}=\frac{\textup{w}}{\textup{M}}$

It is also known that one mole of a gas at STP occupies a volume of 22.4 litres. Thus, if a gas occupies x L at STP, then the number of moles of the gas can be calculated by dividing the actual volume occupied by the gas at STP with the volume occupied by 1 mole of the gas at STP.

Thus, number of moles of a gas =$\frac{\textup{Volume\ occupied\ by\ gas\ at\ STP}}{\textup{Volume\ occupied\ by\ 1\mole\ of\ the\ gas\ at\ STP}}$

The volume of gas and the number of moles of gas at temperature and pressure other than the STP can be related by ideal gas equation, PV = nRT.

Chemical Equations and Stoichiometry

Let the balanced chemical equation we have is MnO2 + 4HCI →  MnCI2 + 2H2O + CI2

The quantitative information drawn from this balanced chemical equation is

The molar ratio in which two reactants (MnO2 and HCI) reacting is 1 : 4.

The molar ratio between two products can also be known i.e. moles of H2O produced would be double the moles of MnCI2 produced.

The initial moles of MnO2 and HCI (to be taken in vessel) for the reaction to occur not necessarily be 1 and 4 respectively or also should not be in the molar ratio of 1:4.

We can start reaction with MnO2 and HCI taken in any molar ratio, but the moles of two reacting will always be in the molar ratio of 1:4.

The balanced chemical equation should follow the law of conservation of mass.

Refer to the following video for stoichiometry

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Let us consider the same chemical systemi.e. MnO2 + 4HCI →  MnCI2 + 2H2O + CI2  with initial composition (in terms of mole) as noMnO2, noHCI, noMnCI2, noH2O, and noCI2.
The noHCI is four times of noMnO2.

When the reaction occurs, these mole numbers change as the reaction be x, then the moles of HCI reacting in the same time interval be 4x since MnO2 and HCI react in the molar ratio of 1 : 4.

Thus, after time t, the composition of the system would be

nMnO2 = noMnO2 – x

nHCI = noHCI – 4x

nMnCI2 = noMnCI2 + x

nH2O = noMnO2 + 2x

nCI2 = noCI2 + x

The algebraic signs, + and – indicates that the reactants are consumed and product are produced.

In general, mole numbers of various species at any time would be given as

nt = noi + voi x

where noi is the initial amount, x is the degree of advancement and vi is the stoichiometric coefficient which will be given a negative sign for reactants and a positive signs for products.

After long time interval from the commencement of reaction i.e., after t time, the composition of the system would be

nMnO2 = 0, nHCI = 0

nMnCI2 = noMnCI2 + noMnO2 = noMnCI2 +

nH2O = noH2O + 2noMnO2 = noH2O +

nCI2 = noCI2 + noMnO2 = noCI2 +

The Limiting Reagent

Let the initial moles of MnO2 and HCI be noMnO2 and noHCI respectively and noHCI ¹ 4noMnO2. Thus, in the given chemical reaction, after ¥ time, one of the reactant will be completely consumed while the other would be left in excess. Thus, the reactant which completely consumed when a reaction goes to completion and which decides the yield of the product is called limiting reagent.

For example, if in the given case noHCI > 4noMnO2, and there is no MnCI2 and H2O in the beginning, then

MnO2           +        4HCI       →           MnCI2        +       CI2       +          2H2O

Initially

noMnO2

noHCI

0

0

0

After time “t”

0

noHCI - 4noMnO2

noMnO2

noMnO2

2noMnO2

Thus, MnO2 is the limiting reagent and the yield of all the products is governed by the amount of MnO2 taken initially.

Similarly, if in the given case noHCI < 4noMnO2 and noMnCI2, CI2 & H2O are present initially, then

MnO2          +           4HCI       →      MnCI2        +     CI2    +      2H2O

Initially

noMnO2

noHCI

0

0

0

After time “t”

noMnO2  –

0

Here, HCI would become limiting reagent & the products yield are decided by the amount of HCI taken initially.

The Yield of Product (Percentage Yield)

Let us suppose that the amount of MnCI2 produced in the last case actually be less than  while the theoretical yield should be . This means that the yield of the product is not 100%. Thus, percentage yield of the product is given as ratio of actual yield by theoretical maximum yield multiplied by 100.

Hence % yield of the product =$\frac{\textup{Actual\ Yield}}{\textup{Theoretical\ Yield}}\times 100$ .

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