UNIT I: Some Basic Concepts of Chemistry
Unit II: Structure of Atom
Unit III: Classification of Elements and Periodicity in Properties
Unit IV: Chemical Bonding and Molecular Structure
Unit V: States of Matter: Gases and Liquids
Unit VI: Chemical Thermodynamics
Unit VII: Equilibrium
Unit VIII: Redox Reaction
Unit IX: Hydrogen
Unit X: s -Block Elements (Alkali and Alkaline Earth Metals)
UNIT XI: P -BLOCK ELEMENTS (CARBON AND BORON FAMILY)
UNIT XII: ORGANIC CHEMISTRY-BASIC PRINCIPLES AND TECHNIQUES
UNIT XIII: HYDROCARBON
UNIT XIV: ENVIRONMENT CHEMISTRY

Stoichiometry and Stoichiometric Calculations

The word ‘stoichiometry’ is derived from two Greek words- ‘stoicheion’ (meaning element) and ‘Metron’ (meaning measure).

Stoichiometry, thus deals with calculation of masses (sometimes volumes also) of the reactants and the products involved in the chemical reaction.

Let us consider the combustion of Methane. A balanced equation for this reaction is as given below.

\operatorname {CH_4}_{(g)}+2\operatorname {O_2}_{(g)}\to \operatorname {CO_2}_{(g)}+2 \operatorname {H_2O}_{(g)}

Here Methane and Dioxygen are called reactants and Carbon Dioxide and Water are called products.

 \operatorname {(g)} indicates that all the reactants and products are gases.

The coeffcients  2 for \operatorname {0_2} and \operatorname {H_2O} are called stoichiometric coeffcients.

Thus, according to the above chemical reaction:

\to One mole of  \operatorname {CH_4}_{(g)} reacts with two Moles of \operatorname {O_2}_{(g)} to give one mole of  \operatorname {CO_2}_{(g)} and two moles of \operatorname {H_2O}_{(g)} .

\to One molecule of \operatorname {CH_4}_{(g)} reacts with  2 molecules of \operatorname {O_2}_{(g)} to give one molecule of  \operatorname {CO_2}_{(g)}  and   2 molecules of  \operatorname {H_2O}_{(g)}.

\to 22.4 \operatorname { L } of  \operatorname {CH_4}_{(g)} reacts  with  44.8 \operatorname { L} of \operatorname {O_2}_{(g)} to give  22.4 \operatorname { L } of  \operatorname {CO_2}_{(g)} and  44.82 \operatorname { L}   of  \operatorname {H_2O}_{(g)}.

\to  16 \operatorname { g} of   \operatorname {CH_4}_{(g)} reacts with  2\times 32 \operatorname { g} of \operatorname {O_2}_{(g)} to have  44 \operatorname { g}  of  \operatorname {CO_2}_{(g)} and  2\times 18 \operatorname { g} of  \operatorname {H_2O}_{(g)}.

From the above relationships;

\text {mass}\rightleftharpoons\text {moles} \rightleftharpoons \text {number of molecules}

 \dfrac{\text {mass}}{\text {Volume}}= \text {Density}

Limiting Reactant / Reagent

Sometimes, in a chemical reaction, the reactants present are not the amount required according to the balanced equation. The amount of products formed then depends upon the reactant which has reacted completely. This reactant reacts completely in the reaction is called Limiting Reactant/Reagent. The reactant which is not consumed completely in the reaction is called an Excess Reactant.

Reactions in Solutions

The concentration of a solution can be expressed in any of the following ways:

1.Mass Percent

It is the mass of the solute in grams per  100 \operatorname { grams }   of the solution.

\operatorname {Mass}\%=\dfrac{\text {Mass of Solute}}{\text {Mass of Solution}} \times 100

2. Volume Percent

It is the number of units of volume of the solute per  100 units of the volume of the solution.

 \operatorname {Volume}\%=\dfrac{\text {Vol. of Solute}}{\text {Vol. of Solution}} \times 100

Parts per Million (PPM) and Parts per Billion (PPB)

It is convenient to measure the concentration of trace quantities of solutes in parts per million and parts per billion. It is independent of temperature.

 \operatorname {PPM}=\dfrac{\text {Mass of Solute Component}}{\text {Total Mass of Solution}} \times 10^6

 \operatorname {PPb}=\dfrac{\text {Mass of Solute Component}}{\text {Total Mass of Solution}} \times 10^8

Normality (N)

The normality of a solution is equal to the number of gram equivalents of solute present per liter of solution.

(I) \operatorname {Normality}=\dfrac{\text {Number of gram equivalent of solute}}{\text {Volume of Solution (lb)}}

(ii) \operatorname {Normality}=\dfrac{\text {Weight of Solute in gm}}{\text {gm equivalent weight of solute}\times \text {volume of solution (ltr)}}

Molarity (M)

The molarity of a solution is equal to the number of moles of the solute present per liter of solution. It can be calculated from the following formulas:

(i)  \operatorname {Molarity}=\dfrac{\text {No.of Moles of Solute (n)}}{\text { Vol. of Solution (in ltr)}}

(ii) If the Molarity and Volume of the solution are changed from   M_1, V_1 to  M_2, V_2 then  M_1V_1=M_2V_2 (Molarity equation)

(iii) In balanced chemical equation, if  n_1 moles of reactant one react with  n_2 moles of reactant two, then

\dfrac{M_1V_1}{N_1}=\dfrac{M_2V_2}{N_2}

(iii) Volume of water added to get a solution of molarity M_2 from V_1 \operatorname {mL} of molarity  M_1 is

V_2-V_1=\left( \dfrac{M_1-M_2}{M_2} \right)V_1

(iv) It two solutions of the same solute are mixed then molarity of resulting solution,

\text{ Molarity}=\dfrac{M_1V_1+M_2V_2}{(V_1+V_2)}

Here, M_1, V_1 are molarity and volume of initial solution and M_2, V_2 are molarity and volume of final solution.

The Relation between Normality and Molarity

\text {Normality} \times \text {Equivalent Mass } = \text {Molarity} \times \text {Molecular Mass}

Molality (m)

Mathematically, Molality can be calculated by

\text {Molality} = \dfrac{\text {No. of Moles of Solute}}{\text {Weight of Solvent (in Kg)}}

Relation between Molarity (M) and Molality (m)

\text {Molality}(m) = \dfrac{\text {Molarity (M)}}{\text {Density}-\dfrac{\text {Molarity}\times \text {Molecular Mass}}{1000}}

Formality (F)

The Formality of a solution may be defined as the number of gram formula masses of ionic solute dissolved per liter of the solution.

\text {F} = \dfrac{\text {No. of gram formula mass of solute}}{\text {Vol. of Solution (in ltr)}}

Mole Fraction (X)

To find the Mole Fraction, divide moles of a constituent (either Solute or Solvent) by the total moles of both the constituents (Solute or Solvent)

Let in a Solution A is Solute and B is Solvent.

Mole fraction  of Solute  X_A=\dfrac{n_A}{n_A+n_B}

 

Mole fraction of Solvent X_B=\dfrac{n_B}{n_A+n_B}

n_A and n_B are number of moles of Solute and Solvent respectively.

X_A+X_B=1

 

 

 

 

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