UNIT 1.3 STOICHIOMETRY

LEC:3                                              STOICHIOMETRY

Before going to discuses Stoichiometry lets discus some some basic concepts

Percentage Composition 

if we have a sample of any compound and we want to check purity of that sample means the given sample contains the same percentage of elements as is present in a pure sample.we can check this by analysis  its percentage of composition of components.

Percentage composition may be expressed by

for example lets take C₆H₁₂O_{6 compound. its molar mass is 180 g  so % composition of each} 

_{Empirical Formula}

An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

If the mass per cent of various elements present in a compound is known, its empirical formula can be determined. Molecular formula can further be obtained if the molar mass is known. 

take a example to discuss this concept

Problem: A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ?

Solution:

step 1: assume mass % as a mass of the atom.

Hydrogen = 4.07 g

carbon = 24.27 g

chlorine = 71.65 g

step 2: convert mass of atom to moles take approximate value.

Hydrogen = 4.07 / 1 = 4

Carbon = 24.27 / 12 = 2

Chlorin = 71.65/35.5 = 2

step 3: divide mole value calculated by smallest value & note simple ratio.

H:C:Cl= 2:1:1

this give the ration of empirical formula of compound.

empirical formula = CH2Cl

step 4: for molecular formula follow these process

mass of empirical formula = 12 + 2 + 35.5 = 49.5

ratio of mass given to the empirical formula = 98.96 / 49.5 = 2

multiply this No to empirical formula ratio to obtain molecular formula

C2H4Cl2.

Stoichiometry and stoichiometry calculation

Stoichiometry deals with the calculation of quantity (mass, volumes, moles, No of particles) of the reactants and the products involved in a chemical reaction.

in this section we are going to study how much reactant required to produce a amount of products, which reactant is limiting reagent (discuss latter), how much product produced etc.

to study in detail let us take a example

CH4 (g)  +  2O2 (g)   →   CO2 (g)  +  2 H2O (g)

First of all Equation should balance, if equation is not balance then first balance it.

Here CH4 and O2 are called reactants and CO2 and H2O are products. in brackets we represent state of compounds solid, liquid and gas.The coefficients 2 for O2 and H2O are called stoichiometric coefficients.They represent the quantity that take part in the reaction or formed in the reaction.

We can relates all compompounds thats take part in chemical reaction in following manners

CH4 (g)     +      2O2 (g)                             ------->                                CO2 (g)       +       2 H2O (g)

1) Mole basis: 1 mole of CH4 reacts with 2 moles of O2 to give 1 mole of CO2 and 2 moles of H2O

2) No of particles basis: 1 molecule of CH4 reacts with 2 molecules of O2 to give 1 molecule of CO2 & 2 molecules of H2O.

3) Volume basis: 22.4 L of CH4  reacts with 44.8 L of O2  to give 22.4 L of CO2  and  44.8 L of H2O.

4) Mass basis: 16 g of CH4  reacts with 2×32 g of O2  to give 44 g of CO2  and 2×18 g of H2O.

we can also relate one quantity to other to solve some specific problem. now let us take a example

Problem: for a given reaction calculate

CH4 (g)  +  2O2 (g)   →   CO2 (g)  +  2 H2O (g)

1) Calculate the amount of water (g) produced by the combustion of 16 g of methane.

2) How many moles of methane are required to produce 22 g CO2 (g) after combustion?

Solution:

1) 16g of methane react with 64 g oxygen to produce 44g of carbon dioxide and 36g of water.

2) 44 g of CO2  produced from 16 g of methane

    22 g of CO2 will produce 8 g of methane means 0.5 moles.

Limiting Reagent

In the reaction, the reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent.

so in this section we have to find first which reactant is limiting reactant and after that we can relate products on the basis of limiting reagent to solve problem.

Problem: 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation.

Solution: 
Step 1: First of all we need a balance chemical equation

N2 (g)   +   3 H2 (g)             ---------------->        2 NH3 (g)

Step 2: Convert all given data to moles.

No of mole of Nitrogen = 50000 / 28 = 1785.7 moles

No of moles of Hydrogen = 10000 / 2 = 5000 moles

Step 3: Check limiting reagent 

1 mole of N2 required 3 moles of H2

so 1785.7 moles of N2 required (1785.7 x 3) 5357.14 moles means we have only 5000 moles of Hydrogen so limiting reagent is Hydrogen

Step 4: Relate compound on the basis of limiting reagent

3 moles of Hydrogen required 2 moles of amonia 

5000 moles of Hydrogen required = (2 / 3) x 5000 = 3333.33 moles of amonia

mass of 3333.33 moles of amonia is = 3333.33 x 17 = 56666.66 g 

56.66 Kg. amonia will produce