Category: Chemistry

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What is Stoichiometry? Balancing Equations

Have you ever noticed the preparation of kesari at your home? In one of the popular methods for the preparation of kesari, the required ingredients to prepare six cups of kesari are as follows.

1. Rava – 1 Cup
2. Sugar – 2 Cups
3. Ghee – \(\frac{1}{2}\) Cup
4. Nuts and Dry Fruits – \(\frac{1}{4}\) Cup

Otherwise,

1 cup rava + 2 cups sugar + \(\frac{1}{2}\) cup ghee + \(\frac{1}{4}\) cup nuts and dry fruits → 6 cups kesari.

From the above information, we will be able to calculate the amount of ingredients that are required for the preparation of 3 cups of kesari as follows

Alternatively, we can calculate the amount of kesari obtained from 3 cups rava as below.

Similarly, we can calculate the required quantity of other ingredients too. We can extend this concept to perform stoichiometric calculations for a chemical reaction. In Greek, stoicheion means element and metron means measure that is, stoichiometry gives the numerical relationship between chemical quantities in a balanced chemical equation.

By applying the concept of stoichiometry, we can calculate the amount of reactants required to prepare a specific amount of a product and vice versa using balanced chemical equation.

Let us consider the following chemical reaction.
C(s) + O₂(g) → CO₂(g)

From this equation, we learnt that 1 mole of carbon reacts with 1 mole of oxygen molecule to form 1 mole of carbon dioxide.

1 mole of C ≡ 1 mole of O₂
≡ 1 mole of CO₂

The symbol ‘≡’ means ‘stoichiometrically 6 cups of Kesari equal to’

Stoichiometric Calculations:

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation in moles. The quantity of reactants and products can be expressed in moles or in terms of mass unit or as volume. These three units are inter convertible.

Let us explain these conversions by considering the combustion reaction of methane as an example. The balanced chemical equation is,

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

Calculations Based on Stoichiometry

1. How many moles of hydrogen is required to produce 10 moles of ammonia?

The balanced stoichiometric equation for the formation of ammonia is

N₂(g) + 3H₂ (g) → 2 NH₃ (g)

As per the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required

∴ to produce 10 moles of ammonia, 3 moles of hydrogen are required

= 15 moles of hydrogen are required.

2. Calculate the amount of water produced by the combustion of 32 g of methane.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

As per the stoichiometric equation,

Combustion of 1 mole (16 g) CH₄ produces 2 moles (2 × 18 = 36 g) of water.

Combustion of 32 g CH₄ produces

3. How much volume of carbon dioxide is produced when 50 g of calcium carbonate is heated completely under standard conditions?

The balanced chemical equation is,

As per the stoichiometric equation,

1 mole (100g) CaCO₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 litres

∴ At STP, 50 g of CaCO₃ on heating produces,

= 11.35 litres of CO₂

4. How much volume of chlorine is required to form 11.2 L of HCl at 273 K and 1 atm pressure?

The balanced equation for the formation of HCl is

H₂(g) + Cl₂(g) → 2 HCl (g)

As per the stoichiometric equation under given conditions,

To produce 2 moles of HCl, 1 mole ofchlorine gas is required

To produce 44.8 litres of HCl, 22.4 litres of chlorine gas are required.

∴ To produce 11.2 litres of HCl,

= 5.6 litres of chlorine are required.

5. Calculate the percentage composition of the elements present in magnesium carbonate. How many kilogram of CO₂ can be obtained by heating 1 kg of 90% pure magnesium carbonate.

The balanced chemical equation is

Molar mass of MgCO₃ is 84 g mol^-1.

84 g MgCO₃ contain 24 g of Magnesium.

∴ 100 g of MgCO₃ contain

= 28.57 g Mg.

i.e. percentage of magnesium
= 28.57 %.

84 g MgCO₃ contain 12 g of carbon

∴ 100 g MgCO₃ contain

= 14.29 g of carbon.

∴ Percentage of carbon
= 14.29 %.

84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

∴ Percentage of oxygen
= 57.14 %.

As per the stoichiometric equation,

84 g of 100 % pure MgCO3 on heating gives 44 g of CO₂.

∴ 1000 g of 90 % pure MgCO₃ gives

= 471.43 g of CO₂
= 0.471 kg of CO₂

Limiting Reagents:

Earlier, we learnt that the stoichiometry concept is useful in predicting the amount of product formed in a given chemical reaction. If the reaction is carried out with stoichiometric quantities of reactants, then all the reactants will be converted into products.

On the other hand, when a reaction is carried out using non-stoichiometric quantities of the reactants, the product yield will be determined by the reactant that is completely consumed. It limits the further reaction from taking place and is called as the limiting reagent. The other reagents which are in excess are called the excess reagents.

Recall the analogy that we used in stoichiometry concept i.e. kesari preparation, As per the recipe requirement, 2 cups of sugar are needed for every cup of rava. Consider a situation where 8 cups of sugar and 3 cups of rava are available (all other ingredients are in excess), as per the cooking recipe, we require 3 cups of rava and 6 cups of sugar to prepare 18 cups of kesari.

Even though we have 2 more cups of sugar left, we cannot make any more quantity of Kesari as there is no rava available and hence rava limits the quantity of Kesari in this case. Extending this analogy for the  chemical reaction in which three moles of sulphur are allowed to react with twelve moles of fluorine to give sulphur hexafluoride.

The balanced equation for this reaction is, S + 3F₂ → SF₆

As per the stoichiometry,

1 mole of sulphur reacts with 3 moles of fluorine to form 1 mole of sulphur hexafluoride and therefore 3 moles of sulphur reacts with 9 moles of fluorine to form 3 moles of sulphur hexafluoride. In this case, all available sulphur gets consumed and therefore it limits the further reaction. Hence sulphur is the limiting reagent and fluorine is the excess reagent. The remaining three moles of fluorine are in excess and do not react.

Urea, a commonly used nitrogen based fertilizer, is prepared by the reaction between ammonia and carbon dioxide as follows.

In a process, 646 g of ammonia is allowed to react with 1.144 kg of CO₂ to form urea.

1. If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent?
2. Calculate the quantity of urea formed and unreacted quantity of the excess reagent.

The balanced equation is

2 NH₃ + CO₂
↓
H₂NCONH₂ + H₂O

Answer:

1. The entire quantity of ammonia is consumed in the reaction. So ammonia is the limiting reagent. Some quantity of CO₂ remains
unreacted, so CO₂ is the excess reagent.

2. Quantity of urea formed

= number of moles of urea formed × molar mass of urea
= 19 moles × 60 g mol^-1
= 1140 g
= 1.14 kg

Excess reagent leftover at the end of the reaction is carbon dioxide.

Amount of carbon dioxide leftver

= number of moles of CO₂ left over × molar mass of CO₂
= 7 moles × 44 g mol^-1
= 308 g.

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Empirical Formula and Molecular Formula

Elemental analysis of a compound gives the mass percentage of atoms present in the compound. Using the mass percentage, we can determine the empirical formula of the compound. Molecular formula of the compound can be arrived at from the empirical formula using the molar mass of the compound.

Empirical formula of a compound is the formula written with the simplest ratio of the number of different atoms present in one molecule of the compound as subscript to the atomic symbol. Molecular formula of a compound is the formula written with the actual number of different atoms present in one molecule as a subscript to the atomic symbol.

Let us understand the empirical formula by considering acetic acid as an example.

The molecular formula of acetic acid is C₂H₄O₂

The ratio of C : H : O is 1 : 2 : 1 and hence the empirical formula is CH₂O.

Determination of Empirical Formula from Elemental Analysis Data:

Step 1:

Since the composition is expressed in percentage, we can consider the total mass of the compound as 100 g and the percentage values of individual elements as mass in grams.

Step 2:

Divide the mass of each element by its atomic mass. This gives the relative number of moles of various elements in the compound.

Step 3:

Divide the value of relative number of moles obtained in the step 2 by the smallest number of them to get the simplest ratio.

Step 4:

(only if necessary) in case the simplest ratios obtained in the step 3 are not whole numbers then they may be converted into whole number by multiplying by a suitable smallest number.

Example:

1. An acid found in Tamarind on analysis shows the following percentage composition: 32 % Carbon; 4 % Hydrogen; 64 % Oxygen. Find the empirical formula of the compound.

The empirical formula is C₂H₃O₃

2. An organic compound present in vinegar has 40 % carbon, 6.6 % hydrogen and 53.4 % oxygen. Find the empirical formula of the compound.

The empirical formula is CH₂O

Molecular formula of a compound is a whole number multiple of the empirical formula. The whole number can be calculated from the molar mass of the compound using the following expression

Calculation of Molecular Formula from Empirical Formula:

Let us understand the calculations of molecular formula from the following example.

Two organic compounds, one present in vinegar (molar mass: 60 g mol^-1), another one present in sour milk (molar mass: 90 g mol^-1) have the following mass percentage composition. C-40%, H-6.6%; O-53.4%. Find their molecular formula.

Since both compounds have same mass percentage composition, their empirical formula are the same as worked out in the example problem no 2. Empirical formula is CH₂O. Calculated empirical formula mass (CH₂O) = 12 + (2×1) + 16 = 30 g mol^-1.

Formula for the compound present in vinegar

∴ Molecular formula = (CH₂O)₂
= C₂H₄O₂ (acitic acid)

Caluculation of molecular formula for the compound present in sour milk.

Molecular formula = (CH₂O)₃
= C₃H₆O₃ (lactic acid)

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Gram Equivalent Mass | Formula, Definition, Diagrams

Similar to mole concept gram equivalent concept is also widely used in chemistry especially in analytical chemistry. In the previous section, we have understood that mole concept is based on molecular mass. Similarly gram equivalent concept is based on equivalent mass.

Definition:

Gram equivalent mass is defined as the mass of an element (compound or ion) that combines or displaces 1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.

Consider the following reaction:

Zn + H₂SO₄ → ZnSO₄ + H₂

In this reaction 1 mole of zinc (i.e. 65.38 g) displaces one mole of hydrogen molecule (2.016 g).

Mass of zinc required to displace 1.008 g hydrogen is

= \(\frac{65.38}{2.016}\) × 1.008
= \(\frac{65.38}{2}\)

The equivalent mass of zinc = 32.69

The gram equivalent mass of zinc = 32.69 g eq^-1

Equivalent mass has no unit but gram equivalent mass has the unit g eq^-1

It is not always possible to apply the above mentioned definition which is based on three references namely hydrogen, oxygen and chlorine, because we can not conceive of reactions involving only with those three references. Therefore, a more useful expression used to calculate gram equivalent mass is given below.

On the basis of the above expression, let us classify chemical entities and find out the formula for calculating equivalent mass in the table below.

Equivalent Mass of Acids, Bases, Salts, Oxidising Agents and Reducing Agents

Mole concept requires a balanced chemical reaction to find out the amount of reactants involved in the chemical reaction while gram equivalent concept does not require the same. We prefer to use mole concept for non-redox reactions and gram equivalent concept for redox reactions.

For example,

If we know the equivalent mass of KMnO₄ and anhydrous ferrous sulphate, without writing balanced chemical reaction we can straightaway say that 31.6 g of KMnO₄ reacts with 152 g of FeSO₄ using gram equivalent concept.

The same can also be explained on the basis of mole concept. The balanced chemical equation for the above mentioned reaction is

10FeSO₄ + 2KMnO₄ + 8H₂SO₄
↓
K₂SO₄ + 2 MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O

i.e. 2 moles (2 × 158 = 316 g) of potassium permanganate reacts with 10 moles (10 × 152 = 1520 g) of anhydrous ferrous sulphate.

∴ 31.6 g KMnO₄ reacts with

\(\frac{1520}{316}\) × 31.6 = 152 g of FeSO₄

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Avogadro’s number | Definition & Units

The total number of entities present in one mole of any substance is equal to 6.022 × 10²³. This number is called Avogadro number which is named after the Italian physicist Amedeo Avogadro who proposed that equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules. Avogadro number does not have any unit.

In a chemical reaction, atoms or molecules react in a specific ratio. Let us consider the following examples

Reaction 1:

C + O₂ → CO₂

Reaction 2:

CH₄ + 2O₂ → CO₂ + 2 H₂O

In the first reaction, one carbon atom reacts with one oxygen molecule to give one carbon dioxide molecule. In the second reaction, one molecule of methane burns with two molecules of oxygen to give one molecule of carbon dioxide and two molecules of water.

It is clear that the ratio of reactants is based on the number of molecules. Even though the ratio is based on the number of molecules it is practically difficult to count the number of molecules.

Because of this reason it is beneficial to use ‘mole’ concept rather than the actual number of molecules to quantify the reactants and the products. We can explain the first reaction as one mole of carbon reacts with one mole of oxygen to give one mole of carbon dioxide and the second reaction as one mole of methane burns with two moles of oxygen to give one mole of carbon dioxide and two moles of water. When only atoms are involved, scientists also use the term one gram atom instead of one mole.

Molar Mass:

Molar mass is defined as the mass of one mole of a substance. The molar mass of a compound is equal to the sum of the relative atomic masses of its constituents expressed in g mol^-1.

Examples:

• Relative atomic mass of one hydrogen atom = 1.008 u
• Molar mass of hydrogen atom = 1.008 g mol^-1
• Relative molecular mass of glucose = 180 u
• Molar mass of glucose = 180 g mol^-1

Molar Volume:

The volume occupied by one mole of any substance in the gaseous state at a given temperature and pressure is called molar volume.

Conditions	Volume occupied by one mole of any gaseous substances (in litre)
273 K and 1 bar pressure (STP)	22.71
273 K and 1 atm pressure	22.4
298 K and 1 atm pressure (Room temperature & Pressure (SATP)	22.47

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Mole Concept – What is a Mole? Related Formulae and Examples

Often we use special names to express the quantity of individual items for our convenience. For example, a dozen roses means 12 roses and one quire paper means 24 single sheets. We can extend this analogy to understand the concept of mole that is used for quantifying atoms and molecules in chemistry. Mole is the SI unit to represent a specific amount of a substance.

To understand the mole concept, let us calculate the total number of atoms present in 12 g of carbon – 12 isotope or molecules in 158.03 g of potassium permanganate, 294.18 g of potassium dichromate and 180 g of glucose.

Table 1.2 Calculation of Number of Entities in One Mole of Substance.

From the calculations we come to know that 12 g of carbon-12 contains 6.022 × 10²³ carbon atoms and same numbers of molecules are present in 158.03 g of potassium permanganate and 294.18 g of potassium dichromate. Similar to the way we use the term ‘dozen’ to represent 12 entities, we can use the term ‘mole’ to represent 6.022 × 10²³ (atoms or molecules or ions)

One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon-12 isotope. The elementary particles can be molecules, atoms, ions, electrons or any other specified particles.

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Molar Mass – Definition, Formula

Similar to relative atomic mass, relative molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit. The relative molecular mass of any compound can be calculated by adding the relative atomic masses of its constituent atoms.

For example,

(i) Relative molecular mass of hydrogen molecule (H₂)
= 2 × (relative atomic mass of hydrogen atom)
= 2 × 1.008 u
= 2.016 u.

(ii) Relative molecular mass of glucose (C₆H₁₂O₆)
= (6 × 12) + ( 12 × 1.008) + (6 × 16)
= 72 + 12.096 + 96
= 180.096 u

Relative Atomic Masses of Some Elements

Element	Relative atomic mass	Element	Relative atomic mass
H	1.008	Cl	35.45
C	12	K	39.10
N	14	Ca	40.08
O	16	Cr	51.99
Na	23	Mn	54.94
Mg	24.3	Fe	55.85
S	32.07	Cu	63.55

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Atomic Mass | Definition, Units & Facts

How much does an individual atom weigh? As atoms are too small with diameter of 10^-10 m and weigh approximately 10^-27 kg, it is not possible to measure their mass directly. Hence it is proposed to have relative scale based on a standard atom.

The C-12 atom is considered as standard by the IUPAC (International Union of Pure and Applied Chemistry), and its mass is fixed as 12 amu (or) u. The amu (or) unified atomic mass unit is defined as one twelfth of the mass of a Carbon-12 atom in its ground state.

i.e. 1 amu (or) 1u ≈ 1.6605 × 10^-27 kg.

In this scale, the relative atomic mass is defined as the ratio of the average atomic mass to the unified atomic mass unit.

Relative atomic mass (A_r)

For example,

Relative atomic mass of hydrogen (A_r)_H

= 1.0078 ≈ 1.008 u.

Since most of the elements consist of isotopes that differ in mass, we use average atomic mass. Average atomic mass is defined as the average of the atomic masses of all atoms in their naturally occurring isotopes. For example, chlorine consists of two naturally occurring isotopes ₁₇Cl³⁵ and ₁₇Cl³⁷ in the ratio 77:23, the average relative atomic mass of chlorine is

= \(\frac{(35×77)+(37×23)}{100}\)
= 35.46 u

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Classification of Matter (Elements, Compounds)

Look around your classroom. What do you see? You might see your bench, table, blackboard, window etc. What are these things made of ? They are all made of matter. Matter is defined as anything that has mass and occupies space. All matter is composed of atoms. This knowledge of matter is useful to explain the experiences that we have with our surroundings.

In order to understand the properties of matter better, we need to classify them. There are different ways to classify matter. The two most commonly used methods are classification by their physical state and by chemical composition as described in the chart.

Physical Classification of Matter:

Matter can be classified as solids, liquids and gases based on their physical state. The physical state of matter can be converted into one another by modifying the temperature and pressure suitably.

Chemical Classification:

Matter can be classified into mixtures and pure substances based on chemical compositions. Mixtures consist of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance. Pure substances are composed of simple atoms or molecules. They are further classified as elements and compounds.

Element:

An element consists of only one type of atom. We know that an atom is the smallest electrically neutral particle, being made up of fundamental particles, namely electrons, protons and neutrons. Element can exist as monatomic or polyatomic units.

Example:

Monatomic unit – Gold (Au), Copper (Cu); Polyatomic unit Hydrogen (H₂), Phosphorous (P₄) and
Sulphur (S₈)

Compound:

Compounds are made up of molecules which contain two or more atoms of different elements.

Example:

Carbon dioxide (CO₂), Glucose (C₆H₁₂O₆), Hydrogen Sulphide (H₂S), Sodium Chloride (NaCl)

Properties of compounds are different from those of their constituent elements. For example, sodium is a shiny metal, and chlorine is an irritating gas. But the compound formed from these two elements, sodium chloride, shows different characteristics as it is a crystalline solid, vital for biological functions.

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Chemistry In Everyday Life – Importance, Examples, Uses

‘Unna unavu, udukka udai, irukka idam’ – in Tamil classical language means food to eat, cloth to wear and place to live. These are the three basic needs of human life. Chemistry plays a major role in providing these needs and also helps us to improve the quality of life.

Chemistry has produced many compounds such as fertilizers, insecticides etc. that could enhance the agricultural production. We build better and stronger buildings that sustain different weather conditions with modern cements, concrete mixtures and better quality steel. We also have better quality fabrics.

Chemistry is everywhere in the world around us. Even our body is made up of chemicals. Continuous biochemical reactions occurring in our body are responsible for human activities. Chemistry touches almost every aspect of our lives, culture and environment.

The world in which we are living is constantly changing, and the science of chemistry continues to expand and evolve to meet the challenges of our modern world. Chemical industries manufacture a broad range of new and useful materials that are used in every day life.

Examples: polymers, dyes, alloys, life saving drugs etc.

When HIV/AIDS epidemic began in early 1980s, patients rarely lived longer than a few years. But now many effective medicines are available to fight the infection, and people with HIV infection have longer and better life.

The understanding of chemical principles enabled us to replace the non eco friendly compounds such as CFCs in refrigerators with appropriate equivalents and increasing number of green processes. There are many researchers working in different fields of chemistry to develop new drugs, environment friendly materials, synthetic polymers etc. for the betterment of the society.

As chemistry plays an important role in our day-to-day life, it becomes essential to understand the basic principles of chemistry in order to address the mounting challenges in our developing country.

Chemistry is everywhere in the world around us. Even our body is made up of chemicals. Continuous bio-chemical reactions occurring in our body are responsible for human activities. Chemistry touches almost every aspect of our lives, culture and environment.

Chemistry is essential for meeting our basic needs of food, clothing, shelter, health, energy, and clean air, water, and soil. Chemical technologies enrich our quality of life in numerous ways by providing new solutions to problems in health, materials, and energy usage.

Food is made from chemicals. Many of the changes you observe in the world around you are caused by chemical reactions. Examples include leaves changing colors, cooking food and getting yourself clean. Knowing some chemistry can help you make day-to-day decisions that affect your life.

The scientific study of the chemical composition of living matter and of the chemical processes that go on in living organisms.

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An Overview of Polymers

The term Polymer is derived from the Greek word ‘polumeres’ meaning “having many parts”. The constitution of a polymer is described in terms of its structural units called monomers. Polymers consists of large number of monomer units derived from simple molecules.

For example: PVC(Poly Vinyl Chloride) is a polymer which is obtained from the monomer vinyl chloride. Polymers can be classified based on the source of availability, structure, molecular forces and the mode of synthesis. The following chart explain different classification of polymers.

Classification of Polymers:

Types of Polymerisation

The process of forming a very large, high molecular mass polymer from small structural units i.e., monomer is called polymerisation. Polymerisation occurs in the following two ways

• Addition polymerisation or chain growth polymerisation
• Condensation polymerisation or step growth polymerisation

Addition Polymerisation

Many alkenes undergo polymerisation under suitable conditions. The chain growth mechanism involves the addition of the reactive end of the growing chain across the double bond of the monomer. The addition polymerisation can follow any of the following three mechanisms depending upon the reactive intermediate involved in the process.

1. Free Radical Polymerisation
2. Cationic Polymerisation
3. Anionic Polymerisation

Free Radical Polymerisation

When alkenes are heated with free radical initiator such as benzyl peroxide, they undergo polymerisation reaction. For example styrene polymerises to polystyrene when it is heated to ionic with a peroxide initiator. The mechanism involves the following steps.

1. Initiation – formation of free radical

2. Propagation Step

The stabilized radical attacks another monomer molecule to give an elongated radical

Chain growth will continue with the successive addition of several thousands of monomer units.

Termination

The above chain reaction can be stopped by stopping the supply of monomer or by coupling of two chains or reaction with an impurity such as oxygen.

Preparation of Some Important Addition Polymers

1. Polythene

It is an addition polymer of ethene. There are two types of polyethylene

• HDPE (High Density Polyethylene)
• LDPE (Low Density polyethylene)

LDPE

It is formed by heating ethene at 200° to 300° C under oxygen as a catalyst. The reaction follows free radical mechanism. The peroxides formed from oxygen acts as a free radical initiator.

It is used as insulation for cables, making toys etc…

HDPE

The polymerization of ethylene is carried out at 373K and 6 to 7 atm pressure using Zeiglar – Natta catalyst [TiCl₄+(C₂H₅)₃Al]. HDPE has high density and melting point and it is used to make bottles, pipe etc..,

Preparation of Teflon (PTFE)

The monomer is tetraflroethylene. When the monomer is heated with oxygen (or) ammonium persulphate under high pressure, Tefln is obtained.

It is used for coating articles and preparing non – stick utensils.

I. Preparation of Orlon (polyacrylonitrile – PAN)

It is prepared by the addition polymerisation of vinylcyanide (acrylonitrile) using a peroxide initiator.

It is used as a substitute of wool for making blankets, sweaters etc…

Condensation Polymerisation

Condensation polymers are formed by the reaction between functional groups an adjacent monomers with the elimination of simple molecules like H₂O, NH₃ etc…. Each monomer must undergo at least two substitution reactions to continue to grow the polymer chain i.e., the monomer must be at least bi functional. Examples: Nylon – 6,6, terylene….

Nylon – 6, 6

Nylon – 6, 6 can be prepared by mixing equimolar adipic acid and hexamethylene – diamine to form a nylon salt which on heating eliminate a water molecule to form amide bonds.

It is used in textiles, manufacture of cards etc…

Nylon – 6

Capro lactam (monomer) on heating at 533K in an inert atmosphere with traces of water gives ∈-v amino carproic acid which polymerises to give nylon – 6

It is used in the manufacture of tyrecards fabrics etc….

II. Preparation of Terylene (Dacron)

The monomers are ethylene glycol and terepathalic acid (or) dimethylterephthalate. When these monomers are mixed and heated at 500K in the presence of zinc acetate and antimony trioxide catalyst, terylene is formed.

It is used in blending with cotton or wool fires and as glass reinforcing materials in safety helmets.

Preparation of Bakelite

The monomers are phenol and formaldehyde. The polymer is obtained by the condensation polymerization of these monomers in presence of either an acid or a base catalyst. Phenol reacts with methanal to form ortho or para hydroxyl methylphenols which on further reaction with phenol gives linear polymer called novolac. Novalac on further heating with formaldehyde undergo cross linkages to form backelite.

Uses:

Navolac is used in paints. Soft backelites are used for making glue for binding laminated wooden planks and in varinishes, Hard backelites are used to prepare combs, pens etc..

Melamine (Formaldehyde Melamine):

The monomers are melamine and formaldehyde. These monomers undergo condensation polymerisation to form melamine formaldehyde resin.

Urea Formaldehyde Polymer:

It is formed by the condensation polymerisation of the monomers urea and formaldehyde.

Co-Polymers:

A polymer containing two or more different kinds of monomer units is called a copolymer. For example, SBR rubber(Buna-S) contains styrene and butadiene monomer units. Co-polymers have properties quite different from the homopolymers.

Natural and Synthetic Rubbers:

Rubber is a naturally occurring polymer. It is obtained from the latex that excludes from cuts in the bark of rubber tree (Ficus elastic). The monomer unit of natural rubber is cis isoprene (2-methyl buta-1,3-diene). Thousands of isoprene units are linearly linked together in natural rubber. Natural rubber is not so strong or elastic. The properties of natural rubber can be modified by the process called vulcanization.

Vulcanization: Cross linking of Rubber

In the year 1839, Charles Good year accidently dropped a mixture of natural rubber and sulphur onto a hot stove. He was surprised to find that the rubber had become strong and elastic. This discovery led to the process that Good year called vulcanization.

Natural rubber is mixed with 3-5% sulphur and heated at 100-150˚C causes cross linking of the cis-1,4-polyisoprene chains through disulphide (-S-S-) bonds. The physical properties of rubber can be altered by controlling the amount of sulphur that is used for vulcanization. In sulphur rubber, made with about 1 to 3% sulphur is sof and stretchy. When 3 to 10% sulphur is used the resultant rubber is somewhat harder but flexible.

Synthetic Rubber:

Polymerisation of certain organic compounds such as buta-1,3-diene or its derivatives gives rubber like polymer with desirable properties like stretching to a greater extent etc., such polymers are called synthetic rubbers.

Preparation of Neoprene:

The free radical polymeristion of the monomer, 2-chloro buta-1,3-diene(chloroprene) gives neoprene.

It is superior to rubber and resistant to chemical action.
Uses: It is used in the manufacture of chemical containers, conveyer belts.

Preparation of Buna-N:

It is a co-polymer of acrylonitrile and buta-1,3-diene.

It is used in the manufacture of hoses and tanklinings.

Preparation of Buna-S:

It is a co-polymer. It is obtained by the polymerisation of buta-1,3-diene and styrene in the ratio 3:1 in the presence of sodium.

Biodegradable Polymers

The materials that are readily decomposed by microorganisms in the environment are called biodegradable. Natural polymers degrade on their own after certain period of time but the synthetic polymers do not. It leads to serious environmental pollution. One of the solution to this problem is to produce biodegradable polymers which can be broken down by soil micro organism.

Examples:

Polyhydroxy butyrate (PHB)
Polyhydroxy butyrate-co-A- hydroxyl valerate (PHBV)
Polyglycolic acid (PGA), Polylactic acid (PLA)
Poly (∈caprolactone) (PCL)

Biodegradable polymers are used in medical field such as surgical sutures, plasma substitute etc… these polymers are decomposed by enzyme action and are either metabolized or excreted from the body.

Preparation of PHBV

It is the co – polymer of the monomers 3 – hydroxybutanoic acid and 3-hydroxypentanoic acid. In PHBV, the monomer units are joined by ester linkages.

Uses:
It is used in ortho paedic devices, and in controlled release of drugs.

Nylon-2-Nylon-6

It is a co – polymer which contains polyamide linkages. It is obtained by the condensation polymersiation of the monomers, glycine and É – amino caproic acid.

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Cleansing Agents Functions and its Types

Soaps and detergents are used as cleansing agents. Chemically soap is the sodium or potassium salt of higher fatty acids. Detergent is sodium salt of alkyl hydrogen sulphates or alkyl benzene sulphonic acids.

Soaps:

Soaps are made from animal fats or vegetable oils. They contain glyceryl esters of long chain fatty acids. When the glycerides are heated with a solution of sodium hydroxide they become soap and glycerol. We have already learnt this reaction under the preparation of glycerol by saponification. Common salt is added to the reaction mixture to decrease the solubility of soap and it helps to precipitate out from the aqueous solution. Soap is then mixed with desired colours, perfumes and chemicals of medicinal importance.

Total Fatty Matter:

The quality of a soap is described in terms of total fatty matter (TFM value). It is defined as the total amount of fatty matter that can be separated from a sample after splitting with mineral acids., Higher the TFM quantity in the soap better is its quality. As per BIS standards, Grade-1 soaps should have 76% minimum TFM, while Grade-2 and 3 must have 70 and 60% , minimum respectively. The other quality parameters are lather, moisture content,mushiness, insoluble matter in alcohol etc..

The Cleansing Action of Soap:

To understand how a soap works as a cleansing agent, let us consider sodium palmitate an example of a soap. The cleansing action of soap is directly related to the structure of carboxylate ions (palmitate ion) present in soap. The structure of palmitate exhibit dual polarity. The hydrocarbon portion is non polar and the carboxyl portion is polar.

The nonpolar portion is hydrophobic while the polar end is hydrophilic. The hydrophobic hydro carbon portion is soluble in oils and greases, but not in water. The hydrophilic carboxylate group is soluble in water. The dirt in the cloth is due to the presence of dust particles intact or grease which stick.

When the soap is added to an oily or greasy part of the cloth, the hydrocarbon part of the soap dissolve in the grease, leaving the negatively charged carboxylate end exposed on the grease surface. At the same time the negatively charged carboxylate groups are strongly attracted by water, thus leading to the formation of small droplets called micelles and grease is flated away from the solid object.

When the water is rinsed away, the grease goes with it. As a result, the cloth gets free from dirt and the droplets are washed away with water. The micelles do not combine into large drops because their surfaces are all negatively charged and repel each other. The cleansing ability of a soap depends upon its tendency to act as a emulsifying agent between water and water insoluble greases.

Detergents:

Synthetic detergents are formulated products containing either sodium salts of alkyl hydrogen sulphates or sodium salts of long chain alkyl benzene sulphonic acids. There are three types of detergents.

Detergents are superior to soaps as they can be used even in hard water and in acidic conditions. The cleansing action of detergents are similar to the cleansing action of soaps.