Separation Technique and practical chemistry

Qualitative and Quantitative Analysis

Separation Techniques

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Why Do We Need to Separate Substances?

Most substances occurring in nature are mixtures. To obtain pure substances, it is necessary to separate the components of the mixtures. For example, seawater is the mixture of water and salt. If we want to use water and salt separately, then we have to separate the salt from the water.

Pure substances have great importance in chemical industries. These are used in laboratories to study their chemical properties.

Components of heterogeneous mixtures can sometimes be separated very easily as the components are physically distinct from each other. For example, methods like handpicking, filtering, sieving etc., can be used to separate the components.

However, to separate the components of a homogeneous mixture, we need to use some special separating techniques.

The commonly used techniques for separating the components of mixtures are as follows:

Evaporation
Centrifugation
Separation of immiscible liquids by a separating funnel
Separation of miscible liquids by:
- Simple distillation
- Fractional distillation
- Sublimation
- Crystallization
- Chromatography

Evaporation

Evaporation is the process in which a liquid is changed into its vapour state by heating. It is used to separate solid substances dissolved in a liquid.

Separation of the coloured component (dye) from blue or black ink

Blue or black ink is prepared by dissolving blue or black dye in water. The dye can be separated from ink by evaporation. Let us perform an activity to do this.

Procedure:

Half fill a beaker with water.
Put a watch glass on top of the beaker.
Add a few drops of blue or black ink on the watch glass.
Now, heat the beaker till a solid mass is obtained in the watch glass.

Observation:

You will observe that a solid residue of the dye is obtained in the watch glass. Ink is a colloidal solution. It is a heterogeneous mixture of dye and water. Heating leads to the evaporation of water. This leaves behind the dye in the watch glass.

Centrifugation

Milk is a heterogeneous mixture of proteins, fats and other nutrients. Cream is separated from milk as a fat-rich substance by the process of centrifugation. The given figure shows this process.

It will be observed that cream collects in the upper layer of milk after centrifugation. Fat (cream) is the lightest component of milk, so it forms a layer on top.

The technique of centrifugation is used in the:

Drying of wet clothes in the spin tub of a washing machine
Extraction of DNA for forensic and experimental purposes
Treatment of wastewater
Processing of sugar and milk

Separation of Two Immiscible Liquids

Examples of immiscible liquid mixtures: Oil and water, Petrol and water

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Separation of Two Miscible Liquids

They can be separated by Simple distillation (for liquids having at least 20−25 K difference in boiling points) and Fractional distillation (for liquids having very similar boiling points). Examples of miscible liquid mixtures are alcohol and water, acetone and water

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Fractional Distillation

This method is used to separate miscible liquids in a mixture. The apparatus used is similar to that used in simple distillation, with the exception of a fractionating column fitted between the condenser and the distillation flask. The fractionating column is packed with glass beads. These provide a large surface area for the hot vapours to cool and condense repeatedly.

Air is a homogeneous mixture of different gases such as oxygen, nitrogen and carbon dioxide. The oxygen cylinders in hospitals and cylinders of carbon dioxide in cars and buildings are prepared by separating individual gases from air.

Aim: To separate a mixture of alcohol, water, salt and sand.

Apparatus required: Beaker, burner, funnel, filter paper, tripod stand, wire gauze and fractionating column

Theory: The given mixture contains alcohol, salt, water and sand. The separation of the mixture depends upon the characteristic properties of its constituents.

Alcohol and salt are soluble in water, but sand is insoluble in water. So, sand can be removed easily by filtration.

The remaining mixture now contains alcohol, salt and water. The boiling points of these are as follows:

Alcohol = 78°C
Salt = 108°C
Water = 100°C

It is evident from the above information that alcohol can be separated from the mixture by simple distillation. During this process, alcohol will boil first and can then be collected in a beaker.

The remaining solution now contains salt dissolved in water which can be separated by evaporation. The water vapours can be condensed back to water, while the salt is left behind in the flask.

Procedure:

Pour the mixture in a beaker.
Take an empty beaker and place the funnel on it.
Cover the inner side of the funnel with filter paper.
Pour the mixture into the funnel, allowing sand to separate from the liquid mixture.
Dry the sand collected on the filter paper.
Pour the remaining solution in a distillation flask and heat to boil.
Vapours of alcohol condense in the condenser and liquid alcohol is collected in the beaker attached to the condenser.
Heat the remaining mixture till water starts evaporating.
Crystals of salt will be left in the flask.

Result: Separation of the mixture is done as follows:

Sand from the mixture: By filtration
Alcohol from the remaining mixture: By distillation
Salt from water: By evaporation

Real life extension: List some of the industrial uses of the separation techniques used in this experiment.

Separation of Gases by Fractional Distillation

Stages involved in separation of the components of air:

Air is first filtered to remove dust particles. Then, it is compressed under high pressure.
This compressed air is then passed through a water condenser to decrease its temperature.
The cold compressed air is now passed through a separator. Here, carbon dioxide separates as dry ice. Due to repeated compression, air becomes cold and turns into liquid.
The liquid air coming out of the separator is passed through an expansion jet into the distillation column. Here, it is warmed slowly. Liquid nitrogen, having the boiling point of −196°C, boils first to form liquid nitrogen gas. This gas is collected from the upper part of the column. Argon, having a boiling point of −186°C, is collected next. Oxygen, having a boiling point of −183°C, is collected last.

Sublimation

Crystals of ammonium chloride and common salt look similar. It is difficult to separate their mixture by ordinary methods. Sublimation is the best way to separate ammonium chloride from common salt. Let us understand this process with the help of an activity.

Procedure:

Take a mixture of common salt and ammonium chloride in a china dish.
Place the dish on a tripod stand and cover it by inverting a funnel over it (as shown in the figure).
The funnel should be plugged on the top with cotton to prevent the vapours of ammonium chloride from escaping into the atmosphere.
Heat the china dish using a burner.
On heating, ammonium chloride sublimes. As a result, ammonium chloride gets separated from common salt and solidifies on the cold walls of the funnel.

Note: For any substance undergoing sublimation, the energy required to convert its solid form into liquid is greater than the energy required to convert the solid form into gas.

Sublimation Printing

It involves using dye sublimation printer to print on a substrate.
The printer uses heat to transfer dye onto the substrate.
Due to the heat, the dye transitions from solid to gas without passing through the intermediate liquid phase.
The dye sublimation printer produces a continuous tone.
The obtained prints are dry and can be used immediately.
The process is clean since no liquid dye is formed.
The coloured ribbons and the heating head must match the size of the substrate.
Only specially coated papers accept sublimated dye.
The sublimated ink is prone to diffusion, so the obtained prints are not sharp.
Once used, the coloured ribbons cannot be used again; so, a lot of dye is wasted.
A negative of the print appears on the coloured ribbons.
Dye sublimation papers and ribbons are very sensitive and can be destroyed by skin oils.
For effective prints, the ribbons need to be free from dust.

Crystallization

We can obtain pure copper sulphate from an impure sample through the process of crystallization. Let us understand this process with the help of an activity.

Procedure:

Take a small amount (5 g) of impure copper sulphate in a china dish.
Dissolve the sample in 10 mL of water.
Filter it to remove impurities.
Evaporate the filtered solution on a water bath till small crystals are formed, indicating that a concentrated solution has been obtained.
Cover the solution with a filter paper and leave it undisturbed at room temperature for the rest of the day.
It will be observed that crystals of pure copper sulphate are obtained and the impurities are left in the solution.

Sedimentation, Decantation and Filtration

Mixtures of mud in water and chalk in water can be separated by using methods such as sedimentation, decantation, and filtration. The definitions of these three methods are given in the following table.

SEDIMENTATION	DECANTATION	FILTRATION
It is the process of settling down of the heavier components present in a mixture.	It is the process of transferring a liquid from one container to another without disturbing the sediments that are present at its bottom.	It is the process of separating the undissolved components from a mixture. It is done by passing the mixture through a material containing fine holes that will allow only one component of the mixture to pass through.

Filtration is a better method than decantation to separate mixtures as some of the solid particles also pass along with the liquid during the process of decantation while filtration allows one to obtain a cleaner liquid. However, decantation is used in cases where recovering the solid substance from the filter paper is difficult.

Sedimentation, Decantation and Filtration

Chromatography

Chromatography can be used to separate the coloured components of a mixture on the basis of the difference in the speeds of the components on chromatograph paper, when dissolved in the same solvent. The adsorbent paper acts as the stationary phase; it carries the components of the mixture on the paper. The mixture acts as the mobile phase and the components get separated. The component which moves slowly (i.e., the less-soluble component) appears as a spot on the lower side of the paper. The component which moves faster (i.e., the more-soluble component) appears as a dot on the higher side of the paper.

Separating coloured components of ink by chromatography

Procedure:

On a thin strip of filter paper, make a boundary at least 3 cm above the lower portion. At the centre of the boundary, pour a drop of ink and let it dry.
Repeat the above action two to three times, letting the drop dry each time, in order to concentrate the mixture. Pour some amount of water in a cylindrical jar.
Place the paper in the jar in a manner that the lower end of the paper dips in water.
Be careful not to dip the ink in water.
Fix the upper end of the paper to the lid of the jar or on the top edge of the jar.
Leave the apparatus undisturbed for some time.

Observation: As the water rises on the filter paper, it carries the drop of ink with itself; after some distance, the ink separates into its constituent colours.

Aim: To separate the green pigment from plant leaves with the help of chromatography.

Theory: Plants survive by making their own food with the help of a green pigment known as chlorophyll. They absorb water from the ground with the help of their roots and carbon dioxide gas from the atmosphere. They use sunlight to convert the absorbed water and carbon dioxide into glucose. Glucose is a sugar which is used by plants as food to supply energy and as the building block for growth. This entire process is named photosynthesis.

Chlorophyll helps in this process of manufacturing food and provides green coloration to plants. In winters, there is often shortage of light and water for photosynthesis to be carried out. During such times, the plants rest and live off the food they stored during summers. When plants stop manufacturing food, chlorophyll starts to disappear and shades of yellow or orange become visible. This is the reason leaves change colour in autumn.

Materials required: Leaves, small beakers, covers lids for beakers, rubbing alcohol, paper coffee filters, shallow pan, hot water, tape, pen, plastic spoon, clock

Procedure:

Collect two to three big leaves from different trees.
Cut each leaf into tiny pieces and place them in a beaker labelled with the name or location of the tree that the leaf came from.
Add rubbing alcohol to the beakers in order to cover the cut pieces of the leaves.
Using a plastic spoon, carefully grind the leaves in the alcohol.
Cover the beakers loosely with lids.
Place the beakers in a shallow tray containing up to 1 inch of hot water.
Keep the beakers in water for at least thirty minutes, until the alcohol is coloured (the darker the better).
Swirl each beaker gently every five minutes. Replace the hot water if it cools off.
Cut out a long thin strip of coffee filter paper for each beaker and label it accordingly.
Remove the beakers from water and uncover them.
Place a strip of filter paper into each beaker such that one end is in the alcohol. Tape the other end to the beaker after bending it around the corner.
The alcohol will travel up the paper, dragging the colours with it.
After thirty to ninety minutes (or longer), the colours will travel to different distances on the paper as the alcohol evaporates.
Different shades of green, and possibly some yellow, orange or red (depending on the type of leaf) will start showing up on the paper.
Remove the strips of filter paper, dry them and then tape them to a piece of plain paper.

Result: List the different colours into which the green pigment got separated.

Real life extension: Find out the applications and uses of chromatography in daily life.

Detection of Elements

An organic compound contains basic elements like carbon, hydrogen, oxygen and some extra elements such as nitrogen, sulphur, phosphorus and halogens. There are many methods that can help in detection of these elements in an organic compound. Let us have a look at them.

Detection of Carbon and Hydrogen

Detected by heating a compound with copper (II) oxide
Carbon present in the compound gets oxidised to CO₂, which is then tested with lime water (lime water turns milky)

Hydrogen present in the compound is oxidised to water, which is then tested with anhydrous copper sulphate (anhydrous copper sulphate turns blue)

Detection of Other Elements (N, S, P and Halogens)

Lassaigne’s Test

Elements (N, S, P and halogens) present in the organic compound are converted from the covalent form to the ionic form by fusing the compound with Na metal.

NaCN, Na₂S and NaX, so formed, are extracted from the fused mass by boiling it with distilled water. The extract obtained is known as sodium fusion extract.
Test for Nitrogen (N)

Sodium fusion extract is boiled with FeSO₄.
It is then acidified with sulphuric acid

Formation of Prussian blue colour confirms the presence of nitrogen.

Test for Sulphur (S)

Sodium fusion extract is acidified with acetic acid and lead acetate is added to it.

Formation of black precipitate of lead sulphide indicates the presence of sulphur.

When sodium fusion extract is treated with sodium nitroprusside, violet colour appears.

Appearance of violet colour indicates the presence of sulphur.

When both N and S are present in an organic compound, sodium thiocyanate (NaSCN) is formed.

NaSCN then gives blood red colour, and not Prussian blue, since there are no free cyanide ions.

Test for Halogens

Sodium fusion extract is acidified with HNO₃ and then treated with AgNO₃.

A white precipitate (soluble in ammonium hydroxide) indicates the presence of chlorine.
A yellowish precipitate (sparingly soluble in ammonium hydroxide) indicates the presence of bromine.
A yellow precipitate (insoluble in ammonium hydroxide) indicates the presence of iodine.

Test for Phosphorus

Compound is heated with an oxidising agent (Sodium peroxide).
Phosphorus is then oxidised to phosphate.
The solution of phosphate is then boiled with HNO₃ and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

Percentage of elements in an organic compound

Once the elements have been detected in an organic compound, it is essential to determine their percentages in the compound to deduce its molecular formula. In the coming section various methods methods are discussed to determine the percentage of elements in organic compound.

Carbon and Hydrogen

Organic compound of known mass is burnt in the presence of excess of oxygen and copper (II) oxide

Carbon and hydrogen (present in the compound) gets oxidised to CO₂ and water respectively.

The mass of water produced can be determined by passing the mixture through a weighed U-tube containing anhydrous CaCl₂.
U-tube containing concentrated solution of KOH absorbs CO₂.
The increase in the mass of CaCl₂ and KOH gives the amounts of water and CO₂, from which the percentages of C and H can be calculated as follows:

Mass of the organic compound = m g

Mass of H₂O = m₁g

Mass of CO₂ = m₂g

∴Percentage of carbon =

And, percentage of hydrogen =

Nitrogen

Dumas Method

When heated with CuO in an atmosphere of CO₂, the nitrogen-containing organic compound yields free nitrogen, CO₂ and H₂O.

The mixture of produced N₂ and CO₂gases is collected over an aqueous solution of KOH so that CO₂ is absorbed by it.
Nitrogen gas is collected in the upper part of the graduated tube.

Mass of the organic compound = m g

Volume of nitrogen collected = V₁ mL

Room temperature = T₁K

Volume of nitrogen at STP =

= V mL (Suppose)

Where,

p₁ = Pressure of nitrogen

V₁ = Volume of nitrogen

p₁= Atmospheric pressure − Aqueous tension

22400 mL N₂ at STP weighs 28g.

V mL N₂ at STP weigh = g

∴Percentage of nitrogen

Kjeldahl’s method

Compound containing nitrogen is heated with conc.H₂SO₄.
Nitrogen gets converted into (NH₄)₂SO₄.
Resulting acidic mixture is heated with an excess of NaOH.
The liberated NH₃ gas is absorbed in an excess of a standard solution of H₂SO₄.
Amount of NH₃ produced is determined by estimating the amount of H₂SO₄ consumed in the reaction, which is calculated by subtracting the un-reacted amount of H₂SO₄ from its total amount.

Calculations:

Mass of the organic compound = m g

Volume of H₂SO₄ of molarity M taken = V mL

Volume of NaOH of molarity M used for titration of excess of H₂SO₄ = V₁mL

V₁mL of NaOH of molarity M

mL of H₂SO₄ of molarity M

Volume of H₂SO₄ of molarity M unused = mL

mL of H₂SO₄ of molarity M

= 2mL of NH₃ solution of molarity M

1000 mL of 1 M NH₃ solution contains 17g NH₃ or 14g of N.

mL of NH₃ solution of molarity M contains

g of N.

∴Percentage of N

This method is not applicable to the compounds containing nitrogen in nitro and azo groups; and to the nitrogen-containing ring (pyridine). This is because the nitrogen present in these compounds does not change to (NH₄)₂SO₄ under these conditions.

Halogens (Carius Method)

An organic compound (of known mass) is heated with fuming HNO₃ in the presence of AgNO₃ (contained in a Carius tube) in a furnace. (shown in the figure)

Halogens present in the compound form the corresponding silver halide (AgX).
AgX is filtered, washed, dried and weighed.
Calculations:

Mass of the organic compound = m g

Mass of AgX formed = m₁g

1 mol of AgX contains 1 mol of ‘X’

Mass of halogen in m₁g of AgX

Percentage of halogen

Sulphur

An organic compound (of known mass) is heated in a Carius tube with sodium peroxide or fuming HNO₃.
Sulphur present in the compound gets oxidised to H₂SO₄, and it gets precipitated as BaSO₄ by adding excess of BaCl₂ solution in water.
The precipitate is filtered, washed, dried and weighed.
Calculations:

Mass of the organic compound taken = m g

Mass of the barium sulphate formed = m₁g

1 mole of BaSO₄ = 233 g of BaSO₄ = 32 g of sulphur

m₁g of BaSO₄ contains g of sulphur.

Percentage of sulphur

Phosphorus

An organic compound (of known mass) is heated with fuming HNO₃.
Phosphorus present in the compound is oxidised to phosphoric acid, which is precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄.12 MoO₃ [by adding NH₃ and ammonium molybdate].
Calculations:

Mass of the organic compound taken = m g

Mass of (NH₄)₃PO₄.12 MoO₃ = m₁g

Molar mass of (NH₄)₃PO₄.12 MoO₃ = 1877 g

Percentage of phosphorus

Oxygen

Usually estimated (in terms of percentage) by the difference between the total percentage-composition (100) and the sum of the percentages of all other elements.
Can be estimated directly as well: If an organic compound (of known mass) is heated in a stream of N₂ gas, it gets decomposed to form O₂ gas and other gaseous products.

- Two moles of CO₂ are obtained from one mole of O₂, i.e., 88 g of CO₂ is obtained if 32 g of O₂ is liberated.
Calculations:

Mass of the organic compound = m g

Mass of CO₂ produced = m₁g

∴m₁ g of CO₂ is obtained from

∴ Percentage of O₂

Qualitative Analysis of Inorganic Salts

The identification of components present in a given sample is known as qualitative analysis. Unlike quantitative analysis, this is concerned only with the identity and not the amount of the substance. In qualitative analysis of an inorganic sample, presence of cations and/or anions is confirmed.

Qualitative analysis involves two kinds of tests. They are:

Dry tests

General preliminary tests
Applicable to solid substances
Carried out before wet analysis
Indicate the presence or absence of ions within a short time interval

Some of the important dry tests are:

Heating
Colour change
Solubility
Flame tests
Spectroscopic tests
Borax bead tests

Wet analysis

It involves dissolving the substance in an appropriate solvent or reagent. The solution is then analysed for the presence of anions and/or cations.

Qualitative analysis of some cations and anions are given below:

Test for Cations

Group	Cation	Group reagent	Reaction	Colour
Zero		Nessler’s reagent	With Nessler’s reagent + OH^-+ [HgI₄]^2- ? NH₂.HgO.HgI	Brown
I	Pb²⁺	dil. HCl	With dil. HCl	White
II A	Cu²⁺	H₂S in presence of dil. HCl	With H₂S	Black
II A	Pb²⁺		With H₂S	Black
II B	As³⁺		With H₂S	Yellow
III	Fe³⁺	NH₄OH in presence of NH₄Cl	With NH₃	Reddish Brown
IV	Co²⁺	H₂S in presence of NH₄Cl	With H₂S	Black
	Ni²⁺		With H₂S	Black
	Mn²⁺		With H₂S	Pink
	Zn²⁺		With H₂S ZnS dissolves in dil HCl ZnS + 2HCl ? ZnCl₂+ H₂S With aq. NH₃	White
V	Ba²⁺	(NH₄)₂ CO₃ in presence of NH₄OH	(NH₄)₂ CO₃solution	White
	Ca²⁺		(NH₄)₂ CO₃solution
	Sr²⁺		(NH₄)₂ CO₃solution
VI	Mg²⁺	No common group reagent	With ammonium solution With NaOH solution	White gelatinous

Test for Anions

Anions of class I
Sulphide ions (S²^-)	Confirmatory test	Add sodium nitroprusside solution to sodium carbonate extract.
	Observation	A purple or violet colour confirms the presence of sulphide ions.
	Reaction
Sulphite ions ()	Confirmatory test	Add BaCl₂ solution to sodium carbonate extract. A white precipitate indicates the presence of, ions in the solution. Filter the solution and add a few drops of KMnO₄ solution, and then acidify with dil. H₂SO₄.
	Observation	Disappearance of pink colour of KMnO₄ confirms the presence of ions.
	Reaction
Carbonate ions ()	Confirmatory test	Add dilute hydrochloric acid to sodium carbonate extract
	Observation	Bubbles of gas that turn lime water milky confirm the presence of carbonate ions.
	Reaction
Nitrite ions ()	Confirmatory test	Dilute water extract with distilled water. Add acetic acid to make the solution acidic. Cool and add a few drops of freshly prepared 0.2 M FeSO₄ solution to this.
	Observation	Appearance of a brown colour confirms the presence of nitrite ions.
	Reaction
Acetate ions ()	Confirmatory test	Take salt and add conc. H₂SO₄ and a few drops of ethyl alcohol to it. Heat the content.
	Observation	Fruity smell confirms the presence of acetate ions.
	Reaction
Anions of class II
Test for halogens
Chloride ions ()	Confirmatory test	Acidify sodium carbonate extract with dil. HNO₃. Add AgNO₃ solution.
	Observation	Formation of a curdy white precipitate, which is soluble in aqueous ammonia, confirms Cl^? ions.
	Reaction
Bromide ions ()	Confirmatory test	Acidify sodium carbonate extract with dil. HNO₃. Add AgNO₃ solution.
	Observation	Formation of a light yellow precipitate, which is partially soluble in aqueous ammonia solution, confirms Br^? ions
	Reaction
Iodide ions ()	Confirmatory test	Acidify sodium carbonate extract with dil. HNO₃. Add AgNO₃ solution.
	Observation	Formation of pale yellow precipitate, which is insoluble in aqueous Ammonia, confirms the presence of I^? ions.
	Reaction
Oxalate ions ()	Confirmatory test	Acidify sodium carbonate extract with dil. acetic acid and add CaCl₂ solution.
	Observation	White precipitate confirms the presence of oxalate ions.
	Reaction
Nitrate ions	Confirmatory test	To water extract, add concentrated sulphuric acid and mix well. Cool the mixture. Add a few drops of saturated solution of FeSO₄ down the side of the test tube.
	Observation	A brown ring will be formed at the zone of contact of the two liquids.
	Reaction
Anions of class III
Sulphate ions ()	Confirmatory test	Acidify sodium carbonate extract with dil. HCl. Add BaCl₂ solution.
	Observation	Appearance of a white precipitate, which is insoluble in conc. HCl and conc. HNO₃_, confirms the presence of ions.
	Reaction
Phosphate ions ()	Confirmatory test	Acidify the sample by adding conc. nitric acid in excess. Treat the solution with ammonium molybdate reagent and warm it.
	Observation	Formation of a yellow crystalline precipitate confirms the presence of phosphate.
	Reaction