Organic Chemistry Purification Methods PDF

very andwhich are 12.7 compounds Organic uantitative seldom mportant are METHODS producedpure; to purify analysis they along when OF a...

very andwhich are 12.7 compounds Organic uantitative seldom mportant are METHODS producedpure; to purify analysis they along when OF are them: PURIFICATION usuallyofisolated with from the the organic contaminated desired natural sourcesor compounds, product. OF with ORGANIC Before insmall order preparedby amounts carrying to COMPOUNDS characterise out of organic other the compounds qualitative them, reactions it is 566 Purification CHEMISTRY-XI The common techniques used for the its nature and also on the nature of thepurification of aparticular compound are based for this purpose are: impurities present in it. The various methods u (i) Sublimation (iü)Crystallisation (iiü)Distillation (iv) Differential extraction (w)Chromatography. ()Sublimation Sublimation is a process of conversion of a solid into gaseous passing through the intermediate liquid state and vice state on heating without versa. Heat Cool Solid Vapour ’ Solid This process is used for the separation of from the non-volatile solids. Iodine, volatile solids, which sublime on heating by this method. camphor, naphthalene, benzoic acid, etc., are purified Sublimate Asbestos sheet Crude substance Fig. 12.1. Sublimation. (iü) Crystallisation Crystallisation is a process of This is the most comnmon waysolidification of a pure substance from its of purifying organic solids. This dissolved state. differences their solubility in a given solvent or a mixture of in method is based upon organic compounds are not fairly soluble in water, other solyents. Since most of the chloroform, ether, etc., are also commonly used for this solvents such as alcohol, acetone, steps given below: purpose. The process involves the 1. Selection of the solvent 2. Preparing the solution 3. Filtration 4. Crystallising 5. Separation and drying of crystal  567 (iii) Distillation Distillation is a process ofconversion of a liquid into vapours by heating followed by condensation of vapours so produced by cooling.The method is used for the purification of 1Buids which boil without decomposition and are associated with non-volatile impurities. The impure liquid is boiled and the vapours, thu8, formed are condensed to get. the pure liquid. Evaporation Condensntion Liquid Vapour >Liquid Types of Distillation: 1. Simple distillation NcER, 2. Fractional distillation 3. Distillation under reduced pressure 4. Steam distillation. Air Condenser To sink Water condenser Impure liquid From tap Pure liquid Fig.12.2. Simple distillation. (iv) Differential Extraction Organic compounds, whether solids or liquids, can be recovered from aqueous solutions by Shaking the solution in a separating funnel with a suitable organie solvent which is Immiscible with water but in which the organic compound is highly soluble. This process 1S known as extraction or solvent extraction. Some commonly employed solvents are, euner, benzene and chloroform. The process is carried out as follows:  568 The aqueous solution is CHEMISTRY--X mixed with a small quantity of the Solvent organic solvent in aseparating layer funnel Fig. 12.3. The funnel is stoppered and its contents shaken Organic for some time when the organic Compound in solvent dissolves out the solute. solvent layer The mixture is now allowed to settle and in this way solvent and Aqueous layer water form two separate layers. Organic The lower aqueous layer is run out (H o Compound in aqueous layer by opening the tap and the solvent layer is collected separately. The whole process may be repeated to remove the solute completely from (a) Before extraction (b) After extraction the aqueous solution. The solute is finally recovered from the Fig. 12.3. Differential extraction. organic solvent by distilling off the latter. (v) Chromatography Chromatography in general, may be described as the ents of a mixture by the differential movement of technique of separating the constitu stationary phase under the influence of mobile phase.individual In fact components through the chromatography derives its name from the word chroma which means colour. Different types of chromatographic techniques are employed depending upon the nature of stationary and mobile phases. Two of the important categories of (A)Adsorption chromatography chromatography are: Solvent (B)Partition chromatography Mixture of compounds A. Adsorption Chromatography This category of chromatography is based on the principal of differential Adsorbent adsorption of various components of the mixture on a (Stationary adsorbent. Some components are more strongly absorbed thansuitable others. phase) When the mobile phase is allowed to move over the stationary phase (adsorbent), the components of the mixture move by varying over the stationary phase. Two main types of distances chromatographic techniques based on the principle of differential adsorption are: Glass (i) Column chromatography Wool (üü)Thin layer chromatography. plug (i)Column involves separation ofChromatography: a mixture over a columnColumn chromatography of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at Fig. 12.4. Colun its lower end. The mixture adsorbed of the adsorbent column packed in a on adsorbent is placed on the top chromatography. glass tube. An appropriate eluent which is aliquid or a mixture ofliquid is allowed to flow downthe column slowly.. Depending upon the degree to which the compounds are The most readily adsorbed substances are retained adsorbed, complete Separation takes place. near the ton and others come down various distances in the column.  This technique is now widely used Solvent Separaion in research laboratories for the added beginning Continued elution nurification of different substances and for the separation of mixtures. For example, a mixture of alcohol and a liquid hydrocarbon may be separated by using alumina as adsorbent and petroleum ether as eluent., (i)Thin Layer Chromatogra phy (TLC): Thin layer chromatography is another type of adsorption chromatog raphy, which involves separation of the substances of a mixture over a thin layer of an adsorbent. Various steps involved are (a) Preparation of TLCplate. Two pencil lines are drawn across the width of the plate about 1 cm each from the top and bottom end. The lower pencil mark First component Second component iscalled the starting line and the upper eluted eluted line is called the finish line or solvent Fig. 12.5. Process of chromatography. line. A thin layer (about 0.2 mm thick)of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable size. The plate is known as thin layer chromatography plate (TLC plate). (ii) Separation process. The solution of the mixture to be separated is applied as asmall spot about 2cm from one end of the TLC plate. The glass plate is then placed in a closed jar containing the solvent (Fig. 12.6). As the solvent in the jar moves up, the componentsof the mixture move up along the plate to different distances depending on their degree of adsorption and separation taking place. The relative adsorption of each component of the mixture is expressed in terms of its retention factor i.e., R, value. Distance moved by the substance from base line R,= Distance moved by the solvent from base line TLC. Plate Developing solvent Fig. 12.6. Development chamber with TLC plate undergoing development. 5/0 R,for compound B= For example, R, for component A = Solvent front A y cm Original 4 spot Xg Cn Starting line Spotted plate Developed plate Fig. 12.7. Thin layer chromatogram. The spots of coloured compounds are visible on TLC plate due to their original colour. Thespots of colourless compounds which are invisible to the eye can be detected by any of the following techniques. (a) U.V light. This method isused for certain organiccompounds which produces fluorescence effect in U.V light. The spots of such substance can be detected by placing under U.V lamp. (b) Use of iodine vapours. In this method, the developed TLC plate is placed in a covered jar containing a few crystals of iodine. Spots of the compounds which absorb iodine will become brown. (c) Chemical methods. This method involves spraying of suitable chemical on the TLCplate. reagents For example, amino acids may be detected by spraying the plate with ninhydrin solution. Similarly, aldehyde/ketones can be detected by spraying the plate with 2, 4-dinitrophenyl-hydrazine. B. Partition Chromatography Partition chromatography is liquid-liquid chromatography unlike TLC and chromatography which represent solid-liguid chromatography. In partition column chromatography, the stationary phase as well as mobile phase are both liquids. Principle. Partition chromatography is based on continuous (distribution) of components of amixture between stationary and mobile differential partitioning common example of partition chromatography is phases. One of the chromatography, a special quality paper known as paper chromatography. ln pape is mainly cellulose, but cellulose does chromatography not constitute stationary paper is used. Paper phase in paper Chromatography is water, phase. The stationdry mobile phase is liquid mixture of two or more trapped or chemically bound to cellulose. 1ne components. substances with water as one 0r e Process. A suitable strip of chromatographic paper is drawn across the width of the paper about 1 or 2 cm from the taken and starting line of components to be separated is bottom, A spot of the mixtue or syringe. The spotted paper is then applied on the starting line with the help of fine capilla suspended in suitable solvent phase. The solvent rises (oramixture (Fig. 12.8.) This solvyent acts as the mobile of solven up the paper bycapillary  PRINCIPLES AND TECHNIQUES action and flows over the 571 spot. different distances depending The stationary and mobile different components of the mixture travel upon their solubility through in (or partitioning between) the Card board Paper strip Jar Solvent Fig.12.8. Paper phases. When the solvent The paper strip so chromatography. reaches the top end, the paper is taken out and is allowed to dry. coloured compounds developed is are visible atknown as a chromatogram. The spots of the separated chromatogram. The different heights from the position of initial spot on the under ultraviolet lightspots of the separated or by the use of an colourless compounds may be observed either thin layer chromatography. appropriate spray reagent as discussed under 12.8 QUALITATIVE ANALYSIS OF ORGANIC 1. Detection of COMPOUNDS Carbon and Hydrogen Asmall quantity of the dry and powdered oxide.The mixture is heated in a well-driedsubstance hard is mixed with freshly glass ignited cupric tube having a bulb at one end. test-tube provided with a delivery in another test tube. The other end of the tube is Carbon present in the compound is dipped under lime water taken turns lime water milky. The hydrogen present in the oxidised to carbon dioxide which condenses on the cooler parts compound is oxidised to turn white anhydrous copper the test-tube or in the bulb of delivery tube.water of which sulphate placed in the bulb into blue. The drops Dry CuO+Organic substance Water drops turn anhydrous CuSO, blue CO, turns lime water milky Fig. 12.9. Detection of carbon and hydrogen.  572 CHEMISTRY-XI C+ 2Cu0 ’ CO, T + 2Cu Co, + Ca(OH), ’ CaCO, + H,0 c Lime water Milky 2H + CuO ’ H,0 2 + Cu CuSO, + 5H,0 ’ CuS0,.5H,0 White Blue 2. Detection of Nitrogen Lassaigne's Test This test is aconfirmatory test for nitrogen in organic compounds. This test involves twe 4 steps: (a) Preparation ofLassaigne's extract. A small piece of sodium metal is heated gently in a fusion tube till it melts to a shining globule. At this stage, a srnall amount of substance is added and the tube is heated strongly. The red hot tube is plunged into distilled water contained in a china dish. The contents are boiled for some time, cooled and then filtered. The filterate is known as sodium extract or Lassaigne's extract. (6) Test for nitrogen. To a part of sodium extract a small amount of a freshl prepared ferrous sulphate solution is added and the drops of ferric chloride solution are then added to thecontents are warmed. A few solution is acidified with dilute hydrochloric acid. Thecontents and of the resulting blue colouration confirms the presence of appearance a prussian nitrogen in the organic compound. Na + C+N ’ NaCN Fuse From organic compound FeSO, + 2NaCN Fe(CN), +Na,SO4 Fe(CN), + 4NaCN. Na,(Fe(CN)|] Sodium ferrocyanide 3Na, (Fe(CN)J + 4FeCl, Fe,(Fe(CN)J + 12NaCl Ferric ferrocyanide (Blue colour) 3. Detection of Sulphur Lassaigne's Test The Lassaigne's extract is divided into two parts and the (a) Lead acetate test. One part of the extract is following tests are performed: lead acetate solution is added. acidified with acetic acid and theu presence of sulphur in the organicFormation of ablack precipitate confirms u Na,S +Pb(CH,CO0), compound. ’ PbS J+ (6) Sodium Black ppt. 2CH,COONa A nitroprusside added to another test. Afew drops of part of the sodium nitroprusside solution are colouration confirms thepresenceLassaigne' of s extract. The appearance of purple So Na,S +Na,[Fe(CN),NO] sulphur. W Sodium Sodium Na,(Fe(CN),NO.S] H extract nitroprusside Purple colouration ORGANIC CHEMISTRY :SOME BASIC PRINCIPLES AND TECHNIQUES 573 Note. If the organic compound contains both nitrogen and sulphur, sodium sulphocyanide may be formed during fusion. This reacts with ferrie ions to form ferric sulphocyanide which gives ablood red colouration. Na + C+N+S NaCNS From organic Sodium compound sulphocyanide 3NaCNS + Fe3+ Fe (CNS), + 3Nat Ferricsulphocyanide (Blood Red) 4. Detection of Halogens Lassaigne's Test Asmall amount of thecompound is fused with metallic sodium as described earlier. The halogen present in the compound combines with sodium to form sodium halide which passes into the sodium extract, X + Na NaX From Sodium (where X may be Cl, Br or I) compound halide A part of the Lassaigne's extract is boiled with dilute nitric acid to expel all the gases if evolved. The silver nitrate solution is added to the resulting solution. If any of the halogen is present, a precipitate is formed. NaX +AgNO, AgX + NaNO, ppt. (a) Awhite precipitate soluble in ammonium hydroxide solution indicates the presence of chloride in the organic compound. NaCl + AgNO ’ AgCl + NaNO, White ppt. (6) A dull yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromide in the organic compound. NaBr + AgNO, ’ AgBr + NaNO, Dull yellow ppt. (c) Abright yellow precipitate, completely insoluble in ammonium hydroxide solution indicates the presence of iodide in the organic compound. Nal + AgNO, Agl + NaNO, Bright yellow ppt. 5. Detection of Phosphorus Avery few organic compounds contain phosphorus. Its presence isdetected by fusing the Compound with an oxidizing mixture of sodium carbonate and potassium nitrate or with Sodium peroxide alone when an alkali phosphate is formed. The fused mass is extracted with water and filtered. The filtrate containing sodium phosphate is heated with conc. HNO, and an excess of ammonium molybdate solution is added. Ayellow ppt. or colouration 1sformed if phosphorus is present. The yellow precipitate isammonium phosphomolybdate, (NH,),|PMo,04ol or (NH),PO,12 MoO,  574 Na,PO,+ 3HNO, H,PO, +3NaNO, H,PO, +12(NH,),Mo0, +21HNO,NH),PO,-12MoO, +21NH,NO, +12H.0 Ammonium Phospho molybdate 12.9 QUANTITATIVE ANALYSIS 1. Estimation of Carbon and Hydrogen-(Liebig's Method) Carbon and hydrogen in the organic compound are estimated together. Principle Aknown mass of the organic compound is heated with dry copper oxide in an atmosphere of air or oxygen free from moisture and carbon dioxide. The carbon and hydrogen of the organic compounds are oxidized to carbon dioxide and water respectively. C+ 2Cu0 ’ CO, + 2Cu 2H + Cu ’ H,0 + Cu CuO Combustion Sample in tube platinum boat pellets Guard tube Furnace Pure dry Anhydrous Potash bulb Oxygen CaCl, Fig. 12.10. Apparatus for the estimation of C and H. NCERT SOLVED PROBLEM 12.20. On complete combustion, 0.246 g of an organic carbon dioxide and 0.101g of water. compound gave 0.198 g of composition of carbon and hydrogen in theDetermine compound. the percentage Solution. Percentage of carbon 12 xWoo, X 100 44× W. VSubstance 12 x 0.198× 100 44 x 0.246 = 21.95% Percentage of hydrogen 2x WH.0 X100 18X WSubstance 2x0.1014 x100 18 x 0.246 = 4.58%.  575 Estimation of Nitrogen There are two methods in use for the (a)Dumnas' Method. estimation of nitrogen. These are: (6) Kjeldahl's Method. (a) Dumas' Method dis method is applicable to Principle nitrogenous compounds: Aknown mass oT the organic compound is heated an atmosphere of carbon stronglywith excess of copper 0xide in dioxide. The carbon and hvdrogen are oxidised and water respectively and nitrogen is set to carbon dioxide produced, it is reduced to free as dinitrogen. If any oxide of nitrogen is dinitrogen by passing over hot reduced copper spiral. The dinitrogen is collected over the concentrated solution of potassium hydroxide and its volume ismeasured at room temperature atmospheric pressure. The chemical reactions and can be represented as: C+ 2H +3Cu0 ’ CO, + H,0 + 3Cu Organic compound 2N + Cu0 ’ N, + oxides of nitrogen Organic compound Oxides of nitrogen + Cu ’ N, + Cu (Reduced copper gauze) The apparatus used consists of: (a) Arrangement for the supply of carbon dioxide (6) Combustion tube (c) Schiffs nitrometer ; an arrangement for the collection of dinitrogen liberated from the organic compound.The KOH solution present in the nitrometer absorbs CO, as well as H,0 produced during combustion. The apparatus used in Dumas' method hasbeen shown in Fig. 12.11. |Note. Since N, is collected over aqueous KOH solution, it is saturated with water vapour. It is, therefore, essential to correct the pressure of aqueous tension i.e., vapour pressure of water. So, Corrected pressure = Atmospheric pressure - Vapour pressure of water Corrected pressure = P - P: Calculations Let the mass of the organic compound taken = Wg Volume of moist N, collected =Ucmß Barometric pressure =P mm Room temperature =TK Pressure of water vapours at TK =p mm Pressure of dry N = (P- p) mm Step I. Toreduce the volume of N, to S.TP. (standard temperature and pressure) V, =v cm, V, =? P, = (P-p), P = 760 mm, T, = T, T, = 273 K  P,V,_PV, T; V, = T,P,V, T, (P-p) (P- p) vx 273 P; Tx760 Step II. Calculation of percentage of nitrogen. 22400 cm of N, at 8.T.P. weigh8 = 28g (G.M.M.) V, cm of N, at S.T.P. weighs 28V2 22400 Nitrogen Coarse CuO KOH solution CuO + Organic compound Reduced CuO gauze copper gauze Furnace CO, Mercury seal Nitrometer Fig. 12.11. Apparatus for estimation of nitrogen by Duma's method. Now amnount of nitrogen present in Wgof organiccompound 28 V 22400 Percentage of N in organic compound 28 V2 100 22400 W NCERT SOLVED PROBLEM 12.21. In Dumas' method for compound gave 50 mL ofestimation of nitrogen, nitrogen collected at 300 0.3g of an organic Ktemperature and 715 mm pressure. Calculate the percentage Composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm) Solution. Volume of nitrogen collected at 300 Kand 715 mm pressure is50mL Actual pressure = 715 - 15 = 700 mn  Volume of nitrogen at STP 273 ×700 × 50 =41.9 mL 300 x 760 22,400 mLof N, at STP weighs = 28 g 41.9 mL of nitrogen weighs 28 x 41.9 22400 Percentage of nitrogen 28x VN, (STP) x 100 22400 x W.Substance 28 x 41.9 x 100 22400 x 0.3 = 17.46% (6) Kjeldahl's Method This method is simpler and convenient. It is largely used for the estimation of nitrogen in food, fertilizers and drugs. The method is, however, not applicable to compounds containing nitrogen in the ring, like pyridine or quinoline, etc, and the compounds, containing nitro (- NO,) and diazo (-N= N-)groups. Principle A known mass of the organic compound is heated with concentrated sulphuric acid. The nitrogen in the organic compound is quantitatively converted into ammnonium sulphate. The resulting liquid is then distilled with excess of sodium hydroxide solution and the ammonia evolved is passed into a known but excess volume of the standard acid (HClor H,SO,). The acid left unused is estimated by titration with some standard alkali. The amount of acid used against ammonia can thus be known and from this, the percentage of nitrogen in the compound can be calculated. Kjeldahl's trap Water outlet Contents of Organic Kjeldahl's flask compound after digestion + + Conc. +CuSO, H,So,y NaOH Water inlet Kjeldahl's flask Known volume of standard acid Fig. 12.12. Apparatus used for estimation of nitrogen by Kjeldahl's method. 578 Chemical reactions involved are: Conc. C+ H +S ’ CO, + H,0 + SO, H,S04 Organic compound Conc. N (NH), S0, H,SO4 (NH), SO4 + 2NaOH ’ Na,SO, +2NH,+ 21,0 2NH, + H,SO, ’ (NH), SO, The apparatus used in Kjeldahl's method consists of heating of organic (a) A long neck round bottom flask known as Kjeldahl's flask. The compound with sulphuric acid is done in this flask. Distillation of (6)Round bottom flask fitted with condenser through Kjeldahl's trap. contents obtained from Kjeldahl's flask with NaOH is carried out in it. (c) Titration flask. The assembly of apparatus is shown in above figure. Limitation of Kjeldahl's method Kjeldahl's method is not used to estimate nitrogen in nitro and azo compound and the compounds in which the N atom is present in a ring e.g., in pyridine. Calculations Let the mass of organiccompound= Wg Volume of standard acid (say HCI)taken =V cm³ Let molarity of acid = M, Let volume of acid unused be Let volume of alkali (say NaOH) of molarityM, used for neutralising unused acid =v, Chemical equation for titration involved is NaOH + HCI ’ NaCl + H,0 unused According to molarity relation, M,U,M,u1 1 or V1 1 M Volume of acid used by ammonia = (V-v,) cm Millimoles of acid used by ammonia = (V - v,) x M, Millimoles of NH, formed = Millimoles of acid used up = (V -u) x M, Mass of NH, formed = Millimoles x 103 x Molar mass = (V-v,) M, x 10-8 x 17g  Mass of nitrogen (V- v) M, × 10x 17× 14 17 =ag (say) :. Percentage of nitrogen ax 100 W Percentage of nitrogen 1.4× M(ncid) Xbasicity of acid x V(acid used) Wsubstance Or 1.4 x N(acid) XV(acid used) Wsubstance NCERT SOLVED PROBLEM 12.22. During estimation of nitrogen present in an organic Kjeldahl's method, the ammonia evolved from 0.5 g of thecompound by compound in Kjeldah>'s estimation of nitrogen, neutralized 10 mL of 1 MH,SO, Find out the percentage of nitrogen in the Solution. 1Mof 10 mL H,S0, = 1 compound. Mof 20mL NH, 1000mL of 1 M ammonia contains = 14 gnitrogen 20 mL of 1 M 14x 20 ammonia contains g nitrogen 1000 14 x 20x 100 Percentage of nitrogen 1000 x 0.5 = 56.0% 3. Estimation of Halogens-Cariusmethod Principle Aknown mass of the organic substance containing halogen is heated with fuming nitric acid and a few crystals of silver nitrate in a sealed tube called Carius tube. The silver halide so formed is separated, washed, dried and weighed. From the weight of silver halide obtained, the percentage of halogen can be calculated. Carbon and hydrogen are oxidized to carbon dioxide and water respectively. The apparatus used has been shown in Fig. 12.13. Carius tube Iron tube HNO, + AgNO, Organic substance Furnace Fig. 12.13. Heating the Carius tube in a furnace for estimation of halogens.  Calculations Let the mass of the organic Substance taken = w g Mass of silver halide (AgX) formed = x g (108 + At. mass of X) parts by mass ofAgX contain halogen = At. mass ofX parts xg by mass of AgX contain halogen At. mass of X 108 + At. maSs of X Hence percentage of halogen At. mass of X 100 108 + At. mass of X W (i) Percentage of chlorine 35.5 100 108 + 35.5 35.5 x WAgc X100 143.5 XWsubstance (iü) Percentage of bromine 80 100 108 + 80 (iüü) Percentage of iodine 127 100 108+ 127 127 x WAcI X l00 235 x Wsubstance Note. Carius method does not give is slightly soluble in nitric acid. The satisfactory results in case of estimation of iodine because Agl results of Carius method have not been very case of highly halogenated aromatic compounds. accurate even in NCERT SOLVED PROBLEM 12,23. In Carius method of estimation of halogen, 0.15 g of an organic gave 0.12g of AgBr. Find Solution. Molar mass out the percentage of bromine in the compound of AgBr= 108 + 80 = 188 g compound 188 g AgBr contains 80 g bromine mol-1 0.12g AgBr contains 80x 0.12 g bromine 188 Percentage of bromine = 80x WAgBr X 100 188 × W. Substance 80 x 0.12 × 100 188x 0.15 =34.04%  A. Estimation of Sulphur and Phosphorus (a) Estimation of Sulphur: Estimation of sulphur in the carried out by Carius nethod. organic compound is Principle Aknown mass of the organic compound is heated with fuming nitric acid in a The sealed tube. sulphur present is quantitativelyconverted into sulphuric acid. Sulphuric acid thus formed is precipitated as barium sulphate by adding barium chloride. The filtered, washed, dried and weighed. From the mass of barium sulphateprecipitate is percentage of sulphur is calculated. The reactions involved can be formed, the represented as : S 2H + HNO, 4 ’ H,SO, Organic From compound HNO, H,SO, +BaCl, ’ BaSO, J+ 2HCI Calculations White ppt. Let the mass of organic substance taken = Wg Mass of barium sulphate (BaSO,) formed = Xg Now, 233 g (G.M.M.)of BaSO, contain S= 32 g 32x xg of BaSO, contain 233 32x Hence,percentage of sulphur 233 x W x 100 32 x WBaso, X100 Percentageof sulphur 233 x WSubstance (6) Estimation of Phosphorus: A known mass of the organic compound is heated with fuming nitric acid. The phosphorus present in the organic compound is oxidised to phosphoricacid (H,PO).The phosphoric acid, thus, formed is treated with magnesia mixture to get the precipitate of magnesium ammonium phosphate (MgNH,PO,). The precipitate is separated, dried and ignited to get magnesium Pyrophosphate (Mg,P,0,). The chemical reactions involved are: P + 3H + 40 HNO3, H,PO, From organic From compound HNO, H,PO, + Magnesia mixture ’ MgNH,PO, 2MgNH,PO, Mg,P,0, +2NH, +H,0 White ppt.  Caleulations compound=Wg the organic Let thenass of obtnined Mass of Mg,P,0, Mg,P,0, contains P = 62 g Now,222g (G.M.M.) of 62 x Igof Mg,P,0,contain P= 222 compound Percentage of P in organic 62 x x 1000 222 W 62 x W, x 100 WMg,P,0, Percentage of Phosphorus = 222 x Wsubstance NCERT SOLVED PROBLEM 12.24. (i) In sulphur estimation,0.157 g of organic compound gave 0.4813 g of BaSO,. What is the percentage of sulphur in organic compound? (ii)0.092 gof organic compound on heating in carius tube and subsequent ignition gave 0.111g of Mg,P,0,. Caleulate the percentage of phosphorus in organic compound. Solution. (i)Mass of BaSO, =0.4813 g Mass of organic compound =0.157g %S = 32 × WBaSO, X 100 233 × VWsubstance 32 x0.4813 × 100 233 x 0.157 = 42.10 (iü)Mass of organic compound = 0.092 g Mass of Mg,P,0, =0.111g % of P 62 x W,Mg,P,07 x 100 222 xWsubstance 62 x 0.111× 100 222 x 0.092 5. =33.69%, Estimationofof Oxygen The Ifthepercentage oxygen in an organic combined percentage of all other comnpound is generally estimated the difference by the difference. gives the.out directly as follows: elements present in the percentage of oxygen. is lessthan 100, A definite mass of The estimation of compound oxygen can also be carried nitrogen gas. The mixture ofan organic compound is when all the oxygen is gaseous converted products decomposed by to carbon containing oxygen is heating in a stream of coke monoxide. This passed lover red-hotpassed gaseous mixture isthen  583 through warm iodine pentoxide (I,0.) when carbon producing iodine. monoxideis oxidised to carbon dioxide heat Organic compound O, + other gaseous 1373 K products 2C + O, ’ 2CO I,0, + 5CO 1, + 500, The percentage of oxygen can be derived from the produced. amount of carbon dioxide or iodine Caleulations Let the mass of organic compound taken = Wg Mass of carbon dioxide =xg 44 g carbon dioxide contains oxygen = 32 g xg carbon dioxide contains 32 x oxygen = 44 Now, from the reaction, CO + 0.CO.. CO,, if amount of oxygen present in CO is 32 parts, the oxygen present in CO is 16 parts..:. Oxygen present in CO= 32 x 16 16x 44 32 44 Percentage of oxygen 16x x 100 16 x Wco, X 100 44W 44 × WSubstance CHN Elemental Analyzer It is an apparatus which is used for a direct determination of the percentages of carbon, hydrogen and nitrogen in an organic compound and automatic experimental techniques. In CHN elemental analyzer a very small quantity of organic sample is required. Therefore,the method is also called microestimation of elements.