CBSE Class 11 Chemistry Notes For Qualitative and Quantitative Analysis Of Organic Compounds

Qualitative Analysis Of Organic Compounds

Since organic compounds are either hydrocarbons or their derivatives, the elements which occur in them are carbon (always present), hydrogen (nearly always present), oxygen (generally present), nitrogen, halogens, sulphur (less commonly present), phosphorus and metals (rarely present). All these elements can be detected by suitable methods.

1. Detection of carbon and hydrogen(C-H)

Detection of carbon and hydrogen Principle:

A small amount of a pure and dry organic compound is strongly heated with dry cupric oxide in a hard glass test tube when carbon present in it is oxidised to carbon dioxide and hydrogen is oxidised to water.

Liberated CO₂ turns lime water milky and the liberated water vapours (H₂O) which get condensed in the bulb of the delivery tube, turn white anhydrous CuSO₄into blue hydrated copper sulphate (CuSO₄-5H₂O)

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Ca(OH)₂ (Lime Water)+ CO₂→CaCO₃↓ (Milky)+ H₂O

CuSO₄ Anhydrous (white)+ 5H₂O→CuSO₄.5H₂O(Blue)

If the organic compound is a volatile liquid or gas, then the vapours of the compound are passed through heated copper oxide and the presence of C02 and H₂O in the liberated gas can be proved in the same way.

If the organic compound contains sulphur in addition to carbon and hydrogen, then sulphur is oxidised to SO₂ which also turns lime water milky due to the formation of calcium sulphite. In that case, the resulting gases are first passed through an acidified solution of potassium dichromate which absorbs SO₂ and then through lime water.

2. Detection of nitrogen

1. Sodalime test

A very small amount of an organic compound is strongly heated with soda lime (NaOH + CaO) In a test tube. The evolution of ammonia having a typical smell Indicates the presence of nitrogen in the compound.

Example:

CH₃CONH₂ (Acetamide)+ [NaOH + CaO]→ CH₃COONa (Sodium acetate)+ NH₃↑

 Sodalime test Limitation:

Organic compounds containing nitrogen as nitro ( — NO₂) or azo (— N =N— ) groups do not evolve NH₃ upon heating with soda lime, i.e., do not give this test.

2. Lassaigne’s test

Preparation of Lassaigne’s extract or sodium extract:

A pea-sized freshly cut dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. A small amount ofthe organic compound is added and the tube is heated strongly for 2-3 minutes till it becomes red hot. The hot tube is then plunged into 10-15 mL of distilled water taken in a mortar. The mixture is then ground thoroughly by a pestle and filtered. The filtrate is known as sodium extract or Lassaigne’s extract.

Test for nitrogen:

The Lassaigne extract is usually alkaline in nature because the excess of metallic sodium reacts with water to form sodium hydroxide. A small amount of freshly prepared FeSO₄ solution is added to a part of sodium extract and the contents are boiled when a light green precipitate of Fe(OH)₂ is obtained.

The mixture is then cooled and acidified with dil.H₂SO₄. The immediate appearance of blue or green colouration or deep blue coloured precipitate of Prussian blue, Fe₄[Fe(CN)₆)₃ indicates the presence of nitrogen.

Chemistry of Lassaigne’s test:

1. When organic compound is fused with metallic Na, carbon and nitrogen present in it combine with Na to form NaCN

2. When the sodium extract is heated with FeSO₄ solution, Fe(OH)₂ is obtained which reacts with sodium cyanide to form sodium ferrocyanide. This is also obtained when FeSO₄ reacts with NaCN.

FeSO₄ + 2NaOH →Fe(OH)₂ + Na₂SO₄

6NaCN + Fe(OH)₂ →Na₄ [Fe(CN)₆ ] (Sodium ferrocyanide) + 2NaOH

FeSO₄ + 6NaCN→Na₂ SO₄ + Na₄ [Fe(CN)₆ ]

Ferrous (Fe²⁺) ions present in hot alkaline solution undergo oxidation by O₂ to give ferric (Fe³⁺) ions. These ferric ions then react with sodium ferrocyanide to produce ferric ferrocyanide (prussian blue)

4Fe(OH)₂ + O₂  + 2H₂O→4 Fe(OH)₃

2Fe(OH)₃ + 3H₂ SO₄ →Fe₂(SO₄ )₃ + 6H₂O

3Na₄ [Fe(CN)₆] + 2Fe₂(SO₄)₃→ Fe₄[Fe(CN)₆]₃  (Prussian Blue)+ 6Na₂SO₄

1. In this test, it is desirable not to add FeCl₃ solution because yellow FeCl₃ causes Prussian blue to appear greenish. For the same reason, the alkaline extract should not be acidified with hydrochloric acid (which produces ferric chloride).

2. When the compound under investigation contains both nitrogen and sulphur, it may combine with sodium during fusion to form sodium thiocyanate (sulphocyanide). This gives blood red colouration with ferric chloride due to the formation of ferric thiocyanate. Thus, Prussian blue is not obtained.

 

⇒ \(\mathrm{Fe}^{3+}+3 \mathrm{NaSCN} \rightarrow \mathrm{Fe}(\mathrm{SCN})_3+3 \mathrm{Na}^{+}\)

3. If fusion is carried out with excess of sodium, the resulting thiocyanate decomposes to give sodium cyanide and sodium sulphide. So, in that case no blood red colouration is visualised.

NaSCN + 2Na→NaCN + Na₂S

However, in that case, black precipitate of FeS is obtained when FeS04 solution is added to sodium extract. The black precipitate dissolves on the addition of dil. H₂SO₄ and the test of nitrogen is then performed with that clear solution.

FeSO₄ + Na₂S→FeS↓ (Black) + Na₂SO₄

FeS + H₂SO₄ →H₂S↑+ FeSQ₄

Lassaigne’s test Limitations:

• Volatile organic compounds if present, escape before reacting when fused with metallic sodium.
• Some nitro compounds may lead to explosion during the fusion process.

3. Detection of sulphur

1. Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen Sulphur present in the compound (which does not contain nitrogen) reacts with sodium metal to form sodium sulphide:

The following tests are then performed with the extract to detect the presence of sulphur

Lead acetate test:

One part ofthe extract is acidified with acetic acid and then a lead acetate solution is added to it. The formation of a black precipitate of lead sulphide (PbS) confirms the presence of sulphur in the compound

Na₂S + Pb(CH₃COO)₂→PbS↓(Black) + 2CH₃COONa

Sodium nitroprusside test:

A few drops of sodium nitroprusside solution are added to another part of the extract. The appearance of a violet or purple colouration confirms the presence of sulphur in the compound.

Na₂S + Na₂[Fe(CN)₅NO](Sodium nitroprusside)→ Sodium sulphonitroprusside Na₄[Fe(CN)₅NOS](Violet)

2. Oxidation test:

The organic compound is fused with a mixture of potassium nitrate and sodium carbonate and as a result, sulphur, if present, gets oxidised to sodium sulphate.

The fused mass is extracted with distilled water and filtered. The filtrate is acidified with dil. HCl and then barium chloride solution is added to it. The formation of a white precipitate insoluble in hydrochloric acid indicates the presence of sulphur in the compound

Na₂SO₄ + BaCl₂→2NaCl + BaSO₄↓ (White)

4. Detection of halogens

1. Beilstein’s test:

A clean and stout Cu-wire flattened at one end is heated in the oxidising flame of a Bunsen burner until it imparts any green or bluish-green colour to the flame. The hot end of the Cu-wire is then touched with the organic compound under investigation and is once again introduced into the flame. The reappearance of green or bluish-green flame due to the formation of volatile copperhalide indicates the presence of halogens in the compound.

Beilstein’s test Limitations:

Many halogen-free compounds,

For example: 

Certain derivatives of pyridine and quinoline, purines, acid amides, urea, thiourea, cyano compounds etc. give this test presumably owing to the formation of volatile copper cyanides.

Therefore, this test is not always trustworthy. It does not indicate which halogen (Cl, Br or I) is present in the organic compound.

The test is not given by fluoro compounds since copper fluoride is non-volatile.

2.  Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen. During fusion, Na combines with the halogen present in the compound to form sodium halide

The extract is then boiled with dilute HNO₃, cooled and a few drops of AgN03 solution are added to it White or yellow precipitation confirms the presence of halogen.

NaX + AgNO₃ → AgX↓ + NaNO₃ [X = Cl,Br,I]

1. Formation of a curdy white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound

NaCl + AgNO₃→AgCU(White) + NaNO₃

AgCl + 2NH₄OH→[Ag(NH₃)₂]Cl(Soluble) + 2H₂O

The formation of a pale yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound.

NaBr + AgNO₃→ AgBr₄-(Pale yellow) + NaNO₃

AgBr + 2NH₄OH → [Ag(NH₃)₂]Br(Soluble) + 2H₂O

The formation of a yellow precipitate insoluble in NH4OH solution indicates the presence of iodine in the organic compound.

Nal + AgNO₃→AgI ↓ (Yellow) + NaNO₃

Function Ol nitric acid:

If the organic compound contains nitrogen and sulphur along with halogens, Lassaigne’s extract contains sodium sulphide (Na₂S) and sodium cyanide (NaCN) along with sodium halide (NaX). These will form a precipitate with silver nitrate solution and hence will interfere with the test.

Na₂S + 2AgNO₃→Ag₂S↓  (Silver sulphide (Black) + 2NaNO₃

NaCN + AgNO₃→ AgCN↓   (Silver cyanide) (white)+ NaNO₃

For this reason, before the addition of AgNO₃ solution to the sodium extract for the detection of halogens, the extract is boiled with dilute HNO₃ which decomposes sodium sulphide and sodium cyanide to vapours of H₂S and HCN respectively

Alternatively, sulphide and cyanide ions can be removed by adding 5% nickel (II) nitrate solution which reÿct? with these ions forming precipitates of nickel (II) sulphide and nickel (II) cyanide

3. Chlorine water test for bromine and iodine:

• A portion of Lassaigne’s extract is boiled with dilute nitric acid or dilute sulphuric acid to decompose NaCN and Na₂S.
• The solution is then cooled, and acidified with dil. H₂SO₄ and a few drops of carbon disulphide or carbon tetrachloride solution are added to it.
• The resulting mixture is then shaken with a few drops of freshly prepared chlorine water and allowed to stand undisturbed for some time.

An orange or brown colouration in the carbon disulphide or carbon tetrachloride layer confirms the presence of bromine, whereas a violet colouration in the layer confirms the presence of iodine in the compound

2NaBr + Cl₂→2NaCl + Br₂ (turns CS₂ or CCl₄ layer orange)

2NaI + Cl₂→2NaCl + I₂ (turns CS₂ or CCl₄ layer violet)

5. Detection of phosphorus

1. The organic compound under investigation is fused with sodium peroxide (an oxidising agent) when phosphorus is oxidised to sodium phosphate

2. The fused mass is then extracted with water, filtered and the filtrate is boiled with concentrated nitric acid. The mixture is then cooled and an excess of ammonium molybdate solution is added to it

3. The appearance of a yellow precipitate or colouration due to the formation of ammonium phosphomolybdate, (NH₄ )₃ PO₄ -12MoO₃, confirms the presence of phosphorus in the compound.

Quantitative Analysis Of Organic Compounds

After detecting the presence of various elements in a particular organic compound, the next step is to determine their respective percentages. This is known as quantitative analysis.

1. Estimation of carbon and hydrogen

Both carbon and hydrogen present in an organic compound are estimated by Liebig’s method.

Liebig’s method

Liebig’s method Principle:

A pure and dry organic compound of known mass is heated strongly with pure and dry copper oxide (CuO) in an atmosphere of air or oxygen-free from carbon dioxide. Both carbon and hydrogen present in the organic compound undergo complete oxidation and get converted into carbon dioxide and water respectively

COthus produced is absorbed in a previously weighed potash bulb containing a strong KOH solution while water produced is absorbed in a previously weighed U-tube containing anhydrous CaCl₂. The U-tube and the bulb are weighed again and from the difference between the two weights, the amount of CO₂ and HaO are determined.

Liebig’s method Procedure:

• The apparatus for the estimation of C and H is The apparatus consists of the following units:
  • Combustion tube,
  • U-tube containing anhydrous CaCl₂ and
  • a bulb containing strong KOH solution.
• The tube is heated l strongly for 2-3 hours till the whole of the organic compound is burnt up.
• After combustion is over, the absorption units (the Utube and the potash bulb) are disconnected and weighed separately

Results and calculations:

Let the mass of organic compound taken = w g. The increase in the mass of the potash bulb, i.e., the mass of CO₂ formed = x g.

The increase in the mass of the U-tube, i.e., the mass of water formed =y g.

Percentage ot carbon: 1 mol of CO₂ (44g) contains 1 gram atom of carbon (12 g).

∴ x g of CO₂ contains = \(\frac{12x}{44}\) g of C

Now, \(\frac{12x}{44}\) g carbon is present in w g organic compound.

∴ The percentage of carbon in the compound

= \(\frac{2 y \times 100}{18 w}=\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Percentage ot hydrogen:  1 mol of water (18g) contains 2 gram-atom hydrogen (2 g).

∴  y g of H₂ O contains = \(\frac{2y}{18}\) g of H

Now, \(\frac{2y}{18}\) hydrogen is present in w g organic compound.

∴ The percentage of hydrogen in the compound

Estimation of C and H by Liebig’s method is suitable for organic compounds containing C, H and O only, but it requires some modifications for compounds containing nitrogen, halogens and sulphur.

1. Compounds containing nitrogen:

During combustion, N present in the organic compound undergoes oxidation to give its oxides (NO, NOz etc.) which are

Also absorbed in an alkali solution along with CO₂. Oxides of nitrogen are decomposed back to nitrogen by placing a copper gauze roll near the exit of the tube. The N₂ so produced is not absorbed by the alkali solution.

2. Compounds containing halogens:

During combustion, halogens present in the organic compound get converted into volatile copper halides which partly decompose to give free halogens. These halogens and volatile copper halides get dissolved in an alkali solution. This can be prevented by placing a silver gauze roll near the exit of the combustion tube. Halogens combine with silver to give non-volatile silver halides.

2Ag + X₂ →2AgX; 2Ag + CuX₂ → 2AgX + Cu

3. Compounds containing only sulphur or sulphur and halogen:

During combustion, elemental sulphur present in the organic compound is oxidised to SO₂ which is absorbed in the potash bulb.

This can be prevented by placing a layer of fused lead chromate near the exit of the tube. SO₂ combines with lead chromate to produce non-volatile lead sulphate.

Lead chromate also reacts with copper halides and free halogens to form lead halides which remain in the combustion tube

Liebig’s method Precautions:

• All the joints ofthe combustion must be air-tight.
• The combustion must be free from CO₂ and water vapour.
• The airflow is controlled in such a way that only 2-3 bubbles are generated per second. A faster flow of air may lead to the formation of carbon monoxide.
• If carbon is deposited on the surface of the combustion mbe, then oxygen instead of air is to be passed at the final stage ofthe process

Example 1: 0.90g of an organic compound on complete combustion yields 2.20 g of carbon dioxide and 0.60 g of water. Calculate the percentages of carbon and hydrogen in the compound.
Answer:

Mass of the organic compound = 0.90 g

Mass of CO₂ formed = 2.2 g & mass of H₂O formed = 0.6 g

Percentage of carbon: 44 g CO₂ contains = 12 g carbon

0.90 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g of carbon

0.20 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g carbon

Percentage of carbon = \(\frac{12 \times 2.20 \times 100}{44 \times 0.90}\)

= 66.67

Percentage of hydrogen:  18 g of H₂O contains = 2g of hydrogen

0.60 g of H₂O contains = \(\frac{2 \times 0.60}{18}\) of hydrogen

0.90 g compound contains = \(\frac{2 \times 0.60}{18}\) g hydrogen

Percentage of hydrogen = \(\frac{2 \times 0.60 \times 100}{18 \times 0.90}\)

= 7.41

2. Estimation ot nitrogen

The following two methods are commonly used for the estimation of nitrogen in an organic compound

1. Duma’s method
2. Kjeldahl’s method.

1. Dumas method

Dumas method Principle:

A weighed amount of the organic compound is heated with an excess of copper oxide in an atmosphere of CO₂. Carbon and hydrogen present in the compound are oxidised to CO₂ and H₂O respectively while nitrogen is set free as nitrogen gas (N₂). If any oxide of nitrogen is formed during this process, it is reduced back to nitrogen by passing over hot reduced copper gauze

When the gaseous mixture thus obtained is passed through a 40% KOH solution taken in a Schiff’s nitrometer, all gases except nitrogen are absorbed by KOH. The volume of nitrogen collected over the KOH solution is noted and from this, the percentage of nitrogen in the compound is calculated

Dumas method Procedure:

• The apparatus used for the estimation of nitrogen by the Dumas method ). It consists of:
  • CO₂ generator combustion tube (a glass tube of diameter 15m/r and length 90 cm) and Schiff’s nitrometer.
  • The combustion tube is packed with [a] coarse CuO which prevents backward diffusion of gases produced,
  • an accurately weighed amount of organic compound of approximately 0.2g mixed with excess of CuO.
  • Coarse CuO and
  • A reduced copper gauze can reduce any oxides of nitrogen back to N₂.
• The Schiff’s nitrometer (connected to the combustion tube) consists of a graduated tube with a reservoir and a tap at the upper end. It has a mercury seal at the bottom to check the backward flow ofthe KOH solution.
• CO₂ generated by heating NaHCO₃ or MgCO₃ is dried by passing through concentrated HS04 and then passed through the combustion tube to displace the air present in the tube.
• The combustion tube is then heated in the furnace. When the combustion is complete, a rapid stream of CO₂ is passed through the tube to sweep away the last traces of N₂.
• The volume of nitrogen collected over the KOH solution in the nitrometer tube is recorded after careful levelling (by making the level of KOH solution both in the tube and reservoir the same).
• The room temperature is recorded and the vapour pressure of the KOH solution at the experimental temperature is noted in the table.

Results and calculations:

Let the mass of the organic compound taken = w g, the volume of N₂ gas collected = V mL, the atmospheric pressure = P mm, the room temperature = t°C and the vapour pressure of KOH solution at t°C =f mm. Hence, the pressure of dry N₂ gas = (P- f) mm.

By applying the gas equation = \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Volume of N₂ at STP, = \(V_2=\frac{P_1 V_1 T_2}{P_2 T_1}=\frac{(P-f) \times V \times 273}{760(273+t)}\)

Percentage of nitrogen: 22400 mL of N₂ gas at STP will weigh = 28 g.

2 mL of N₂ gas at STP will weigh = \(\frac{28 \times V_2}{22400} \mathrm{~g}\)

Percentage of nitrogen:

= \(\frac{\text { Mass of nitrogen }}{\text { Mass of the compound }} \times 100\)

= \(\frac{28}{22400} \times \frac{V_2}{w} \times 100\)

The amount of nitrogen estimated by this method is 0.2% higher than the actual quantity. This is because a fraction of the small amount of air (nitrogen) entrapped in copper oxide gets deposited in the nitrometer

Example:

In Dumasg of an method organic for compound, the estimation gave 31.7 of nitrogen, mL of nitrogen at 14°C and 758 mm pressure. Calculate the percentage of nitrogen in the compound (vapour pressure of water at 14°C = 12 mm.
Answer:

If the volume ofnitrogen at STP is V mL, then

V = \(\frac{31.7(758-12)}{(273+14)} \times \frac{273}{760}\)

= 29. 6 mL

Now, at STP 22400 mL of nitrogen will weigh = 28 g

∴ 29.6 mL nitrogen at STP will weigh = \(\) = 0.037%

∴  Percentage of nitrogen = \(=\frac{0.037 \times 100}{0.1877}\)

= 19.71

2. Kjeldahl’s method

Kjeldahl’s method Principle:

A known mass of a nitrogenous organic compound is digested (heated strongly) with concentrated H₂SO₄ in the presence of a little potassium sulphate (to increase the boiling point of H₂SO₄) and a little CuSO₄ (a catalyst) when the nitrogen present in the compound is quantitatively converted into ammonium sulphate. The resulting mixture is heated with excess of NaOH solution and the ammonia evolved is passed into a known but excess volume of a standard acid (HCl or H₂SO₄ ).

The excess acid left after the neutralisation of ammonia is estimated by titration with some standard alkali

2NaOH + H₂SO₄  →Na₂SO₄ + 2H₂O

Kjeldahl’s Method Procedure

•  The organic compound (0.5-5 g) is heated with a cone. H₂SO₄ in a Kjeldahl’s flask which is a long-necked round-bottomed flask with a loose stopper. A small amount of potassium sulphate and a few drops of mercury (or a little CuS ) are added.
• The reaction mixture is heated for 2-3 hours when carbon and hydrogen present in the compound are oxidised to CO₂ and H₂O respectively while nitrogen is quantitatively converted to ammonium sulphate.
• CO₂ and water vapours escape through the loose stopper while ammonium sulphate remains in the flask.
• The Kjeldahl’s flask is then cooled and the contents of the flask are transferred to a round-bottomed flask.
• The mixture is diluted with water and excess caustic soda solution (40%) is added.
• The flask is then connected to a Liebig condenser through Kjeldahl’s tap. The lower end of the condenser is dipped in a known volume of standard acid (N/10 HCl or H₂SO₄ ) taken in a conical flask.
• The flask is then heated. The evolved ammonia gas is absorbed in the acid solution.

Results and calculations:

Let the mass of the organic compound = w g, volume of acid taken = V₁ mL, normality of acid solution x(N) and volume of x(N) alkali required to neutralise unused acid = V₂ mL.

Now, V₂ mL  × (N) alkali = V₂ mL × (N) acid.

∴ The quantity of acid used for neutralising ammonia = VjmL x(N) acid solution- V2 mL x(N) acid solution = (V₁– V₂) mL x(N) acid solution = (V₁– V₂)mL x(N) ammonia solution.

Now, 1000 mL (N) ammonia solution contains 17 g of NH₃ or 14 g of nitrogen.

Therefore, (V₁–  V₂)mL × (N)

Ammonia solution contains = \(\frac{14 \times\left(V_1-V_2\right) x}{1000}\) g of nitrogen

This amount of nitrogen is present in w g compound

%  of nitrogen = \(\mathrm{n}=\frac{14 \times\left(V_1-V_2\right) x \times 100}{1000 \times w}=\frac{1.4\left(V_1-V_2\right) x}{w}\)

= \(\frac{1.4 \times \text { Vol. of acid used } \times \text { Normality of the acid used }}{\text { Mass of the compound taken }}\)

Kjeldahl’s method  Limitations:

• Nitrogen present in pyridine, quinoline, diazo compound, azo compound, and nitro compound, cannot be converted into ammonium salts by this method.
• So this method does not apply to such compounds. Thus although Dumas’s method applies to all compounds, Kjeldahl’s method has limited applications.

The percentage of nitrogen estimated in this method is not correct.

Kjeldahl’s method Utility:

This experiment can be performed quite easily and quickly. So in those cases where correct estimation of nitrogen content is not necessary

For example: Fertiliser, soil, food¬ stuff etc.), this method has its application.

This method is not troublesome and hence the possibility of error can be minimised by repeating the experiment several times.

Example:  0.257 g of a nitrogenous organic compound was analysed by Kjeldahl’s method and ammonia evolved was absorbed in 50 mL of (N/10) H₂SO₄. The unused acid required 23.2 mL of (N/10) NaOH solution for complete neutralisation. Determine the percentage of nitrogen in the compound.
Answer:

Acid taken = 50 mL (N/10) H₂SO₄ solution

= 5 mL(N) H₂SO₄ solution

Alkali required to neutralise the excess acid

= 23.2 mL (N/10) NaOH = 2.32 mL (N) NaOH

Now, 2.32 mL(N) NaOH = 2.32 mL(N) H₂SO₄ solution

The acid required to neutralise the ammonia evolved

= 5 mL (N) H₂SO₄– 2.32 mL (N) H₂SO₄ solution

= 2.68 mL (N) H₂SO₄ s 2.68 mL (N) NH₃

[v KmL(N) H₂SO₄S VmL(N) ammonia]

1000 mL (N) ammonia solution contains 14g nitrogen

So, 2.68mL (N) ammonia solution contains \(\frac{14 \times 2.68}{1000}\) g nitrogen

So, 0.257g compound contains = \(\frac{14 \times 2.68}{1000}\) g nitrogen

Percentage of nitrogen = \(=\frac{14 \times 2.68 \times 100}{1000 \times 0.257}\)

= 14.6

3. Estimation of halogens: Carius method

Halogens Carius method Principle:

An organic compound containing halogen of known mass is heated with fuming nitric acid and a few crystals of silver nitrate. The halogen present in the compound is converted into the corresponding silver halide (AgX). From the mass of the compound taken and that of the silver halide formed, the percentage of halogen in the compound can be calculated.

Carius method Procedure:

• About 5 mL of fuming nitric acid and 2 g of silver nitrate crystals are placed in a hard glass tube (Carius tube) of about 50 cm in length and 2 cm in diameter.
• A small amount of accurately weighed organic compound is taken in a small tube and the tube is placed carefully into the Carius tube in such a way that nitric acid does not enter the tube.
• The Carius tube is then sealed. Now, it is placed in an outer iron jacket and heated in a furnace at 550-560 K for about six hours
• As a result, C, H and S present in the compound are oxidised to CO₂, H₂O and H₂SO₄ respectively. The halogen present in the compound gets converted into silver halide which is precipitated.
• The tube is cooled and when the sealed capillary end is heated in a burner’s flame, a small hole is formed through which the gases escape.
• The capillary end is now cut off and the precipitate of silver halide is filtered, washed, dried and weighed

Results & calculations:

Let, the mass of organic compound taken = w g and the mass of silver halide (AgX) formed—x g. Now, 1 mol of AgX contains 1 gram-atom of X (X = Cl, Br or I ), i.e., (108 + atomic mass of X) g of AgX contain (atomic mass of X)g of X

∴ xg AgX contain = \(\frac{\text { atomic mass of } X}{(108+\text { atomic mass of } X)}\) X JCg of X

This amount of halogen (X) is present in w g compound.

Percentage of halogen in the compound

= \(\frac{\text { atomic mass of } \mathrm{X}}{(108+\text { atomic mass of } \mathrm{X})} \times \frac{x}{w} \times 100\)

1. Percentage of chlorine (atomic mass = 35.5 )

= \(\frac{35.5}{(108+35.5)} \times \frac{x}{w} \times 100\)

= \(\frac{35.5}{143.5} \times \frac{x}{w} \times 100\)

2. Percentage of bromine (atomic mass = 80 )

= \(\frac{80}{(108+80)} \times \frac{x}{w} \times 100\)

= \(\frac{80}{188} \times \frac{x}{w} \times 100\)

3. Percentage of iodine (atomic mass = 127 )

= \(\frac{127}{(108+127)} \times \frac{x}{w} \times 100\)

= \(\frac{127}{235} \times \frac{x}{w} \times 100\)

Example: In the Carius method, 0.256 g of an organic compound containing bromine gave 0.306 g of AgBr. Find out the percentage of bromine in the compound.
Answer:

The mass of the organic compound taken = 0.256 g and the mass of AgBr formed = 0.306 g.

Now, 1 mol of AgBr = 1 gram-atom of Br

or, (108 + 80) g or 188 g of AgBr = 80 gofBr

i.e., 188 g of AgBr contains = 80 g of bromine

0.306 g of Ag Br contain \(\frac{80}{188} \times 0.306\) g of bromine

This amount of bromine is present in 0.256g compound

Percentage6 of bromine = \(\frac{80}{188} \times \frac{0.306}{0.256} \times 100\)

= 50.9

4. Estimation of sulphur: Carius method

Sulphur Carius method Principle:

When an organic compound containing sulphur is heated with fuming nitric acid in a sealed tube (Carius tube), sulphur is quantitatively oxidised to sulphuric acid. It is then precipitated as barium sulphate by adding a barium chloride solution. The precipitate is then filtered, washed, dried and weighed. From the amount of BaSO₄ formed, the percentage of sulphur is calculated.

Resultsandcalculations: Let the mass of organic compound

= w g and the mass of barium sulphate formed = xg

Now, 1 mol of BaSO₄ contains 1 gram-atom of S, i.e.,

(137 + 32 + 64) g or, 233 g BaO₄ contain 32 g sulphur.

∴ x g of BaSO₄ contain \(\frac{32}{233}\) × x g of sulphur

So, this amount of sulphur is present in the w g compound.

∴ Percentage of sulphur = \(\frac{32}{233} \times \frac{x}{w} \times 100\)

Example: In sulphur estimation by the Carius method, 0.79 g of an organic compound gave 1.164 g of barium sulphate. Calculate the percentage of sulphur in the compound.
Answer:

The mass of the organic compound taken = 0.79 g and the mass of BaSO₄ obtained = 1.164 g.

Now, 1 mol of BaSO₄ = 1 gram-atom of S or,

233 g of BaSO₄= 32 g of i.e.,

233 g of BaSO₄ contain = 32 g ofsulphur.

1.164 g of BaSO₄ contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

So, 0.79 g compound contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

Percentage of sulphur =  \(\frac{32}{233} \times \frac{1.164}{0.79} \times 100\)

= 20.24

5. Estimation of phosphorus: Carius method

Phosphorus Carius method principle:

An organic compound (containing P) of known mass is heated with fuming nitric acid in a sealed tube (Carius tube). Phosphorus present in the compound is oxidised to phosphoric acid which is precipitated as ammonium phosphomolybdate by heating it with concentrated nitric acid and then adding ammonium molybdate.

Ammonium phosphomolybdate (Yellow)

The precipitate of ammonium phosphomolybdate thus obtained is filtered, washed, dried and weighed.

Alternative method:

The phosphoric acid is precipitated as magnesium ammonium phosphate (MgNH₄PO₄) by the addition of magnesia mixture (a solution containing MgCl₂, NH₄Cl and NH₄OH ).

The precipitate is filtered, washed, dried and then ignited to give magnesium pyrophosphate (Mg₂P₂O)

From the mass of magnesium pyrophosphate, the percentage of phosphorus in the compound can be easily calculated.

Results and calculations:

Let the mass of the organic compound taken = wg and the mass of ammonium phosphomolybdate formed = xg.

1 mol (NH₄)₃PO₄-12MoO₃ contains 1 gram-atom of P.

or, 3(14 + 4) + 31 + 4 × 16 + 12(96 + 3 × 16) = 1877g of

(NH₄)₃PO₄. 12MOO₃ = 31 g of P, i.e., 1877 g of

(NH₄)₃PO₄ . 12MoO₃ contain 31 g of phosphorus

xg of (NH₄)₃PO₄ -12MoO₃ contain \(\)

xg of phosphorus. This amount of phosphorus is present in w g of the compound.

Percentage of phosphorus = \(\frac{31}{1877} \times \frac{x}{w} \times 100\)

= \(\frac{31}{1877} \times \frac{\text { Mass of ammonium phosphomolybdate }}{\text { Mass of the compound taken }} \times 100\)

Alternative calculation: Let the mass of Mg2P20? formed

= xg. Now, 1 mol of (Mg₂P₂O) contains 2 gram-atom of P

or, (24 × 2 + 31 × 2 + 16 × 7) = 222 g of (Mg₂P₂O)

i.e., 222g of (Mg₂P₂O) contains 62 g of phosphorus.

x g of (Mg₂P₂O)contain \(\frac{62}{222} \times x\) g of phosphorus

w g organic compound contain \(\frac{62}{222} \times x\) xg phosphorus

Percentage of phosphorus = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Example: 0.35 g of an organic compound containing phosphorus gave 0.65 g magnesium pyrophosphate, Mg₂P₂O in the Carius method. Calculate the percentage of phosphorus in the given compound.
Answer:

The mass of the organic compound taken = 0.35 g and the mass of Mg₂P₂O formed = 0.65 g.

Now, 1 mol of Mg₂P₂O contains 2 gram-atom of P.

Or, (2 ×  24 + 2 × 31 + 16 × 7) g of Mg₂P₂O= 2 × 31 g of P

i.e., 222 g of Mg₂P₂O contains 62 g of phosphorus

0.65 g Mg₂P₂Ocontain \(\frac{62}{222} \times 0.65\) x 0.65 g of phosphorus

This amount of P is present in a 0.35 g compound.

Percentage of phosphorus = \(\frac{62}{222} \times \frac{0.65}{0.35} \times 100\)

6. Estimation of oxygen

Percentage of oxygen = 100 – (sum of the percentages of all other elements). However, the oxygen content of a compound can be estimated directly as follows:

Oxygen Principle:

An organic compound of known mass is decomposed by heating in a stream of nitrogen gas. The mixture of the gaseous products including oxygen is passed over red-hot coke when all the oxygen combines with carbon to form carbon monoxide. The mixture is then passed through warm iodine pentoxide (I₂O₅ ) when CO undergoes oxidation to CO₂ and iodine is liberated.

From the knowledge of the mass of iodine liberated or C02 produced, the percentage of oxygen in the compound can be calculated easily

Results and calculations: Let, the mass of the organic compound taken = wg and the mass of CO₂ formed =xg. We find that each mole of oxygen liberated from the compound will produce 2 mol of CO₂

x g. of CO₂ is obtained from \(\frac{32}{88} \times x=\frac{16}{44} \times\) = x x g of O₂

This amount of oxygen is present in w g ofthe compound.

Percentage of oxygen = \(\frac{16}{44} \times \frac{x}{w} \times 100\)

1. Percentage of carbon (Liebig’s method)

= \(\frac{12}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

2. Percentage of carbon (Liebig’s method)

= \(\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

3. Percentage of nitrogen (Dumas method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

4. Percentage of nitrogen (KjeldahTs method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

5. Percentage of halogen (Carius method)

= \(\frac{\text { At. mass of } X}{(108+\text { At. mass of } X)} \times \frac{\text { Mass of AgX }}{\text { Mass of the compound taken }} \times 100\)

X = Cl(35.5) , Br(80) or 1(127)

6. Percentage of sulphur (Carius method)

= \(\frac{32}{233} \times \frac{\text { Mass of } \mathrm{BaSO}_4 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

7. Percentage of phosphorus (Carius method)

\(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

8. Percentage of oxygen

= \(\frac{16}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of compound taken }} \times 100\).