Chemistry Qualitative and quantitative analysis of organic compounds

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=> Qualitative and quantitative analysis of organic compounds

### QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS

The elements present in organic compounds are carbon and hydrogen. In addition to these, they may also contain oxygen, nitrogen, sulphur, halogens and phosphorus.

### Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heating the compound with copper(II) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue).

color{red}(C + 2 CuO overset(Delta)→ 2Cu + CO_2)

color{red}(2H + CuO overset(Delta)→ Cu + H_2O)

color{red}(CO_2 + Ca(OH)_2 → CaCO_3 ↓ + H_2O)

color{red}(5H_2O + underset("White")(CuSO_4) → underset("Blue")(CuSO_4) .5 H_2O)

### Detection of Other Elements

Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by color{green}("Lassaigne’s test"). The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:

color{red}(Na + C + N overset(Delta)→ NaCN)

color{red}(2Na + S overset(Delta)→ Na_2 S)

color{red}(Na + X overset(Delta)→ Na X)

(color{red}(X = Cl, Br) or color{red}(I)) color{red}(C, N, S) and color{red}(X) come from organic compound.

Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled
water. This extract is known as sodium fusion extract.

color{green}("(A) Test for Nitrogen")

The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanoferrate(II) to produce iron(III) hexacyanoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.

color{red}(6CN^(-) + Fe^(2+) → [Fe(CN)_6 ]^(4-))

color{red}(3[Fe(CN)_6]^(4-) + 4Fe^(3+) overset(x H_2O)→ underset("Prussian blue")(Fe_4[Fe(CN)_6]_3 . x H_2O)

color{green}("(B) Test for Sulphur")

(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.

color{red}(S^(2-) + Pb^(2+) → underset("Black")(PbS))

(b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur.

color{red}(S^(2-) + [ Fe(CN)_5 NO]^(2-) → underset("Violet")([Fe(CN)_5 NOS]^(4-)))

In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red
colour and no Prussian blue since there are no free cyanide ions.

color{red}(Na +C + N + S → Na SCN)

color{red}(Fe^(3+) + SCN^(-) → underset("Blood red")([ Fe(SCN)]^(2+)))

If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give
their usual tests.

color{red}(NaSCN + 2Na → NaCN + Na_2S)

color{green}("(C) Test for Halogens")

The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.

color{red}(X^(-) + Ag^(+) → AgX)

color{red}(X) represents a halogen color{red}(– Cl, Br) or color{red}(I).

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens.

color{green}("(D) Test for Phosphorus")

The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

color{red}(Na_3PO_4+ 3HNO_3 → H_3PO_4+ 3 NaNO_3)

color{red}(H_3PO_4 + underset("Ammonium molybdate")(12(NH_4)_2 MoO_4) + 21 HNO_3 → underset("Ammonium phosphomolybdate")((NH_4)_3 PO_4 .12 MoO_3) + 21 NH_4NO_3 + 12 H_2O)

### QUANTITATIVE ANALYSIS

The percentage composition of elements present in an organic compound is determined by the methods based on the following principles:

### Carbon and Hydrogen

Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.

color{red}(C_xH_y + (x+y/4) O_2 → xCO_2 + (y/2) H_2O)

The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Fig.12.14). The

increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated.

Let the mass of organic compound be color{red}(m g), mass of water and carbon dioxide produced be color{red}(m_1) and color{red}((m_2 g) respectively;

color{red}("Percentage of carbon =" (12 xx m_2 xx 100)/(44 xx m))

color{red}("Percentage of hydrogen " = (2 xx m_1 xx 100)/(18 xx m))

Q 3214656559

On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound.

Solution:

Percentage of carbon  = ( 12 xx 0.198 xx 100)/(44xx0.246)

 = 21.95%

Percentage of hydrogen  = ( 2xx0.1014xx100)/(18xx0.246)

 = 4.58%

### Nitrogen

There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.

(i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.

color{red}(C_x H_y N_z + (2x+y/2) CuO → xCO_2 + y/2 H_2O + z/2 N_2 + (2x+y/2) Cu)

Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig.12.15).

Let the mass of organic compound color{red}(= m g) Volume of nitrogen collected color{red}(= V_1 mL) Room temperature color{red}(= T_1K)

Volume of nitrogen at STP  color{red}(= (p_1 V_1 xx 273)/(760 xx T_1))

(Let it be color{red}(V mL)) Where color{red}(p_1) and color{red}(V_1) are the pressure and volume of nitrogen, color{red}(p_1) is different from the atmospheric pressure at which nitrogen gas is collected. The value of color{red}(p_1) is obtained by the relation;

color{red}(p_1=) Atmospheric pressure – Aqueous tension color{red}(22400 mL N_2) at STP weighs 28 g.

color{red}(V m L N_2) at STP weighs color{red}( = ( 28 xx V)/(22400) g)

color{red}("Percentage of nitrogen " = ( 28xxV xx 100)/(22400 xx m))

(ii) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Fig. 12.16). The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.

color{red}("Organic compound" + H_2SO_4 → (NH_4)_2 SO_4 overset(2NaOH)→ Na_2SO_4 +2NH_3 + 2H_2O)

color{red}(2NH_3 + H_2SO_4 → (NH_4)_2 SO_4)

Let the mass of organic compound taken color{red}(= m g)

Volume of color{red}(H_2SO_4) of molarity, color{red}(M), taken color{red}(= V mL)

Volume of color{red}(NaOH) of molarity, color{red}(M), used for titration of excess of color{red}(H_2SO_4 = V_1 mL V_1mL) of color{red}(NaOH) of molarity color{red}(M = V_1 /2 mL) of color{red}(H_2SO_4) of molarity color{red}(M)

Volume of color{red}(H_2SO_4) of molarity color{red}(M) unused color{red}(= (V - (V_1)/2) mL)

color{red}((V- (V_1)/2) mL) of color{red}(H_2SO_4) of molarity color{red}(M)

color{red}(= 2(V-(V_1)/2) mL) of color{red}(NH_3) solution of molarity color{red}(M).

color{red}(1000 mL) of color{red}(1 M NH_3) solution contains 17g color{red}(NH_3) or 14 g of color{red}(N)

color{red}((2(V-(V_1)/2) mL) of color{red}(NH_3) solution of molarity color{red}(M) contains:

color{red}((14 xxM xx 2 (V - (V_1)/2))/(1000) g N)

Percentage of color{red}(N = (14 xx M xx 2 ( V - (V_1)/2))/(1000) xx 100/m)

color{red}( = ( 1.4 xx M xx2 (V - (V_1)/2))/m)

Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Q 3224756651

In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure.
Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)

Solution:

Volume of nitrogen collected at 300K and 715mm pressure is 50 mL
Actual pressure = 715-15 =700 mm

Volume of nitrogen at STP  = ( 273 xx 700 xx 50)/(300 xx 760)

 = 41.9 mL

22,400 mL of N_2 at STP weighs = 28 g

"41.9 mL of nitrogen weighs"  = ( 28 xx41.9)/(22400) g

"Percentage of nitrogen" = (28 xx 41.9 xx 100)/(22400 xx 0.3)

 = 17.46 %
Q 3244756653

During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H_2SO_4. Find out the percentage of nitrogen in the compound.

Solution:

1 M of 10 mL H_2SO_4=1M of 20 mL NH_3

1000 mL of 1M ammonia contains 14 g nitrogen

20 mL of 1M ammonia contains

(14 xx 20)/(1000) g nitrogen

Percentage of nitrogen  = (14 xx 20 xx100)/(1000xx 0.5) = 56.0%

### Halogens

Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.12.17) in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed. Let the mass of organic

compound taken color{red}(= m g)
Mass of color{red}(AgX) formed color{red}(= m_1 g)
1 mol of color{red}(AgX) contains 1 mol of color{red}(X)
Mass of halogen in color{red}(m_1g) of color{red}(AgX)

color{red}(= ("atomic mass of" X xx m_1 g)/("molecular mass of" AgX))

color{red}("Percentage of halogen" = (" atomic mass of " X xx m_1 xx 100)/("molecular mass of" AgX xx m))

Q 3264756655

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

Solution:

Molar mass of AgBr = 108 + 80 = 188 g mol^(-1)

188 g AgBr contains 80 g bromine

0.12 g AgBr contains (80 xx 0.12)/(188) g bromine

Percentage of bromine  = ( 80 xx 0.12 xx 100)/(188 xx 0.15)

 = 34.05%

### Sulphur

A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.

Let the mass of organic compound taken color{red}(= m g)

and the mass of barium sulphate formed color{red}(= m_1g)

1 mol of color{red}(BaSO_4 = 233g BaSO_4 = 32g "sulphur")

color{red}(m_1 g BaSO_4 "contains" (32 xx m_1)/(233) g " sulphur")

color{red}("Percentage of sulphur" = (32xxm_1 xx 100)/(233 xx m))
Q 3284756657

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Solution:

Molecular mass of BaSO_4 = 137+32+64 = 233g

233 g BaSO_4 contains 32 g sulphur

0.4813 g BaSO_4 contains (32 xx 0.4813)/(233) g sulphur

Percentage of sulphur  = ( 32 xx 0.4813 xx 100)/(233 xx 0.157)

 = 42.10%

### Phosphorus

A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, color{red}((NH_4)_3 PO_4.12MoO_3), by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as color{red}(MgNH_4PO_4) by adding magnesia mixture which on ignition yields color{red}(Mg_2P_2O_7).

Let the mass of organic compound taken color{red}(= m g) and mass of ammonium phospho molydate color{red}(= m_1g)

Molar mass of color{red}((NH_4)_3PO_4.12MoO_3 = 1877 g)

Percentage of phosphorus  color{red}(= ( 31 xx m_1 xx 100)/(1877 xx m) %)

If phosphorus is estimated as color{red}(Mg_2P_2O_7)

Percentage of phosphorus color{red}( = (62 xx m_1 xx 100)/(222 xx m)%)

where, color{red}(222 u) is the molar mass of color{red}(Mg_2P_2O_7, m), the mass of organic compound taken, color{red}(m_1),
the mass of color{red}(Mg_2P_2O_7) formed and 62, the mass of two phosphorus atoms present in the
compound color{red}(Mg_2P_2O_7.)

### Oxygen

The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (color{red}(I_2O_5)) when carbon monoxide is oxidised to carbon dioxide producing iodine.

color{red}("Compound" overset("heat")→ O_2 + " other gaseous products")

color{red}(2C+ O_2 overset(1373K)→ 2CO) ] \ \ \ \ 5 \ \ \ \ \ \(A)

color{red}(I_2O_5 + 5CO → I_2 + 5 CO_2) ] \ \ \ \ \ \ 2 \ \ \ \ \ (B)

On making the amount of color{red}(CO) produced in equation (A) equal to the amount of color{red}(CO) used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.

Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.

Let the mass of organic compound taken be color{red}(m g)

Mass of carbon dioxide produced be color{red}(m_1 g)

therefore color{red}(m_1 g) carbon dioxide is obtained from color{red}(( 32 xx m_1)/(88) g O_2)

color{red}(therefore "Percentage of oxygen" = ( 32 m_1 xx 100)/(88 m) %)

The percentage of oxygen can be derived from the amount of iodine produced also.

Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time.