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

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)`

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))`

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.

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))`

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))`

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.)`

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.