Quantitative Analysis - Continued

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Estimation of Halogens

The amount of halogen in an organic compound is estimated by Carius method.

Principle

Carius method involves conversion of the halogen present in the organic compound into the corresponding silver halide. From the mass of the organic compound taken and the mass of silver halide formed, the percentage of halogen in the compound can be calculated.

Method

A small known mass (0.2 to 0.3 g) of the given organic compound is heated with fuming nitric acid and a few crystals of silver nitrate at about 200°C in a sealed tube (called Carius tube) for 5 to 6 hours. Carbon and hydrogen are oxidized to carbon dioxide and water while halogen forms a precipitate of silver halide (AgX). After filtration and washing, the precipitate of silver halide is dried and weighed.

Fig:16.18 A Carius tube for the estimation of halogens

Calculations

Let,

The mass of the compound taken be = W g

Mass of AgX formed = W₁ g

From stoichiometry,

Ag = X

1 mol 1 mol

1 molecular mass 1 atomic mass

Therefore,

Estimation of Sulphur

Sulphur present in any organic compound is also estimated by Carius method.

Principle

Sulphur present in any organic compound is oxidized to sulphuric acid by

digesting with fuming nitric acid. Sulphuric acid so formed is then quantitatively precipitated as barium sulphate by the addition of excess barium chloride. The precipitate of barium sulphate is filtered, washed, dried and weighed.

Process

A known mass of the organic compound is heated strongly with fuming

HNO₃ in a Carius tube for about 2 hours. The contents of the Carius tube are cooled and treated with excess of barium chloride (BaCl₂) solution to precipitate SO₄^2- ions present in the solution as BaSO₄. The precipitate of barium sulphate (BaSO₄) is filtered, washed and dried.

Calculations

Let,

The mass of the organic compound be = W g

Mass of the precipitate of BaSO₄ = W₁ g

From stoichiometry,

BaSO_{4 =} S

1 mol 1 mol

1 molecular mass 1 atomic mass

233 g 32 g

Therefore,

Then,

Estimation of Phosphorus

The amount of phosphorus present in any organic compound is determined by Carius method.

Principle

The phosphorus present in an organic compound is oxidized to phosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipitated as MgNH₄PO₄, which on ignition is converted into Mg₂P₂O₇. From the mass of Mg₂P₂O₇, the mass of phosphorus present in the sample of the organic compound can be calculated.

Process

A known mass (0.2 - 0.3 g) of the organic substance is heated with fuming nitric acid in a Carius tube for about 2 hours. The phosphorus of the organic compound gets converted into phosphoric acid. The total phosphate content of the solution is reacted with magnesia mixture (mixture of a solution containing magnesium chloride, ammonium chloride and a little ammonia) to precipitate it as MgNH₄PO₄. The precipitate of MgNH₄PO₄ is filtered, washed, dried and then ignited to give magnesium pyrophosphate (Mg₂P₂O₇).

Calculations

Let,

The mass of the organic compound taken be = W g

Mass of Mg₂P₂O₇ obtained = W₁ g

From stoichiometry,

Mg₂P₂O₇ = 2P

1 mol          2 mol

1 x molecular mass 2 x atomic mass

222 g 2 x 31 g = 62 g

Therefore,

Then,

Estimation of Oxygen

There is no direct method for the estimation of oxygen present in any organic compound. The percentage of oxygen in an organic compound is usually obtained by subtracting the sum of the percentages of all other elements from 100.

% of oxygen in an organic compound = 100 - (Sum of the % of all other elements present in the compound)

Problems

1. An organic compound weighing 0.244 g on combustion with dry oxygen produces 0.616 g of CO₂ and 0.108 g of H₂O. Determine the percentage composition of the compound

Solution

Mass of the organic compound taken = 0.244 g

Mass of CO₂ formed = 0.616 g

Mass of H₂O formed = 0.108 g

We know that CO₂ = C and H₂O = 2H44 g   12 g   18 g     2g

(iii) Percentage of O in the compound = 100 - (68.85 + 4.92) = 26.23

The percentage composition of the compound is, C = 68.85, H = 4.92 and O = 26.23.

2. An organic substance weighing 0.35 g was Kjeldahlised and the ammonia obtained was passed into 100 mL of N/5 H₂SO₄. The excess acid required 154 mL of N/10 NaOH for neutralization. Calculate the percentage of nitrogen in the compound.

Solution

The volume of H₂SO₄ neutralised by N/10 NaOH can be obtained as follows:

N_acid x V_acid = N_alkali x V_alkali

Therefore,

Volume of N/5 H₂SO₄ used for neutralizing NH₃= (100 - 77) mL = 23 mL

Then, Percentage of nitrogen in the sample =

3. An organic compound weighs 0.316 g after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound.(Atomic masses: Ba = 137, S = 32, O = 16, C = 12, H = 1).

Solution

Mass of the substance taken = 0.316 g

Mass of BaSO₄ formed = 0.466 g

From stoichiometry,

BaSO₄ = S (molecular mass of the BaSO₄ = 137+ 32 + 64 = 233)233       32

Then,