Estimation and Determination

Estimation and Determination Notes

Estimation and Determination

Determination of Molecular Mass of Acids by Silver Salt Method

This method is based on the fact that they form insoluble silver salts, which upon heating decompose to leave a residue of metallic silver.
Procedure: The unknown acid is dissolved in water and treated with a slight excess of ammonium hydroxide. The excess of ammonia is boiled off. To this sufficient quantity of silver nitrate is added when a white precipitate of silver salt is obtained. The precipitate is separated by filtration, washed successively with water, alcohol and ether and dried in the steam oven. About 0.2 g of the dry silver salt is weighed into a crucible and ignited until the decomposition is complete. Ignition is repeated until the crucible with the residue of silver has attained constant weight. The molecular mass of the acid is then calculated from the mass of the silver salt taken and the mass of the residue of metallic silver obtained from it.
Weight of unknown carboxylic acid = W g
Weight of silver salt of that acid = X1 g
Weight of metallic silver = X2 g
i.e. X2 g of silver are obtained from X1 g of silver salt
Then 108 g of silver are obtained from = (X1/X2) X 108 g of silver salt
Thus, molar mass of silver salt of carboxylic acid = (X1/X2) X 108 g
For monocarboxylic acid, molar mass = (molar mass of Ag-salt) - (atomic mass of Ag) + (Atomic mass of H)
= (108 X1/X2) − 108 + 1
= (108 X1/X2) − 107
For polybasic carboxylic acid of basicity (n), molar mass =n(Molar mass of Ag-salt) - n(atomic mass of Ag) + n(atomic mass of H)
= (n108 X1/X2) − 108n + 1n
= (n108 X1/X2) − 107n
Molar mass of Acid = [wt. of Ag salt/wt. of Ag) X 108 − 107] Basicity of Acid

Q. About 0.759 g of silver salt of a dibasic acid was ignited when a residue of 0.463 g of metallic silver was left. Calculate the molecular mass of the acid.

Solution: 0.463 g of metallic silver is left by silver salt = 0.759 g
216 g of metallic silver is left by silver salt =( 0.759/0.463) X 216 = 354 g
so, molecular mass of dibasic acid = (mol. wt of Ag salt − 2) X 108 + 2 = 140 g

Determination of Molecular Weight of a Base by H2PtCl6 Method

Most of the bases combine with hydrochloroplatinic acid to form insoluble salts known as platinichlorides. These double salts may be represented by the general formula B2H2PtCl6. where B2 stands for one equivalent of the base.
On ignition, H2PtCl6 decomposes to leave a residue of Pt.
B2H2PtCl6 → Pt
Knowing the weight of B2H2PtCl6 taken and the Pt left as residue, the molecular weight of the base can be calculated.
Let 'x' gm be the weight of the B2H2PtCl6 taken and 'a' of the Pt residue.
Since one molecule of B2H2PtCl6 contains one Pt atom. So, 195 gm Pt will be left as residue by one molecular weight of the H2PtCl6.
But, a gm of Pt left by x gm of B2H2PtCl6
So, 195 gm of Pt will left by (x/a)195 gm
Hence, molecular weight of B2H2PtCl6 = (x/a)195
The equivalent weight of the base (B) =
(B2H2PtCl6 − H2PtCl6)/2
or, (mol. wt of H2PtCl6 − 410)/2
or, [(x/a)195 − 410]/2
If the acidity of the base is 'n', the molecular weight of the base-
[(x/a)195 − 410]/(n/2)

Detection of Nitrogen

Sodalime Test

The given substance is mixed with double the amount of sodalime and heated in a test tube. The vapour of ammonia evolved show the presence of nitrogen.

Sodium Test (Lassaigne's Test)

This is a golden test for the detection of nitrogen in all classes of nitrogenous compounds. It involves the following steps:
The substance is heated strongly with Na metal
            Na + C + N → NaCN
The water extract of the fused mass is boiled with ferrous sulphate solution.
            FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4
            6NaCN + Fe(OH)2 → Na4[Fe(CN)6] + 2NaOH
To the cooled solution is then added a little ferric chloride solution and excess of concentrated HCl.
            Na4[Fe(CN)6] + 4FeCl3 → Fe4[Fe(CN)6]3 + 12NaCl
                                                        prussian blue
The formation of prussian blue or green coloration confirms the presence of nitrogen in given organic compound.

Detection of Halogens:

Sodium Test:

Upon fussion with sodium, the halogens in the organic compound are converted to the corresponding sodium halides. Thus,
          Cl + Na → NaCl
          Br + Na → NaBr
          I + Na → NaI
Acidify a portion of sodium extract with dilute HNO3 and add to it AgNO3 solution.
White ppt. soluble in ammonia indicates Chlorine
Yellowish ppt. sparingly soluble in ammonia indicates Bromine
Yellow ppt. insoluble in ammonia indicates Iodine.

Detection of Sulphur

Sodalime Test:

If sulphur is present in the given organic compound, upon fussion with Na reacts to form sodium sulphide.
          2Na + S → Na2S
Thus, the sodium extract obtained from the fused mass may be tested as:
A. To a portion, add freshly prepared sodium nitropruside solution. A deep violet coloration indicates the presence of sulphur.
B. Acidify a second portion of the extract with acetic acid and then add lead acetate solution. A black ppt. of lead sulphide confirm the presence of sulphur.
          Pb(CH3COO)2 + N2S → PbS + 2CH3COONa
            Lead Acetate               Black ppt.

Detection of Phosphorus

The solid substance is heated with an oxidizing agent such as concentrated nitric acid or a mixture of sodium carbonate and potassium nitrate. The phosphorus present in the substance is thus oxidized to phosphate. the residue is extracted with water, boiled with some nitric acid and then a hot solution of ammonium molybdate is added to it in excess. A yellow coloration or ppt. indiccates the presence of phosphurs.

Estimation of Nitrogen

Kjeldahl Method

This method is based on the fact that when an organic compound containing nitrogen is heated with concentrated sulphuric acid, the nitrogen in it is quantitatively converted into ammonium sulphate. The resultant liquid is then treated with excess of alkali and the liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia is determined by finding the ammount of acid neutralized by back titration with some statndard alkali.
Let the weight of organic substance be x gm.
V volume of N HCl is required for complete neutralization of NH3 evolved.
V ml. N HCl ≡ V ml. N NH3
1000ml N NH3 contains 17 gm of NH3 or 14 gm N
Ammount of N present in V mol of N NH3 = (14/1000). V . N gm
% of nitrogen = wt. of nitrogen (100/wt. of substance)
                       = (14/1000).V.N.(100/x) = 1.4VN/x
where: N = Normality and V = Volume of the acid used.

Estimation of Sulphur

Carius Method

Known weight of organic compound is heated with fuming nitric acid in carius tube, sulphur present in organic compound oxidized to sulphuric acid. The whole solution is then transfered in a beaker and BaCl2 solution in excess is added to it for complete ppt. The ppt. of BaSO4 is then washed, dried and ignited to obtain the constant wt. of BaSO4. Now by knowing the weight of organic compound and BaSO4, the percentage of sulphur in the organic compound is easily calculated.
          Organic Compound + HNO3 → CO2 + H2 + H2SO4
          H2SO4 + BaCl2 → BaSO4 + 2HCl
Let wt. of an organic compound = a gm
wt. of BaSO4 = b gm

molecular weight of BaSO4 = 233
∵ 233 gm BaSO4 contains 32 gm S.
∴ b gm BaSO4 contains (32 X b)/233 gm S.
∵ a gm of organic compound contains (32 X b)/233 gm S.
∴ 100 gm of organic compound contains (32 X b X 100)/233 X a gm S.
∴ % of Sulphur in organic compound = (32 X b X 100)/233 X a