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Organic Chemistry – Some Basic Principles and Techniques - Qualitative and Quantitative Analysis

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Qualitative Analysis: Detection of Carbon and Hydrogen is done by heating the compound with CuOCuO, converting Carbon to CO2CO_2 (turns lime water milky) and Hydrogen to H2OH_2O (turns anhydrous CuSO4CuSO_4 blue).

Lassaigne's Test: Used for detection of Nitrogen, Sulphur, and Halogens. The organic compound is fused with metallic Sodium to convert covalent bonds into ionic bonds (NaCNNaCN, Na2SNa_2S, NaXNaX).

Detection of Nitrogen: The sodium fusion extract is boiled with FeSO4FeSO_4 and acidified with concentrated H2SO4H_2SO_4. Formation of Prussian blue color due to Fe4[Fe(CN)6]3xH2OFe_4[Fe(CN)_6]_3 \cdot xH_2O confirms Nitrogen.

Detection of Sulphur: Sodium fusion extract treated with Sodium Nitroprusside gives a violet color ([Fe(CN)5NOS]4[Fe(CN)_5NOS]^{4-}).

Quantitative Estimation of Carbon and Hydrogen (Liebig's Method): A known mass of compound is burnt in the presence of excess Oxygen and CuOCuO. Carbon is oxidized to CO2CO_2 (absorbed in KOHKOH solution) and Hydrogen to H2OH_2O (absorbed in anhydrous CaCl2CaCl_2).

Dumas Method for Nitrogen: The compound is heated with CuOCuO in an atmosphere of CO2CO_2, yielding N2N_2 gas. The volume of N2N_2 is measured at STPSTP.

Kjeldahl's Method for Nitrogen: The compound is digested with concentrated H2SO4H_2SO_4 to form (NH4)2SO4(NH_4)_2SO_4. On heating with NaOHNaOH, NH3NH_3 is evolved and absorbed in a known volume of standard acid. Note: This method is not applicable to compounds containing Nitrogen in nitro/azo groups or in the ring (e.g., Pyridine).

Carius Method for Halogens: The compound is heated with fuming HNO3HNO_3 and AgNO3AgNO_3. Halogen is converted to AgXAgX precipitate, which is filtered, washed, dried, and weighed.

📐Formulae

% of Carbon=1244×Mass of CO2 formedMass of organic compound×100\% \text{ of Carbon} = \frac{12}{44} \times \frac{\text{Mass of } CO_2 \text{ formed}}{\text{Mass of organic compound}} \times 100

% of Hydrogen=218×Mass of H2O formedMass of organic compound×100\% \text{ of Hydrogen} = \frac{2}{18} \times \frac{\text{Mass of } H_2O \text{ formed}}{\text{Mass of organic compound}} \times 100

% of Nitrogen (Dumas method)=2822400×VSTP of N2Mass of organic compound×100\% \text{ of Nitrogen (Dumas method)} = \frac{28}{22400} \times \frac{V_{STP} \text{ of } N_2}{\text{Mass of organic compound}} \times 100

% of Nitrogen (Kjeldahl method)=1.4×N×VMass of organic compound\% \text{ of Nitrogen (Kjeldahl method)} = \frac{1.4 \times N \times V}{\text{Mass of organic compound}} (where NN and VV are normality and volume of acid used to neutralize NH3NH_3)

% of Halogen=Atomic mass of XMolecular mass of AgX×Mass of AgX formedMass of organic compound×100\% \text{ of Halogen} = \frac{\text{Atomic mass of } X}{\text{Molecular mass of } AgX} \times \frac{\text{Mass of } AgX \text{ formed}}{\text{Mass of organic compound}} \times 100

% of Sulphur=32233×Mass of BaSO4 formedMass of organic compound×100\% \text{ of Sulphur} = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4 \text{ formed}}{\text{Mass of organic compound}} \times 100

💡Examples

Problem 1:

In Carius method of estimation of halogen, 0.150.15 g of an organic compound gave 0.120.12 g of AgBrAgBr. Find the percentage of bromine in the compound. (Atomic mass: Ag=108Ag = 108, Br=80Br = 80)

Solution:

Mass of compound (mm) = 0.150.15 g; Mass of AgBrAgBr (m1m_1) = 0.120.12 g. Molar mass of AgBr=108+80=188AgBr = 108 + 80 = 188 g/mol. % of Br=80188×0.120.15×100=34.04%\% \text{ of } Br = \frac{80}{188} \times \frac{0.12}{0.15} \times 100 = 34.04\%.

Explanation:

The formula for the percentage of halogen is applied using the mass of the silver halide precipitate formed relative to the initial mass of the sample.

Problem 2:

0.300.30 g of an organic compound gave 5050 mL of Nitrogen collected at 300300 K and 715715 mm pressure in Dumas method. Calculate the percentage of Nitrogen. (Aqueous tension at 300300 K = 1515 mm)

Solution:

Pressure of dry N2=71515=700N_2 = 715 - 15 = 700 mm. Volume at STP (VSTPV_{STP}) = P1V1T0P0T1=700×50×273760×300=41.9\frac{P_1 V_1 T_0}{P_0 T_1} = \frac{700 \times 50 \times 273}{760 \times 300} = 41.9 mL. % of N=2822400×41.90.30×100=17.46%\% \text{ of } N = \frac{28}{22400} \times \frac{41.9}{0.30} \times 100 = 17.46\%.

Explanation:

First, the volume of N2N_2 gas is corrected to STP conditions by subtracting aqueous tension from the total pressure and then applying the combined gas law. Then, the percentage formula for the Dumas method is used.

Qualitative and Quantitative Analysis Revision - Class 11 Chemistry CBSE