Easy Practice — Set 1

Quick Recap — Purification Methods

• Crystallisation: separates a solid using differences in solubility; sublimation for solids that vaporise directly (camphor, naphthalene).
• Distillation: simple (large b.p. gap), fractional (close b.p.), steam (steam-volatile, water-immiscible, e.g. aniline), reduced-pressure (compounds that decompose at their b.p.).
• Differential (solvent) extraction: partition between two immiscible solvents; chromatography: adsorption (column, TLC) or partition (paper); $R_f=\dfrac{\text{distance by solute}}{\text{distance by solvent}}$.
• Lassaigne's (sodium-fusion) test detects N, S and halogens; a sharp melting point indicates purity.

Beyond-NCERT JEE Formulae

Estimation of the elements is the quantitative heart of this chapter and precisely what JEE Main rewards here. Each block tells you when to use it and flags the trap examiners like to set. Atomic masses used throughout: C = 12, H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Br = 80, I = 127, Ag = 108, Ba = 137.

1. Carbon and hydrogen — Liebig combustion

When to use: a known mass of the compound is burnt in a stream of oxygen over hot CuO; the water is trapped in anhydrous $CaCl_2$ (or $Mg(ClO_4)_2$) and the carbon dioxide in KOH. You weigh the mass GAINED by each absorption tube.

• %C $=\dfrac{12}{44}\times\dfrac{\text{mass CO}_2}{\text{mass sample}}\times100$
• %H $=\dfrac{2}{18}\times\dfrac{\text{mass H}_2\text{O}}{\text{mass sample}}\times100$
• Oxygen is never burnt-estimated directly; take it by difference, %O = 100 - (sum of all the other percentages).

[JEE Tip] The factor $\dfrac{12}{44}$ appears because 44 g of carbon dioxide carries only 12 g of carbon, and $\dfrac{2}{18}$ because 18 g of water carries just 2 g of hydrogen. Trap: the numerator mass is the mass of CO2 (or H2O) actually COLLECTED in the tube, never the mass of the sample.

2. Nitrogen — Kjeldahl and Dumas

When to use Kjeldahl: nitrogen bonded as an amine or amide, which digestion with hot concentrated $H_2SO_4$ turns into $(NH_4)_2SO_4$ and then, on adding alkali, into ammonia.

• %N $=\dfrac{1.4\times N_{acid}\times V_{acid}}{\text{mass sample}}$, with $V_{acid}$ in mL and $N_{acid}$ the normality of the standard acid neutralised.
• Back-titration form (a measured EXCESS of acid is taken, then the leftover is titrated with standard base): %N $=\dfrac{1.4\,(N_{acid}V_{acid}-N_{base}V_{base})}{\text{mass sample}}$.

When to use Dumas: the universal method, and the ONLY one for nitrogen held in a ring, nitro, azo or diazo group; the nitrogen is swept out as $N_2$ gas over CuO and its volume measured.

• %N $=\dfrac{28}{22400}\times\dfrac{V_{N_2}\text{ at STP}}{\text{mass sample}}\times100$

[JEE Tip] The constant 1.4 already folds in "14 g of nitrogen per equivalent, per mL of 1 N acid" (that is, $14\times\tfrac{1}{1000}\times100$). For $H_2SO_4$ remember normality $=2\times$ molarity — the single most common Kjeldahl slip. Kjeldahl FAILS for pyridine, nitro and azo nitrogen, so reach for Dumas there. In Dumas, if the gas is collected over water, subtract the aqueous tension from the pressure before reducing the volume to STP.

3. Halogens (Carius) and sulphur

When to use: the compound is heated with fuming $HNO_3$ (plus $AgNO_3$ for a halogen) in a sealed Carius tube; the halogen drops out as silver halide, while sulphur is oxidised to sulphate and weighed as $BaSO_4$.

• %X $=\dfrac{\text{at.mass }X}{\text{mol.mass }AgX}\times\dfrac{\text{mass }AgX}{\text{mass sample}}\times100$, with AgCl = 143.5, AgBr = 188, AgI = 235.
• %S $=\dfrac{32}{233}\times\dfrac{\text{mass BaSO}_4}{\text{mass sample}}\times100$
• Phosphorus (beyond NCERT): it is oxidised to $H_3PO_4$ and weighed as $Mg_2P_2O_7$ (molar mass 222), so %P $=\dfrac{62}{222}\times\dfrac{\text{mass Mg}_2\text{P}_2\text{O}_7}{\text{mass sample}}\times100$.

[JEE Tip] Every one of these ratios is simply (mass of the wanted element locked inside the precipitate) divided by (molar mass of that precipitate). Fluorine cannot be estimated this way, because AgF is soluble and never precipitates.

4. Silver-salt method — equivalent weight of an acid

When to use: an unknown organic acid whose pure silver salt can be prepared and then ignited, leaving a residue of metallic silver.

• Silver replaces one acidic hydrogen per equivalent (a mass change of $108-1=107$), so the salt's equivalent weight is $(E+107)$ and $\dfrac{\text{mass Ag}}{\text{mass salt}}=\dfrac{108}{E+107}$. Rearranging gives the working formula:
• Equivalent weight $E=\dfrac{108\times\text{mass of silver salt}}{\text{mass of Ag residue}}-107$.

[JEE Tip] This returns the EQUIVALENT weight; multiply by the basicity to reach the molar mass (a dibasic acid has $M=2E$). Memorise 107 as $108-1$. The mirror-image trick for an organic base ignites its chloroplatinate salt to leave platinum.

5. Molar mass — Victor Meyer and Rast (camphor)

When to use Victor Meyer: a volatile liquid or low-boiling solid; a weighed drop is flash-vaporised and pushes out an equal volume of air, collected over water.

• Molar mass $M=\dfrac{\text{mass of substance}\times22400}{V\text{ (vapour at STP, in mL)}}$; reduce the collected volume to STP, and strip the aqueous tension, before substituting.

When to use Rast: the molar mass of a non-volatile solute, using molten camphor as the solvent.

• $\Delta T_f=K_f\times m$ leads to molar mass $M=\dfrac{1000\times K_f\times w_2}{w_1\times\Delta T_f}$, where $w_2$ is the solute mass, $w_1$ the camphor mass in grams, and $K_f\approx40$ K kg/mol for camphor.

[JEE Tip] Camphor is the chosen Rast solvent precisely because its $K_f$ (about 40) is enormous, so even a pinch of solute gives a large, easily read depression. Both methods hand you the MOLAR mass; pair it with the empirical formula from the percentage composition through $n=\dfrac{\text{molar mass}}{\text{empirical-formula mass}}$ to fix the molecular formula.

Solved Examples — Beyond-NCERT Formulae

Example 1 — %C, %H and the molecular formula (Liebig). On complete combustion, 0.23 g of a compound of C, H and O gave 0.44 g of carbon dioxide and 0.27 g of water. Find its percentage composition and molecular formula (molar mass = 46).

• Carbon: $\dfrac{12}{44}\times\dfrac{0.44}{0.23}\times100=52.17$, so %C = 52.17.
• Hydrogen: $\dfrac{2}{18}\times\dfrac{0.27}{0.23}\times100=13.04$, so %H = 13.04.
• Oxygen by difference: %O = 100 - 52.17 - 13.04 = 34.78.
• Divide each percentage by its atomic mass, then by the smallest, to get the mole ratio:

$$C:H:O=\dfrac{52.17}{12}:\dfrac{13.04}{1}:\dfrac{34.78}{16}=4.35:13.04:2.17=2:6:1$$

Answer: %C = 52.17, %H = 13.04, %O = 34.78; the empirical formula $C_2H_6O$ has mass 46, equal to the molar mass, so the molecule is $C_2H_6O$ (ethanol).

Example 2 — %N by Kjeldahl (back-titration). The ammonia liberated from 0.70 g of a compound was absorbed in 50 mL of 0.5 N $H_2SO_4$; the unreacted acid then required 30 mL of 0.5 N NaOH. Find %N.

• Acid actually neutralised by the ammonia $=(N_{acid}V_{acid})-(N_{base}V_{base})=(0.5\times50)-(0.5\times30)=10$ milli-equivalents.
• Substitute into the Kjeldahl relation (percentage nitrogen):

$$\dfrac{1.4\times10}{0.70}=20.0$$

Answer: %N = 20.0.

Example 3 — %N by Dumas (reduce the gas to STP first). By Dumas' method 0.42 g of a compound gave 58 mL of nitrogen collected over water at 300 K and 760 mm Hg; the aqueous tension at 300 K is 26.7 mm. Find %N.

• Pressure of the dry nitrogen alone = 760 - 26.7 = 733.3 mm.
• Reduce the volume to STP:

$$V_{STP}=58\times\dfrac{733.3}{760}\times\dfrac{273}{300}=50.93\text{ mL}$$

• Now apply the Dumas relation (percentage nitrogen):

$$\dfrac{28}{22400}\times\dfrac{50.93}{0.42}\times100=15.16$$

Answer: %N = 15.2, consistent with aniline, $C_6H_5NH_2$. Forgetting the aqueous-tension and STP corrections is the built-in trap.

Example 4 — %Cl by the Carius method. 0.45 g of an organic chloride, heated with fuming $HNO_3$ and $AgNO_3$ in a Carius tube, gave 0.574 g of AgCl. Find %Cl (AgCl = 143.5).

• Percentage chlorine:

$$\dfrac{35.5}{143.5}\times\dfrac{0.574}{0.45}\times100=31.56$$

Answer: %Cl = 31.56, matching chlorobenzene, $C_6H_5Cl$.

Example 5 — %S from the barium sulphate precipitate. 0.20 g of a sulphur compound, oxidised in a Carius tube and precipitated as barium sulphate, yielded 0.466 g of BaSO4. Find %S (BaSO4 = 233).

• Percentage sulphur:

$$\dfrac{32}{233}\times\dfrac{0.466}{0.20}\times100=32.0$$

Answer: %S = 32.0.

Example 6 — Molar mass by the Victor Meyer method. 0.29 g of a volatile liquid, flash-vaporised, displaced 112 mL of air measured at STP. Find its molar mass.

• The moles of vapour are $\dfrac{112}{22400}$, so the molar mass is:

$$M=\dfrac{0.29\times22400}{112}=58\text{ g/mol}$$

Answer: M = 58 g/mol (acetone, $C_3H_6O$).

Example 7 — Equivalent weight by the silver-salt method. 0.60 g of the pure silver salt of a dibasic acid, on ignition, left 0.36 g of metallic silver. Find the equivalent weight and molar mass of the acid (Ag = 108).

• Apply the silver-salt relation:

$$E=\dfrac{108\times0.60}{0.36}-107=180-107=73$$

• The acid is dibasic, so its molar mass is $M=2\times73=146$.

Answer: equivalent weight = 73, molar mass = 146 (adipic acid, $C_6H_{10}O_4$).

Example 8 — Molar mass by the Rast (camphor) method. 0.30 g of a non-volatile solute in 20 g of molten camphor ($K_f=40$ K kg/mol) lowered the freezing point by 3.0 K. Find the molar mass of the solute.

• Apply the Rast relation:

$$M=\dfrac{1000\times40\times0.30}{20\times3.0}=\dfrac{12000}{60}=200\text{ g/mol}$$

Answer: M = 200 g/mol.