10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
Steam distillation relies on the principle that when two immiscible liquids (like water and an organic compound) are heated together, the mixture boils when the sum of their individual (partial) vapour pressures equals atmospheric pressure — this allows the organic compound to distil over at a temperature significantly below its own normal boiling point, since water's vapour pressure contributes substantially to reaching this combined total.
Example: Steam distillation is commonly used to extract and purify essential oils (like eucalyptus oil or clove oil) from plant materials, allowing extraction at a much lower, non-decomposing temperature.
In Lassaigne's test, the organic compound is fused with sodium metal, converting any covalently bound nitrogen into sodium cyanide ($NaCN$).
The resulting Lassaigne's extract (containing $NaCN$, if nitrogen was present) is then treated with freshly prepared ferrous sulfate solution, followed by acidification — if nitrogen is present, a characteristic Prussian blue (or green) coloration forms, due to the formation of ferric ferrocyanide, confirming nitrogen's presence in the original compound.
Adsorption chromatography separates components based on differential adsorption onto a solid stationary phase (like silica gel or alumina) — components with stronger adsorption affinity move more slowly through the system.
Partition chromatography separates components based on their differential distribution (partitioning) between two immiscible liquid phases — a stationary liquid phase (often held on an inert solid support) and a mobile liquid phase — components partition differently between these two liquid phases based on their relative solubility in each.
Crystallisation exploits the fact that a compound's solubility in a given solvent typically increases with temperature — a hot, saturated (or slightly supersaturated) solution of the impure solid is prepared, then allowed to cool slowly.
As the solution cools, the solubility limit decreases, causing the desired compound (present in larger quantity) to selectively crystallise out of solution, since impurities (present in much smaller quantities) generally remain dissolved in the mother liquor (or are more effectively excluded from the forming, ordered crystal lattice) — the resulting purified crystals can then be separated by filtration, leaving the impurities behind in solution.
In Duma's method, the organic compound is heated with copper(II) oxide in an atmosphere of carbon dioxide gas — this combustion converts the compound's nitrogen content into free nitrogen gas ($N_2$), while carbon and hydrogen are oxidised to $CO_2$ and water respectively.
The resulting gas mixture is passed through a concentrated potassium hydroxide (KOH) solution, which absorbs the $CO_2$ (both the excess $CO_2$ used as a carrier gas, and any $CO_2$ formed from combustion), leaving only the pure nitrogen gas, whose volume is then measured directly and used to calculate the percentage of nitrogen originally present in the compound.
In Carius method, a known mass of the organic compound is heated with fuming nitric acid and silver nitrate in a sealed tube (Carius tube) — this treatment oxidatively decomposes the compound, converting any covalently bound halogen into the corresponding silver halide precipitate ($AgCl$, $AgBr$, or $AgI$).
This precipitate is filtered, washed, dried, and precisely weighed — since the stoichiometric relationship between the organic compound and the silver halide is known, the mass of silver halide obtained allows direct calculation of the percentage of halogen originally present in the sample.
A pure organic solid typically melts sharply, over a very narrow temperature range (often within 1-2°C), at a specific, well-characterised melting point value.
The presence of even small amounts of impurity typically causes the melting point to both decrease and occur over a noticeably broader (widened) temperature range — this depression and broadening of the melting point is a simple, reliable, and widely used practical indicator for assessing the purity of a given organic solid sample, without requiring more sophisticated analytical instrumentation.
Qualitative analysis aims to determine which elements are present in a compound, without measuring their exact amounts. Example: Lassaigne's test, which confirms the presence (or absence) of nitrogen, sulfur, and halogens.
Quantitative analysis aims to determine the precise percentage/amount of each element present in the compound. Example: Liebig's method for carbon and hydrogen estimation, which precisely measures the mass of $CO_2$ and $H_2O$ produced upon combustion to calculate exact percentage composition.
Silica gel and alumina are polar, porous solid materials with a very high surface area, capable of adsorbing organic molecules onto their surface through various intermolecular interactions (like hydrogen bonding or dipole-dipole interactions), depending on each molecule's specific polarity/functional groups.
Different compounds within a mixture, having different degrees of polarity or different functional groups, are adsorbed onto these stationary phases with differing strengths — this differential adsorption strength is precisely the basis for their effective separation as the mobile phase (solvent) carries them through the column at different rates, making silica gel and alumina particularly effective, widely available, and cost-effective stationary phase materials for this technique.
A pure liquid compound characteristically boils at one specific, constant temperature (at a given, fixed atmospheric pressure) throughout the entire distillation/boiling process.
The presence of an impurity (a dissolved solute, or another miscible liquid) typically causes the boiling point to either be elevated (for a non-volatile impurity, as in colligative boiling point elevation) or to occur over a range of varying temperatures (if the impurity is itself volatile, distilling alongside the main compound at differing rates) — this deviation from a single, sharp, constant boiling point is a simple, practical indicator used to assess the purity of a liquid organic sample.
10 questions · 5 marks each · Detailed step-by-step solutions
Percentage of carbon: $\%C = \dfrac{12}{44}\times\dfrac{\text{mass of }CO_2}{\text{mass of compound}}\times100 = \dfrac{12}{44}\times\dfrac{0.44}{0.2}\times100$
$\%C = \dfrac{12}{44}\times2.2\times100 = 0.2727\times2.2\times100 = \mathbf{60\%}$
Percentage of hydrogen: $\%H = \dfrac{2}{18}\times\dfrac{\text{mass of }H_2O}{\text{mass of compound}}\times100 = \dfrac{2}{18}\times\dfrac{0.18}{0.2}\times100$
$\%H = 0.1111\times0.9\times100 = \mathbf{10\%}$
Moles of $N_2$ collected $= \dfrac{44.8}{22400} = 0.002\text{ mol}$
Mass of nitrogen $= 0.002\times28 = 0.056\text{ g}$ (since each mole of $N_2$ contains 28 g of nitrogen)
$\%N = \dfrac{\text{mass of nitrogen}}{\text{mass of compound}}\times100 = \dfrac{0.056}{0.3}\times100$
$\%N = \mathbf{18.67\%}$
$\%Br = \dfrac{80}{188}\times\dfrac{\text{mass of }AgBr}{\text{mass of compound}}\times100$
$\%Br = \dfrac{80}{188}\times\dfrac{0.2868}{0.15}\times100$
$\%Br = 0.4255\times1.912\times100 \approx \mathbf{81.35\%}$
Moles of HCl used (= moles of $NH_3$ liberated, since they react 1:1): $n_{HCl} = M\times V = 1\times0.020 = 0.02\text{ mol}$
Since 1 mole of $NH_3$ corresponds to 1 mole of N: moles of N = 0.02 mol
Mass of N $= 0.02\times14 = 0.28\text{ g}$
$\%N = \dfrac{0.28}{0.5}\times100 = \mathbf{56\%}$
$\%S = \dfrac{32}{233}\times\dfrac{\text{mass of }BaSO_4}{\text{mass of compound}}\times100$
$\%S = \dfrac{32}{233}\times\dfrac{0.466}{0.2}\times100$
$\%S = 0.1373\times2.33\times100 \approx \mathbf{32\%}$
%C: $\dfrac{12}{44}\times\dfrac{0.22}{0.15}\times100 = 0.2727\times1.467\times100 \approx 40\%$
%H: $\dfrac{2}{18}\times\dfrac{0.09}{0.15}\times100 = 0.1111\times0.6\times100 = 6.67\%$
%O (by difference, since only C, H, O present): $100 - 40 - 6.67 = \mathbf{53.33\%}$
Summary: C = 40%, H = 6.67%, O = 53.33%
$\%Cl = \dfrac{35.5}{143.5}\times\dfrac{\text{mass of }AgCl}{\text{mass of compound}}\times100$
$\%Cl = \dfrac{35.5}{143.5}\times\dfrac{0.3675}{0.25}\times100$
$\%Cl = 0.2474\times1.47\times100 \approx \mathbf{36.37\%}$
Moles of $N_2 = \dfrac{22.4}{22400} = 0.001\text{ mol}$
Mass of nitrogen $= 0.001\times28 = 0.028\text{ g}$
$\%N = \dfrac{0.028}{0.1}\times100 = \mathbf{28\%}$
$\%Cl = \dfrac{35.5}{143.5}\times\dfrac{0.5735}{0.3}\times100 = 0.2474\times1.9117\times100 \approx 47.29\%$
Mass of Cl per mole of compound $= \dfrac{47.29}{100}\times85 \approx 40.2\text{ g}$
Number of Cl atoms per molecule $= \dfrac{40.2}{35.5} \approx \mathbf{1.13 \approx 1\text{ atom}}$
(Rounding to the nearest whole number, as expected for a real molecular formula, this compound contains 1 chlorine atom per molecule.)
Total moles of HCl taken $= 0.5\times0.050 = 0.025\text{ mol}$
Moles of excess HCl (neutralised by NaOH) $= 0.5\times0.020 = 0.01\text{ mol}$
Moles of HCl actually reacted with $NH_3$ = Total − Excess $= 0.025 - 0.01 = 0.015\text{ mol}$
This equals moles of nitrogen: mass of N $= 0.015\times14 = 0.21\text{ g}$
$\%N = \dfrac{0.21}{0.7}\times100 = \mathbf{30\%}$