Q.In the estimation of sulphur by the Carius method, 0.468 g of an organic sulphur compound gave 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Concept and Intuition

In the Carius method for phosphorus, the organic compound is oxidatively decomposed (fuming $\mathrm{HNO_3}$), converting all its phosphorus to phosphate, which is precipitated as $\mathrm{MgNH_4PO_4}$ and then ignited to the stable, easily-weighed pyrophosphate $\mathrm{Mg_2P_2O_7}$. Since the stoichiometric relationship between the original compound's phosphorus and the final $\mathrm{Mg_2P_2O_7}$ is fixed (2 P atoms end up in every 1 formula unit of $\mathrm{Mg_2P_2O_7}$), weighing the pyrophosphate lets you back-calculate the mass, and hence percentage, of phosphorus in the original sample.

Step-by-Step Solution

1. Molar mass of $\mathrm{Mg_2P_2O_7}$: $2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222\ \mathrm{g/mol}$.
2. Moles of $\mathrm{Mg_2P_2O_7}$ obtained: $\dfrac{0.444\ \mathrm{g}}{222\ \mathrm{g/mol}} = 0.002\ \mathrm{mol}$.
3. Each formula unit of $\mathrm{Mg_2P_2O_7}$ contains 2 phosphorus atoms (it comes from combining 2 $\mathrm{MgNH_4PO_4}$ units on ignition): moles of P $= 2 \times 0.002 = 0.004\ \mathrm{mol}$.
4. Mass of phosphorus $= 0.004\ \mathrm{mol} \times 31\ \mathrm{g/mol} = 0.124\ \mathrm{g}$. …

Concept and Intuition

In combustion analysis, all the carbon in the sample ends up in the CO₂ produced, and all the hydrogen ends up in the H₂O produced. Whatever mass is left over (total sample mass minus C and H masses) must be oxygen, since the compound contains only C, H, and O.

Step-by-Step Solution

1. Mass of C $=0.942\times\dfrac{12}{44}=0.257$ g $\Rightarrow$ moles C $=0.257/12=0.0214$.
2. Mass of H $=0.231\times\dfrac{2}{18}=0.0257$ g $\Rightarrow$ moles H $=0.0257/1=0.0257$.
3. Mass of O $=0.42-0.257-0.0257=0.1374$ g $\Rightarrow$ moles O $=0.1374/16=0.00859$.
4. Divide by the smallest (0.00859): C : H : O $\approx2.49:2.99:1.00\approx2.5:3:1$. …

Concept and Intuition

In Carius estimation of sulfur, the organic compound is oxidised so that all its sulfur precipitates as $BaSO_4$. Since 1 mole of $BaSO_4$ (233 g) corresponds to 1 mole of S (32 g), the mass of sulfur in the sample is $\dfrac{32}{233}\times(\text{mass of }BaSO_4)$, and dividing by the sample mass and multiplying by 100 gives the percentage of sulfur.

Step-by-Step Solution

1. Molar mass of $BaSO_4 = 137 + 32 + 4(16) = 233$ g/mol.
2. Formula: $\%S = \dfrac{32}{233}\times \dfrac{\text{mass of }BaSO_4}{\text{mass of compound}}\times 100$.
3. Substitute known values: $8.89 = \dfrac{32 \times 0.233}{233 \times x}\times 100$.
4. Compute numerator: $32 \times 0.233 \times 100 = 745.6$. …

Concept and Intuition

In combustion analysis, all the carbon in the sample ends up in $\mathrm{CO_2}$ and all the hydrogen ends up in $\mathrm{H_2O}$. Knowing the masses of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ produced lets us back-calculate the mass of C and H originally present. Any remaining mass (sample mass − mass C − mass H) must be oxygen, since the compound is stated to contain only C, H, and O.

Step-by-Step Solution

1. Moles of $\mathrm{CO_2} = \dfrac{0.46}{44} = 0.010455$ mol $\Rightarrow$ mass of C $= 0.010455\times12 = 0.12545$ g.
2. %C $= \dfrac{0.12545}{0.2}\times100 = 62.7\%$.
3. Moles of $\mathrm{H_2O} = \dfrac{0.25}{18} = 0.013889$ mol $\Rightarrow$ mass of H $= 0.013889\times2 = 0.027778$ g. …

Concept and Intuition

Direct nitration of aniline (without protecting the -NH₂ group) partially protonates the amine in the strongly acidic nitrating mixture, so the reaction proceeds through a mix of free aniline (activating, ortho/para-directing) and anilinium ion (deactivating, meta-directing) pathways, giving a characteristic product distribution: ortho ≈ 51%, para ≈ 47%, meta ≈ 2%. So B here is p-nitroaniline. Diazotisation of a primary aromatic amine converts -NH₂ to a diazonium salt (-N₂⁺Cl⁻) at low temperature, and reacting this diazonium salt with CuCN/KCN (a Sandmeyer-type reaction) substitutes -N₂⁺ with a -CN (nitrile) group, releasing N₂ gas.

Step-by-Step Solution

1. Aniline + HNO₃/H₂SO₄ at 288 K → 51% ortho-nitroaniline (A), 47% para-nitroaniline (B), 2% meta-nitroaniline (C).
2. B (p-nitroaniline) $\xrightarrow{NaNO_2/HCl,\ 273-278K}$ p-nitrobenzenediazonium chloride. …

Concept and Intuition

Two consecutive dehydrohalogenations of a vicinal dihalide (using strong bases, the second — alc. KOH being insufficient for the resulting vinyl halide — needing $NaNH_2$) build a triple bond, converting a dibromide into an alkyne. Terminal alkynes undergo Markovnikov (Hg²⁺/H⁺-catalyzed) hydration to give methyl ketones exclusively, via an enol intermediate that tautomerizes.

Step-by-Step Solution

1. Starting material $\mathrm{CH_3-CHBr-CH_2Br}$ has vicinal (adjacent-carbon) bromines.
2. Alc. KOH, $\Delta$ performs the first E2 elimination, forming a bromoalkene (e.g. $\mathrm{CH_2=CBr-CH_3}$).
3. $NaNH_2$, a much stronger base, performs the second elimination on the less-reactive vinylic C–H, giving the alkyne: $A = \mathrm{CH_3-C \equiv CH}$ (propyne). …

Concept and Intuition

Empirical formula determination converts mass percentages to moles (using atomic masses), then finds the simplest whole-number mole ratio by dividing every mole value by the smallest one.

Step-by-Step Solution

1. Assume 100 g of compound: 72.7 g C, 13.1 g H, 14.1 g N.
2. Moles of C $= 72.7/12 = 6.058$.
3. Moles of H $= 13.1/1 = 13.1$.
4. Moles of N $= 14.1/14 = 1.007$.
5. Divide each by the smallest (1.007): C $\approx 6.02$, H $\approx 13.01$, N $= 1.00$.
6. Simplest ratio C:H:N = 6:13:1, giving empirical formula $\mathrm{C_6H_{13}N}$. …

Concept and Intuition

An empirical formula is found by converting the given mass percentages (or mass ratio) of each element into moles (dividing by atomic mass), then finding the simplest whole-number ratio of those mole values. Carbon's atomic mass (12) is much larger than hydrogen's (1), so even though the mass ratio favours carbon (4:1), the mole ratio can favour hydrogen once the division by atomic mass is done.

Step-by-Step Solution

1. Take a convenient basis: assume 100 g total sample, split as 80 g C and 20 g H (matching the 4:1 mass ratio).
2. Moles of C: $\dfrac{80}{12} = 6.67$ mol.
3. Moles of H: $\dfrac{20}{1} = 20$ mol.
4. Divide both by the smaller value (6.67): C: $\dfrac{6.67}{6.67}=1$; H: $\dfrac{20}{6.67}=3$. …

Step 1 - moles of carbon. In photosynthesis $O_2$ is released in a $1:1$ mole ratio with the $CO_2$ consumed ($6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$):

Step 2 - moles of hydrogen.

Mass check: $85.7 + 14.4 \approx 100\ g$ (matches the sample).

Step 3 - empirical formula.

$C : H = 7.14 : 14.4 \approx 1 : 2 \;\Rightarrow\; \text{empirical } CH_2.$ …

Concept and Intuition

In a metal-deficient non-stoichiometric compound like $Cu_{1.8}S$, the metal

site vacancies are electronically compensated by some of the remaining metal

ions being oxidized to a higher charge state. Here, sulfide is always $S^{2-}$,

so the total positive charge contributed by all copper ions (a mix of $Cu^+$

and $Cu^{2+}$) must sum to exactly $+2$ per formula unit, even though there

are only $1.8$ (not 2) copper atoms present.

Step-by-Step Solution

1. Let the amount of $Cu^{2+}$ be $y$ mol, so $Cu^{+}$ present $= (1.8-y)$ mol (total copper $=1.8$).
2. Total positive charge must balance the $-2$ charge of one $S^{2-}$: $2(y) + 1(1.8-y) = 2$.
3. Simplify: $2y + 1.8 - y = 2 \Rightarrow y + 1.8 = 2 \Rightarrow y = 0.2$. …

Concept and Intuition

In combustion analysis, all the carbon in the sample ends up as $CO_2$ and all the hydrogen ends up as $H_2O$. Since we know the molar mass fraction of C in $CO_2$ (12/44) and of H in $H_2O$ (2/18), we can back-calculate the mass of each element in the original sample, then express these as percentages of the sample mass.

Step-by-Step Solution

1. Mass of C $= \dfrac{12}{44}\times 0.535 = 0.1459\ g$. $\%C = \dfrac{0.1459}{0.765}\times100 = 19.07\%$.
2. Mass of H $= \dfrac{2}{18}\times 0.13 = 0.01444\ g$. $\%H = \dfrac{0.01444}{0.765}\times100 = 1.888\%$. …