JEE Challenger
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ChemMedAdv 2023 • Paper 1

Calculate Weight of Tetrameric Cyclic Silicone Product from Hydrolysis

The stoichiometric reaction of 516 g516\text{ g} of dimethyldichlorosilane with water results in a tetrameric cyclic product X\mathbf{X} in 75%75\% yield. The weight (in g\text{g}) of X\mathbf{X} obtained is ___.

[Use, molar mass (g mol−1\text{g mol}^{-1}): H=1\text{H} = 1, C=12\text{C} = 12, O=16\text{O} = 16, Si=28\text{Si} = 28, Cl=35.5\text{Cl} = 35.5]

Official Numerical Answer222

Topics & Concepts

p-Block Elementsp-Block ElementsStoichiometry

Step-by-Step Solution

To determine the weight of the tetrameric cyclic silicone product X\mathbf{X} obtained, we follow a step-by-step stoichiometric analysis:

1. Molar Mass of Dimethyldichlorosilane, (CH3)2SiCl2(\text{CH}_3)_2\text{SiCl}_2

Using the given atomic masses:

  • Si=28 g mol−1\text{Si} = 28\text{ g mol}^{-1}
  • C=2×12=24 g mol−1\text{C} = 2 \times 12 = 24\text{ g mol}^{-1}
  • H=6×1=6 g mol−1\text{H} = 6 \times 1 = 6\text{ g mol}^{-1}
  • Cl=2×35.5=71 g mol−1\text{Cl} = 2 \times 35.5 = 71\text{ g mol}^{-1}

Molar mass of (CH3)2SiCl2=28+24+6+71=129 g mol−1\text{Molar mass of } (\text{CH}_3)_2\text{SiCl}_2 = 28 + 24 + 6 + 71 = 129\text{ g mol}^{-1}

2. Number of Moles of Dimethyldichlorosilane

Moles of (CH3)2SiCl2=Given massMolar mass=516 g129 g mol−1=4 moles\text{Moles of } (\text{CH}_3)_2\text{SiCl}_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{516\text{ g}}{129\text{ g mol}^{-1}} = 4\text{ moles}

3. Hydrolysis and Formation of Tetrameric Cyclic Product X\mathbf{X}

Hydrolysis of dimethyldichlorosilane yields dimethylsilanediol, which undergoes condensation polymerization to yield the cyclic tetramer X\mathbf{X}, octamethylcyclotetrasiloxane [(CH3)2SiO]4[(\text{CH}_3)_2\text{SiO}]_4:

4(CH3)2SiCl2+4H2O⟶[(CH3)2SiO]4+8HCl4(\text{CH}_3)_2\text{SiCl}_2 + 4\text{H}_2\text{O} \longrightarrow [(\text{CH}_3)_2\text{SiO}]_4 + 8\text{HCl}

From the stoichiometry of the reaction, 4 moles4\text{ moles} of (CH3)2SiCl2(\text{CH}_3)_2\text{SiCl}_2 theoretically produce 1 mole1\text{ mole} of the cyclic tetramer X\mathbf{X}.

4. Molar Mass of the Tetrameric Cyclic Product X\mathbf{X}

The monomeric unit of silicone is −[(CH3)2SiO]−-[(\text{CH}_3)_2\text{SiO}]-: Molar mass of (CH3)2SiO=28+16+(2×12)+(6×1)=74 g mol−1\text{Molar mass of } (\text{CH}_3)_2\text{SiO} = 28 + 16 + (2 \times 12) + (6 \times 1) = 74\text{ g mol}^{-1}

Since product X\mathbf{X} is tetrameric, its formula is [(CH3)2SiO]4[(\text{CH}_3)_2\text{SiO}]_4: Molar mass of X=4×74 g mol−1=296 g mol−1\text{Molar mass of } \mathbf{X} = 4 \times 74\text{ g mol}^{-1} = 296\text{ g mol}^{-1}

5. Theoretical and Actual Yield Calculation

  • Theoretical mass of X\mathbf{X} (for 100% yield from 4 moles4\text{ moles} of monomer): Theoretical mass=1 mole×296 g mol−1=296 g\text{Theoretical mass} = 1\text{ mole} \times 296\text{ g mol}^{-1} = 296\text{ g}

  • Actual mass of X\mathbf{X} (at 75%75\% yield): Actual mass=296 g×75100=296 g×0.75=222 g\text{Actual mass} = 296\text{ g} \times \frac{75}{100} = 296\text{ g} \times 0.75 = 222\text{ g}

Final Answer: The weight of X\mathbf{X} obtained is 222.

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