Problem 102 - Chemistry and Chemical Reactivity

Question

Cloth can be waterproofed by coating it with a silicone layer. This is done by exposing the cloth to \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}\) vapor. The silicon compound reacts with OH groups on the cloth to form a waterproofing film (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) of \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right]_{n},\) where \(n\) is a large integer number. $$\begin{aligned}&n\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}+2 n \mathrm{OH}^{-} \rightarrow\\\&&2 n \mathrm{Cl}^{-}+n \mathrm{H}_{2} \mathrm{O}+\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right]_{n} \end{aligned}$$ The coating is added layer by layer, with each layer of \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right]_{n}\) being 0.60 nm thick. Suppose you want to waterproof a piece of cloth that is 3.00 square meters, and you want 250 layers of waterproofing compound on the cloth. What mass of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}\) do you need?

Step-by-Step Solution

Answer

Approximately 0.785 grams of \((\mathrm{CH}_{3})_{2} \mathrm{SiCl}_{2}\) are needed.

Step 1: Calculate the Total Volume of Waterproofing

First, we need to calculate how much volume the 250 layers of coating will occupy. Each layer of the compound is 0.60 nm thick. To find the total thickness in meters, multiply the number of layers by the thickness of each layer: \[ \text{Total thickness} = 250 \times 0.60 \text{ nm} = 150 \text{ nm} = 150 \times 10^{-9} \text{ m} = 1.50 \times 10^{-7} \text{ m}\] The total volume of the coating will then be the area of the cloth multiplied by the total thickness:\[ \text{Volume} = 3.00 \text{ m}^2 \times 1.50 \times 10^{-7} \text{ m} = 4.50 \times 10^{-7} \text{ m}^3 \]

Step 2: Calculate the Mass of the Waterproofing Film

Using the density of the waterproofing film, calculate the total mass. The density is given as 1.0 g/cm³, which is equivalent to 1000 kg/m³:\[ \text{Mass} = \text{Density} \times \text{Volume} = 1000 \text{ kg/m}^3 \times 4.50 \times 10^{-7} \text{ m}^3 = 4.50 \times 10^{-4} \text{ kg} \]Converting to grams:\[ 4.50 \times 10^{-4} \text{ kg} = 0.45 \text{ g} \]

Step 3: Determine the Molar Mass of the Waterproofing Compound

Calculate the molar mass of \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO} \). The molar masses of the individual atoms are as follows: - C: 12.01 g/mol- H: 1.01 g/mol- Si: 28.09 g/mol- O: 16.00 g/molUsing this info, calculate:\[ \text{Molar Mass} = 2(12.01) + 6(1.01) + 28.09 + 16.00 = 74.16 \text{ g/mol} \]

Step 4: Calculate Moles of Waterproofing Film

Determine the moles of \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO} \) required:\[ \text{Moles} = \frac{ \text{Mass of } \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}}{ \text{Molar Mass} } = \frac{0.45 \text{ g}}{74.16 \text{ g/mol}} \approx 0.00606 \text{ mol} \]

Step 5: Calculate Moles of \((\mathrm{CH}_{3})_{2} \mathrm{SiCl}_{2}\) Required

From the chemical equation:\[ n(\mathrm{CH}_3)_2 \mathrm{SiCl}_2 + 2n \mathrm{OH}^{-} \rightarrow 2n \mathrm{Cl}^{-} + n \mathrm{H}_2\mathrm{O} + (\left(\mathrm{CH}_3\right)_2\mathrm{SiO})_n \]1 mole of \((\mathrm{CH}_3)_2 \mathrm{SiO})_n\) corresponds to 1 mole of \((\mathrm{CH}_3)_2 \mathrm{SiCl}_2\). So, you need 0.00606 moles of \((\mathrm{CH}_3)_2 \mathrm{SiCl}_2\).

Step 6: Determine the Molar Mass of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}\)

Calculate the molar mass of \((\mathrm{CH}_3)_2 \mathrm{SiCl}_2\):- Si: 28.09 g/mol- C: 12.01 g/mol- H: 1.01 g/mol- Cl: 35.45 g/molCalculate the total:\[ \text{Molar Mass} = 2(12.01) + 6(1.01) + 28.09 + 2(35.45) = 129.61 \text{ g/mol} \]

Step 7: Calculate the Mass of \((\mathrm{CH}_3)_2 \mathrm{SiCl}_2\) Required

The mass needed for the 0.00606 moles of \((\mathrm{CH}_3)_2 \mathrm{SiCl}_2\) is calculated as follows:\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.00606 \text{ mol} \times 129.61 \text{ g/mol} \approx 0.785 \text{ g} \]

Key Concepts

Understanding Chemical Reactions in Silicone Waterproofing

Silicone waterproofing relies on a fascinating chemical reaction. When cloth is exposed to dimethyldichlorosilane \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2} \) vapor, it reacts with hydroxide ions (OH-) on the cloth's surface. This reaction is important because it forms a durable waterproof layer called polysiloxane \( \left(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right)_{n} \).

The balanced equation for this reaction is:
\[ n\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2} + 2n \mathrm{OH}^{-} \rightarrow 2n \mathrm{Cl}^{-} + n \mathrm{H}_{2} \mathrm{O} + \left(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right)_{n} \]

*Key Points of the Reaction:*

• Each molecule of dimethyldichlorosilane reacts with two hydroxide ions.
• This produces chloride ions (\( 2n \mathrm{Cl}^{-} \)), water (\( n \mathrm{H}_{2} \mathrm{O} \)), and the waterproofing film \( \left(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right)_{n} \).
• The layers are formed gradually, making the coating effective and long-lasting.

Understanding this chemical process is crucial, as each step contributes to creating the protective layer that keeps fabrics dry.

Molar Mass Calculation of Compounds

Molar mass calculations are essential for determining how much of a substance we need in a reaction. Calculating the molar mass involves adding up the atomic masses of all atoms in a compound.

*Let's look at two important compounds in our problem:*

• **\( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO} \):**
  - Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
  - Hydrogen (H): 1.01 g/mol × 6 = 6.06 g/mol
  - Silicon (Si): 28.09 g/mol
  - Oxygen (O): 16.00 g/mol
  - **Total Molar Mass:** 74.16 g/mol
  
• **\( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2} \):**
  - Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol
  - Hydrogen (H): 1.01 g/mol × 6 = 6.06 g/mol
  - Silicon (Si): 28.09 g/mol
  - Chlorine (Cl): 35.45 g/mol × 2 = 70.90 g/mol
  - **Total Molar Mass:** 129.61 g/mol
  

Performing these calculations accurately ensures that we have the right amount of each compound for the process, guaranteeing the effectiveness of the waterproofing treatment.

Connecting Density and Volume in Calculations

Density is a critical concept that connects mass and volume. In the context of silicone waterproofing, understanding density helps to calculate how much coating can be applied.

*Applying Density to Calculate Mass:*

• **Density of the Waterproofing Film:** 1.0 g/cm³ or 1000 kg/m³
• **Volume Calculation:**
  - The thickness of each layer is 0.60 nm, so for 250 layers, the total thickness is 150 nm.
  - Convert this to meters: 150 nm = 1.50 × 10^{-7} m.
  - The area of the cloth is 3.00 m².
  - **Total Volume:** 3.00 m² × 1.50 × 10^{-7} m = 4.50 × 10^{-7} m³
  
• **Mass Calculation:**
  - Using the density formula: Mass = Density × Volume
  - Mass = 1000 kg/m³ × 4.50 × 10^{-7} m³ = 4.50 × 10^{-4} kg
  - Convert to grams: 4.50 × 10^{-4} kg = 0.45 g
  

Calculating these quantities accurately allows for the correct application of silicone layers, ensuring a reliable waterproof barrier is achieved across the fabric surface.