What amount of bromine will be required to convert 2 g of phenol into 2, 4, 6-tribromophenol?
(Given molar mass in g mol\(^{-1}\) of C, H, O, Br are 12, 1, 16, 80 respectively)
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- 10.22 g
- 6.0 g
- 4.0 g
- 20.44 g
The Correct Option is A
Approach Solution - 1
To determine the amount of bromine required to convert 2 g of phenol into 2,4,6-tribromophenol, we need to consider the balanced chemical reaction and the stoichiometry involved. Let's go through the steps:
- The reaction between phenol \( \text{C}_6\text{H}_5\text{OH} \) and bromine (\( \text{Br}_2 \)) to form 2,4,6-tribromophenol \( \text{C}_6\text{H}_2\text{Br}_3\text{OH} \) is as follows:
\(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\) - Determine the molar masses:
- Molar mass of phenol (\( \text{C}_6\text{H}_5\text{OH} \)):
Carbon (C): 12 × 6 = 72 g/mol
Hydrogen (H): 1 × 6 = 6 g/mol
Oxygen (O): 16 × 1 = 16 g/mol
Total = 72 + 6 + 16 = 94 g/mol - Molar mass of bromine (\( \text{Br}_2 \)):
Bromine (Br): 80 × 2 = 160 g/mol
- Molar mass of phenol (\( \text{C}_6\text{H}_5\text{OH} \)):
- Determine the moles of phenol:
\(n_{\text{phenol}} = \frac{2 \text{ g}}{94 \text{ g/mol}} \approx 0.0213 \text{ mol}\) - According to the balanced equation, 1 mole of phenol reacts with 3 moles of bromine. Therefore, the moles of bromine required will be:
\(n_{\text{Br}_2} = 3 \times 0.0213 = 0.0639 \text{ mol}\) - Calculate the mass of bromine needed:
\(m_{\text{Br}_2} = 0.0639 \text{ mol} \times 160 \text{ g/mol} = 10.224 \text{ g}\)
Therefore, the amount of bromine required is approximately 10.22 g, corresponding to the correct option: 10.22 g. This calculation confirms that option \(10.22 \text{ g}\) is correct.
Approach Solution -2
To determine the amount of bromine needed to convert 2 g of phenol into 2,4,6-tribromophenol, we follow this procedure:
Step 1: Calculate the molar mass of phenol (C6H5OH)
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of phenol = 6(12) + 6(1) + 16 = 94 g/mol
Step 2: Calculate the number of moles of phenol
Number of moles = mass / molar mass = 2 g / 94 g/mol ≈ 0.0213 mol
Step 3: Determine the reaction and molar mass of tribromophenol
Reaction: C6H5OH + 3 Br2 → C6H2Br3OH + 3 HBr
Molar mass of Br2 = 2(80) = 160 g/mol
Step 4: Calculate bromine required for the reaction
Bromine required per mol of phenol = 3 mols of Br2
Number of moles of Br2 = 3 × 0.0213 mol = 0.0639 mol
Mass of Br2 = moles × molar mass = 0.0639 mol × 160 g/mol = 10.224 g
Thus, the amount of bromine required is 10.22 g.
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