Question

What volume of 1-butanol is needed to convert 1.30 grams of

benzhydrol to n-butyl benzhydryl ether?

Answer 1

`\large \boxed{\text{0.646 mL}}`

Step-by-step explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

MM: 74.12 184.24

CH₃CH₂CH₂CH₂OH + (C₆H₅)₂CHOH ⟶ Product

m/g: 1.30

ρ/g·mL⁻¹: 0.8098

1. Moles of benzhydrol (BH)

`\text{Moles of BH } =\text{1300 mg BH} * \frac{\text{1 mmol BH}}{\text{184.24 mg BH}} =\text{7.056 mmol BH}`

2. Moles of butan-1-ol (BuOH)

The molar ratio is 1 mmol BuOH:1 mmol BH.

`\text{Moles of BuOH}=\text{7.056 mmol BH} * \frac{\text{1 mmol BuOH}}{ \text{1 mmol BH}} = \text{7.056 mmol BuOH}`

3. Mass of BuOH

`m = \text{7.056 mmol BuOH} * \frac{\text{74.12 mg BuOH}}{\text{1 mmol BuOH}} = \text{523.0 mg BuOH}`

4. Volume of BuOH

`V = \text{0.5230 g BuOH} * \frac{\text{1 mL BuOH}}{\text{0.8098 g BuOH}} = \text{0.646 mL BuOH}\\\\\text{You must use $\large \boxed{\textbf{0.646 mL}}$ of butan-1-ol}`