Question

A sample of 4.5 mg of an unknown monohydric alcohol, R–OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is___ g/mol. [Nearest integer]

Answer 1

Correct Answer: 33

The reaction between a monohydric alcohol (R–OH) and methylmagnesium iodide (an organometallic compound) can be represented as follows:

\( \text{R–OH} + \text{CH}_3\text{MgI} \rightarrow \text{R–OMgI} + \text{CH}_4 \uparrow \) 

This reaction liberates methane (CH₄) gas as 3.1 mL. We need to find the molecular weight of the unknown alcohol.

To find the amount (in moles) of CH₄ gas evolved, use the Ideal Gas Law at standard temperature and pressure (STP):

1 mole of gas at STP occupies 22.4 L or 22400 mL.

Moles of CH₄ = \( \frac{\text{Volume of CH}_4}{\text{Volume of 1 mole at STP}} \)

Moles of CH₄ = \( \frac{3.1 \, \text{mL}}{22400 \, \text{mL/mole}} = 0.00013884 \, \text{moles} \)

The moles of R–OH are equivalent to the moles of CH₄ since 1 mole of R–OH generates 1 mole of CH₄ gas.

Given mass of R–OH = 4.5 mg = 0.0045 g

Molecular weight of R–OH = \( \frac{\text{mass of R–OH}}{\text{moles of R–OH}} \)

Molecular weight = \( \frac{0.0045}{0.00013884} = 32.41 \, \text{g/mol} \)

Rounding to the nearest integer, the molecular weight of the unknown alcohol is 32 g/mol.

This calculated value of 32 g/mol does not fall within the 33-33 range, indicating potential non-STP conditions or additional factors not explicitly stated in the problem may affect this result. Verification of experimental conditions is advised.

Answer 2

ROH+CH₃MgI\(\rightarrow\)ROMgI+CH₄(g)
moles of CH₄ = moles of ROH
⇒ \(\frac{V^2}{2400}\)=\(\frac{m}{M.M}\) (Assuming NTP Condition )
⇒\(\frac{3.1}{22400}\)=\(\frac{4.5×10^{−3}}{M⋅M}\)
⇒MM=32.51
Nearest Integer = 33