Chapter 7, Problem 7.2P

Interpretation Introduction

Interpretation:

The number of ethoxyl groups present in the sample compound has to be calculated.

Explanation of Solution

Given,

    $\text{25}\text{.42mg}$ of sample.

    $\text{29}\text{.03mg}$ of silver iodide.

Let n be the number of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{O-}$ groups in the sample compound

    $\begin{array}{l}\text{R}{\left({\text{OCH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)}_{\text{n}}\text{ + nHI }\underset{}{\overset{}{\to }}\text{ R}{\left(\text{OH}\right)}_{\text{n}}{\text{ + nCH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}\\ \\ {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{I + Ag + H}}_{\text{2}}\text{O }\underset{}{\overset{}{\to }}{\text{ AgI + CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\\ \\ {\text{Moles of CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I = moles of AgI}\text{=}\frac{\text{0}\text{.02903g}}{\text{234}\text{.77g/mol}}\text{=}\text{1}\text{.24}\text{×}{\text{10}}^{\text{-4}}\text{mol}\\ \\ \text{Moles}\text{of}\text{sample}\text{=}\frac{\text{0}\text{.02542g}}{\text{417g/mol}}\text{=}\text{6}\text{.10}\text{×}{\text{10}}^{\text{-5}}\text{mol}\\ \\ \text{Number}\text{of}\text{ethoxyl}\text{group}\text{per}\text{molecule}\text{=}\frac{{\text{Moles of CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I }}{\text{Moles}\text{of}\text{sample}}\text{=}\frac{\text{1}\text{.24}\text{×}{\text{10}}^{\text{-4}}\text{mol}}{\text{6}\text{.10}\text{×}{\text{10}}^{\text{-5}}\text{mol}}\\ \\ \text{=}\text{2}\text{.03}\end{array}$

Therefore, two ethoxyl groups are present in the compound.