Chapter 11, Problem 11.61P

Interpretation Introduction

Interpretation:

The percentage yield of following reaction has to be determined.

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\left(aq\right)+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\left(aq\right)\stackrel{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\to }{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_2{\text{CH}}_3\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Percent yield is ratio of actual yield by theoretical yield. Actual yield is amount of actual product formed in chemical reaction.

Theoretical yield is amount of product formed from reactant according to stoichiometry in balanced chemical reaction.

The expression for percent yield is as follows:

  $\text{Percent yield}\left(\%\right)=\left(\frac{\text{actual yield}}{\text{theoreticalyield}}\right)\left(100\right)$