Chapter 3, Problem 3.112P

Interpretation Introduction

Interpretation:

The percent yield for the formation of zirconium oxalate is to be calculated.

Concept introduction:

The expression to calculate the moles of a solute when given mass and molecular mass of compound are given is as follows:

  $\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The formula to calculate the percentage yield is as follows:

  $\text{Percentage yield}=\left(\frac{\text{actual yield}}{\text{theoretical yield}}\right)100\%$

The formula to calculate moles is as follows:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$        (1)