Chapter 9, Problem 9.45FU

(a)

Interpretation Introduction

Interpretation: The mass of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{Cl}$ form from $2.2\times {10}^2\text{ kg}$ of naphthol should be calculated.

Concept Introduction: The number of moles is calculated using the formula:

  $\text{number of moles = }\frac{\text{given mass}}{\text{Molar Mass}}$

Answer to Problem 9.45FU

  $2.2\times {10}^2\text{ kg}$ of naphthol forms $\text{ 315}\text{.3 kg}$ of $\text{C11H7O2Cl}$ .

Explanation of Solution

Given:

Mass of naphthol = 2.2 x 10² kg

Molar mass of naphthol = 144.17 g/mol

Equality:

1 kg = 1000 g

1 mole napthol = 144.17 g

1 mole napthol = 1 mole $\text{C11H7O2Cl}$

1 mole $\text{C11H7O2Cl}$ = 206.62 g $\text{C11H7O2Cl}$

  $\begin{array}{c}\text{2}{\text{.2 x 10}}^{\text{2}}\text{ kg x }\frac{\text{1000 g}}{\text{1 kg}}\text{ x }\frac{\text{1 mole napthol}}{\text{144}\text{.17 g napthol}}\text{x}\frac{\text{1 mole C11H7O2Cl}}{\text{1 mole napthol}}\text{x}\frac{\text{206}\text{.62 g C11H7O2Cl}}{\text{1 mol C11H7O2Cl}}\text{x}\frac{\text{1kg}}{\text{1000 g}}\\ \text{= 315}\text{.3 kg}\end{array}$

(b)

Interpretation Introduction

Interpretation: The theoretical yield of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{Cl}$ should be calculated.

Concept Introduction: The number of moles is calculated using the formula:

  $\text{number of moles = }\frac{\text{given mass}}{\text{Molar Mass}}$

Answer to Problem 9.45FU

The theoretical yield of $\text{C11H7O2Cl}$ is $\text{ 143}\text{.4 g}$ .

Explanation of Solution

Given:

Mass of napthol = 100 g

Molar mass of napthol = 144.17 g/mol

Mass of phosgene = 100 g

Molar mass of phosgene = 98.92 g/mol

Equality:

1 mole of phosgene = 98.92 g of phosgene

1 mole napthol = 144.17 g of napthol

Convert mass to moles;

  $\text{100 g phosgene x }\frac{\text{1molephosgene}}{\text{98}\text{.92 g of phosgene}}\text{= 1}\text{.01 mole}$

  $\text{100 g napthol x }\frac{\text{1mole napthol}}{\text{144}\text{.17 g of napthol}}\text{= 0}\text{.694 mole}$

The molar ratio of naphthol and phosgene is 1:1 thus, limiting reagent can be determined by just seeing the moles of both reactants.

Since moles of napthol are less so it must be limiting reagent and will determine the amount of product.

1 mole napthol = 1 mole $\text{C11H7O2Cl}$

1 mole $\text{C11H7O2Cl}$ = 206.62 g $\text{C11H7O2Cl}$

Hence $\text{0}\text{.694 moles napthol x }\frac{\text{1mole C11H7O2Cl}}{\text{1 mole napthol}}\text{ x}\frac{\text{206}\text{.62 g C11H7O2Cl}}{\text{1 mole C11H7O2Cl}}\text{= 143}\text{.4 g C11H7O2Cl}$

(c)

Interpretation Introduction

Interpretation: The percent yield should be calculated if the actual yield of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}{\text{O}}_{\text{2}}\text{Cl}$ is 115 g.

Concept Introduction: The actual yield of the product is the amount of product which is formed at the end of reaction experimentally whereas theoretical yield represents the amount of product calculated by expecting that 100 % of reactant changes to product. The percent yield can be shown as:

  $\text{Percent yield = }\frac{\text{Actual yield}}{\text{Theoretical yield}}\text{ x 100}$

Answer to Problem 9.45FU

The percent yield of $\text{C11H7O2Cl}$ is 80.2 %.

Explanation of Solution

Given:

Actual yield of $\text{C11H7O2Cl}$ = 115 g

Theoretical yield = 143.4 g

  $\text{Percent yield = }\frac{\text{Actual yield}}{\text{Theoretical yield}}\text{ x 100}$

Plug the values;

  $\text{Percent yield = }\frac{\text{115 g}}{\text{143}\text{.4 g}}\text{ x 100 = 80}\text{.2 \%}$