Chapter 5, Problem 5.47UKC

(a)

Interpretation Introduction

Interpretation:

The molar mass of linolenic acid has to be calculated.

Answer to Problem 5.47UKC

The molar mass of linolenic acid is $\text{278}\text{.4}\text{g/mol}\text{.}$

Explanation of Solution

The formula weight of ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}$ is calculated as,

  $\begin{array}{l}\text{30}\text{H}\text{atoms×1}\text{.008}\text{amu}\text{=}\text{30}\text{.24}\text{amu}\\ \text{18}\text{C}\text{atoms×12}\text{.01}\text{amu}\text{=}\text{216}\text{.18}\text{amu}\\ \text{2}\text{O}\text{atoms×16}\text{.00}\text{amu}\text{=}\underset{\_}{\text{32}\text{.00}\text{amu}}\\ \text{Formula weight}\text{of}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}\text{=278}\text{.42}\text{amu}\end{array}$

Since, the formula weight of ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}$ is $\text{278}\text{.4}\text{amu,}$ the molar mass of linolenic acid is $\text{278}\text{.4}\text{g/mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The balanced reaction of linolenic acid with hydrogen has to be written.

Answer to Problem 5.47UKC

The balanced reaction of linolenic acid with hydrogen is,

  ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+3H}}_{\text{2}}\stackrel{}{\to }{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$

Explanation of Solution

The given reaction is,

  ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+3H}}_{\text{2}}\stackrel{}{\to }{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$

Hydrogen atoms are unbalanced in the reaction.

Place the coefficient $\text{3}$ before hydrogen in the reactant side to balance hydrogen.

The balanced reaction of linolenic acid with hydrogen is,

  ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+3H}}_{\text{2}}\stackrel{}{\to }{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$

Atoms in the reactant side: $\text{36}\text{H's,18}\text{C'S}$ and $\text{2}\text{O's}\text{.}$

Atoms in the product side: $\text{36}\text{H's,18}\text{C'S}$ and $\text{2}\text{O's}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The balanced reaction of linolenic acid with oxygen has to be written.

Answer to Problem 5.47UKC

The balanced reaction of linolenic acid with oxygen is,

  ${\text{2 C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+49 O}}_{\text{2}}\stackrel{}{\to }\text{36}{\text{CO}}_{\text{2}}\text{+30}{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The given reaction is,

  ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

The reaction is unbalanced.

Place the coefficient $\text{2}$ to balance carbon before ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}$ in the reactant side and place the coefficient $\text{32}$ to balance carbon before ${\text{CO}}_{\text{2}}$ in the product side.

Place the coefficient $\text{2}$ to balance oxygen before ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}$ in the reactant side.  Place the coefficient $\text{49}$ to balance oxygen before ${\text{O}}_{\text{2}}$ in the reactant side and place the coefficient $\text{32}$ to balance oxygen before ${\text{CO}}_{\text{2}}$ in the product side and place the coefficient $\text{30}$ to balance oxygen before ${\text{H}}_{\text{2}}\text{O}$ in the product side.

Place the coefficient $\text{2}$ to balance hydrogen before ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}$ in the reactant side.  place the coefficient $\text{30}$ to balance hydrogen before ${\text{H}}_{\text{2}}\text{O}$ in the product side.

The balanced reaction of linolenic acid with oxygen is,

  ${\text{2 C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+49 O}}_{\text{2}}\stackrel{}{\to }\text{36}{\text{CO}}_{\text{2}}\text{+30}{\text{H}}_{\text{2}}\text{O}$

Atoms in the reactant side: $\text{36}\text{C's,60}\text{H'S}$ and $\text{102}\text{O's}\text{.}$

Atoms in the product side: $\text{36}\text{C's,60}\text{H'S}$ and $\text{102}\text{O's}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The number of grams of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ formed from $\text{10}\text{.0}\text{g}$ of linolenic acid has to be given.

Answer to Problem 5.47UKC

The number of grams of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ formed from $\text{10}\text{.0}\text{g}$ of linolenic acid is $\text{10}\text{.2}\text{g}\text{.}$

Explanation of Solution

Given,

Grams of linolenic acid are $\text{10}\text{.0}\text{g}\text{.}$

The balanced chemical equation is,

  ${\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}{\text{+3H}}_{\text{2}}\stackrel{}{\to }{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$

The moles of the reactant linolenic acid is calculated as,

  $\text{10}\cancel{\text{g}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}}{\text{278}\text{.4}\cancel{\text{g}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}}}\text{=0}\text{.0359}\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}$

The moles of the product ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ is calculated as,

  $\text{0}\text{.0359}\cancel{\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}}{\text{1}\cancel{\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{30}}{\text{O}}_{\text{2}}}}\text{=0}\text{.0359}\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$

The number of grams of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ is calculated as,

  $\text{0}\text{.0359}\cancel{\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}}\text{×}\frac{\text{284}\text{.48}\text{g}{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}}{\text{1}\cancel{\text{mol}{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}}}\text{=10}\text{.2}\text{g}{\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$

The number of grams of ${\text{C}}_{\text{18}}{\text{H}}_{\text{36}}{\text{O}}_{\text{2}}$ formed from $\text{10}\text{.0}\text{g}$ of linolenic acid is $\text{10}\text{.2}\text{g}\text{.}$