Chapter 15, Problem 15.110P

(a)

Interpretation Introduction

Interpretation:

Compound A containing $\text{C, H and O}$ in branched structure is optically active.  A $\text{0}\text{.500 g}$ of sample burns in excess ${\text{O}}_{\text{2}}$ to yield $\text{1}{\text{.25 g of CO}}_{\text{2}}$ and $\text{0}{\text{.613 g of H}}_{\text{2}}\text{O}$.  The empirical formula of the compound has to be determined.

Explanation of Solution

Given,

• $\text{1}{\text{.25 g of CO}}_{\text{2}}$
• $\text{0}{\text{.613 g of H}}_{\text{2}}\text{O}$
• Total sample = $\text{0}\text{.500 g}$.

From the given mass, moles can be calculated.

For Carbon:

  $\begin{array}{l}\text{Moles of C = }\left(\text{1}{\text{.25 g CO}}_{\text{2}}\right)\left(\frac{{\text{1 mol CO}}_{\text{2}}}{\text{440}{\text{.1 g CO}}_{\text{2}}}\right)\left(\frac{\text{1 mol C}}{{\text{1 mol CO}}_{\text{2}}}\right)\\ \\ \text{ = 0}\text{.0284026 mol C}\end{array}$

For Hydrogen:

  $\begin{array}{l}\text{moles of H = }\left(\text{0}{\text{.613 g H}}_{\text{2}}\text{O}\right)\left(\frac{{\text{1 mol H}}_{\text{2}}\text{O}}{\text{18}{\text{.02 g H}}_{\text{2}}\text{O}}\right)\left(\frac{\text{2 mol H}}{{\text{1 mol H}}_{\text{2}}\text{O}}\right)\\ \\ \text{ = 0}\text{.0680355 mol H}\end{array}$

The mass of carbon and hydrogen can be calculated from the moles obtained.

For carbon:

  $\begin{array}{l}\text{Mass of C (g) = }\left(\text{0}\text{.0284026 mol C}\right)\left(\frac{\text{12}\text{.01 g C}}{\text{1 mol C}}\right)\\ \\ \text{ = 0}\text{.341115 g C}\end{array}$

For hydrogen:

  $\begin{array}{l}\text{Mass of H (g) = }\left(\text{0}\text{.0680355 mol H}\right)\left(\frac{\text{1}\text{.008 g H}}{\text{1 mol H}}\right)\\ \\ \text{ = 0}\text{.06857978 g H}\end{array}$

The mass of oxygen can be determined by subtracting the mass of carbon and hydrogen from the total mass of $\text{C, H and O}$.

  $\begin{array}{l}\text{mass of O (g) = 0}\text{.500 g (C,}\text{}\text{H,O)-(0}\text{.341115 g C + 0}\text{.06857978 g H)}\\ \\ \text{ = 0}\text{.0903052 g O}\end{array}$

The mass of the oxygen is $\text{0}\text{.0903052 g}$.

By using the mass of oxygen, moles can be determined.

  $\begin{array}{l}\text{moles of O = }\left(\text{0}\text{.0903052 g O}\right)\left(\frac{\text{1 mol O}}{\text{16}\text{.00 g O}}\right)\\ \\ \text{ = 0}\text{.005644075 mol O}\end{array}$

The moles of $\text{C, H and O}$ are divided by smallest number of moles.

Carbon:

  $\frac{\text{0}\text{.0284026 mol C}}{\text{0}\text{.005644075 mol O}}\text{ = 5}$

Hydrogen:

  $\frac{\text{0}\text{.0680355 mol H}}{\text{0}\text{.005644075 mol O}}\text{ = 12}$

Oxygen:

  $\frac{\text{0}\text{.005644075 mol O}}{\text{0}\text{.005644075 mol O}}\text{ = 1}$

The empirical formula can be given as ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$.

(b)

Interpretation Introduction

Interpretation:

When $\text{0}\text{.225 g}$ of compound A vaporizes at $\text{755 torr}$ and ${\text{97}}^{\text{0}}\text{C}$, the vapour occupies $\text{78}\text{.0 mL}$.  The molecular formula of compound A has to be determined.

Concept Introduction:

The equation used to find the molecular weight of the compound is

  $\begin{array}{l}\text{PV = }\frac{\text{mRT}}{\text{mass}}\\ \\ \text{mass = }\frac{\text{mRT}}{\text{PV}}\end{array}$

Where, $\begin{array}{l}\text{P = pressure}\\ \\ \text{V = volume}\\ \\ \text{R = gas constant}\\ \\ \text{T = temperature}\\ \\ \text{m = mass of compound}\end{array}$

Explanation of Solution

The molecular weight of the compound can be calculated as

Given,

• pressure = $\text{755 torr}$
• temperature = ${\text{97}}^{\text{0}}\text{C}$
• volume = $\text{78}\text{.0 mL}$
• total mass = $\text{0}\text{.225 g}$

  $\begin{array}{l}\text{mass = }\frac{\text{mRT}}{\text{PV}}\\ \\ \text{ = }\frac{\left(\text{0}\text{.225g}\right)\left(\text{0}\text{.0821}\frac{\text{L}\text{.atm}}{\text{mol}\text{.K}}\right)\left(\left(\text{273+97}\right)\text{K}\right)}{\left(\text{755 torr}\right)\left(\text{78}\text{.0 mL}\right)}\left(\frac{\text{1 mL}}{{\text{10}}^{\text{-3}}\text{ L}}\right)\left(\frac{\text{760 torr}}{\text{1 atm}}\right)\\ \\ \text{ = 88}\text{.206 = 88}\text{.2 g/mol}\end{array}$

The empirical formula weight can be calculated.

Empirical formula: ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$

Empirical mass: $\text{88}\text{.15 g/mol}$.

As the molecular weight and empirical mass of nearly same, the molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$.

(c)

Interpretation Introduction

Interpretation:

On careful oxidation, Compound A yields a ketone.  The compound A has to be drawn and named.  The chiral centre has to be circled.

Concept Introduction:

Carbonyl compounds are formed by the oxidation of alcohols.  Ketone is formed on oxidation of secondary alcohol.

Explanation of Solution

The compound A is a secondary alcohol as it gives a ketone as product on oxidation.  The structure of the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$ can be

The chiral centre in the structure is the carbon bonded to four different groups, ${\text{-CH}}_{\text{3}}{\text{, -CH}}_{\text{2}}\text{, -OH and H}$.