Chapter 22, Problem 160IP

(a)

Interpretation Introduction

Interpretation: The empirical formula and the molecular formula of the given helicene are to be calculated. The balanced chemical reaction for the combustion of helicene is to be stated.

Concept introduction: Helicines are defined as the polycyclic aromatic compounds in which the aromatic ring is annulated to provide helically shaped molecule. The molecular formula of any organic compound is determined by using empirical formula when the percent of each element is given in the compound.

To determine: The empirical formula of the given helicene.

Answer to Problem 160IP

Answer

The empirical formula of the given helicene is ${\text{C}}_{\text{5}}{\text{H}}_{\text{3}}$.

Explanation of Solution

Explanation

The empirical formula of the given helicene is ${\text{C}}_{\text{5}}{\text{H}}_{\text{3}}$.

Helicene is an aromatic compound so it must contain carbon and hydrogen atoms.

Given

Weight of ${\text{CO}}_{\text{2}}$ is $0.50\text{63 g}$.

Weight of compound is $0.145\text{0 g}$.

The weight of carbon is calculated by using the formula,

$\text{Weight of carbon}={\text{Weight of CO}}_{\text{2}}\times \frac{\text{Molecular weight of carbon}}{{\text{Molecular weightof CO}}_{\text{2}}}$

The molar mass of carbon is $12\text{g/mol}$.

The molar mass of carbon dioxide is $44\text{g/mol}$.

Substitute the value of weight of ${\text{CO}}_{\text{2}}$, molecular weight of carbon and ${\text{CO}}_{\text{2}}$ to calculate the weight of carbon in the above formula.

$\begin{array}{c}\text{Weight of carbon}={\text{Weight of CO}}_{\text{2}}\times \frac{\text{Molecular weight of carbon}}{{\text{Molecular weightof CO}}_{\text{2}}}\\ =0.506\text{3 g}\times \frac{\text{12 g/mol}}{4\text{4 g/mol}}\\ =0.138\text{0 g}\end{array}$

The weight of carbon is $0.13\text{80 g}$.

Therefore the weight of hydrogen is calculated as,

$\begin{array}{c}\text{Weight of hygrogen}=\text{Weight of compound}-\text{Weight of carbon}\\ =0.145\text{0 g}-0.138\text{0 g}\\ =0.00\text{7 g}\end{array}$

Therefore the number of moles of carbon and hydrogen is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the value of the given mass and the molar mass, to calculate the number of moles of carbon and hydrogen in the above equation.

For carbon,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{0.\text{1380}\text{g}}{\text{12 }\text{g/mol}}\\ =0.01\text{15 mol}\end{array}$

For hydrogen,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{0.007\text{g}}{1\text{g/mol}}\\ =0.007\text{mol}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent.

For carbon $\left(\text{C}\right)$,

$\text{C}=\frac{0.0115\text{mol}}{0.007\text{mol}}=1.642$

For hydrogen $\left(\text{H}\right)$,

$\text{H}=\frac{0.007\text{mol}}{0.007\text{mol}}=1$

By multiplying each with 3 we get the whole number as,

For carbon

$1.642\times 3=4.926\approx 5$.

For hydrogen

$1\times 3=3$.

Hence, the empirical formula of the compound is ${\text{C}}_5{\text{H}}_3$.

(b)

Interpretation Introduction

Interpretation: The empirical formula and the molecular formula of the given helicene are to be calculated. The balanced chemical reaction for the combustion of helicene is to be stated.

Concept introduction: Helicines are defined as the polycyclic aromatic compounds in which the aromatic ring is annulated to provide helically shaped molecule. The molecular formula of any organic compound is determined by using empirical formula when the percent of each element is given in the compound.

To determine: The molecular formula of the given helicene.

Answer to Problem 160IP

Answer

The molecular formula of the given helicene is ${\text{C}}_{\text{20}}{\text{H}}_{\text{12}}$.

Explanation of Solution

Explanation

The molecular formula of the given helicene is ${\text{C}}_{\text{20}}{\text{H}}_{\text{12}}$.

Given

Molality is $0.017\text{5 m}$.

Weight of solvent is $12.\text{5 g}$.

Weight of solute is $0.09\text{38 g}$.

Therefore the molecular weight of solute is calculated by the given expression.

$\text{Molality}\left(m\right)=\frac{\text{Weight of solute }\times \text{1000}}{\text{Molecular Weight of solute }\times \text{Weight of solvent}}$

Substitute the values of weight of solute, molality and weight of solvent in the above expression.

$\begin{array}{c}\text{Molality}\left(m\right)=\frac{\text{Weight of solute }\times \text{1000}}{\text{Molecular Weight of solute }\times \text{Weight of solvent}}\\ \text{Molecular Weight}=\frac{\text{Weight ofsolute}\times 1000}{\text{Molality}\times \text{Weight ofsolvent}}\\ =\frac{0.093\text{8 g}\times 1000}{0.017\text{5 m}\times 12.\text{5 g}}\\ =248.\text{8 g}\end{array}$

Now, The empirical mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{3}}$ is $\text{63 g}$. Then number of atoms ( $n$) is calculated by using the formula,,

$n=\frac{\text{Molecular weight}}{\text{Empirical weight}}$

Substitute the value of molecular and empirical weight in the above expression.

$\begin{array}{c}n=\frac{\text{Molecular weight}}{\text{Empirical weight}}\\ =\frac{248.8}{63}\\ =4\end{array}$

Therefore the molecular formula is calculated as,

$\begin{array}{c}\text{Molecular formula}={\left(\text{Empirical formula}\right)}_n\\ ={\left({\text{C}}_{\text{5}}{\text{H}}_{\text{3}}\right)}_4\\ ={\text{C}}_{\text{20}}{\text{H}}_{\text{12}}\end{array}$

Hence, the molecular formula is ${\text{C}}_{\text{20}}{\text{H}}_{\text{12}}$

(c)

Interpretation Introduction

Interpretation: The empirical formula and the molecular formula of the given helicene are to be calculated. The balanced chemical reaction for the combustion of helicene is to be stated.

Concept introduction: Helicines are defined as the polycyclic aromatic compounds in which the aromatic ring is annulated to provide helically shaped molecule. The molecular formula of any organic compound is determined by using empirical formula when the percent of each element is given in the compound.

To determine: The balanced chemical reaction for the combustion of helicene.

Answer to Problem 160IP

Answer

The balanced chemical reaction for the combustion of helicene is,

${\text{C}}_{\text{20}}{\text{H}}_{\text{12}}+{\text{ 23O}}_{\text{2}}\underset{}{\to }2{\text{0CO}}_{\text{2}}+{\text{ 6H}}_{\text{2}}\text{O}$.

Explanation of Solution

Explanation

The balanced chemical reaction for the combustion of helicene is shown below.

The balanced reaction for the combustion of helicene is,

${\text{C}}_{\text{20}}{\text{H}}_{\text{12}}+{\text{ 23O}}_{\text{2}}\underset{}{\to }2{\text{0CO}}_{\text{2}}+{\text{ 6H}}_{\text{2}}\text{O}$