Chapter 10, Problem 120QRT

(a)

Interpretation Introduction

Interpretation:

The molecular formula of the hydrocarbon has to be derived.

Concept Introduction:

The compounds that contain carbon and hydrogen atoms are known as hydrocarbon compounds.  There are two classes of hydrocarbon compounds which are saturated and unsaturated hydrocarbons.

Saturated hydrocarbon compound has only one type that is alkane.  Unsaturated hydrocarbon compound are of two types: alkenes and alkynes.  The alkane contains only single bond between the carbon chains.

Explanation of Solution

The number of moles of hydrocarbon is calculated by the formula shown below.

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

The mass of hydrocarbon is $\text{1}\text{.58 g}$.

The molar mass of hydrocarbon is $\text{40}\text{.1 g/mol}$.

Substitute the values in the above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{\text{1}\text{.58 g}}{\text{40}\text{.1 g/mol}}\\ =0.039\text{mol}\end{array}$

The volume of $\text{1 mole}$ of hydrogen gas at STP is $\text{22}\text{.4 L}$.

Thus, the number of moles of hydrogen occupied by $\text{1}\text{.77 L}$ of hydrogen is calculated as shown below.

    $\begin{array}{c}\text{Moles}\text{of hydrogen}=\frac{\text{1}\text{.77 L}}{\text{22}\text{.4 L}}\times \text{1 mole}\\ =0.079\text{0 mole}\end{array}$

The moles of hydrogen per moles of hydrocarbon are calculated as shown below.

    $\begin{array}{c}\text{Moles}\text{of hydrogen per mole of hydrocarbon}=\frac{\text{Moles of hydrogen}}{\text{Moles of hydrocarbon}}\\ =\frac{\text{0}\text{.079}}{\text{0}\text{.0394}}\\ =2.00\end{array}$

Thus, one mole of hydrocarbon reacts with two moles of hydrogen.  This indicates the presence of two double or one triple bond in the hydrocarbon.

The mass percent of carbon is $\text{90 \%}$.

The mass percent of hydrogen is $\text{10 \%}$.

The mass of carbon is calculated as shown below.

    $\begin{array}{c}\text{Mass of carbon}=1.5\text{8 g}\times \frac{90}{100}\\ =1.422\text{ g}\end{array}$

The mass of hydrogen is calculated as shown below.

    $\begin{array}{c}\text{Mass of hydrogen}=1.5\text{8 g}\times \frac{10}{100}\\ =0.158\text{ g}\end{array}$

The number of moles of carbon is calculated by the formula shown below.

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

The mass of carbon is $\text{1}\text{.422 g}$.

The molar mass of carbon is $\text{12}\text{.01 g/mol}$.

Substitute the values in the above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{\text{1}\text{.422 g}}{\text{12}\text{.01 g/mol}}\\ =0.1184\text{mol}\end{array}$

The number of moles of hydrogen is calculated by the formula shown below.

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

The mass of hydrogen is $\text{0}\text{.158 g}$.

The molar mass of hydrogen n is $\text{1}\text{.0079 g/mol}$.

Substitute the values in the above formula.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{\text{0}\text{.158 g}}{\text{1}\text{.0079 g/mol}}\\ =0.1568\text{mol}\end{array}$

The molar ratio of $\text{C}:{\text{H}}_{\text{2}}$ can be obtained by dividing the number of moles with the lowest number of mole that is $0.1184\text{mol}$.

  $\begin{array}{c}\text{C}:\text{H}=\frac{0.1184\text{mol}}{0.1184\text{mol}}:\frac{0.1568\text{mol}}{0.1184\text{mol}}\\ \text{C}:\text{H}=1:1.32\\ \text{C}:\text{H}\approx 3:4\end{array}$

Thus, the molecular formula of the hydrocarbon is ${\text{C}}_{\text{3}}\text{H}{\text{4}}_{}$.

(b)

Interpretation Introduction

Interpretation:

The structural formula for two possible isomers has to be drawn.

Concept Introduction:

The molecules which have same molecular formula and same connectivity of atoms but different arrangement of atoms in space are known as stereoisomers.  Stereocentre is the region or atom in a molecule due to which molecule is showing stereoisomerism.  The interchange of groups at stereocentre generates stereoisomer.

Explanation of Solution

The molecular formula of the hydrocarbon is ${\text{C}}_{\text{3}}\text{H}{\text{4}}_{}$.  This means two double bond or one triple bond must be present in the hydrocarbon. Thus, the structural formulas for two possible isomers are shown below.

Figure 1

(c)

Interpretation Introduction

Interpretation:

The balanced chemical equation has to be used to show the bromination and hydrogenation reaction for one of the isomers.

Concept Introduction:

The addition of a bromine atom in the given compound is known as bromination.  Bromination occurs through an electrophilic substitution reaction.  Bromine atom acts as an electrophile which causes the formation of sigma bond in the reaction.

Explanation of Solution

Propyne undergoes bromination to form tetrabromo product. The balanced chemical equation is shown below.

Figure 2

Propyne undergoes hydrogenation to form propane. The balanced chemical equation is shown below.

Figure 3