Q153CP - Chemistry: Molecular Nature Of Matter And Change

Question

Carboxylic acids react with alcohol to form esters, which are found in all plants and animals. Some are responsible for the flavours and odours of fruits and flowers. What is the percent by mass of carbon in each of the following esters?

Name	Formula	Odour
Isoamyl isovalerate	C4H9COOC5H11	Apple
Amyl butyrate	C3H7COOC5C11	Apricot
Isoamyl acetate	CH3COOC5H11	Banana
Ethyl butyrate	C3H7COOC2H5	pineapple

 

Step-by-Step Solution

Answer

The mass percentage of carbon in each of the esters can be found as,

Isoamylisovalerate- $\mathbf{69}\mathbf{.}\mathbf{7}\mathbf{\%}$

Amyl butyrate- $\mathbf{68}\mathbf{.}\mathbf{29}\mathbf{\%}$

Isoamyl acetate- $\mathbf{64}\mathbf{,}\mathbf{56}\mathbf{\%}$

Ethyl butyrate-  $\mathbf{62}\mathbf{.}\mathbf{02}\mathbf{\%}$

Step 1: Finding the molecular mass for isoamyl isovalerate

This can be simply equated to${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{Y}}{\mathbf{O}}_{\mathbf{Z}}\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{,}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{20}}{\mathbf{O}}_{\mathbf{2}}$

The atomic mass is given as,

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{ }\mathbf{u}$

On adding the atomic masses to get molecular mass,

$$\begin{gathered} \mathbf{molecularmass}\mathbf{=}\mathbf{10}\mathbf{C}\mathbf{+}\mathbf{20}\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{O} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{10}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{20}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{16}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{172}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{amu} \end{gathered}$$

The percentage of carbon is 

$$\begin{gathered} \mathbf{percentage}\mathbf{=}\frac{\mathbf{120}\mathbf{.}\mathbf{1}}{\mathbf{172}\mathbf{.}\mathbf{3}}\mathbf{\times }\mathbf{100} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{69}\mathbf{.}\mathbf{7}\mathbf{\%} \end{gathered}$$

Step 2: Finding the molecular mass for Amyl butyrate

This can be simply equated to${\mathbf{C}}_{\mathbf{x}}{\mathbf{H}}_{\mathbf{y}}{\mathbf{O}}_{\mathbf{z}}\mathbf{ }\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{,}{\mathbf{C}}_{\mathbf{9}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{O}}_{\mathbf{2}}$

The atomic mass is given as,

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{ }\mathbf{u}$

On adding the atomic masses to get molecular mass,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{=}\mathbf{9}\mathbf{C}\mathbf{+}\mathbf{18}\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{O} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{9}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{18}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{16}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{158}\mathbf{.}\mathbf{27}\mathbf{ }\mathbf{amu} \end{gathered}$$

The percentage of carbon is 

$$\begin{gathered} \mathbf{percentage}\mathbf{=}\frac{\mathbf{108}\mathbf{.}\mathbf{09}}{\mathbf{158}\mathbf{.}\mathbf{27}}\mathbf{\times }\mathbf{100} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{68}\mathbf{.}\mathbf{29}\mathbf{\%} \end{gathered}$$

Step 3: Finding the molecular mass for Isoamyl acetate

This can be simply equated to${\mathbf{C}}_{\mathbf{x}}{\mathbf{H}}_{\mathbf{y}}{\mathbf{O}}_{\mathbf{z}}\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{,}\mathbf{ }{\mathbf{C}}_{\mathbf{7}}{\mathbf{H}}_{\mathbf{14}}{\mathbf{O}}_{\mathbf{2}}$

The atomic mass is given as,

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{ }\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{u}$

On adding the atomic masses to get molecular mass,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{7}\mathbf{C}\mathbf{+}\mathbf{14}\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{O} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{7}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{14}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{16}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{130}\mathbf{.}\mathbf{21}\mathbf{amu} \end{gathered}$$

The percentage of carbon is 

$$\begin{gathered} \mathbf{Percentage}\mathbf{=}\frac{\mathbf{84}\mathbf{.}\mathbf{07}}{\mathbf{130}\mathbf{.}\mathbf{21}}\mathbf{\times }\mathbf{100} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{64}\mathbf{.}\mathbf{56}\mathbf{\%} \end{gathered}$$

Step 4: Finding the molecular mass for Ethyl butyrate

This can be simply equated to${\mathbf{C}}_{\mathbf{x}}{\mathbf{H}}_{\mathbf{y}}{\mathbf{O}}_{\mathbf{z}}\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{,}\mathbf{ }{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{2}}$

The atomic mass is given as,

$\mathbf{C}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{u}\mathbf{,}\mathbf{O}\mathbf{=}\mathbf{16}\mathbf{u}$

On adding the atomic masses to get molecular mass,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{6}\mathbf{C}\mathbf{+}\mathbf{12}\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{O} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{6}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{12}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{16}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{116}\mathbf{.}\mathbf{18}\mathbf{ }\mathbf{amu} \end{gathered}$$

The percentage of carbon is 

$$\begin{gathered} \mathbf{Percentage}\mathbf{=}\frac{\mathbf{72}\mathbf{.}\mathbf{06}}{\mathbf{116}\mathbf{.}\mathbf{18}}\mathbf{\times }\mathbf{100} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{62}\mathbf{.}\mathbf{02}\mathbf{\%} \end{gathered}$$