The aim of this question is to learn the **calculations** involved in **molar proportions** of chemical compounds.

This is a **three-step process**. First, we find the **molar mass of the given compound** using its chemical formula. Second, we find the **no. of moles contained in unit volume** using the following formula:

\[ 1 \ mL \ of \ chemical \ compound \ = \ y \ mole \]

Third, we find the **no. of moles contained in the given volume** using the following formula:

\[ x \ mL \ of \ chemical \ compound \ = \ x \ \times \ y \ mole \]

## Expert Answer

The **chemical formula of acetic anhydride** is:

\[ C_4 H_6 O_3 \]

Since:

\[ \text{Molar mass of carbon } = \ 12 \ amu \]

\[ \text{Molar mass of hydrogen } = \ 1 \ amu \]

\[ \text{Molar mass of oxygen } = \ 16 \ amu \]

So:

\[ \text{Molar mass of acetic anhydride } = \ 4 \ ( \ 12 \ ) \ + \ 6 \ ( \ 1 \ ) \ + \ 3 \ ( \ 16 \ ) \ amu \]

\[ \Rightarrow \text{Molar mass of acetic anhidride } = \ 48 \ + \ 6 \ + \ 48 \ amu \]

\[ \Rightarrow \text{Molar mass of acetic anhydride } = \ 102 \ amu \]

This means that:

\[ 1 \ mole \ of \ acetic \ anhydride \ = \ 102 \ g \]

Alternatively:

\[ 102 \ g \ of \ acetic \ anhydride \ = \ 1 \ mole \]

\[ \boldsymbol{ \Rightarrow 1 \ g \ of \ acetic \ anhydride \ = \ \dfrac{ 1 }{ 102 } \ mole } \ … \ … \ … \ (1) \]

Given:

\[ \text{ Density of acetic anhydride } = \ 1.08 \ \frac{ g }{ mL } \]

Which means:

\[ 1 \ mL \ of \ acetic \ anhydride \ = \ 1.08 \ g \]

Using the value of $ 1 \ g $ from the **equation (1)**:

\[ 1 \ mL \ of \ acetic \ anhydride \ = \ \times \ 1.08 \ \times \ \dfrac{ 1 }{ 102 } \ mole \]

\[ \Rightarrow 1 \ mL \ of \ acetic \ anhydride \ = \ \dfrac{ 1.08 }{ 102 } \ mole \]

\[ \boldsymbol{ \Rightarrow 1 \ mL \ of \ acetic \ anhydride \ = \ 0.010588 \ mole } \ … \ … \ … \ (2) \]

For $ 5 \ mL $ quantity, **equation (2)** becomes:

\[ 5 \ mL \ of \ acetic \ anhydride \ = \ 5 \ \times \ 0.010588 \ mole \]

\[ \boldsymbol{ \Rightarrow 5 \ mL \ of \ acetic \ anhydride \ = \ 0.053 \ mole } \]

## Numerical Result

\[ 5 \ mL \ of \ acetic \ anhydride \ = \ 0.05294 \ mole \]

## Example

For the** same chemical compound**, Calculate the **no. of moles** contained in $ 1 \ L $.

Recall **equation (2)**:

\[ 1 \ mL \ of \ acetic \ anhydride \ = \ 0.010588 \ mole \]

For $ 1 \ L \ = \ 1000 \ mL $:

\[ 1000 \ mL \ of \ acetic \ anhydride \ = \ 1000 \ \times \ 0.010588 \ mole \ = \ 10.588 \ mole \]