Q142CP - Chemistry: Molecular Nature Of Matter And Change

Question

Succinic acid (below) is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid, and calculate the mass percent of each element.

                                               

Step-by-Step Solution

Answer

The molecular formula is ${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{4}}$  an empirical formula ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}$ . The molecular mass is 118.10 amu. The mass percentage is found as,  $\mathbf{40}\mathbf{.}\mathbf{68}\mathbf{\%}\mathbf{C}\mathbf{,}\mathbf{5}\mathbf{.}\mathbf{13}\mathbf{\%}\mathbf{H}$ and $\mathbf{54}\mathbf{.}\mathbf{19}\mathbf{\%}\mathbf{O}$

Step 1: Finding the molecular formula

From the Structure given, we can say le has 4 C-atoms, 6 H-atoms and 4-oxygen atoms. Thus, the molecular formula for succinic acid is ${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{4}}$.

Step 2: Finding the empirical formula

By dividing the subscripts by a greater common factor, we will get the empirical formula.

 ${\mathbf{C}}_{\frac{\mathbf{4}}{\mathbf{2}}}{\mathbf{H}}_{\frac{\mathbf{6}}{\mathbf{2}}}{\mathbf{O}}_{\frac{\mathbf{4}}{\mathbf{2}}}\mathbf{=}{\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{4}}$

Thus, the empirical formula for succinic acid is ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{2}}$

Step 3: Finding the molecular mass

The atomic masses of elements present can be 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 mass,

$$\begin{gathered} \mathbf{Molecular}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{4}\mathbf{\times }\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{+}\mathbf{6}\mathbf{\times }\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{+}\mathbf{4} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\times }\mathbf{atomic}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{O} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{4}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{6}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{\left(}\mathbf{16}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{118}\mathbf{.}\mathbf{10}\mathbf{amu} \end{gathered}$$

Step 4: Finding the mass percentage

On solving for the mass percentage for each element,

$$\begin{gathered} \mathbf{\%}\mathbf{massC}\mathbf{=}\frac{\mathbf{4}\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{)}}{\mathbf{118}\mathbf{.}\mathbf{10}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{68}\mathbf{\%}\mathbf{C} \\ \mathbf{\%}\mathbf{massH}\mathbf{=}\frac{\mathbf{6}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{)}}{\mathbf{118}\mathbf{.}\mathbf{10}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{13}\mathbf{\%}\mathbf{H} \\ \mathbf{\%}\mathbf{massO}\mathbf{=}\frac{\mathbf{4}\mathbf{\left(}\mathbf{16}\mathbf{\right)}}{\mathbf{118}\mathbf{.}\mathbf{10}}\mathbf{\times }\mathbf{100}\mathbf{\%} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{54}\mathbf{.}\mathbf{19}\mathbf{\%}\mathbf{O} \end{gathered}$$