Q144CP - Chemistry: Molecular Nature Of Matter And Change

Question

Which of the following models represent compounds having the same empirical formula? What is the molecular mass of this common empirical formula?

                                                                  

Step-by-Step Solution

Answer

B,C and D have same empirical formula, i.e.,${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{O}$ .

Step 1: Finding the molecular formula of each compound

The molecular formula can be find out by looking at the structures,

         $$\begin{gathered} \mathbf{A}\mathbf{.}\mathbf{ }{\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{10}}{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{B}\mathbf{.}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{O} \\ \mathbf{C}\mathbf{.}\mathbf{ }{\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{D}\mathbf{.}\mathbf{ }{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{3}} \\ \mathbf{E}\mathbf{.}\mathbf{ }{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{2}} \end{gathered}$$

Step 2: Finding the empirical formula

Empirical formula of a chemical compound is the simplest whole number ratio of atoms present in a compound.

 

Then identify the empirical formula,

      $\mathbf{A}\mathbf{)}\mathbf{ }{\mathbf{C}}_{\frac{\mathbf{4}}{\mathbf{2}}}{\mathbf{H}}_{\frac{\mathbf{10}}{\mathbf{2}}}{\mathbf{O}}_{\frac{\mathbf{2}}{\mathbf{2}}}\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{O}$

        $\mathbf{B}\mathbf{)}$It already present in simplest whole number so empirical formula is same as molecular formula i.e.,       ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{O}$.

       $$\begin{gathered} \mathbf{C}\mathbf{)}\mathbf{ }{\mathbf{C}}_{\frac{\mathbf{4}}{\mathbf{2}}}{\mathbf{H}}_{\frac{\mathbf{18}}{\mathbf{2}}}{\mathbf{O}}_{\frac{\mathbf{2}}{\mathbf{2}}}\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{O} \\ \mathbf{D}\mathbf{)}\mathbf{ }\mathbf{ }{\mathbf{C}}_{\frac{\mathbf{6}}{\mathbf{3}}}{\mathbf{H}}_{\frac{\mathbf{12}}{\mathbf{3}}}{\mathbf{O}}_{\frac{\mathbf{3}}{\mathbf{3}}}\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{O} \end{gathered}$$