Question

Write the molecular formula of each of the following compound whose:

(b) empirical formula$={\mathrm{C}}_3{\mathrm{H}}_4{\mathrm{NO}}_3$, molecular weight$=408$.

Answer 1

Step 1: Given information.

• Empirical formula$={\mathrm{C}}_3{\mathrm{H}}_4{\mathrm{NO}}_3$
• Molecular weight$=408$

Step 2: Calculation of empirical formula weight

• The empirical formula weight for ${\mathrm{C}}_3{\mathrm{H}}_4{\mathrm{NO}}_3$ is calculated as follows:

$\begin{array}{rcl}\mathrm{Empirical}\mathrm{formula}\mathrm{weight}& =& \left(12\times 3+1\times 4+14+16\times 3\right)\mathrm{g}\\ & =& 102\mathrm{g}\end{array}$

Step 3: Calculation of value of $\mathit{n}$

• Determine the value of $n$ as follows:

$\begin{array}{rcl}n& =& \frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}}\\ & =& \frac{408}{102}\\ & =& 4\end{array}$

Step 4: Determination of molecular formula

• The molecular formula is determined as follows:

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{formula}& =& {\left(\mathrm{Empirical}\mathrm{formula}\right)}_n\\ & =& {\left({\mathrm{C}}_3{\mathrm{H}}_4{\mathrm{NO}}_3\right)}_4\\ & =& {\mathrm{C}}_{12}{\mathrm{H}}_{16}{\mathrm{N}}_4{\mathrm{O}}_{12}\end{array}$

Therefore, the molecular formula of the compound is ${\mathrm{C}}_{12}{\mathrm{H}}_{16}{\mathrm{N}}_4{\mathrm{O}}_{12}$.