Chapter 3, Problem 3.109QE

Interpretation Introduction

Interpretation:

The empirical formula of the compound has to be determined for the compound that contains only carbon, hydrogen, oxygen and nitrogen.  Combustion of a $\text{1}\text{.48g}$ sample in excess ${\text{O}}_{\text{2}}$ yields $\text{2}\text{.60g}$${\text{CO}}_{\text{2}}$ and $\text{0}\text{.799g}$${\text{H}}_{\text{2}}\text{O}$.  And a separate experiment of that same compound shows that a $\text{2}\text{.43g}$ sample contain $\text{0}\text{.340g}$$\text{N}$.

Explanation of Solution

This sample yields $\text{2}\text{.60g}$${\text{CO}}_{\text{2}}$ and $\text{0}\text{.799g}$${\text{H}}_{\text{2}}\text{O}$.

One mole of ${\text{CO}}_{\text{2}}$ has a mass of $\text{44}\text{.01g}$ and one mole of ${\text{H}}_{\text{2}}\text{O}$ has a mass of $\text{18}\text{.02g}$.

Then,

    ${\text{Moles of H}}_2\text{O = 0}{\text{.799g H}}_2\text{O × }\left(\frac{{\text{1mol H}}_2\text{O}}{\text{18}{\text{.02g H}}_2\text{O}}\right)\text{= 0}{\text{.044 mol H}}_2\text{O}$

    ${\text{moles of CO}}_{\text{2}}\text{ = 2}{\text{.60 g CO}}_{\text{2}}\text{ × }\left(\frac{{\text{1mol CO}}_{\text{2}}}{\text{44}{\text{.01g CO}}_{\text{2}}}\right)\text{= 0}{\text{.059mol CO}}_{\text{2}}$.

From the formulas of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$, one mole of ${\text{H}}_{\text{2}}\text{O}$ has two moles of hydrogen and one mole of ${\text{CO}}_{\text{2}}$ has one mole carbon.  These relationships are used to convert moles of each compound into moles of each element.

    $\text{Moles H = 0}{\text{.044 H}}_2\text{O × }\left(\frac{\text{2 mol H}}{\text{1 mol H}}\right)\text{= 0}\text{.088 mol H}$.

    $\text{Moles C = 0}{\text{.059 mol CO}}_2\text{ × }\left(\frac{\text{1mol C}}{{\text{1 mol CO}}_2}\right)\text{= 0}\text{.059 mol C}$.

Use the molar mass of hydrogen and carbon to calculate the mass of each present in sample.

    $\text{Mass H = 0}\text{.088 mol H ×}\left(\frac{\text{1}\text{.01gH}}{\text{1molH}}\right)\text{ = 0}\text{.088 g H}$.

    $\text{Mass C = 0}\text{.059 mol C ×}\left(\frac{\text{12}\text{.01gC}}{\text{1mol C}}\right)\text{ = 0}\text{.708 g C}$.

    $\begin{array}{l}{\text{Mass}}_{\text{sample}}{\text{= mass}}_{\text{C}}{\text{+ mass}}_{\text{H}}{\text{+ mass}}_{\text{O}}\\ \\ {\text{mass}}_{\text{O}}\text{= 1}\text{.48g - (0}\text{.088 g H + 0}\text{.708 g C) = 0}\text{.683 g}\end{array}$

    $\text{Amount O = 0}\text{.683 g O × }\left(\frac{\text{1mol O}}{\text{15}\text{.999 g O}}\right)\text{= 0}\text{.0426 mol O}$.

    $\text{Amount N = 0}\text{.340 g N × }\left(\frac{\text{1mol N}}{\text{14}\text{.0067 g N}}\right)\text{= 0}\text{.024 mol N}$.

This calculation will yield the relative number of moles of each element.  To convert into integers, divide each by the smallest number 0.024.

    $\text{Amount H = 0}\text{.088mol H }÷\text{ 0}\text{.024 = 3}\text{.66 mol H}$.

    $\text{Amount C = 0}\text{.059 mol C }÷\text{ 0}\text{.024 = 2}\text{.45 mol C}$

    $\text{Amount O = 0}\text{.0426 mol O }÷\text{ 0}\text{.024 = 1}\text{.77 mol O}$.

    $\text{Amount N = 0}\text{.024 mol N }÷\text{ 0}\text{.024 = 1 mol N}$.

The empirical formula of the compound is ${\text{C}}_{\text{4}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{N}$.