Chapter 10, Problem 3STP

Interpretation Introduction

Interpretation:

The empirical formula for ethanol using graph should be predicted.

Concept introduction:

The representation of atoms of a compound in simple whole number ratio is known as empirical formula.

The rules for determining empirical formula are given as follows:

• Determine the mass of elements in given compound.
• Calculate number of moles using molar mass of each compound or element.
• Divide each number of moles by smallest number of mole value calculated in second step.
• Round the value calculated in step 3 to nearest whole number.

Answer to Problem 3STP

The empirical formula for ethanol is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$.

Explanation of Solution

From the graph, considering 100 g mass, mass of carbon = 52.2 g, mass of hydrogen = 13.0 g and mass of oxygen = 34.8 g.

Now, calculating the moles of carbon, hydrogen, and oxygen using their molar mass as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{\text{52}\text{.2}\text{g}}{\text{12}\text{.0107}\text{g/mol}}\\ {\text{n}}_{\text{C}}=4.35\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{13}\text{.0}\text{g}}{\text{1}\text{.008}\text{g/mol}}\\ {\text{n}}_{\text{H}}=12.9\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{\text{34}\text{.8}\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=2.175\text{mol}\end{array}$

Now, dividing each mole value by smallest mole value as follows:

$\begin{array}{c}{\text{n}}_{\text{c}}=\frac{4.35}{2.175}\\ =2\\ {\text{n}}_{\text{H}}=\frac{12.9}{2.175}\\ =6\\ {\text{n}}_{\text{O}}=\frac{2.175}{2.175}\\ =1\end{array}$

Thus, the empirical formula of ethanol is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$.

Hence, the correct option is (C) that is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$.