Chapter 10, Problem 2STP

Interpretation Introduction

Interpretation:

The molecular formula for butanoic acid should be predicted.

Concept introduction:

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

Chemical formula or molecular formula represents the type of atoms and number of atoms present in a molecule using numerical subscripts and atomic symbol.

The rules for determining empirical formula is 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 2STP

The molecular formula for butanoic acid is ${\text{C}}_4{\text{H}}_8{\text{O}}_{\text{2}}$.

Explanation of Solution

Assuming the percentage is given by mass that means 54.5 g, 9.1 g, and 36.4 g of carbon, hydrogen, and oxygen is present in butanoic acid.

Mass of carbon = 54.5 g

Mass of hydrogen = 9.1 g

Mass of oxygen = 36.4 g

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

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{54.3\text{g}}{12.0107\text{g/mol}}\\ {\text{n}}_{\text{C}}=4.5\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{9}\text{.1}\text{g}}{\text{1}\text{.008}\text{g/mol}}\\ {\text{n}}_{\text{H}}=9.03\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{\text{36}\text{.4}\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=2.3\text{mol}\end{array}$

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

$\begin{array}{c}{\text{n}}_{\text{C}}=\frac{4.5}{2.3}\\ =2\\ {\text{n}}_{\text{H}}=\frac{9.03}{2.3}\\ =4\\ {\text{n}}_{\text{O}}=\frac{2.3}{2.3}\\ =1\end{array}$

2. Thus, the empirical formula of hydrazine is ${\text{C}}_2{\text{H}}_{\text{4}}\text{O}$.

Now, n that is the ratio of given molar mass and empirical formula mass is calculated as follows:

$\text{n}=\frac{\text{Molecular}\text{formula}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}$

The empirical formula mass is calculated as follows:

$\begin{array}{c}\text{Empirical}\text{}\text{formula}\text{mass}=\left(2\times \text{C}+4\times \text{H}+1\times \text{O}\right)\text{g/mol}\\ =\left(2\times 12.0107+4\times 1.008+1\times 15.999\right)\text{g/mol}\\ \text{Empirical}\text{}\text{formula}\text{mass}=44.0524\text{g/mol}\end{array}$

Calculation of n as follows:

$\begin{array}{c}\text{n}=\frac{\text{Molecular}\text{formula}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}\\ =\frac{\text{88}\text{.1}\text{g/mol}}{\text{44}\text{.0524}\text{g/mol}}\\ =1.99\\ \text{n}=2\end{array}$

Now, the number obtained will be multiplied to empirical formula as follows:

$\begin{array}{c}\text{Molecular}\text{formula}={\text{C}}_2{\text{H}}_4\text{O}\times 2\\ ={\text{C}}_4{\text{H}}_8{\text{O}}_{\text{2}}\end{array}$

Thus, the molecular formula for butanoic acid is ${\text{C}}_4{\text{H}}_8{\text{O}}_{\text{2}}$.