Chapter 10, Problem 1STP

Interpretation Introduction

Interpretation:

Whether the molecular formula, empirical formula, molar mass or chemical properties for acetaldehyde and butanoic acid are same needs to be determined, using the given graph.

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 1STP

Empirical formula for acetaldehyde and butanoic acid is same.

Explanation of Solution

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

1. Mass of carbon = 54.5g

Mass of hydrogen = 9.1g

Mass of oxygen = 36.4g

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

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{54.5\text{g}}{\text{12}\text{.0107}\text{g/mol}}\\ {\text{n}}_{\text{C}}=4.5\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{9}\text{.1}\text{g}}{1.008\text{g/mol}}\\ {\text{n}}_{\text{H}}=9.03\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{36.4\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=2.3\text{mol}\end{array}$

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

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

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

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

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

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:

$\begin{array}{c}\text{n}=\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}\\ =\frac{\text{44}\text{.05}\text{g/mol}}{\text{44}\text{.0524}\text{g/mol}}\\ =0.99\\ \text{n}=1\end{array}$

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

$\begin{array}{c}\text{Molecular}\text{formula}={\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}\times \text{1}\\ ={\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}\end{array}$

Now, the empirical formula and molecular formula for butanoic acid is calculated as follows:

4. Mass of carbon = 54.5g

Mass of hydrogen = 9.1g

Mass of oxygen = 36.4g

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

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{54.5\text{g}}{\text{12}\text{.0107}\text{g/mol}}\\ {\text{n}}_{\text{C}}=4.5\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{9}\text{.1}\text{g}}{1.008\text{g/mol}}\\ {\text{n}}_{\text{H}}=9.03\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{36.4\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=2.3\text{mol}\end{array}$

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

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

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

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

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

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:

$\begin{array}{c}\text{n}=\frac{\text{Molar}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}\\ =\frac{\text{88}\text{.11}\text{g/mol}}{\text{44}\text{.05}\text{g/mol}}\\ \text{n}=2\end{array}$

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

$\begin{array}{c}\text{Molecular}\text{formula}={\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}\times 2\\ ={\text{C}}_{\text{4}}{\text{H}}_8{\text{O}}_2\end{array}$

The molar mass of acetaldehyde and butanoic acid is 44.05 and 88.11 g/mol respectively.

Also, acetaldehyde is a compound having -CHO as a functional group and butanoic acid is having -COOH as a functional group. Thus, the chemical properties of both will be different.

The empirical formula for acetaldehyde and butanoic acid is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$. Thus, both will have same empirical formula.