Chapter 10.3, Problem 49LC

Interpretation Introduction

Interpretation: The empirical and molecular formula of a compound that contains 58.8% C, 9.8% H, and 31.4% O and has a molar mass, 102 g/mol should be calculated.

Concept Introduction: Molecular formula depicts the exact number of atoms in a molecule of a compound whereas an empirical formula depicts the minimum integer subscripts of atoms in a molecule of a compound.

Answer to Problem 49LC

The empirical and molecular formula of a compound that contains 58.8% C, 9.8% H, and 31.4% O and has a molar mass, 102 g/mol is ${\text{C}}_5{\text{H}}_{10}{\text{O}}_2$ .

Explanation of Solution

The percent by mass of each element in the compound represents the percent composition as:

  $\text{percent by mass of an element = }\frac{\text{mass of an element}}{\text{mass of the compound}}\times 100\%$

So, 100.0 g of the compound contains 58.8 g C, 9.8 g H, and 31.4 g O.

By using the molar mass of an element, the ratio of masses can be changed to the ratio of moles by using the following conversion factor:

  $\text{mass of element (g) }\times \text{ molar mass of element (g/mol) = mol of element}$ …. (1)

The molar mass of $\text{C, H}$ and $\text{O}$ is 12.0 g/mol, 1.0 g/mol, and 16.0 g/mol respectively.

The percent by mass of $\text{C, H}$ and $\text{O}$ can be converted to moles of $\text{C, H}$ and $\text{O}$ using the relation (1) as:

  $\begin{array}{l}\text{58}\text{.5 g C }\times \frac{1\text{ mol C}}{12.0\text{ g C}}\text{ = 4}\text{.9 mol C}\\ \text{9}\text{.8 g H }\times \frac{1\text{ mol H}}{1.0\text{ g H}}\text{ = 9}\text{.8 mol H}\\ \text{31}\text{.4 g O }\times \frac{1\text{ mol O}}{16.0\text{ g O}}\text{ = 1}\text{.9 mol O}\end{array}$

The mole ratio of $\text{C, H}$ and $\text{O}$ is ${\text{C}}_{\text{4}\text{.9}}{\text{H}}_{\text{9}\text{.8}}{\text{O}}_{\text{1}\text{.9}}$ .

Now, each molar quantity is divided by the smaller number of moles:

  $\begin{array}{l}\frac{\text{4}\text{.9 mol C}}{1.9}=\text{ 2}\text{.58 mol C}\\ \frac{\text{9}\text{.8 mol H}}{1.9}=\text{ 5}\text{.16 mol H}\\ \frac{\text{1}\text{.9 mol O}}{1.9}\text{ = 1}\text{.0 mol O}\end{array}$

Multiply each mole ratio by 2 to obtain the mole ratio in the whole number:

  $\begin{array}{l}\text{C = 2}\text{.58 }\times \text{ 2 }\simeq \text{ 5}\\ \text{H = 5}\text{.16 }\times \text{ 2 }\simeq \text{ 10}\\ \text{O = 1 }\times \text{ 2 }=\text{ 2}\end{array}$

Thus, the empirical formula of a compound is ${\text{C}}_5{\text{H}}_{10}{\text{O}}_2$ .

Now, to determine the molecular formula of the compound, the empirical formula mass of the compound is calculated as:

  $\begin{array}{l}{\text{efm of C}}_5{\text{H}}_{10}{\text{O}}_2\text{ = 5(12}\text{.0 g/mol) + 10(1}\text{.0 g/mol) + 2(16}\text{.0) g/mol}\\ {\text{efm of C}}_5{\text{H}}_{10}{\text{O}}_2\text{ = 102 g/mol}\end{array}$

Now, the molar mass is divided by the empirical formula mass as:

  $\frac{102\text{ g/mol}}{\text{102 g/mol}}\text{ = 1}$

Multiply the subscripts of the empirical formula, ${\text{C}}_5{\text{H}}_{10}{\text{O}}_2$ with 1:

  ${\text{C}}_5{\text{H}}_{10}{\text{O}}_2\times 1{\text{ = C}}_5{\text{H}}_{10}{\text{O}}_2$

Hence, the molecular formula of a compound that contains 58.8% C, 9.8% H, and 31.4% O and has a molar mass of 102 g/mol is ${\text{C}}_5{\text{H}}_{10}{\text{O}}_2$ .