Chapter 10, Problem 79A

(a)

Interpretation Introduction

Interpretation: To determine the molecular formula from the empirical formula.

Concept Introduction: Empirical formula would be a term used to describe a chemical formula that displays the most basic ratio of the components in a compound instead of the overall number of atoms in the molecule.

Answer to Problem 79A

The molecular formula will be ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$

Explanation of Solution

The given empirical formula is ${\text{CH}}_{\text{2}}\text{O}$ .

The molar mass of the compound is 90 g/mol

Molar mass of the empirical formula is 30 g/mol.

Molecular formula can be determined by using the formula:

  $\begin{array}{c}\text{n=}\frac{\text{molar}\text{mass}}{\text{molar}\text{mass}\text{of}\text{empirical}\text{formula}}\\ \text{=}\frac{\text{90}\text{g/mol}}{\text{30}\text{g/mol}}\\ =3\end{array}$

The relation between molar mass and empirical formula can be written as:

  $\begin{array}{c}\text{Molecular}\text{formula=(empirical}\text{formula}{\text{)}}_{\text{n}}\\ {\text{=(CH}}_{\text{2}}{\text{O)}}_{\text{3}}\\ {\text{=C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}\end{array}$

Therefore, the molecular formula will be ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$

(b)

Interpretation Introduction

Interpretation: To determine the molecular formula from the empirical formula.

Concept Introduction: Empirical formula would be a term used to describe a chemical formula that displays the most basic ratio of the components in a compound instead of the overall number of atoms in the molecule.

Answer to Problem 79A

The molecular formula will be ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$

Explanation of Solution

The given empirical formula is $\text{HgCl}$ .

The molar mass of the compound is 472.2 g/mol

Molar mass of the empirical formula is 272 g/mol.

Molecular formula can be determined by using the formula:

  $\begin{array}{c}\text{n=}\frac{\text{molar}\text{mass}}{\text{molar}\text{mass}\text{of}\text{empirical}\text{formula}}\\ \text{=}\frac{472.2\text{g/mol}}{272\text{g/mol}}\\ =2\end{array}$

The relation between molar mass and empirical formula can be written as:

  $\begin{array}{c}\text{Molecular}\text{formula=(empirical}\text{formula}{\text{)}}_{\text{n}}\\ {\text{=(HgCl)}}_2\\ {\text{=Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}\end{array}$

Therefore, the molecular formula will be ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$