Chapter 10.4, Problem 65PP

Interpretation Introduction

Interpretation:

The molecular formula of given compound with the help of percent composition of C, H and O elements and molar mass of 110.0 g/mol needs to be determined.

Concept introduction:

Empirical formula is the simplest formula of any organic or inorganic compound that represents the simple ratio of all atoms present in the molecule. It can be calculated with the help of elemental composition of molecule. Certain steps must be used to get the empirical formula:

• Consider mass % as mass in grams and calculate moles of element with the help of molar mass
• Calculate the moles of each element in least whole number
• Write the number of each atom as subscript to write the empirical formula.

Answer to Problem 65PP

Molecular formula = ${\text{C}}_{\text{6}}{\text{H}}_6{\text{O}}_{\text{2}}$

Explanation of Solution

Given information:

C = 65.45 %

O = 29.09 %

H = 5.45 %

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.007 g/mol

Molar mass of O = 15.99 g/mol

Calculate moles of each part:

$\begin{array}{l}\text{Moles of C = }\frac{\text{65}\text{.45 }}{\text{12}\text{.01 }}\text{= 5}\text{.449}\\ \text{Moles of O = }\frac{\text{29}\text{.09}}{\text{15}\text{.99 }}\text{= 1}\text{.819}\\ \text{Moles of H = }\frac{\text{5}\text{.45}}{\text{1}\text{.007 }}\text{= 5}\text{.412}\end{array}$

Divide by least moles to get the whole number:

$\begin{array}{l}\text{Moles of C = }\frac{\text{ 5}\text{.449}}{\text{ 1}\text{.819}}\text{= 3}\\ \text{Moles of O = }\frac{\text{ 1}\text{.819}}{\text{ 1}\text{.819}}\text{= 1}\\ \text{Moles of H = }\frac{\text{ 5}\text{.412}}{\text{ 1}\text{.819}}\text{= 3}\end{array}$

Thus the empirical formula must be ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{O}$.

Mass of empirical formula = ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{O=(3}\times \text{12}\text{.01)+(3}\times \text{1}\text{.007)+15}\text{.99=55}\text{.04}$

Molar mass = 110.0 g/mol

$\begin{array}{l}\text{n = }\frac{\text{molar mass}}{\text{mass of empirical formula}}\text{=}\frac{\text{110}\text{.0g/mol }}{\text{55}\text{.04 g/mol}}\text{=2}\\ {\text{Molecular formula = C}}_{\text{3}}{\text{H}}_{\text{3}}\text{O}\times 2={\text{C}}_{\text{6}}{\text{H}}_6{\text{O}}_{\text{2}}\end{array}$

Conclusion

Molecular formula = ${\text{C}}_{\text{6}}{\text{H}}_6{\text{O}}_{\text{2}}$