Chapter 10.4, Problem 61PP

Interpretation Introduction

Interpretation:

Using the percentage of the elements, the empirical formula of aspirin needs to be determined.

Concept introduction:

Empirical formula is the chemical formula having the simplest atoms ratio in a compound. Percentage of elements denotes percentage of mass per $100\text{ gram}$ sample. One can find the moles then using the mass − mole conversion where

$\text{number of moles =mass}\times \frac{\text{1 mole}}{\text{molar mass}}$

The ratio of the moles of the compound is found. Divide both the components by the smallest mole. The numbers are then multiplied by a smallest number which will give a ratio of whole numbers.

Answer to Problem 61PP

Empirical formula of aspirin is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$

Explanation of Solution

Given information:

$\begin{array}{l}60\cdot 00\text{\% carbon}\\ 4\cdot 44\text{\% hydrogen}\\ 35\cdot 56\text{\% oxygen}\end{array}$

Assuming the mass percentage for a $100\text{ gram}$ sample, thus,

Mass of carbon $=\frac{60\cdot 00}{100}\times 100\text{ grams = 60 grams C}$

Mass of hydrogen $=\frac{4\cdot 44}{100}\times 100\text{ grams = 4}\cdot 44\text{ grams H}$

Mass of oxygen $=\frac{\text{35}\cdot 56}{100}\times 100\text{ grams = 35}\cdot 56\text{ grams O}$

Using mass − mole conversion, one find number of moles of each element.

Molar mass of carbon is $=12\cdot 01\text{ grams/mol}$

Molar mass of hydrogen is $=1\cdot 008\text{ grams/mol}$

Molar mass of oxygen $=16\text{ grams/mol}$

Moles of carbon $\text{=60 g}\times \frac{1\text{ mole}}{12\cdot 01\text{ g}}=4\cdot 996\text{ moles C}$

Moles of hydrogen $=4\cdot 44\times \frac{\text{1 mole}}{1\cdot 008\text{ g}}\cong 4\cdot 405\text{ moles H}$

Moles of oxygen $=35\cdot 56\times \frac{\text{1 mole}}{16\text{ g}}=2\cdot 223\text{ moles O}$

Now divide the values by the smallest mole value:

$\begin{array}{l}\frac{4\cdot 996}{2\cdot 223}\text{moles C =2}\cdot 25\text{ mole C}\\ \frac{4\cdot 405}{2\cdot 223}\text{moles H =1}\cdot 98\simeq 2\text{ moles H}\\ \frac{2\cdot 223}{2\cdot 223}\text{ moles O = 1 mole O}\end{array}$

Hence the ratio will be $\left[2\cdot 25\text{ moles C}\right]:\left[\text{2 moles H}\right]:\left[1\text{ mole O}\right]$

In order to get a ratio of whole number, multiply both the number by $4$

$\begin{array}{l}\left[4\times 2\cdot 25\text{ moles C}\right]:\left[\text{4}\times \text{2 moles H}\right]:\left[4\times 1\text{ mole O}\right]\\ 9\text{ moles C: 8 moles H: 4 moles O}\end{array}$

Hence the empirical formula of the compound is ${\text{C}}_{\text{9}}{\text{H}}_{\text{8}}{\text{O}}_{\text{4}}$.