Chapter 10.4, Problem 59PP

Interpretation Introduction

Interpretation:

Using the percentage of the elements, the empirical formula that will be formed by the given elements 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 use 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 59PP

${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$

Explanation of Solution

Given information:

$\begin{array}{l}\text{35}\cdot \text{98\% aluminium}\\ \text{64}\cdot \text{02\% sulfur}\end{array}$

Assuming the mass percentage for a $100\text{ gram}$ sample, one can see that given mass of aluminum $=35\cdot 98\text{ grams}$. This can be shown as follows:

$\frac{35\cdot 98}{100}\times 100\text{ grams = 35}\cdot \text{98 grams}$ and $\frac{64\cdot 02}{100}\times 100\text{ grams = 64}\cdot 02\text{ grams}$

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

Molar mass of aluminum $=26\cdot 98\text{ grams}$

Molar mass of sulfur $=32\cdot 07\text{ grams}$

Moles of aluminum $=35\cdot 98\times \frac{\text{1 mole}}{26\cdot 98}\text{=1}\cdot 334\text{ moles}$

Moles of sulfur $=64\cdot 02\times \frac{\text{1 mole}}{32\cdot 07}\text{=1}\cdot 996\text{ moles}$

Now, divide the values by the smallest mole value. Hence, divide both the numbers by $\text{1}\cdot 334$ as follows:

$\begin{array}{l}\frac{\text{1}\cdot 334}{\text{1}\cdot 334}\text{moles Al =1 mole Al}\\ \frac{1\cdot 996}{1\cdot 334}\text{moles S =1}\cdot 496\cong 1\cdot 5\text{ mole S}\end{array}$

Now, multiply both the number to get a whole number ratio by a simplest number. Here, the ratio $1\text{ mole Al : 1}\cdot \text{5 mole S}$

In order to get a ratio of whole numbers , multiply $2$ on both sides

$\begin{array}{l}2\times 1\text{ mole Al:2}\times 1\cdot 5\text{ moles S}\\ \text{2 moles Al: 3 moles S}\end{array}$

Hence the empirical formula of the compound is ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$