Chapter 10.4, Problem 69SSC

Interpretation Introduction

Interpretation:

The empirical formula of the oxide which contains 0.545 g Al and 0.485 g O 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 69SSC

The empirical formula must be ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Explanation of Solution

Given information:

Mass of Al = 0.545 g

Mass of O = 0.485 g

Molar mass of Al = 26.98 g/mol

Molar mass of O = 15.99 g/mol

Calculate moles of Al and O:

$\begin{array}{l}\text{Al = }\frac{\text{0}\text{.545g}}{\text{26}\text{.98g/mol}}\text{= 0}\text{.020 moles}\\ \text{O = }\frac{\text{0}\text{.485 g}}{\text{15}\text{.99 g/mol}}\text{= 0}\text{.030 moles}\end{array}$

Divide with least moles to get the whole number:

$\begin{array}{l}\text{Al = }\frac{\text{0}\text{.020 moles}}{\text{0}\text{.020 moles}}\text{= 1}\times \text{2=2}\\ \text{O = }\frac{\text{0}\text{.030 moles}}{\text{0}\text{.020 moles}}\text{= 1}\text{.5}\times \text{2=3}\end{array}$

Thus the empirical formula must be ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$

Conclusion

The empirical formula must be ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$