Chapter 10.4, Problem 68SSC

Interpretation Introduction

Interpretation: The empirical formula of a compound of iron and oxygen which yields $\text{174}\text{.86 g of Fe and 75}\text{.14 g of O}$ needs to be determined.

Concept introduction:

Empirical formula is the simplest ratio in a molecule. The ratio of empirical formula and molecular formula is a whole number.

Answer to Problem 68SSC

The empirical formula is $F{e}_2{O}_3$.

Explanation of Solution

Given information:

The mass of iron and oxygen elements is given as follows:

Iron $=174.86\text{ g}$

Oxygen $=75.14\text{ g}$

Calculation:

To determine the empirical formula, first calculate the number of moles of Fe and O as follows:

$n=\frac{m}{M}$

Molar mass of Fe and O is 55.84 g/mol and 16 g/mol. Thus, number of moles will be:

$\begin{array}{l}\text{For Iron = 174}\text{.86 }\times \frac{\text{One mole Fe}}{\text{55}\text{.84 g of Fe}}=\text{ 3}\text{.131 mol}\\ \text{For Oxygen = 75}\text{.14 }\times \frac{{\text{One mole of O}}_2}{{\text{16 g of O}}_2}\text{= 4}\text{.69 mol}\end{array}$

Dividing by smallest number of moles,

$\begin{array}{l}\text{Iron = }\frac{3.131}{3.131}=1\\ \text{Oxygen}=\frac{4.69}{3.131}=1.49\approx 1.5\end{array}$

The formula will be $Fe{O}_{\frac{3}{2}}$ which is equal to $F{e}_2{O}_3.$

Conclusion

Therefore, the empirical formula is $F{e}_2{O}_3.$