Chapter 10.4, Problem 64PP

Interpretation Introduction

Interpretation:

The empirical formula of the compound needs to be determined.

Concept introduction:

The empirical formula tells about the simplest ratio of atoms of present in the molecule. It does not give the actual number of atoms present in the molecule.

Answer to Problem 64PP

The empirical formula of the compound is ${\text{K}}_{\text{2}}\text{O}$.

Explanation of Solution

Given information:

An oxide of potassium is decomposed in $19\cdot 55{\text{g K and 4g O}}_2$

Calculation: The mass of K and O is 19.55 g and 4.00 g respectively. The molar mass of K and O is 39.1 g/mol and 16 g/mol respectively.

The number of moles of K and O can be calculated as follows:

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

Thus,

$\begin{array}{l}{n}_K=\frac{19.55\text{ g}}{39.1\text{ g/mol}}=0.5\text{ mol}\\ n_O=\frac{4\text{ g}}{16\text{ g/mol}}=0.25\text{ mol}\end{array}$

The ratio of number of moles of K and O will be:

$\begin{array}{l}K:O=0.5:0.25\\ =\frac{0.5}{0.25}:\frac{0.25}{0.25}\\ =2:1\end{array}$

Thus, the empirical formula is ${\text{K}}_{\text{2}}\text{O}$.

Conclusion

Therefore, the empirical formula of the compound is ${\text{K}}_{\text{2}}\text{O}$.