Chapter 6, Problem 61A

Interpretation Introduction

Interpretation:

The empirical formula of the oxide needs to be predicted.

Concept Introduction:

The simplest positive integer ratio of the atoms present in the chemical compound is known as empirical formula.

Answer to Problem 61A

The empirical formula of the compound is HgO.

Explanation of Solution

Given information:

Mass of mercury = 4.01 g

Mass of oxide of mercury = 4.33 g

The calculation of mass of oxygen in the compound is shown below:

Mass of oxygen = mass of oxide compound − mass ogf mercury

= 4.33 g − 4.01 g

= 0.32 g

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{Hg}\text{=}\frac{\text{mass}\text{of}\text{Hg}}{\text{molar}\text{mass}\text{of}\text{Hg}}\\ \text{=}\frac{\text{4}\text{.01}\text{g}}{\text{200}\text{.59}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.0199}\text{mol}\\ \simeq \text{0}\text{.02}\text{mol}\\ \text{moles}\text{of oxygen}\text{=}\frac{\text{mass}\text{of}\text{oxygen}}{\text{molar}\text{mass}\text{of}\text{oxygen}}\\ \text{=}\frac{\text{0}\text{.32}\text{g}}{\text{16}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.02}\text{mol}\end{array}$

Molar ratio is shown below:

Hg:O = 0.02 : 0.02

= 0.02/0.02 : 0.02/0.02

= 1:1

Conclusion

The empirical formula of the compound is HgO