Question

When the reddish-brown mercury(II) oxide, \(\mathrm{Hg} \mathrm{O}\), is heated, it decomposes to its elements, liquid mercury metal and oxygen gas: (a) What is the molar mass of \(\mathrm{Hg} \mathrm{O}\) ? (b) What is the molar mass of \(\mathrm{Hg}\) ? (c) If \(2.00 \mathrm{~g} \mathrm{Hg} \mathrm{O}\) is decomposed to \(\mathrm{Hg}\), predict the mass of the pure Hg metal produced.

Short answer

The molar mass of \(\mathrm{HgO}\) is 216.59 g/mol; the molar mass of \(\mathrm{Hg}\) is 200.59 g/mol; if 2.00 g of \(\mathrm{HgO}\) is decomposed, it will produce 1.85 g of \(\mathrm{Hg}\).

Step-by-step solution

Calculating the molar mass of \(\mathrm{Hg} \mathrm{O}\)

Firstly look up the atomic masses of mercury and oxygen in the periodic table. Multiply this by the number of atoms in \(\mathrm{HgO}\) and then add them together to find the molar mass. For instance, the atomic mass of mercury(Hg) is approximately 200.59 g/mol and the atomic mass of oxygen(O) is approximately 16.00 g/mol. Therefore, the molar mass of HgO is \(200.59 \,g/mol + 16.00 \,g/mol = 216.59 \,g/mol\).

Calculating the molar mass of \(\mathrm{Hg}\)

From the periodic table, it's known that the atomic mass of mercury is approximately 200.59 g/mol. Given that there's only one atom of mercury in \(\mathrm{Hg}\), the molar mass thus equals 200.59 g/mol.

Predicting the mass of the pure Hg metal produced

Find the number of moles of \(\mathrm{HgO}\) decomposed by dividing the given mass by its molar mass thus, \(2.00 \,g/\,216.59 \,g/mol = 0.00924 \,mol\). Since the balanced equation shows the relationship to be 1:1 between \(\mathrm{HgO}\) and \(\mathrm{Hg}\), there will be the same number of moles of \(\mathrm{Hg}\) produced. Multiply this by the molar mass of \(\mathrm{Hg}\) to find the mass, so \(0.00924 \,mol \times 200.59 \,g/mol = 1.85 g\).

Key concepts

Mercury(II) oxide decomposition

Mercury(II) oxide, denoted as \(\mathrm{HgO}\), undergoes decomposition upon heating, breaking down into its basic components, liquid mercury (\(\mathrm{Hg}\)) and oxygen gas (\(\mathrm{O_2}\)). This process is a classic example of a decomposition reaction in chemistry, where one compound transforms into two or more simpler substances.

In this specific reaction, \(\mathrm{HgO}\) is unique due to its reddish-brown color, which distinguishes it from many other commonly encountered compounds. When heated, \(\mathrm{Hg}\) emerges as a silver-colored liquid, while \(\mathrm{O_2}\) is released as a gas. This reaction is not only interesting visually but also key in the field of chemical studies as it illustrates fundamental principles, such as stoichiometry, which we will explore in the next section. To summarize, mercury(II) oxide decomposition encapsulates the transformation from compound to elemental forms under thermal influence.

Chemical stoichiometry

Chemical stoichiometry is the study of quantitative relationships or ratios of reactants and products in a chemical reaction. It is crucial for understanding how much of each substance is consumed or produced. This concept is widely applicable in scientific research and practical applications like chemical manufacturing and laboratory experiments.

In the decomposition of \(\mathrm{HgO}\), stoichiometry helps us understand the one-to-one molar ratio between \(\mathrm{HgO}\) and \(\mathrm{Hg}\). The balanced chemical equation for this reaction is:

• \(2 \mathrm{HgO} \rightarrow 2 \mathrm{Hg} + \mathrm{O_2}\)

This tells us that two moles of \(\mathrm{HgO}\) will produce two moles of \(\mathrm{Hg}\) and one mole of \(\mathrm{O_2}\). Understanding stoichiometry not only helps in predicting the amounts of products formed but also ensures that the preservation of mass is maintained throughout the reaction. It is a foundational concept in chemistry that allows for precise calculations and predictions in chemical processes.

Predicting mass of products

After understanding the role of stoichiometry in reactions, predicting the mass of products becomes significantly easier. In our example with mercury(II) oxide, the key is to use the molar masses correctly.

• First, we identified the molar mass of \(\mathrm{HgO}\) as 216.59 g/mol.
• We then found the molar mass of \(\mathrm{Hg}\) to be 200.59 g/mol.

Given an initial amount of \(2.00 \, \mathrm{g}\) of \(\mathrm{HgO}\), the number of moles of \(\mathrm{HgO}\) decomposed can be calculated using its molar mass, yielding approximately \(0.00924\, \mathrm{mol}\). Since the reaction's stoichiometry shows a 1:1 mole ratio between \(\mathrm{HgO}\) and \(\mathrm{Hg}\), \(0.00924\, \mathrm{mol}\) of \(\mathrm{Hg}\) is produced. Multiply the moles by the molar mass of \(\mathrm{Hg}\) to find the mass of the product: \(0.00924\, \mathrm{mol} \times 200.59\, \mathrm{g/mol} = 1.85\, \mathrm{g}\).

By developing a firm grasp on these calculations, students can predict the mass of products from any decomposing reactions with accuracy, enhancing their understanding and capability in chemistry.