Problem 85 - Chemistry An Atoms First Approach

Question

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. Heating 0.4172 g of the other compound results in a mass loss of 0.016 g. Determine the empirical formula of each compound.

Step-by-Step Solution

Answer

The empirical formulas of the two binary compounds of mercury and oxygen are HgO and Hg2O.

Step 1: Determine the mass of oxygen in each compound

For the first compound: Mass of mercury = 0.6018 g Since the initial mass is 0.6498 g, the mass of oxygen is: Mass of oxygen = 0.6498 g - 0.6018 g = 0.048 g For the second compound: Mass loss due to heating = 0.016 g (this is the mass of oxygen that escaped) So, the mass of mercury = 0.4172 g - 0.016 g = 0.4012 g

Step 2: Convert mass to moles

Using the molar mass of mercury (Hg) = 200.59 g/mol and oxygen (O) = 16.00 g/mol to convert masses to moles: For the first compound: Moles of Hg = 0.6018 g Hg * (1 mol Hg / 200.59 g Hg) = 0.00300044 mol Hg Moles of O = 0.048 g O * (1 mol O / 16.00 g O) = 0.003 mol O For the second compound: Moles of Hg = 0.4012 g Hg * (1 mol Hg / 200.59 g Hg) = 0.00199932 mol Hg Moles of O = 0.016 g O * (1 mol O / 16.00 g O) = 0.001 mol O

Step 3: Find the simplest ratio of moles

For the first compound: Divide the moles of both elements by the smallest value (0.003) to find the simplest ratio: Ratio_Hg = 0.00300044 mol Hg / 0.003 mol = 1.000133 Ratio_O = 0.003 mol O / 0.003 mol = 1 So, the empirical formula for the first compound is HgO. For the second compound: Divide the moles of both elements by the smallest value (0.001) to find the simplest ratio: Ratio_Hg = 0.00199932 mol Hg / 0.001 mol = 1.99932 Ratio_O = 0.001 mol O / 0.001 mol = 1 The ratio of Hg and O is approximately 2:1. Therefore, the empirical formula for the second compound is Hg2O.

Key Concepts

Understanding Moles

The concept of moles is central in chemistry as it helps us link the mass of a substance to the number of atoms or molecules it contains. One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles. This could be atoms, molecules, or ions. By using the molar mass of an element, which is the mass of one mole of its atoms in grams, we can convert the mass of a sample to moles.
For example, the molar mass of mercury (Hg) is 200.59 g/mol, which means one mole of mercury atoms weighs 200.59 grams. Similarly, the molar mass of oxygen (O) is 16.00 g/mol. In our exercise, once we have the mass of mercury and oxygen in grams, we convert those to moles using their respective molar masses. This conversion assists in determining how many moles of each element are present, essential for finding the empirical formula.

The Importance of Mass Conversion

Mass conversion is a crucial step when working with chemical elements and compounds in problems like this. It involves changing mass values into moles, enabling us to compare ratios of elements in compounds. To convert the mass of a substance to moles, we utilize the formula: \( \text{moles} = \frac{\text{mass in grams}}{\text{molar mass (g/mol)}} \).
For instance, converting 0.6018 g of mercury to moles, given the molar mass of mercury is 200.59 g/mol, involves dividing the mass by the molar mass. This gives us the number of moles, a fundamental step to establish the empirical formula of the compounds. This process also ensures accuracy in calculations, as it transitions measurements from the macroscopic scale (grams) to the atomic scale (moles).

Chemical Decomposition in Practice

Chemical decomposition is a reaction where a compound breaks down into simpler substances. In this scenario, heating causes the binary compounds of mercury and oxygen to decompose, resulting in elemental mercury and oxygen gas. This is indicative of a decomposition reaction, which is crucial in both laboratory and industrial chemical processes.
During decomposition, the loss of mass is often due to the release of a gaseous component, in this case, oxygen. By understanding the mass lost when the compound is heated, we can determine the quantity of oxygen released and use this to calculate the percentage composition of the compound. This provides the foundation for further calculations to find the compound's empirical formula.

Exploring Binary Compounds

Binary compounds are composed of two different elements. In our exercise, we are dealing with binary compounds of mercury and oxygen. These compounds are significant because their properties and reactions can tell us a lot about their composition and the chemical behavior of each element.
By decomposing the binary compounds of mercury and oxygen, we were able to measure the amount of each element present. This information allows us to determine the mass of individual elements, convert these to moles, and finally establish their empirical formula. Understanding the concept of binary compounds provides insight into how these elements combine to form stable compounds, highlighting stoichiometric ratios that lead to distinctive chemical properties.