Problem 42 - General Chemistry: Principles and Modern Applications

Question

Joseph Priestley, a British chemist, was credited with the discovering oxygen in \(1774 .\) In his experiments, he generated oxygen gas by heating \(\mathrm{HgO}(\mathrm{s}) .\) The other product of the decomposition reaction is \(\mathrm{Hg}(1) .\) What volume of wet \(\mathrm{O}_{2}(\mathrm{g})\) is obtained from the decomposition of \(1.0 \mathrm{g} \mathrm{HgO}(\mathrm{s}),\) if the gas is collected over water at \(25^{\circ} \mathrm{C}\) and a barometric pressure of \(756 \mathrm{mmHg} ?\) The vapor pressure of water is \(23.76 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The volume of wet O2 obtained from the decomposition of 1.0g HgO is approximately 58mL. However, the actual volume of dry O2 will be less than this due to the presence of water vapor.

Step 1: Molar Mass of HgO

First, determine the molar mass of HgO by adding the atomic masses of Hg and O, which is \(202.59 g/mol + 16.00 g/mol = 218.59 g/mol\).

Step 2: Find the moles of HgO

Next, find the number of moles of HgO in 1g by using the formula \(\frac{mass}{molar\: mass}\). Substituting the given values, we have \(\frac{1g}{218.59g/mol} = 0.00457moles\).

Step 3: Moles of O2 produced

The balanced chemical reaction of the decomposition of HgO is \(2HgO \rightarrow 2Hg + O2\). From the balanced equation, we know that 2 moles of HgO produce 1 mole of O2. Therefore, 0.00457 moles of HgO would produce half as many moles of O2, which is \(0.00457 * 0.5 = 0.002285 moles\).

Step 4: Use the Ideal Gas Law

Use the ideal gas law \(PV = nRT\) to calculate the volume of the gases at the given conditions. Remember the gas is collected over water, so the pressure of the gas will be the barometric pressure minus the vapor pressure of the water. Thus, the pressure of the gas is \(756mmHg - 23.76mmHg = 732.24mmHg\). Converting that to atmospheres gives \(732.24mmHg \times \frac{1atm}{760mmHg} = 0.96374atm\). \(T\) (in Kelvin) here is \(25°C + 273.15 = 298.15K\) and \(R = 0.0821 L.atm/K.mol\). Substituting these values into the ideal gas law equation gives \(V = \frac{nRT}{P} = \frac{0.002285mol \times 0.0821 L.atm/K.mol \times 298.15K}{0.96374atm} = 0.058 L or 58 mL\). So, the collected O2 will occupy 58mL under the given conditions.

Step 5: Account for Water Vapor Pressure

The measured volume of O2 is not correct because it contains water vapor. Therefore the volume of dry O2 will be less than 58mL. The correct volume of dry O2 can't be determined without more information, specifically, the vapor pressure of water at the temperature of the gas in the eudiometer (gas collection tube).

Key Concepts

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. This process often requires energy, such as heat, light, or electricity, to take place. In the context of the given exercise, the decomposition reaction involves mercury(II) oxide (HgO) which, when heated, decomposes to form mercury (Hg) and oxygen gas (O_2). This reaction can be represented by the equation:

• \(2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2\)

This balanced equation shows that two moles of HgO yield one mole of O_2 gas and two moles of mercury(Hg). Here, the decomposition reaction is crucial because it indicates the molar ratio between reactants and products, allowing us to predict how much O_2 will be produced from a given amount of HgO. Understanding this concept helps in calculating the outcomes of reactions, especially when dealing with chemical equations involving mass and volume relationships. The decomposition reaction is an example of how matter can change from one form to another, illustrating one of the core principles of chemistry.

Ideal Gas Law

In chemistry, the Ideal Gas Law is an important equation used to relate the pressure, volume, temperature, and number of moles of a gas. It is represented by the equation:

• \(PV = nRT\)

Here, \(P\) stands for pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. This law is particularly useful when you need to calculate the volume of gas produced in a reaction, as seen in the exercise.When applying the Ideal Gas Law to the decomposition reaction of HgO, it's important to remember that the gas (oxygen in this case) is collected over water, which means the pressure of the gas includes the vapor pressure of water. Therefore, the actual pressure of the oxygen gas must be adjusted by subtracting the water vapor pressure from the barometric pressure.After adjusting the pressure, you can substitute the known values into the Ideal Gas Law formula to find the volume of the gas. This process emphasizes the importance of considering environmental factors, such as water vapor presence, when using gas laws to solve chemical equations. Understanding the Ideal Gas Law allows chemists to make accurate predictions about gas behavior under various conditions.

Molar Mass Calculation

Determining the molar mass of a compound is a key step in many chemical calculations as it allows for the conversion between mass and moles. The molar mass is calculated by summing the atomic masses of all the atoms present in a molecule of the compound. In the exercise, calculating the molar mass of mercury(II) oxide (HgO) is essential for determining how much O_2 is produced from a given mass of HgO.

• First, identify the atomic masses: Mercury (Hg) has an atomic mass of 202.59 g/mol, and Oxygen (O) has an atomic mass of 16.00 g/mol.
• Adding these atomic masses gives \(202.59 + 16.00 = 218.59\) g/mol for HgO.

To find the number of moles of HgO from a specific mass, you use the formula:

• \(\text{Moles} = \frac{\text{mass}}{\text{molar mass}}\)

This calculation forms the foundation for further computations in chemistry, such as finding the quantities of substances involved in reactions. By understanding how to calculate molar mass and convert it into moles, students acquire a vital tool for analyzing and predicting the outcomes of chemical reactions.