Question

The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

Short answer

The mass percent of MgO in the mixture is approximately 78.2 %.

Step-by-step solution

Calculate moles of CO2 initially present

We are given the initial pressure of CO2, the volume of the container, and the temperature. Using the Ideal Gas Law, we can calculate the moles of CO2 initially present: \(PV = nRT\) Where, P = Pressure in atm = 740 Torr × (1 atm / 760 Torr) = 0.9737 atm V = Volume = 3.00 L R = Gas constant = 0.0821 L atm/mol K T = Temperature in Kelvin = 20°C + 273.15 = 293.15 K \[n_{CO_2\, initial} = \frac{PV}{RT} = \frac{(0.9737)(3.00)}{(0.0821)(293.15)} \approx 0.1190 \,mol\]

Calculate moles of CO2 remaining after the reaction

Similarly, we can calculate the moles of CO2 remaining after the reaction using the Ideal Gas Law and the final conditions: P( final) = 390 Torr × (1 atm / 760 Torr) = 0.5132 atm \[n_{CO_2\, final} = \frac{PV}{RT} = \frac{(0.5132)(3.00)}{(0.0821)(293.15)} \approx 0.0637 \,mol\]

Calculate moles of CO2 reacted with MgO

The difference between the initial and final moles of CO2 is equal to the moles of CO2 that reacted with MgO: \[n_{CO_2\, reacted} = n_{CO_2\, initial} - n_{CO_2\, final} = 0.1190 - 0.0637 = 0.0553 \, mol\]

Determine moles of MgO

From the balanced reaction, we know the stoichiometry of CO2 and MgO is 1:1. Therefore, the moles of MgO must be equal to the moles of CO2 reacted: \[n_{MgO} = n_{CO_2\, reacted} = 0.0553 \,mol\]

Calculate mass of MgO in the mixture

To calculate the mass of MgO in the mixture, multiply the number of moles of MgO by its molar mass: Molar mass of MgO = 24.31 g/mol (Mg) + 16.00 g/mol (O) = 40.31 g/mol Mass of MgO = n_{MgO} × Molar mass of MgO = 0.0553 mol × 40.31 g/mol ≈ 2.23 g

Calculate the mass percent of MgO in the mixture

Now that we know the mass of MgO in the mixture, we can find the mass percent by dividing the mass of MgO by the total mass of the mixture (2.85 g) and multiplying by 100%: Mass percent of MgO = (Mass of MgO / Total mass) × 100 % = (2.23 g / 2.85 g) × 100 % ≈ 78.2 % The mass percent of MgO in the mixture is approximately 78.2 %.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a helpful formula for understanding the relationship between pressure, volume, and temperature of a gas. This relationship is captured by the equation \(PV = nRT\). Here, \(P\) represents the pressure of the gas, \(V\) is the volume it occupies, \(n\) is the number of moles of the gas, \(R\) is the ideal gas constant, and \(T\) is the measured temperature in Kelvin.

The Ideal Gas Law is particularly useful when you need to calculate the number of moles of a gas, as shown in our exercise.

• First, remember to always convert temperatures to Kelvin by adding 273.15 to the Celsius measurement.
• Also, when dealing with pressure, ensure it is converted to atmospheres (atm), especially if originally given in units like torr. Use the conversion factor \(1 \text{ atm} = 760 \text{ torr}\).

This law assumes ideal behavior, meaning it works best under conditions of low pressure and high temperature, where gases act most ideally.

Mass Percent Calculation

Mass percent is a quick way to express the concentration of a compound in a mixture. It gives us the mass of a specific component compared to the total mass of the mixture, presented as a percentage.
To calculate mass percent, use the formula:
\[\text{Mass percent} = \left( \frac{\text{Mass of component}}{\text{Total mass of mixture}} \right) \times 100\%\]
In this exercise, we determined the mass of \(MgO\) in the mixture to find the mass percent of \(MgO\). Calculate \(MgO\)'s mass percent by taking the mass of \(MgO\) that reacted (which we found from stoichiometry) divided by the total sample mass, then multiply by 100.

This gives us useful insight into the composition of the mixture, which in this problem turned out to be approximately \(78.2\% \ MgO\).

Stoichiometry

Stoichiometry revolves around the concept of the conservation of mass. It helps in understanding the quantitative relationships in chemical reactions, allowing us to calculate the amounts of reactants needed or products formed.
In the context of the exercise, we used stoichiometry to relate moles of \(CO_2\) that reacted to moles of \(MgO\). Since we know from the balanced equation that 1 mole of \(MgO\) reacts with 1 mole of \(CO_2\), the moles of each involved in the reaction are equal.

• First, determine the moles of a reagent (here \(CO_2\)) from the initial conditions using the Ideal Gas Law.
• Then, find the number of moles left over after the reaction finishes.
• The difference tells you how much has reacted, and stoichiometry lets you know if more than one reactant or product is involved。

Through stoichiometry, we discovered the mass of \(MgO\) that actually reacted, leading to the calculation of its mass percent.

Chemical Reactions

Chemical reactions describe the transformation of substances, depicted by chemical equations. In our exercise, \(MO(s) + CO_2(g) \rightarrow MCO_3(s)\) showcases a solid oxide reacting with carbon dioxide to form a carbonate.
The products and reactants in a chemical equation are interconnected, representing the conservation of mass and energy.

• A balanced chemical equation ensures that the number of atoms for each element in the reaction remains the same pre- and post-reaction.
• This balance is crucial for precise stoichiometric calculations.Understanding these reactions helps in predicting the outcome of mixing substances under certain conditions.

In this problem, only \(MgO\) was considered to react with \(CO_2\), despite the presence of \(CuO\), reflecting selective reactiveness that can often be observed in chemical processes under specific conditions.