Question

The oxides of Group \(2 \mathrm{~A}\) metals (symbolized by \(\mathrm{M}\) 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 \(\mathrm{MgO}\) in the mixture? Assume that only the \(\mathrm{Mg} \mathrm{O}\) reacts with \(\mathrm{CO}_{2}\).

Short answer

To find the mass percent of MgO in the mixture, we can follow these steps: 1. Calculate the moles of CO₂ reacted using the given initial and final pressures and the ideal gas law. The difference in moles of CO₂ is \(Δn_{CO₂} = n_{initial} - n_{final}\). 2. The moles of MgO in the sample are equal to the moles of CO₂ reacted: \(n_{MgO} = Δn_{CO₂}\). 3. Calculate the mass of MgO in the sample: \(mass_{MgO} = n_{MgO} \times Molar\_mass_{MgO}\). 4. Calculate the mass percent of MgO in the mixture: \(mass\%_{MgO} = \frac{mass_{MgO}}{mass_{total}} \times 100\%\).

Step-by-step solution

Calculate moles of CO₂ reacted

We know the initial pressure (740 torr) and final pressure (390 torr) of CO₂ in the container. To find the moles reacted, we can use the ideal gas law. We calculate the difference in moles, as only the reacted CO₂ affects the pressure change. First, we convert the pressures into atmospheres, as the ideal gas law requires pressure in atm. \(P_{initial}\) = \(\frac{740}{760}\) atm \(P_{final}\) = \(\frac{390}{760}\) atm Now, we can use the ideal gas law to find the initial and final moles of CO₂: \(PV = nRT\) Where: P is the pressure V is the volume n is the moles R is the gas constant (0.0821 L atm/mol K) T is the temperature in Kelvin We know the volume (3 L) and temperature (20°C = 293 K) of the container. So, we can now calculate the initial and final moles of CO₂: \(n_{initial} = \frac{P_{initial}V}{RT}\) \(n_{final} = \frac{P_{final}V}{RT}\) And the difference in moles of CO₂: \(Δn_{CO₂} = n_{initial} - n_{final}\)

Calculate moles of MgO in the sample

Now that we have the moles of CO₂ that reacted, we can use stoichiometry to find the moles of MgO in the sample. We know that one mole of MgO reacts with one mole of CO₂: \(MgO + CO₂ \to MgCO₃\) Hence, the moles of MgO in the sample are equal to \(Δn_{CO₂}\) moles. \(n_{MgO} = Δn_{CO₂}\)

Calculate the mass of MgO in the sample

We know the moles of MgO and the molar mass of MgO (40.30 g/mol). So, we can find the mass of MgO in the sample: \(mass_{MgO} = n_{MgO} \times Molar_mass_{MgO}\)

Calculate the mass percent of MgO in the mixture

Finally, we can calculate the mass percent of MgO in the mixture using the mass of MgO and the given total mass of the mixture (2.85 g): \(mass\%_{MgO} = \frac{mass_{MgO}}{mass_{total}} × 100\%\)

Key concepts

Ideal Gas Law

Understanding the ideal gas law is crucial when studying gaseous reactions. It is a mathematical relationship that allows us to calculate the amount of gas (in moles) involved in a reaction given the pressure, volume, and temperature of the system. The formula is expressed as
\[ PV = nRT \]
where P stands for pressure, V represents volume, n is the number of moles, R is the ideal gas constant, and T is the absolute temperature in Kelvin. In this exercise, the ideal gas law helps determine the change in moles of CO₂ gas once it reacts. This change in moles is directly proportional to the change in pressure, assuming the volume and temperature remain constant. Consequently, the gas law allows us to explore the extent of the reaction by comparing the pressures before and after the reaction.

Molar Mass

Molar mass is an indispensable concept in chemistry, defining the mass of one mole of a substance (in grams per mole). It can be determined by summing the atomic masses of all the atoms in a molecule, which are listed in the periodic table. In the context of the exercise, the molar mass of MgO, which comprises one magnesium atom and one oxygen atom, is necessary to translate moles of MgO into grams.
This conversion is vital as it bridges the gap between the microscopic scale of atoms and molecules and the macroscopic world we can measure in labs. By multiplying the number of moles of MgO with its molar mass
\[ mass_{MgO} = n_{MgO} \times Molar_mass_{MgO} \]
we obtain the actual mass of MgO that reacted with CO₂, providing insight into the composition of the original mixture.

Reaction Stoichiometry

Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the balanced chemical equation, which reflects the conservation of mass and the mole ratio of substances involved in the reaction. According to the stoichiometry of the given reaction
\[ MgO + CO_2 \to MgCO_3 \]
one mole of MgO reacts with one mole of CO₂. This 1:1 mole ratio allows us to directly equate the moles of reacted CO₂, derived from the ideal gas law, to the moles of MgO in the sample. This straightforward ratio underscores the convenience and power of stoichiometry in predicting the amounts of substances required or produced in a chemical reaction, and is especially helpful in this exercise for determining the proportion of MgO that reacted.

Percent Composition

Percent composition is an expression of the concentration of a particular component in a mixture, based on its mass relative to the total mass of the mixture. It is commonly used in analytical chemistry to describe the purity of substances or to calculate the empirical formula of a compound. The formula for the mass percent of a component is
\[ mass\% = \frac{mass_{component}}{mass_{total}} \times 100\% \]
In our exercise, we use percent composition to find out how much MgO, by mass, is present in the original 2.85-g sample of MgO and CuO. Mass percent is an invaluable figure in real-world applications such as quality control in manufacturing processes, determination of nutritional content in food, and more broadly, whenever a composition needs to be quantified. Empowering students to use percent composition calculations can significantly enhance their practical understanding of the subject.