Question

The oxides of Group 2A 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 -g sample containing only MgO and CuO is placed in a \(3.00-\mathrm{L}\) 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 \(\mathrm{MgO}\) reacts with \(\mathrm{CO}_{2}\) .

Short answer

The mass percent of MgO in the mixture can be calculated in the following steps: 1. Calculate the initial moles of CO2: \(n = \frac{PV}{RT} = \frac{(740/760)(3.00)}{0.0821(293.15)} \approx 0.116\,\text{mol}\) 2. Calculate the moles of CO2 after reaction: \(n = \frac{PV}{RT} = \frac{(390/760)(3.00)}{0.0821(293.15)} \approx 0.063\,\text{mol}\) 3. Calculate moles of CO2 that reacted: \(0.116 - 0.063 = 0.053\,\text{mol}\) 4. Calculate moles of MgO that reacted: \(0.053\,\text{mol}\) (from the 1:1 reaction ratio) 5. Convert moles of MgO to mass: \(0.053\,\text{mol} \times 40.3\, \frac{\text{g}}{\text{mol}} \approx 2.14\,\text{g}\) 6. Calculate mass percent of MgO: \(\frac{2.14}{2.85} \times 100 \approx 75.1\%\) The mass percent of MgO in the mixture is approximately \(75.1\%\).

Step-by-step solution

Calculate the initial moles of CO2

We can use the ideal gas law formula: \[ PV = nRT \] where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature. Given, initial pressure (P) = 740 Torr = 740/760 atm (1 atm = 760 Torr) Volume (V) = 3.00 L Temperature (T) = 20 + 273.15 = 293.15 K Gas constant (R) = 0.0821 L.atm/(K.mol) Now, we can solve for n (moles of CO2). \[ n = \frac{PV}{RT} \]

Moles of CO2 after reaction

After the reaction, the pressure inside the flask is 390 Torr = 390/760 atm. Using the same volume and temperature, we can calculate the moles of CO2 remaining after the reaction using the ideal gas law formula.

Calculate moles of CO2 that reacted

Subtracting the moles of CO2 after the reaction from the initial moles of CO2, we can find the moles of CO2 that reacted.

Calculate moles of MgO that reacted

From the given reaction, \[ \mathrm{MO} + \mathrm{CO}_{2} \longrightarrow \mathrm{MCO}_{3} \] As we can see, the balanced reaction ratio is 1:1 for MgO and CO2. Therefore, the moles of MgO that reacted are equal to the moles of CO2 that reacted.

Convert moles of MgO to mass

We can convert moles of MgO to its mass using the molar mass of MgO (40.3 g/mol): \[ \text{Mass of }MgO = \text{moles of }MgO \times \text{molar mass of }MgO \]

Calculate mass percent of MgO

Finally, we can calculate the mass percent of MgO in the 2.85-g sample using the formula: \[ \text{Mass \% of }MgO = \frac{\text{mass of }MgO}{\text{total mass of the sample}} \times 100 \]

Key concepts

Group 2A Metals

Group 2A metals, also known as alkaline earth metals, include beryllium, magnesium, calcium, strontium, barium, and radium. These metals are located in the second column of the periodic table and share some intriguing chemical characteristics.
Their oxides, like magnesium oxide (MgO) and calcium oxide (CaO), are typically ionic and form by the reaction of metal with oxygen.
These oxides often react with carbon dioxide to form carbonates, which is the focus of the typical reaction seen in chemistry problems related to stoichiometry:

• They usually have higher melting and boiling points compared to Group 1 metals.
• They react with water (though not as vigorously as Group 1) to form alkaline solutions, hence the name "alkaline earth metals."

In our problem, the key player is magnesium oxide (MgO), which reacts with carbon dioxide to form magnesium carbonate (MgCO\(_3\)) according to the simplified reaction equation: \[ \text{MO} + \text{CO}_2 \rightarrow \text{MCO}_3 \] Here, "M" symbolizes a Group 2A metal, highlighting its interaction with CO\(_2\). Understanding these properties allows us to predict the behaviors and products of reactions involving these metals.

Gas Laws

Understanding the behavior of gases is crucial in chemistry, especially when dealing with chemical reactions that involve gaseous reactants or products.
The ideal gas law is particularly helpful in calculating various properties of gases involved in chemical reactions. It is expressed as: \[ PV = nRT \] where:

• P is the pressure of the gas, often converted from units like Torr to atmospheres for consistency.
• V is the volume of the gas in liters.
• n is the number of moles of the gas.
• R is the ideal gas constant, which is often given as 0.0821 L.atm/(K.mol).
• T is the temperature in Kelvin, which requires conversion from Celsius.

In our context, initial conditions in the tank like pressure and temperature allow us to calculate how much carbon dioxide is present initially versus after the reaction completes. These calculations give crucial information about the quantities of reactants and products, aiding in solving the stoichiometry of the reaction between MgO and CO\(_2\).

Stoichiometry

Stoichiometry helps us understand and predict the quantities of substances consumed and produced in a chemical reaction.
It relies on balanced chemical equations to provide a quantitative relationship between reactants and products.
For the reaction of the oxides of Group 2A metals with carbon dioxide, stoichiometry can be applied to find out how much of each reactant is needed or used. In this problem, the reaction is as follows: \[ \text{MgO} + \text{CO}_2 \rightarrow \text{MgCO}_3 \] This equation indicates a 1:1 molar ratio between MgO and CO\(_2\).
By calculating the moles of CO\(_2\) that reacted, we can immediately deduce the moles of MgO that must have reacted, since they react in equal amounts.

This foundational stoichiometric relationship is essential in determining the mass of MgO that is a part of the original sample, enabling us to move towards calculating its mass percentage.

Mass Percentage Calculation

Mass percentage is a way of expressing the concentration of a component in a mixture or compound.
It represents how much of that component is present compared to the total mass.
This is useful when analyzing mixtures to determine the makeup of different elements or compounds quantitatively. Here is how you generally calculate mass percent:\[ \text{Mass \,\%} = \left( \frac{\text{mass of component}}{\text{total mass of mixture}} \right) \times 100 \] For our exercise, once we've established how much MgO reacted (based on stoichiometric calculations using the ideal gas law), we can convert those moles to mass (using MgO's molar mass).
This mass is then plugged into the mass percentage formula, comparing it against the total initial mass of the sample.
This method highlights not only how to find out the proportion of MgO in a complex sample but also reinforces the necessity of understanding the molar relationships and mass conversions in chemistry. Knowing this technique is essential for anyone tackling chemistry problems involving mixtures and reactions.