Question

A chemist weighed out \(5.14 \mathrm{g}\) of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50\) -\(\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at \(30.0^{\circ} \mathrm{C}\) and \(750 .\) torr. After the reaction to form \(\mathrm{BaCO}_{3}(s)\) and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was \(230 .\) torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

Short answer

The mass percentages of BaO(s) and CaO(s) in the mixture are approximately 50.6% and 25.1%, respectively.

Step-by-step solution

Calculate moles of CO2 before and after the reaction

To find the initial and final moles of CO₂, we 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 ideal gas constant, and T is the temperature in Kelvin. Given are initial pressure (750 torr), final pressure (230 torr), temperature (30°C), and volume (1.50 L). First, convert temperature from Celsius to Kelvin: \(T = 30 + 273.15 = 303.15 K\) Next, convert pressure from torr to atm, since the gas constant R is given in atm: \(P_{initial} = \frac{750}{760} \approx 0.9868 atm\), \(P_{final} = \frac{230}{760} \approx 0.3026 atm\) Now, rearrange the ideal gas law formula to solve for n: \(n = \frac{PV}{RT}\). Calculate the initial and final moles of CO₂: \(n_{initial} = \frac{(0.9868)(1.50)}{(0.08206)(303.15)} \approx 0.06034 mol\) \(n_{final} = \frac{(0.3026)(1.50)}{(0.08206)(303.15)} \approx 0.01845 mol\)

Calculate moles of BaO and CaO using stoichiometry

The total decrease in moles of CO₂ represents equal moles of BaO(s) and CaO(s), as there is a 1:1 stoichiometric ratio in their reactions. Therefore, subtract the final number of moles from the initial number of moles to find the moles of BaO(s) and CaO(s) participating in the reaction: \(0.06034 - 0.01845 \approx 0.04189 mol\) Let x be the number of moles of BaO(s) and y be the number of moles of CaO(s). Their sum will be equal to 0.04189 mol: \(x + y = 0.04189\)

Convert moles to grams and find mass percentages

To find the mass of BaO(s) and CaO(s) in the sample, multiply the moles of each by their respective molar masses: \(137.33x + 56.08y = 5.14\), where x and y are in moles, and the equation states the mass of the sample in grams, 5.14 g. Now we have a system of two equations with two unknowns: (1) \(x + y = 0.04189\) (2) \(137.33x + 56.08y = 5.14\) Multiply Equation (1) by 56.08 and subtract from Equation (2) to isolate x: \(x = \frac{5.14 - 56.08(0.04189)}{137.33 - 56.08} \approx 0.01891 \mathrm{mol}\) Substitute x back into Equation (1) to find y: \(y = 0.04189 - 0.01891 \approx 0.02298 \mathrm{mol}\) Now, convert these moles to grams by multiplying by their respective molar masses: Mass of BaO(s) = \(0.01891 \times 137.33 \approx 2.6 g\) Mass of CaO(s) = \(0.02298 \times 56.08 \approx 1.29 g\) Finally, calculate the mass percentages of BaO(s) and CaO(s) in the mixture: % BaO(s) = \(\frac{2.6}{5.14} \times 100 \approx 50.6\% \) % CaO(s) = \(\frac{1.29}{5.14} \times 100 \approx 25.1\% \) Thus, the mass percentages of BaO(s) and CaO(s) in the mixture are approximately 50.6% and 25.1%, respectively.

Key concepts

Stoichiometry and the Ideal Gas Law

Stoichiometry problems often require a deep understanding of the ideal gas law, which is expressed as the equation: \(PV = nRT\). In this law, \(P\) stands for pressure, \(V\) stands for volume, \(n\) stands for the number of moles, \(R\) is the universal gas constant, and \(T\) represents temperature in Kelvin. This equation allows scientists to calculate the amount of gas in moles based on the measured pressure, volume, and temperature.

When dealing with mixtures and reactions, as in the problem provided, it's critical to apply the ideal gas law to find the change in the number of moles of gas before and after the reaction. This change is directly related to the stoichiometry of the reaction, which in turn allows calculation of the quantities of the reactants. It's crucial to always convert pressures to the same units as the gas constant (usually atmospheres, atm) and temperatures to Kelvin for accurate calculations.

To improve understanding of this topic, students should practice converting units of pressure and temperature, as well as become comfortable with rearranging the ideal gas law formula to solve for the unknown variable.

Moles Calculation and Stoichiometry

Calculating moles is a foundational skill in chemistry, especially in stoichiometry problems. The concept of a mole allows chemists to count particles (atoms, molecules, etc.) by weighing them. One mole is defined as the amount of substance that contains as many particles as there are atoms in 12 grams of pure carbon-12, which is approximately \(6.022 \times 10^{23}\) particles (Avogadro's number).

In the context of our problem, calculating the moles consumed or produced in a chemical reaction involves using the stoichiometric coefficients from the balanced chemical equations. These coefficients indicate the proportions of reactants and products involved. For instance, a 1:1 ratio in the reaction between CO₂ and the metal oxides (BaO and CaO) implies that a mole of CO₂ reacts with a mole of BaO or CaO each. By determining the difference between the initial and final moles of CO₂, we can derive the moles of BaO and CaO that reacted.

Students can improve their skills in mole calculations by practicing with a variety of chemical equations and by always ensuring that they're using the correct molar masses of substances and maintaining proper unit consistency throughout the calculation.

Mass Percent Composition in Mixtures

Mass percent composition is another significant concept in chemistry, particularly for stoichiometry problems involving mixtures. It expresses the mass of each component as a percentage of the total mass of the mixture. It is a useful concept for identifying the purity of a substance or the concentration of components in mixtures.

In the given problem, after calculating the moles of BaO and CaO, we convert these to grams and then find their mass percentages relative to the total mass of the original mixture. To convert moles to grams, we multiply the number of moles by the molar mass of the compound. The mass percent is then found by dividing the mass of the individual component by the total mass and multiplying by 100. This calculation provides a quantifiable expression of composition that can help in analyzing and comparing mixtures.

Understanding mass percent composition is crucial for students, not only for solving textbook problems but also for practical applications like preparing solutions in a lab or analyzing nutritional information on food labels. Engaging in exercises that involve determining mass percent composition from actual lab data or everyday products can enhance student proficiency.