Determining Atomic Weight Of Metal M In M2(SO4)3 A Stoichiometry Problem
In this comprehensive article, we will delve into a fascinating chemistry problem that involves determining the atomic weight of a metal M in the compound M2(SO4)3. This problem is a classic example of stoichiometry and gravimetric analysis, where we use the mass of a precipitate formed in a chemical reaction to determine the amount of a specific element in the original compound. To solve this problem, we will carefully analyze the reaction between the metal sulfate M2(SO4)3 and barium chloride (BaCl2), which results in the formation of barium sulfate (BaSO4), a highly insoluble precipitate. By using the mass of BaSO4 obtained and the stoichiometry of the reaction, we can accurately calculate the molar mass of the metal sulfate and, subsequently, the atomic weight of the metal M.
The core concept we will be exploring is the conservation of mass, a fundamental principle in chemistry. This principle states that mass is neither created nor destroyed in a chemical reaction. Thus, the amount of sulfate ions (SO4^2-) in the original metal sulfate compound will be directly related to the amount of barium sulfate precipitate formed. This relationship is crucial for our calculations and will allow us to establish a clear pathway to determine the unknown atomic weight. Additionally, we will reinforce our understanding of molar mass, stoichiometry, and gravimetric analysis, making this article an invaluable resource for students and enthusiasts of chemistry. Let’s begin this step-by-step journey to unravel the mystery of metal M and its atomic weight.
A metal M forms the sulfate M2(SO4)3. A 0.596 g sample of the sulfate reacts with excess BaCl2 to give 1.220 g BaSO4. What is the atomic weight of M? (Atomic weights: S = 32, Ba = 137.3, Cl = 35.5)
A) 27
B) 70
C) 54
D) 23
Step 1: Write the Balanced Chemical Equation
To begin solving this problem, it's essential to first write the balanced chemical equation for the reaction between the metal sulfate M2(SO4)3 and barium chloride (BaCl2). This step is crucial because it establishes the stoichiometric relationships between the reactants and products, which are necessary for accurate calculations. The balanced equation is:
M2(SO4)3 + 3 BaCl2 → 3 BaSO4 + 2 MCl3
This equation tells us that one mole of M2(SO4)3 reacts with three moles of BaCl2 to produce three moles of barium sulfate (BaSO4) and two moles of metal chloride (MCl3). The key relationship here is between M2(SO4)3 and BaSO4, as we are given the mass of BaSO4 produced and need to relate it back to the metal M.
Understanding the balanced equation is paramount because it provides the mole ratios necessary to convert between the mass of BaSO4 and the moles of M2(SO4)3. Without a balanced equation, the stoichiometric calculations would be inaccurate, leading to an incorrect final answer. This step highlights the importance of paying close attention to the fundamental principles of chemical reactions and stoichiometry.
Step 2: Calculate Moles of BaSO4
Next, we need to calculate the number of moles of BaSO4 formed in the reaction. This step is essential as it allows us to connect the mass of the precipitate obtained experimentally to the molar quantities involved in the reaction. To do this, we use the molar mass of BaSO4, which can be calculated from the atomic weights of barium (Ba), sulfur (S), and oxygen (O). The atomic weights are:
- Ba = 137.3 g/mol
- S = 32.0 g/mol
- O = 16.0 g/mol
The molar mass of BaSO4 is calculated as:
Molar mass BaSO4 = 137.3 + 32.0 + (4 × 16.0) = 233.3 g/mol
Now, we can calculate the number of moles of BaSO4 using the given mass of 1.220 g:
Moles BaSO4 = Mass BaSO4 / Molar mass BaSO4 = 1.220 g / 233.3 g/mol ≈ 0.00523 mol
This calculation is a crucial step in gravimetric analysis, where the mass of a precipitate is used to determine the amount of a specific substance. The accuracy of this step directly affects the final result, emphasizing the importance of precise measurements and calculations.
Step 3: Determine Moles of M2(SO4)3
Using the balanced chemical equation from Step 1, we can now determine the number of moles of M2(SO4)3 that reacted. The balanced equation shows that 1 mole of M2(SO4)3 produces 3 moles of BaSO4. Therefore, we can use this stoichiometric ratio to convert the moles of BaSO4 calculated in Step 2 to moles of M2(SO4)3.
The stoichiometric ratio is:
(1 mol M2(SO4)3) / (3 mol BaSO4)
So, the moles of M2(SO4)3 are:
Moles M2(SO4)3 = Moles BaSO4 × (1 mol M2(SO4)3) / (3 mol BaSO4) = 0.00523 mol × (1/3) ≈ 0.00174 mol
This conversion is a pivotal step in solving the problem. By correctly applying the stoichiometric ratio, we link the experimentally obtained quantity of BaSO4 back to the original compound, M2(SO4)3. This step exemplifies the power of stoichiometry in quantitative chemical analysis, allowing us to relate the amounts of different substances involved in a reaction.
Step 4: Calculate the Molar Mass of M2(SO4)3
Now that we know the moles of M2(SO4)3 and the mass of the sample (0.596 g), we can calculate the molar mass of M2(SO4)3. The molar mass is given by the formula:
Molar mass = Mass / Moles
So, the molar mass of M2(SO4)3 is:
Molar mass M2(SO4)3 = 0.596 g / 0.00174 mol ≈ 342.53 g/mol
This calculation provides a critical piece of information needed to determine the atomic weight of the metal M. The molar mass of M2(SO4)3 represents the total mass of the compound per mole, and it includes the masses of the metal M, sulfur (S), and oxygen (O). By subtracting the known masses of sulfur and oxygen, we can isolate the mass contributed by the metal M.
This step underscores the importance of understanding the relationship between mass, moles, and molar mass. It showcases how molar mass acts as a bridge between macroscopic measurements (mass) and microscopic quantities (moles), allowing us to work with chemical substances in a quantitative manner.
Step 5: Determine the Atomic Weight of Metal M
The molar mass of M2(SO4)3 consists of the combined atomic weights of two M atoms, three sulfur (S) atoms, and twelve oxygen (O) atoms. We know the atomic weights of sulfur and oxygen, and we have calculated the molar mass of M2(SO4)3. Thus, we can set up an equation to solve for the atomic weight of M.
The equation is:
2 × (Atomic weight of M) + 3 × (Atomic weight of S) + 12 × (Atomic weight of O) = Molar mass of M2(SO4)3
Substituting the known values:
2 × (Atomic weight of M) + 3 × (32.0 g/mol) + 12 × (16.0 g/mol) = 342.53 g/mol
2 × (Atomic weight of M) + 96.0 g/mol + 192.0 g/mol = 342.53 g/mol
2 × (Atomic weight of M) + 288.0 g/mol = 342.53 g/mol
Now, we can solve for the atomic weight of M:
2 × (Atomic weight of M) = 342.53 g/mol - 288.0 g/mol
2 × (Atomic weight of M) = 54.53 g/mol
Atomic weight of M = 54.53 g/mol / 2 ≈ 27.27 g/mol
This final calculation is the culmination of all the previous steps. By systematically working through the stoichiometry of the reaction and using the molar mass of M2(SO4)3, we have successfully isolated and determined the atomic weight of the metal M. The result, approximately 27.27 g/mol, is very close to the atomic weight of aluminum (Al), which is 27 g/mol. Therefore, the metal M is likely aluminum.
Step 6: Final Answer
The calculated atomic weight of metal M is approximately 27.27 g/mol. Looking at the options provided, the closest answer is:
A) 27
Therefore, the atomic weight of metal M is 27.
In this detailed exploration, we successfully determined the atomic weight of metal M in the compound M2(SO4)3. By meticulously following the principles of stoichiometry and gravimetric analysis, we navigated through the problem step by step. We began by writing a balanced chemical equation, which provided the crucial stoichiometric relationships between the reactants and products. This foundation allowed us to accurately convert the mass of BaSO4 precipitate to the moles of M2(SO4)3.
The process involved calculating the moles of BaSO4 formed, using its molar mass, and then determining the moles of M2(SO4)3 that reacted, based on the stoichiometric ratio from the balanced equation. We then calculated the molar mass of M2(SO4)3 using the mass of the sample and the moles of M2(SO4)3. Finally, by subtracting the known masses of sulfur and oxygen from the molar mass of M2(SO4)3, we were able to isolate and calculate the atomic weight of the metal M, which was approximately 27 g/mol.
This comprehensive solution highlights the importance of several key concepts in chemistry. Stoichiometry plays a pivotal role in relating the quantities of reactants and products in a chemical reaction, and gravimetric analysis allows us to quantitatively determine the amount of a substance by measuring the mass of a precipitate. The molar mass concept serves as a bridge between macroscopic measurements (mass) and microscopic quantities (moles), enabling us to perform accurate calculations.
The step-by-step approach used in this solution exemplifies the systematic problem-solving strategies essential in chemistry. By breaking down the problem into smaller, manageable steps, we can tackle complex questions with confidence. This method not only leads to the correct answer but also enhances our understanding of the underlying chemical principles. The final answer of 27 g/mol strongly suggests that the metal M is aluminum, which further solidifies the accuracy of our calculations.
In conclusion, this exercise serves as a valuable learning experience, reinforcing the importance of stoichiometry, gravimetric analysis, and the application of molar mass in determining the atomic weights of elements in chemical compounds. It showcases the power of quantitative analysis in chemistry and the elegance of solving chemical problems through a systematic and logical approach.
