Problem 150 - Chemistry and Chemical Reactivity

Question

The elements A and Z combine to produce two different compounds: \(\mathrm{A}_{2} \mathrm{Z}_{3}\) and \(\mathrm{AZ}_{2}\). If 0.15 mol of \(\mathrm{A}_{2} \mathrm{Z}_{3}\) has a mass of \(15.9 \mathrm{g}\) and \(0.15 \mathrm{mol}\) of \(\mathrm{AZ}_{2}\) has a mass of \(9.3 \mathrm{g}\), what are the atomic weights of A and Z?

Step-by-Step Solution

Answer

The atomic weights are: A = 26 g/mol, Z = 18 g/mol.

Step 1: Determine the Molar Mass of Compounds

We know 0.15 mol of \(A_2Z_3\) weighs 15.9 g, and 0.15 mol of \(AZ_2\) weighs 9.3 g. Calculate the molar mass for each compound by dividing the total mass by the number of moles. \(\text{Molar mass of } A_2Z_3 = \frac{15.9 \text{ g}}{0.15 \text{ mol}} = 106 \text{ g/mol}\) \(\text{Molar mass of } AZ_2 = \frac{9.3 \text{ g}}{0.15 \text{ mol}} = 62 \text{ g/mol}\)

Step 2: Write Equations for Molar Masses

Use the formulas \( A_2Z_3 \) and \( AZ_2 \) to express the molar masses in terms of the atomic weights of A and Z.For \(A_2Z_3\):\[ 2M_A + 3M_Z = 106 \]For \(AZ_2\):\[ M_A + 2M_Z = 62 \]

Step 3: Solve the System of Equations

We must solve the system of equations to find \( M_A \) and \( M_Z \).Equation 1: \[ 2M_A + 3M_Z = 106 \]Equation 2: \[ M_A + 2M_Z = 62 \]Multiply Equation 2 by 2:\[ 2M_A + 4M_Z = 124 \]Subtract Equation 1 from this result to eliminate \( M_A \):\[ (2M_A + 4M_Z) - (2M_A + 3M_Z) = 124 - 106 \]\[ M_Z = 18 \]

Step 4: Find the Atomic Weight of A

Use the value of \( M_Z \) found in Step 3 and replace it back into Equation 2:\[ M_A + 2(18) = 62 \]\[ M_A + 36 = 62 \]\[ M_A = 62 - 36 \]\[ M_A = 26 \]

Step 5: Verify the Solution

Substitute \( M_A = 26 \) and \( M_Z = 18 \) back into Equation 1 to verify the correctness of the solution:\[ 2(26) + 3(18) = 52 + 54 = 106 \]Since both sides equal 106, our values for \( M_A \) and \( M_Z \) are correct.

Key Concepts

Molar Mass Calculations

Molar mass is a fundamental concept in chemistry which allows us to convert between the mass of a compound and the amount of substance (in moles). To calculate the molar mass of a compound, we divide the given mass of a substance by the number of moles present. In the given exercise, two compounds, \(A_2Z_3\) and \(AZ_2\), have their molar masses calculated.
The formula used is:
\[ \text{Molar Mass} = \frac{\text{Mass of Substance (g)}}{\text{Moles of Substance}} \]
For \(A_2Z_3\): 0.15 mol has a mass of 15.9 g. Thus, its molar mass is:
\( \frac{15.9 \text{ g}}{0.15 \text{ mol}} = 106 \text{ g/mol} \).
Similarly, for \(AZ_2\): 0.15 mol has a mass of 9.3 g, leading to:
\( \frac{9.3 \text{ g}}{0.15 \text{ mol}} = 62 \text{ g/mol} \).
Mastering molar mass calculations is essential for many stoichiometry problems as it serves as the bridge between mass given and the amount in moles, facilitating further analysis.

System of Equations

Often in chemistry, we encounter problems where we need to solve for unknowns using equations derived from chemical formulas. A system of equations emerges naturally in such contexts.
In this exercise, the molar mass equations for the compounds \(A_2Z_3\) and \(AZ_2\) are set up using the atomic weights.
The equations are:
For \(A_2Z_3\): \[ 2M_A + 3M_Z = 106 \] For \(AZ_2\): \[ M_A + 2M_Z = 62 \] By solving these simultaneously, we find the values for \(M_A\) and \(M_Z\).
This process involves standard algebraic manipulation: first aligning the equations, then using methods like substitution or elimination (here elimination is used), to find the solution.

• Multiply Equation 2 by 2.
• Subtract the result from Equation 1 to eliminate \(M_A\).

Systems of equations are a powerful tool in stoichiometry for deriving unknown values in terms of known weights or masses.

Atomic Weight Determination

Understanding atomic weight is crucial for identifying elements in a compound. Given the molar mass equations calculated, we aim to determine the atomic weights, \(M_A\) and \(M_Z\).
Using the system of equations set up from the molar mass calculations, solving for \(M_Z\) first gives an atomic weight of 18.
Substituting this into one of the original equations yields \(M_A = 26\). Finally, the solution is verified by checking if these values satisfy the other equation: \[ 2(26) + 3(18) = 106 \] Both sides equal 106, demonstrating that the atomic weights are correct.
Determining the correct atomic weights allows chemists to analyze elements quantitatively, contributing to precise chemical calculations and understanding of compound composition.

• First find \(M_Z\).
• Substitute to find \(M_A\).
• Verify the solution.

Understanding these calculations is key to mastering more complex stoichiometry problems.