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 masses of \(\mathrm{A}\) and \(\mathrm{Z} ?\)

Short answer

The atomic masses are \( m_A = 26 \) and \( m_Z = 18 \).

Step-by-step solution

Calculate Molar Mass of Compounds

Start by calculating the molar mass of each compound from the given molar amounts and masses. For \( \mathrm{A}_2 \mathrm{Z}_3 \), the molar mass (\( M_{\mathrm{A}_2 \mathrm{Z}_3} \)) is calculated as follows: \[ M = \frac{15.9 \text{ g}}{0.15 \text{ mol}} = 106 \text{ g/mol} \] Similarly, for \( \mathrm{AZ}_2 \), the molar mass (\( M_{\mathrm{AZ}_2} \)) is: \[ M = \frac{9.3 \text{ g}}{0.15 \text{ mol}} = 62 \text{ g/mol} \]

Establish Equations for Molar Masses

Using the molar masses, establish equations. For \( \mathrm{A}_2 \mathrm{Z}_3 \), \( 2m_A + 3m_Z = 106 \). For \( \mathrm{AZ}_2 \), \( m_A + 2m_Z = 62 \). Here, \( m_A \) is the atomic mass of \( \mathrm{A} \) and \( m_Z \) is the atomic mass of \( \mathrm{Z} \).

Solve the Equation System

Solve the system of equations. Let's multiply the second equation by 2: \( 2m_A + 4m_Z = 124 \). Now subtract the first equation from this new equation to eliminate \( 2m_A \):\[ (2m_A + 4m_Z) - (2m_A + 3m_Z) = 124 - 106 \] Simplifying gives: \( m_Z = 18 \). Substitute \( m_Z = 18 \) back into the second equation: \( m_A + 2(18) = 62 \), yielding \( m_A = 26 \).

Verify the Solution

Verify that both equations hold true with \( m_A = 26 \) and \( m_Z = 18 \). Substitute into the first equation: \( 2(26) + 3(18) = 106 \). Substitute into the second equation: \( 26 + 2(18) = 62 \). Both equations are satisfied, confirming the solution.

Key concepts

Molar Mass Calculation

Understanding molar mass calculation is crucial to solving problems related to chemistry. The molar mass of a compound is the mass of one mole of its molecules. It is measured in grams per mole (g/mol).

To find the molar mass of a compound, we use the formula:
\[M = \frac{m}{n}\]where:

• \(M\) is the molar mass,
• \(m\) is the mass of the sample in grams,
• \(n\) is the amount of substance in moles.

Let's apply this to the chemical compounds given in the problem. For the compound \(\mathrm{A}_2 \mathrm{Z}_3\), with a mass of 15.9 g for 0.15 mol, the molar mass is:
\[106\ g/mol = \frac{15.9\ g}{0.15\ mol}\]Similarly, for \(\mathrm{AZ}_2\), the mass of 9.3 g for 0.15 mol will give a molar mass:
\[62\ g/mol = \frac{9.3\ g}{0.15\ mol}\]The molar mass helps us understand the proportions and composition of elements within a compound. Calculating the molar mass helps identify the individual atomic masses within the compounds involved.

Chemical Compounds

A chemical compound is a substance composed of two or more different elements chemically bonded together. In our problem, the compounds \(\mathrm{A}_2 \mathrm{Z}_3\) and \(\mathrm{AZ}_2\) illustrate how elements A and Z combine in different ratios.

The subscripts in a chemical formula represent the number of atoms of each element in a single molecule of the compound:

• In \(\mathrm{A}_2 \mathrm{Z}_3\), there are 2 atoms of A for every 3 atoms of Z.
• In \(\mathrm{AZ}_2\), there is 1 atom of A for every 2 atoms of Z.

The different formulas show the versatility of how these elements can bond, forming different compounds with distinct properties. Understanding such formulas is a fundamental skill in chemistry that helps relate the structure and properties of compounds.

System of Equations

Solving a system of equations is essential to finding solutions where multiple factors or conditions must be satisfied simultaneously. In our exercise, the molar masses provide a system of equations derived from the compound formulas.

The equations given by the molar masses are:

• For \(\mathrm{A}_2 \mathrm{Z}_3\): \(2m_A + 3m_Z = 106\)
• For \(\mathrm{AZ}_2\): \(m_A + 2m_Z = 62\)

Here, \(m_A\) is the atomic mass of element A and \(m_Z\) is the atomic mass of element Z. Solving this system requires algebraic manipulation to isolate one variable.

To do this, multiply the second equation by 2 to align coefficients of \(m_A\):
\[2m_A + 4m_Z = 124\]Subtract the first equation from this new equation to eliminate \(2m_A\), and find:
\[m_Z = 18\]Substituting \(m_Z = 18\) back into the simpler, second equation resolves the atomic mass for A:
\[m_A = 26\]These calculated values need to satisfy both original equations to ensure correctness, thus confirming the solution.