Question

Suppose that \(\Lambda\) and \(\mathrm{B}\) are two clements which form compounds \(\mathrm{B}_{2} \Lambda_{3}\) and \(\mathrm{B}_{2} \Lambda\), respectively. If \(0.05 \mathrm{~mole}\) of \(\mathrm{B}_{2} \Lambda_{3}\) weighs \(9.0 \mathrm{~g}\) and \(0.10 \mathrm{~mole}\) of \(\mathrm{B}_{2} \Lambda\) weighs \(10 \mathrm{~g}\), then the atomic weight of \(\Lambda\) and \(\mathrm{B}\), respectively is (1) 30 and 40 (2) 40 and 30 (3) 20 and 5 (4) 15 and 20

Short answer

The atomic weights are \(M_\Lambda = 40\) and \(M_B = 30\).

Step-by-step solution

- Write down given information

For compound \( \text{B}_2 \Lambda_3 \), \(0.05\) moles weigh \(9.0 \text{g}\). For compound \( \text{B}_2 \Lambda \), \(0.10\) moles weigh \(10 \text{g}\).

- Calculate the molar mass of compounds

The molar mass of \( \text{B}_2 \Lambda_3 \) is calculated as follows: \( \text{molar mass} = \frac{9.0 \text{g}}{0.05 \text{moles}} = 180 \text{g/mol} \). For \( \text{B}_2 \Lambda \), the molar mass is \( \text{molar mass} = \frac{10 \text{g}}{0.10 \text{moles}} = 100 \text{g/mol} \).

- Set up the system of equations

For \( \text{B}_2 \Lambda_3 \), \( 2M_B + 3M_\Lambda = 180 \). For \( \text{B}_2 \Lambda \), \( 2M_B + M_\Lambda = 100 \).

- Solve the system of equations

Subtract the second equation from the first equation: \(2M_B + 3M_\Lambda - (2M_B + M_\Lambda) = 180 - 100 \). Simplifying, \(2M_\Lambda = 80 \), thus \(M_\Lambda = 40\). Substitute \(M_\Lambda \) back into the second equation: \(2M_B + 40 = 100 \). Solving for \(M_B \), \(2M_B = 60\), thus \(M_B = 30\).

Key concepts

Molecular Mass Calculation

Molecular mass is the sum of the atomic masses of all atoms in a molecule. It's also called molecular weight. To find it, you need the atomic masses of individual elements.
Here's the process:

• Identify the number of each type of atom in the molecule.
• Multiply the atomic mass of each element by the number of atoms of that element in the molecule.
• Add the results together to get the total molecular mass.

In our problem, we calculated the molecular masses of two compounds:
1. For compound \( \text{B}_2 \Lambda_3 \), we used the data \( 9.0 \text{g} \) for \( 0.05 \text{moles} \), which resulted in a molecular mass of \( 180 \text{g/mol} \).

2. For \( \text{B}_2 \Lambda \), we used \( 10 \text{g} \) for \( 0.10 \text{moles} \), giving a molecular mass of \( 100 \text{g/mol} \). Getting these figures helps us in further determining the atomic weights of the elements involved.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It relies on the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. Here's how it works:

• Use balanced chemical equations to find the ratios of the reactants and products.
• Use these ratios to convert between moles of different substances in the reaction.

In our exercise, we used stoichiometry when we set up the system of equations:

For \( \text{B}_2 \Lambda_3 \), the equation is \( 2M_B + 3M_\Lambda = 180 \).
For \( \text{B}_2 \Lambda \), it's \( 2M_B + M_\Lambda = 100 \).

We then solved these equations step by step to determine the atomic weights of \( \Lambda \) and \( B \). By subtracting the second equation from the first one, we simplified to find the weight of the elements.

Chemical Equations

A chemical equation represents a chemical reaction. It shows the reactants transforming into products using their chemical formulas. Balancing chemical equations ensures the conservation of mass, as the number of atoms for each element remains the same before and after the reaction.
Here are the steps to balance a chemical equation:

• Write down the unbalanced equation.
• Count the number of atoms of each element in both reactants and products.
• Add coefficients to balance the number of atoms of each element on both sides of the equation.

In the context of our exercise, we used chemical equations to represent the compounds. By knowing the molecular formulas \( \text{B}_2 \Lambda_3 \) and \( \text{B}_2 \Lambda \), and calculating their molecular masses, we set up balanced equations to eventually determine the atomic weights of \( \Lambda \) and \( B \).