Elements and combine to form two distinct compounds, and , illustrating the Law of Multiple Proporti — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
Elements $A$ and $B$ combine to form two distinct compounds, $B_2A_3$ and $B_2A$, illustrating the Law of Multiple Proportions. Analytical data reveals that $0.05\text{ mole}$ of $B_2A_3$ weighs exactly $9.0\text{ g}$ and $0.10\text{ mole}$ of $B_2A$ weighs exactly $10.0\text{ g}$. Calculate the exact atomic weight of element $A$.
💡 Solution & Explanation
Let the atomic weights of $A$ and $B$ be $a$ and $b$ respectively. Molar mass of $B_2A_3 = \frac{9.0\text{ g}}{0.05\text{ mol}} = 180\text{ g/mol}$. Thus, $2b + 3a = 180$ (Equation 1). Molar mass of $B_2A = \frac{10.0\text{ g}}{0.10\text{ mol}} = 100\text{ g/mol}$. Thus, $2b + a = 100$ (Equation 2). Subtracting Equation 2 from Equation 1: $(2b + 3a) - (2b + a) = 180 - 100 \Rightarrow 2a = 80 \Rightarrow a = 40$. The atomic weight of element A is $40$.