Empirical (simplest) formula problem in stoichiometry:\nA compound contains 50% by mass of element A (atomic weight = 10) and 50% by mass of element B (atomic weight = 20). Determine the simplest whole-number formula of the compound.

Introduction / Context:
This question tests the foundational stoichiometry skill of converting mass percentages and atomic weights into an empirical (simplest) formula. Empirical formulas are widely used in analytical chemistry, materials science, and biochemical analyses to represent the lowest whole-number ratio of atoms present in a compound.

Given Data / Assumptions:

• Mass composition: A = 50% by mass; B = 50% by mass.
• Atomic weight: A = 10; B = 20.
• Empirical formula requires the simplest whole-number mole ratio.
• Assume a 100 g sample for convenience (a standard approach).

Concept / Approach:
Convert each element’s mass to moles, then reduce to the smallest whole-number ratio. The empirical formula expresses this ratio directly. If necessary, multiply by small integers (2, 3, etc.) to remove fractional subscripts while maintaining the same atomic ratio.

Step-by-Step Solution:
Assume 100 g total: mass of A = 50 g; mass of B = 50 g.Moles of A = 50/10 = 5.00 mol.Moles of B = 50/20 = 2.50 mol.Mole ratio A:B = 5.00 : 2.50 = 2 : 1.Write simplest formula from the ratio: A2B.

Verification / Alternative check:
Check proportional masses implied by A2B: Using AWs, mass fraction A = (210) / (210 + 1*20) = 20/40 = 0.5 = 50%. Mass fraction B = 20/40 = 50%. The composition matches, verifying the result.

Why Other Options Are Wrong:
AB3 implies A:B = 1:3, inconsistent with the 2:1 mole ratio.A2B3 implies 2:3, also not 2:1.AB2 implies 1:2, the inverse of the correct ratio.

Common Pitfalls:
Forgetting to convert mass to moles; reducing ratios incorrectly; leaving fractional subscripts without clearing them by multiplying through; assuming molecular formula instead of empirical when only percent composition is provided.

Final Answer:
A2B