Atomic counting in a sample of ammonia (NH3):\nGiven 8.5 g of NH3, determine the total number of atoms present, expressed as X × 10^23. Choose the closest value for X.

Introduction / Context:
Counting atoms in a given mass of a compound is a staple skill in chemical stoichiometry. This problem asks for the total number of atoms in a known mass of ammonia (NH3), reported in the convenient scientific form X × 10^23. Solving it reinforces conversions among grams, moles of molecules, and finally moles of atoms using Avogadro’s constant.

Given Data / Assumptions:

• Molar mass of NH3 = 17 g/mol (14 for N and 1 each for 3 H).
• Sample mass m = 8.5 g NH3.
• Each NH3 molecule contains 4 atoms (1 N + 3 H).
• Avogadro’s constant NA ≈ 6.022 × 10^23 entities per mole.

Concept / Approach:
The workflow is: grams → moles of molecules → moles of atoms → number of atoms. First compute moles of NH3 from its molar mass. Multiply by the number of atoms per molecule to get moles of atoms. Finally, convert moles of atoms to number of atoms using NA. Report the result as X × 10^23 and pick X appropriately.

Step-by-Step Solution:

Verification / Alternative check:
If the mass were 17 g (one mole) of NH3, there would be 4 × 6.022 × 10^23 = 24.088 × 10^23 atoms. Halving the mass to 8.5 g halves the count to 12.044 × 10^23, matching the calculation.

Why Other Options Are Wrong:

• 9.03, 6.02, 3.01, 1.204: These arise from common slip-ups such as forgetting the 4 atoms per molecule, using molecules instead of atoms, or mis-scaling by factors of 10.

Common Pitfalls:
Not multiplying by the number of atoms per molecule, rounding too early, or misreading the requested format (X × 10^23 rather than × 10^24). Work symbolically, then round only at the end.

Final Answer:
12.04 × 10^23 atoms