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To solve this problem, let's break it down step-by-step.

1. **Determine the number of moles of \( \text{AlF}_3 \):**
   - Given:
     - Mass of \( \text{AlF}_3 \) sample = 65.3 grams
     - Molar mass of \( \text{AlF}_3 \) = 84.0 g/mol
   - Formula:
     \[
     \text{moles of AlF}_3 = \frac{\text{mass}}{\text{molar mass}}
     \]
   - Calculation:
     \[
     \text{moles of AlF}_3 = \frac{65.3}{84.0} \approx 0.78 \text{ moles}
     \]

2. **Determine the number of formula units of \( \text{AlF}_3 \):**
   - Avogadro's number (the number of formula units per mole) is \( 6.022 \times 10^{23} \).
   - Formula:
     \[
     \text{formula units} = \text{moles} \times 6.022 \times 10^{23}
     \]
   - Calculation:
     \[
     \text{formula units} = 0.78 \times 6.022 \times 10^{23} \approx 4.70 \times 10^{23}
     \]

**Answers (rounded to the nearest hundredth):**
- Moles of \( \text{AlF}_3 \): **0.78**
- Formula units of \( \text{AlF}_3 \): **4.70E23**

Would you like further clarification on any part of this calculation?

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**Related Questions:**
1. How would the calculations change if the sample mass was doubled?
2. What would the number of moles be for a different molar mass?
3. How is Avogadro's number derived, and why is it used here?
4. How does rounding affect precision in chemical calculations?
5. What is the significance of significant figures in these calculations?

**Tip:** Always double-check units to ensure the correct application of formulas in chemistry problems.

In a 65.3 gram sample of aluminum fluoride, there are _____ moles of AlF₃ and _____ formula units. Aluminum fluoride has a molar mass of 84.0 g/mol. (use 'E' to represent 10^x). Round your answers to the nearest 100th.

Find the number of moles and formula units in a 65.3 g sample of aluminum fluoride (AlF₃) with a molar mass of 84.0 g/mol. This problem guides you through stoichiometric calculations using Avogadro's number and is suitable for high school chemistry students.

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