Notes on Converting Formula Units, Moles, and Mass (NaNO3)

Key concepts

Avogadro's number sets the scale between moles and individual entities: $N_A = 6.022 \times 10^{23} \ \frac{\text{entities}}{\text{mol}}.$
Formula unit vs molecule:
- For ionic compounds (like NaNO₃), the basic counting unit is a formula unit, not a discrete molecule.
- Analogy: counting entities in a generic sense (e.g., a dozen donuts vs a dozen cars) shows that the identity of the unit does not affect the mole concept.
Molar mass (molar mass, M): mass per mole of a substance (units: g/mol). For NaNO₃, the molar mass is approximately $M{NaNO3} \approx 85 \ \frac{g}{\text{mol}}.$
Dimensional analysis / unit cancellation:
- When converting between quantities, place units so that they cancel between numerator and denominator, yielding the desired unit (moles or grams).
Direct conversion caveat:
- You cannot directly convert from a number of entities to mass; you must first go through moles (entities → moles via Avogadro's number) and then to mass via molar mass.
Sig figs (significant figures):
- Use the number of sig figs from the given data to determine the precision of the final answer.
- Example: 4.63 × 10³⁴ has 3 sig figs; results should reflect the appropriate number of sig figs (e.g., 3 sig figs for the first example).
Practical takeaway:
- If the question asks for moles, you do not need molar mass; if it asks for mass, you need molar mass.
Summary formulas:
- Entities → moles: $n = \frac{N}{N_A}.$
- Moles → mass: $m = n \; M.$
- Molar mass: $M = \sum (\text{atomic mass} \times \text{counts})$ (for NaNO₃, roughly 85 g/mol).

Example 1: Formulas units to moles (NaNO₃)

Given: 4.63 × 10³⁰ formula units of NaNO₃.
- Note: NaNO₃ is ionic; counting units are formula units, not molecules.
Goal: find the number of moles.
Conversion factor (dimensional analysis):
- $\frac{1 \ \text{mol}}{N_A \ \text{units}} = \frac{1 \text{ mol}}{6.022 \times 10^{23} \text{ units}}.$
Calculation (units cancel to leave moles):
- $n = \frac{4.63 \times 10^{24} \ \text{units}}{1} \times \frac{1 \ \text{mol}}{6.022 \times 10^{23} \ \text{units}} \approx 7.69 \ \text{mol}.$
Sig figs:
- The given data 4.63 has 3 sig figs; the result is reported with 3 sig figs: $n \approx 7.69 \ \text{mol}.$
Key point: Identity of the formula unit is irrelevant for the mole count; 1 mole always contains $N_A = 6.022 \times 10^{23}$ formula units.
Note on molar mass: Not needed here since the question asks for moles, not mass.
Quick recap formula used:
- $n = \frac{N}{N_A}$ where N is the number of formula units.

Example 2: Formulas units to mass via moles (NaNO₃)

Given: 3.7 × 10¹²¹ formula units of NaNO₃.
Step 1: Convert entities to moles:
- $n = \frac{3.7 \times 10^{21} \ \text{units}}{6.022 \times 10^{23} \ \text{units/mol}} \approx 6.14 \times 10^{-3} \ \text{mol}.$
Step 2: Convert moles to mass using molar mass $M{NaNO3} \approx 85 \ \frac{\text{g}}{\text{mol}}.$
- The conversion factor for mass: $1 \ \text{mol} \to 85 \ \text{g}$ , i.e., $\frac{85 \ \text{g}}{1 \text{mol}}.$ (Place 85 g on top and 1 mol on bottom in the dimensional analysis.)
Calculation for mass:
- $m = n \times M = \left(\frac{3.7 \times 10^{21} \ \text{units}}{6.022 \times 10^{23} \ \text{units/mol}}\right) \times (85 \ \frac{\text{g}}{\text{mol}}) \approx 0.522 \ \text{g}.$
Sig figs:
- The initial value 3.7 × 10²¹ has 2 sig figs; final mass should reflect 2 sig figs → $m \approx 0.52 \ \text{g}.$ (0.522 rounds to 0.52 with two sig figs.)
Rationale for steps:
- Start with entities, convert to moles via Avogadro's number, then convert to mass via molar mass.
- Align units so that grams are the desired unit at the end; cancel moles along the way.
Practice takeaway:
- If you only need mass, include molar mass in your calculation; if you only need moles, molar mass is not required.

Worked recap and general approach

Core workflow for problems like these:
- Step 1: Identify what is given (entities/formula units). Determine desired quantity (moles or mass).
- Step 2: Convert entities to moles using $N_A = 6.022 \times 10^{23} \ \frac{\text{entities}}{\text{mol}}$ :
- $n = \frac{N}{N_A}.$
- Step 3: If mass is requested, convert from moles to mass using $M$ (the molar mass with units g/mol):
- $m = n \times M.$
- Step 4: Apply appropriate significant figures based on the given data.
- Step 5: Take note that for ionic compounds, the counting unit is a formula unit; the method is the same as for molecules, but terminology differs.
Quick reference values used:
- Avogadro's number: $N_A = 6.022 \times 10^{23} \frac{\text{entities}}{\text{mol}}.$
- Molar mass of NaNO₃: $M{NaNO3} \approx 85 \frac{\text{g}}{\text{mol}}.$
Real-world relevance:
- These conversions are foundational in analytical chemistry and material science whenever you need to relate microscopic counts (ions, molecules) to macroscopic mass or amount of substance.