Avogadro's number, empirical formula, molecular formula, and percent composition

Avogadro's Number and Mass–Mole Relationship

• Micro vs Macro: Atoms/molecules are microscopic; moles/grams are macroscopic.

• Atomic Mass (A): Average mass of an element (in amu). Molar mass (in g/mol) numerically equals atomic mass (in amu).

  • Example: Hydrogen: A $\approx 1.008\ \text{amu}$ per atom; $M_{\text{H}} = 1.008\ \text{g/mol}$ per mole.

• The periodic table provides atomic masses for calculations.

• One mole of any substance contains Avogadro's number of particles ($N_A = 6.022 \times 10^{23}$).

Molecular & Molar Mass

• Molecular mass: Sum of atomic masses in amu for a molecule.

• Molar mass: Same numerical value as molecular mass, but in g/mol.

  • Example (SO$_2$): Molecular mass = $64.072\ \text{amu}$; Molar mass = $64.07\ \text{g/mol}$.

• Key Formulas:

  • Moles ($n$) from mass ($m$) and molar mass ($M$): $n = \frac{m}{M}$.

  • Number of particles ($N$) from moles ($n$) and Avogadro's number ($N<em>A$): $N = n N</em>A$.

Mass to Moles to Atoms Workflow

• Process: Start with mass (g) $\rightarrow$ convert to moles (n) $\rightarrow$ convert to molecules (N) via $N<em>A$ $\rightarrow$ determine total atoms via atomicity (e.g., $4\ \text{H}$ atoms in CH$4$).

  • Example (25.6g Urea CH$4$N$2$O):

  • $M_{\text{urea}} \approx 60.06\ \text{g/mol}$

  • $n \approx 0.4267\ \text{mol}$

  • Molecules $\approx 2.57 \times 10^{23}$

  • Hydrogen atoms $\approx 1.03 \times 10^{24}$.

Percent Composition by Mass

• Determines the mass fraction of each element in a compound.

• Formula: \text{%X} = \frac{\text{moles of X} \times \text{Molar Mass of X}}{\text{Molar Mass of Compound}} \times 100\%

  • Example (Ethanol, C$2$H$6$O):

  • $M \approx 46.068\ \text{g/mol}$

  • \text{%C} \approx 52.14\%, \text{%H} \approx 13.30\%, \text{%O} \approx 34.56\%.

• Check: Percentages should sum to $\approx 100\%$.

Empirical & Molecular Formulas

• Empirical Formula: Simplest whole-number ratio of atoms in a compound.

  • Procedure (from % Composition, assuming 100g):

  1. Convert percent (g) of each element to moles.

  2. Divide all mole values by the smallest mole value to get ratios.

  3. Multiply ratios by a small integer if needed to obtain whole numbers.

  • Example (Ascorbic acid: C 40.92%, H 4.58%, O 54.5%): Leads to C$3$H$4$O$_3$.

• Molecular Formula: Actual number of atoms; a multiple of the empirical formula.

  • If molar mass (M) is known: $n = \frac{M}{M<em>{\text{emp}}}$ (where $M</em>{\text{emp}}$ is empirical formula mass).

  • Molecular Formula = Empirical Formula $\times$ $n$.

  • Example: Empirical C$2$H$4$Br (emp. mass $\approx 107.95\ \text{g/mol}$), actual M $\approx 215.9\ \text{g/mol}$. Here, $n \approx 2$, so Molecular Formula = C$4$H$8$Br$_2$.