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 (NA): N = n NA.
Mass to Moles to Atoms Workflow
Process: Start with mass (g) \rightarrow convert to moles (n) \rightarrow convert to molecules (N) via NA \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):
Convert percent (g) of each element to moles.
Divide all mole values by the smallest mole value to get ratios.
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{\text{emp}}} (where M{\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$_2$.