Chemistry Notes: Empirical Formula, Molarity & Stoichiometry

Empirical Formula from Percent Composition

• Assume a 100 g sample for percent composition: sample mass $m_{sample}=100\text{ g}$
• Given: $m<em>C=49.29\text{ g},\ m</em>H=9.653\text{ g},\ m<em>N=19.16\text{ g}$; compute $m</em>O = 100 - (m<em>C + m</em>H + m_N) = 21.897\text{ g}$
• Convert to moles (using molar masses: C=12, H=1, N=14, O=16):
  • $n_C = \frac{49.29}{12} = 4.104\,\text{mol}$
  • $n_H = \frac{9.653}{1} = 9.653\,\text{mol}$
  • $n_N = \frac{19.16}{14} = 1.368\,\text{mol}$
  • $n_O = \frac{21.897}{16} = 1.368\,\text{mol}$
• Smallest mole value: $n_{min}=1.368\,\text{mol}$
• Compute mole ratios by dividing each by the smallest:
  • $\frac{n<em>C}{n</em>{min}} = \frac{4.104}{1.368} \approx 3.0$
  • $\frac{n<em>H}{n</em>{min}} = \frac{9.653}{1.368} \approx 7.06 \approx 7$
  • $\frac{n<em>N}{n</em>{min}} = 1$
  • $\frac{n<em>O}{n</em>{min}} = 1$
• Empirical formula: $\mathrm{C<em>3H</em>7NO}$
• Notes:
  • Empirical formula is the simplest whole-number ratio; molecular formula is a multiple of the empirical formula.
  • To get the molecular formula, divide the molar mass of the compound by the molar mass of the empirical formula.

Molarity, Dilution, and Stoichiometry

• Molarity basics: $M = \frac{n}{V}$
• Dilution (volume-change) relation: $M<em>1 V</em>1 = M<em>2 V</em>2$
• Stoichiometry with non-1:1 ratios: for a reaction $a\,A + b\,B \rightarrow \text{products}$, the relationship is $a\,M<em>A V</em>A = b\,M<em>B V</em>B$
  • If solving for an unknown molarity, use: $M<em>B = \frac{a\,M</em>A\,V<em>A}{b\,V</em>B}$
  • If the reaction is 1:1, this reduces to $M<em>A V</em>A = M<em>B V</em>B$
• Example (acid–base neutralization): $\mathrm{Ca(OH)<em>2 + 2\ HCl \rightarrow CaCl</em>2 + 2\ H_2O}$
  • Two moles of HCl react with 1 mole of Ca(OH)₂ (stoichiometric coefficients 2:1)
  • When solving for the molarity of an unknown solution, incorporate the coefficients: e.g., if solving for $M<em>{HCl}$ given volumes, use $M</em>{HCl} = \frac{2\,M<em>{Ca(OH)</em>2}\,V<em>{Ca(OH)</em>2}}{2\,V<em>{HCl}} = \frac{M</em>{Ca(OH)<em>2}\,V</em>{Ca(OH)<em>2}}{V</em>{HCl}}$
• Practical tip: read the question carefully to determine whether to include the stoichiometric coefficients; ICE tables are often not needed for these grams-to-moles problems

Quick Reference: Key Formulas

• Percent yield (conceptual): $\%<br /> Yield = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$
• Empirical vs molecular formulas: empirical is the simplest ratio; molecular formula = (empirical formula) × n where n is a whole number; determine n from molar masses

Preview: Lewis Structures (brief)

• Lewis structure shows valence electrons and how atoms bond
• Bond-line drawings are a simplified representation used in organic chemistry
• This topic is introduced gradually and will be reinforced in future sessions