Stoichiometry & Chemical Formulas

Stoichiometry Review

• Grams to Moles Conversion (O$_2$ molecule)

  • Problem: Convert $1.35$ grams of oxygen gas ($O_2$) to moles of oxygen gas.

  • Key Concept: Moles are the "currency of chemistry," requiring knowledge of molar mass.

  • Molar Mass of O$<em>2$: Each oxygen atom (O) weighs approximately $16 ext{ g/mol}$. Since $O</em>2$ is diatomic, one mole of $O_2$ weighs $2 imes 16 ext{ g/mol} = 32 ext{ g/mol}$.

  • Calculation: Divide the given mass by the molar mass.
    $1.35 ext{ g O}<em>2 imes \frac{1 ext{ mol O}</em>2}{32 ext{ g O}_2}$

  • Self-Correction: If units do not cancel correctly in the dimensional analysis, flip the conversion factor.

• Grams to Moles of Atoms Conversion (O atoms from O$_2$ molecule)

  • Problem: Convert $1.25$ grams of oxygen gas ($O_2$) to moles of oxygen atoms (O).

  • Step 1: Convert grams of O$<em>2$ to moles of O$</em>2$ (as above).
    $1.25 ext{ g O}<em>2 imes \frac{1 ext{ mol O}</em>2}{32 ext{ g O}_2}$

  • Step 2: Convert moles of O$_2$ to moles of O atoms.

    • The chemical formula $O<em>2$ indicates that there are $2$ oxygen atoms for every $1$ molecule of $O</em>2$. Therefore, there are $2$ moles of oxygen atoms for every $1$ mole of $O_2$ molecules.

  • Full Calculation:
    $1.25 ext{ g O}<em>2 imes \frac{1 ext{ mol O}</em>2}{32 ext{ g O}<em>2} imes \frac{2 ext{ mol O}}{1 ext{ mol O}</em>2}$

Molecular vs. Empirical Formulas

• Introduction to Formulas

  • A chemical formula (e.g., for ionic compounds) tells us the ratio of elements present. For molecular compounds, we differentiate between molecular and empirical formulas.

• Molecular Formula

  • Definition: Represents the exact number of each type of atom in a single molecule of a compound.

  • Example: Hydrogen peroxide has a molecular formula of $H<em>2O</em>2$, indicating two hydrogen atoms and two oxygen atoms per molecule.

  • Significance: Different molecular formulas (e.g., $H<em>2O$ for water vs. $H</em>2O_2$ for hydrogen peroxide) represent entirely different substances with distinct properties.

• Empirical Formula

  • Definition: Represents the lowest whole number ratio of elements in a compound.

  • For Ionic Compounds: The formula unit is always the empirical formula (e.g., NaCl).

  • For Molecular Compounds: It may or may not be the same as the molecular formula.

  • Example: For hydrogen peroxide ($H<em>2O</em>2$), the lowest whole number ratio of H to O is $1:1$, so its empirical formula is HO.

Determining Empirical and Molecular Formulas from Mass Percent Composition

• Practical Application

  • In research labs (e.g., developing new drugs) or forensic science (CSI), instruments are used to determine the composition of unknown compounds.

  • Instruments:

    1. Elemental Analyzer: Determines the mass percentage of various elements (e.g., C, H, O, N) in a compound.

    2. Mass Spectrometer ("Mass Spec"): Determines the molar mass of the compound.

  • Goal: Convert mass percent data into an empirical formula, and then use the molar mass to find the molecular formula.

• Example Calculation: Find the empirical and molecular formula for a compound with:

  • $54.55\% ext{ Carbon (C)}$

  • $9.09\% ext{ Hydrogen (H)}$

  • $36.36\% ext{ Oxygen (O)}$

  • Known Molar Mass: $88 ext{ g/mol}$

  • (Note: Sometimes percentage for one element might be omitted, requiring you to calculate it by subtracting other percentages from $100\%$.)

• Steps to Determine the Empirical Formula:

  1. Assume a $100 ext{ g}$ Sample: This simplifies conversion from percent to grams.

     • $54.55 ext{ g C}$

     • $9.09 ext{ g H}$

     • $36.36 ext{ g O}$

  2. Convert Grams of Each Element to Moles (using molar masses from the periodic table):

     • Carbon: $54.55 ext{ g C} imes \frac{1 ext{ mol C}}{12.01 ext{ g C}} = 4.54 ext{ mol C}$

     • Hydrogen: $9.09 ext{ g H} imes \frac{1 ext{ mol H}}{1.008 ext{ g H}} = 9.02 ext{ mol H}$

     • Oxygen: $36.36 ext{ g O} imes \frac{1 ext{ mol O}}{16.00 ext{ g O}} = 2.27 ext{ mol O}$

  3. Divide All Mole Values by the Smallest Mole Value to find the simplest whole-number ratio:

     • The smallest mole value is $2.27 ext{ mol O}$.

     • Carbon: $\frac{4.54 ext{ mol C}}{2.27 ext{ mol O}} \approx 2 ext{ mol C} ) (ratio of Carbon to Oxygen)</p></li><li><p><strong>Hydrogen</strong>:$ \frac{9.02 ext{ mol H}}{2.27 ext{ mol O}} \approx 3.97 ext{ mol H} \approx 4 ext{ mol H} ) (ratio of Hydrogen to Oxygen)

     • Oxygen: $\frac{2.27 ext{ mol O}}{2.27 ext{ mol O}} = 1 ext{ mol O} ) (ratio of Oxygen to Oxygen)</p></li><li><p><strong>Resulting Empirical Formula</strong>:$C2H4O$</p></li></ul></li><li><p><strong>Address Non-Whole Number Ratios (if necessary)</strong>:</p><ul><li><p>If, after division, the ratios are not whole numbers but are close to simple fractions, multiply all ratios by an integer to convert them to whole numbers.</p></li><li><p><strong>Fractional Multipliers</strong>:</p><ul><li><p>If a ratio ends in$.5$(e.g.,$2.5$), multiply all ratios by$2$(e.g.,$2.5:1$becomes$5:2$).</p></li><li><p>If a ratio ends in$.333$or$.666$(e.g.,$1.333$), multiply all ratios by$3$(e.g.,$1.333:1$becomes$4:3$).</p></li><li><p>If a ratio ends in$.25$or$.75$(e.g.,$4.25$), multiply all ratios by$4$(e.g.,$4.25:1$becomes$17:4$).</p></li></ul></li><li><p><em>Instructor's Note</em>: Homework problems might present more complex fractions (e.g.,$.2$requiring a$5x$multiplier). For tests, simpler multipliers ($2, 3, 4$) are typical.</p></li></ul></li></ol></li><li><p><strong>Steps to Determine the Molecular Formula</strong>:</p><ol><li><p><strong>Calculate the Empirical Formula Mass (EFM)</strong>:</p><ul><li><p>For the empirical formula$C2H4O$:</p><ul><li><p>EFM =$(2 imes 12.01 ext{ g/mol C}) + (4 imes 1.008 ext{ g/mol H}) + (1 imes 16.00 ext{ g/mol O})$</p></li><li><p>EFM =$24.02 + 4.032 + 16.00 = 44.052 ext{ g/mol}$(approximately$44 ext{ g/mol}$)</p></li></ul></li></ul></li><li><p><strong>Determine the Multiplier (n)</strong>:</p><ul><li><p>Divide the given Molar Mass (MM) by the Empirical Formula Mass (EFM).</p></li><li><p>$n = \frac{ ext{Molar Mass}}{ ext{Empirical Formula Mass}} = \frac{88 ext{ g/mol}}{44 ext{ g/mol}} = 2$</p></li></ul></li><li><p><strong>Multiply the Subscripts in the Empirical Formula by$n$</strong>:</p><ul><li><p>Molecular Formula =$C{(2 imes 2)}H{(4 imes 2)}O{(1 imes 2)} = C4H8O2$</p></li></ul></li></ol></li><li><p><strong>Test-Taking Strategy (Multiple Choice)</strong>:</p><ul><li><p>Once you have the empirical formula, you know the molecular formula must be a whole-number multiple of it (e.g.,$C2H4O$,$C4H8O2$,$C6H{12}O3$$, etc.).

     • On a multiple-choice exam, calculate the molar mass for the options that are integer multiples of the empirical formula. The option whose molar mass matches the given molar mass is the correct answer. Distractors will include formulas that are not multiples of the empirical formula.

     • The core concept is that the molecular formula reflects a consistent multiple of the empirical formula's atomic ratio.