Empirical Formulas, Molecular Formulas, and Combustion Analysis

Foundations of Chemical Formulas and Compounds

• Chemical formulas represent the elemental composition and exact atomic ratios of substances:

  • Sucrose ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$): Molar mass of $342.297\,\text{g\,mol}^{-1}$.
  • Water ($\text{H}_2\text{O}$): Standard triatomic covalent molecule.
  • Aluminum ($\text{Al}$): Molar mass of $26.982\,\text{g\,mol}^{-1}$.
  • Sodium chloride ($\text{NaCl}$): Formula mass of $58.44\,\text{g\,mol}^{-1}$.

• Chemical formulas are subdivided into two main categories based on the information provided:

  • Empirical Formulas: Express the simplest, lowest whole-number ratio of atoms present in a compound.
  • Molecular Formulas: Express the exact number of atoms of each element present in a single molecule of a covalent compound.

Empirical Formulas

• An empirical formula contains the simplest whole-number ratio of atoms in a chemical compound.

• The actual chemical formula (molecular formula) is either identical to the empirical formula or a whole-number multiple of it.

• General Procedure for Determining Empirical Formulas from Mass Data:

  1. Convert percent composition data directly to mass data by assuming a $100.0\,\text{g}$ sample size.
  2. Convert the mass of each element into moles of each element using its respective molar mass.
  3. Divide the calculated mole quantities of all elements by the smallest calculated mole value to establish a provisional ratio.
  4. If the resulting numbers are not whole numbers, multiply all values by an appropriate integer factor to convert decimals to whole numbers.
  5. Write the resulting integer values as subscripts following each respective element symbol.

• Conversion of Decimal Subscripts to Whole Numbers:

  • When dividing by the smallest mole value yields decimal subscripts, multiply all subscripts by the specified factor based on the decimal ending:

*   Decimal ending $.20$: Fractional equivalent is $\frac{1}{5}$; multiply all subscripts by $5$.
*   Decimal ending $.25$: Fractional equivalent is $\frac{1}{4}$; multiply all subscripts by $4$.
*   Decimal ending $.33$: Fractional equivalent is $\frac{1}{3}$; multiply all subscripts by $3$.
*   Decimal ending $.40$: Fractional equivalent is $\frac{2}{5}$; multiply all subscripts by $5$.
*   Decimal ending $.50$: Fractional equivalent is $\frac{1}{2}$; multiply all subscripts by $2$.
*   Decimal ending $.67$: Fractional equivalent is $\frac{2}{3}$; multiply all subscripts by $3$.
*   Decimal ending $.75$: Fractional equivalent is $\frac{3}{4}$; multiply all subscripts by $4$.

• Worked Example 1: Determination from Mass Data
  • Problem: A $20.0\,\text{g}$ sample contains $8.00\,\text{g C}$, $1.34\,\text{g H}$, and $10.7\,\text{g O}$. Determine its empirical formula.
  • Step 1: Convert elemental masses to moles:

$8.00\,\text{g C} \times \frac{1\,\text{mol C}}{12.011\,\text{g C}} = 0.666\,\text{mol C}$

$1.34\,\text{g H} \times \frac{1\,\text{mol H}}{1.008\,\text{g H}} = 1.33\,\text{mol H}$

$10.7\,\text{g O} \times \frac{1\,\text{mol O}}{15.999\,\text{g O}} = 0.669\,\text{mol O}$

*   Step 2: Express as a preliminary non-integer formula:

$\text{C}_{0.666}\text{H}_{1.33}\text{O}_{0.669}$

*   Step 3: Convert ratios to whole numbers by dividing each value by the smallest calculated mole value ($0.666$):

$\text{C}: \frac{0.666}{0.666} = 1.00$

$\text{H}: \frac{1.33}{0.666} = 1.99 \approx 2.00$

$\text{O}: \frac{0.669}{0.666} = 1.00$

*   Result: The mole ratio is $1:2:1$, producing the empirical formula $\text{CH}_2\text{O}$.

• Worked Example 2: Empirical Formula Conversions (Example 3.45)

  • Determining empirical formulas from non-integer or reducible formulas:
    • $\text{BO}_{1.5}$: Multiplying all subscripts by $2$ yields $\text{B}_2\text{O}_3$.
    • $\text{C}_6\text{H}_6$: Dividing all subscripts by $6$ yields $\text{CH}$.
    • $\text{Na}_{0.5}\text{P}_{0.25}\text{O}$: Multiplying all subscripts by $4$ yields $\text{Na}_2\text{PO}_4$.
    • $\text{C}_3\text{H}_{2.67}\text{O}_{1.33}$: Recognizing $.67$ as $\frac{2}{3}$ and $.33$ as $\frac{1}{3}$, multiply all subscripts by $3$ to yield $\text{C}_9\text{H}_8\text{O}_4$.
  • Verification of existing empirical formulas:
    • $\text{H}_2\text{O}$: Subscript ratio $2:1$ is fully reduced; it is an empirical formula.
    • $\text{NaCl}$: Subscript ratio $1:1$ is fully reduced; it is an empirical formula.

• Worked Example 3: Determination from Percent Composition (Example 3.48)

  • Problem: Determine the empirical formula of a compound composed of $43.7\%$ P and $56.3\%$ O by mass.
  • Step 1: Assume a $100.0\,\text{g}$ sample, yielding $43.7\,\text{g P}$ and $56.3\,\text{g O}$.
  • Step 2: Convert mass to moles:

$43.7\,\text{g P} \times \frac{1\,\text{mol P}}{30.974\,\text{g P}} = 1.411\,\text{mol P}$

$56.3\,\text{g O} \times \frac{1\,\text{mol O}}{15.999\,\text{g O}} = 3.519\,\text{mol O}$

*   Preliminary subscript representation: $\text{P}_{1.411}\text{O}_{3.519}$.
*   Step 3: Divide both mole values by the smaller mole value ($1.411$):

$\text{P}: \frac{1.411}{1.411} = 1.00$

$\text{O}: \frac{3.519}{1.411} = 2.494 \approx 2.50$

*   Step 4: Multiply subscripts by $2$ to convert $.50$ into a whole number:

$1.00 \times 2 = 2$

$2.50 \times 2 = 5$

*   Result: The empirical formula is $\text{P}_2\text{O}_5$.

• Empirical Formulas in Chemical Classification:
  • Ionic Compounds: Most ionic compounds are defined and written in terms of their empirical formulas (e.g., $\text{NaCl}$). However, certain polyatomic ions contain even atomic ratios that do not simplify to the empirical ratio (e.g., peroxide $\text{O}_2^{2-}$, mercury(I) $\text{Hg}_2^{2+}$, oxalate $\text{C}_2\text{O}_4^{2-}$, and peroxydisulfate $\text{S}_2\text{O}_8^{2-}$).
  • Covalent Compounds: Empirical formulas represent simplified ratios and are utilized directly to calculate molecular formulas when molar mass is known.

Molecular Formulas

• The molecular formula defines the actual number of atoms of each element in a single molecule of a covalent compound.

  • Example: One molecule of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) contains exactly $6$ carbon atoms, $12$ hydrogen atoms, and $6$ oxygen atoms.

• Relationship to Empirical Formula:

  • The molecular formula is always an integer multiple ($n$) of the empirical formula.
  • Compounds can share identical empirical formulas and percent compositions while possessing distinct molecular formulas, physical structures, and chemical properties.

*   Acetylene ($\text{C}_2\text{H}_2$) and Benzene ($\text{C}_6\text{H}_6$) both possess the empirical formula $\text{CH}$ and identical mass percent composition, yet they are distinct molecules with different molecular formulas.

• Calculation of Molecular Formula:
  • Step 1: Calculate the molar mass of the empirical formula.
  • Step 2: Determine the integer multiplying factor ($n$):

$n = \frac{\text{molar mass of the compound}}{\text{molar mass of the empirical formula}}$

*   Step 3: Multiply all subscripts in the empirical formula by $n$:

$\text{Molecular Formula} = n \times (\text{Empirical Formula})$

• Worked Example 1: Styrene (Example 3.50)
  • Problem: Styrene is a molecular compound used in polystyrene production. Its empirical formula is $\text{CH}$, and its molar mass is $104\,\text{g\,mol}^{-1}$. Find its molecular formula.
  • Step 1: Molar mass of empirical formula $\text{CH}$:

$12.011\,\text{g\,mol}^{-1} + 1.008\,\text{g\,mol}^{-1} = 13.019\,\text{g\,mol}^{-1} \approx 13.02\,\text{g\,mol}^{-1}$

*   Step 2: Calculate multiplying factor $n$:

$n = \frac{104\,\text{g\,mol}^{-1}}{13.02\,\text{g\,mol}^{-1}} = 7.99 \approx 8$

*   Step 3: Apply multiplier to formula:

$\text{Molecular Formula} = 8 \times (\text{CH}) = \text{C}_8\text{H}_8$

• Worked Example 2: Complete Determination from Elemental Mass (Example 3.51)
  • Problem: A $72.33\,\text{g}$ sample contains $28.90\,\text{g C}$, $4.86\,\text{g H}$, and $38.57\,\text{g O}$. The compound has a molar mass of $60.05\,\text{g\,mol}^{-1}$. Determine its molecular formula.
  • Step 1: Convert mass of each element to moles:

$28.90\,\text{g C} \times \frac{1\,\text{mol C}}{12.011\,\text{g C}} = 2.406\,\text{mol C}$

$4.86\,\text{g H} \times \frac{1\,\text{mol H}}{1.008\,\text{g H}} = 4.82\,\text{mol H}$

$38.57\,\text{g O} \times \frac{1\,\text{mol O}}{15.999\,\text{g O}} = 2.411\,\text{mol O}$

*   Unsimplified ratio: $\text{C}_{2.406}\text{H}_{4.82}\text{O}_{2.411}$.
*   Step 2: Divide by the smallest mole value ($2.406$):

$\text{C}: \frac{2.406}{2.406} = 1.00$

$\text{H}: \frac{4.82}{2.406} = 2.00$

$\text{O}: \frac{2.411}{2.406} = 1.00$

*   Empirical formula: $\text{CH}_2\text{O}$.
*   Step 3: Calculate empirical formula mass:

$12.011\,\text{g\,mol}^{-1} + 2(1.008\,\text{g\,mol}^{-1}) + 15.999\,\text{g\,mol}^{-1} = 30.026\,\text{g\,mol}^{-1}$

*   Step 4: Divide molar mass by empirical formula mass:

$n = \frac{60.05\,\text{g\,mol}^{-1}}{30.026\,\text{g\,mol}^{-1}} = 2.000$

*   Step 5: Multiply empirical formula subscripts by $2$:

$\text{Molecular Formula} = 2 \times (\text{CH}_2\text{O}) = \text{C}_2\text{H}_4\text{O}_2$

Combustion Analysis

• Combustion analysis is a laboratory technique used to determine the empirical formula of organic compounds containing carbon, hydrogen, and optionally oxygen or other elements.

• Experimental Principles and Process:

  • A sample of known mass is burned completely in a stream of pure oxygen ($\text{O}_2$) inside a combustion chamber.
  • All carbon in the sample is quantitatively converted into carbon dioxide ($\text{CO}_2$).
  • All hydrogen in the sample is quantitatively converted into water ($\text{H}_2\text{O}$).
  • The combustion gas stream ($\text{H}_2\text{O} + \text{CO}_2 + \text{O}_2$) flows through sequential absorber tubes:
    1. First chamber: Contains a selective absorbent for $\text{H}_2\text{O}$.
    2. Second chamber: Contains a selective absorbent for $\text{CO}_2$
  • The mass gain of each absorber corresponds to the mass of $\text{H}_2\text{O}$ and $\text{CO}_2$ produced.

• Mathematical Equations for Mass Determination:
  • Mass of Carbon in sample (Equation 3.4):

$\text{mass of C} = \text{mass of CO}_2 \times \frac{1\,\text{mol CO}_2}{44.009\,\text{g CO}_2} \times \frac{1\,\text{mol C}}{1\,\text{mol CO}_2} \times \frac{12.011\,\text{g C}}{1\,\text{mol C}}$

*   Mass of Hydrogen in sample (Equation 3.5):

$\text{mass of H} = \text{mass of H}_2\text{O} \times \frac{1\,\text{mol H}_2\text{O}}{18.015\,\text{g H}_2\text{O}} \times \frac{2\,\text{mol H}}{1\,\text{mol H}_2\text{O}} \times \frac{1.008\,\text{g H}}{1\,\text{mol H}}$

*   Mass of Oxygen in sample (by difference):

$\text{mass of O} = \text{mass of total sample} - (\text{mass of C} + \text{mass of H})$

• Worked Example 1: Butyl Butyrate (Example 3.53)
  • Problem: A $2.50\,\text{g}$ sample of butyl butyrate (containing C, H, and O; molar mass = $144.21\,\text{g\,mol}^{-1}$) yields $6.10\,\text{g CO}_2$ and $2.50\,\text{g H}_2\text{O}$. Determine the molecular formula.
  • Step 1: Calculate mass of C and H from combustion products:

$\text{mass of C} = 6.10\,\text{g CO}_2 \times \frac{1\,\text{mol CO}_2}{44.009\,\text{g CO}_2} \times \frac{1\,\text{mol C}}{1\,\text{mol CO}_2} \times \frac{12.011\,\text{g C}}{1\,\text{mol C}} = 1.66\,\text{g C}$

$\text{mass of H} = 2.50\,\text{g H}_2\text{O} \times \frac{1\,\text{mol H}_2\text{O}}{18.015\,\text{g H}_2\text{O}} \times \frac{2\,\text{mol H}}{1\,\text{mol H}_2\text{O}} \times \frac{1.008\,\text{g H}}{1\,\text{mol H}} = 0.28\,\text{g H}$

*   Step 2: Calculate mass of O by subtracting C and H from original sample mass:

$\text{mass of O} = 2.50\,\text{g sample} - (1.66\,\text{g C} + 0.28\,\text{g H}) = 0.56\,\text{g O}$

*   Step 3: Convert masses of C, H, and O to moles:

$\text{C}: 1.66\,\text{g C} \times \frac{1\,\text{mol C}}{12.011\,\text{g C}} = 0.138\,\text{mol C}$

$\text{H}: 0.280\,\text{g H} \times \frac{1\,\text{mol H}}{1.008\,\text{g H}} = 0.278\,\text{mol H}$

$\text{O}: 0.56\,\text{g O} \times \frac{1\,\text{mol O}}{15.999\,\text{g O}} = 0.035\,\text{mol O}$

*   Unsimplified ratio: $\text{C}_{0.138}\text{H}_{0.278}\text{O}_{0.035}$.
*   Step 4: Divide moles by the smallest mole quantity ($0.035$):

$\text{C}: \frac{0.138}{0.035} = 3.94 \approx 4$

$\text{H}: \frac{0.278}{0.035} = 7.94 \approx 8$

$\text{O}: \frac{0.035}{0.035} = 1.0$

*   Empirical formula: $\text{C}_4\text{H}_8\text{O}$.
*   Step 5: Determine multiplying factor $n$ using empirical formula mass:

$\text{Empirical formula mass} = 4(12.011\,\text{g\,mol}^{-1}) + 8(1.008\,\text{g\,mol}^{-1}) + 15.999\,\text{g\,mol}^{-1} = 72.106\,\text{g\,mol}^{-1}$

$n = \frac{144.21\,\text{g\,mol}^{-1}}{72.106\,\text{g\,mol}^{-1}} = 2.00$

*   Step 6: Calculate molecular formula:

$\text{Molecular Formula} = 2 \times (\text{C}_4\text{H}_8\text{O}) = \text{C}_8\text{H}_{16}\text{O}_2$

• Worked Example 2: Hydrocarbon Combustion Analysis
  • Problem: A $1.75\,\text{g}$ sample of a compound containing only carbon and hydrogen yields $5.39\,\text{g CO}_2$ and $2.48\,\text{g H}_2\text{O}$ upon combustion analysis.
  • Step 1: Calculate mass of C and H:

$\text{mass of C} = 5.39\,\text{g CO}_2 \times \frac{1\,\text{mol CO}_2}{44.009\,\text{g CO}_2} \times \frac{1\,\text{mol C}}{1\,\text{mol CO}_2} \times \frac{12.011\,\text{g C}}{1\,\text{mol C}} = 1.471\,\text{g C}$

$\text{mass of H} = 2.48\,\text{g H}_2\text{O} \times \frac{1\,\text{mol H}_2\text{O}}{18.015\,\text{g H}_2\text{O}} \times \frac{2\,\text{mol H}}{1\,\text{mol H}_2\text{O}} \times \frac{1.008\,\text{g H}}{1\,\text{mol H}} = 0.2775\,\text{g H}$

*   Verify total mass: $1.471\,\text{g C} + 0.2775\,\text{g H} = 1.7485\,\text{g} \approx 1.75\,\text{g}$ (confirms compound contains no oxygen).
*   Step 2: Convert masses to moles:

$\text{moles of C} = \frac{1.471\,\text{g}}{12.011\,\text{g\,mol}^{-1}} = 0.1225\,\text{mol C}$

$\text{moles of H} = \frac{0.2775\,\text{g}}{1.008\,\text{g\,mol}^{-1}} = 0.2753\,\text{mol H}$

*   Step 3: Divide by smallest mole value ($0.1225$):

$\text{C}: \frac{0.1225}{0.1225} = 1.00$

$\text{H}: \frac{0.2753}{0.1225} = 2.247 \approx 2.25$

*   Step 4: Convert decimal subscript ($.25$) to whole number by multiplying all subscripts by $4$:

$\text{C}: 1.00 \times 4 = 4$

$\text{H}: 2.25 \times 4 = 9$

*   Resulting Empirical Formula: $\text{C}_4\text{H}_9$ (Empirical formula mass = $57.116\,\text{g\,mol}^{-1}$).
*   Step 5: Determine molecular formula based on specified molar mass ranges:
    *   If molar mass is given as approximately $114.23\,\text{g\,mol}^{-1}$ or in the range $100.00\text{--}130.00\,\text{g\,mol}^{-1}$:

$n = \frac{114.23\,\text{g\,mol}^{-1}}{57.116\,\text{g\,mol}^{-1}} = 2.00 \implies \text{Molecular Formula} = \text{C}_8\text{H}_{18}$

    *   If molar mass is in the range $150.00\text{--}200.00\,\text{g\,mol}^{-1}$:

$n = \frac{171.35\,\text{g\,mol}^{-1}}{57.116\,\text{g\,mol}^{-1}} = 3.00 \implies \text{Molecular Formula} = \text{C}_{12}\text{H}_{27}$