4.3 Determination of Percent Composition, Empirical Formulas, and Molecular Formulas

Chemical Formulas and the Identity of Compounds

• The chemical identity of a compound is defined by its elemental makeup.
• Chemical formulas are the most succinct and efficient way of representing this elemental makeup.
• When a laboratory scientist works with an unknown compound, the first experimental step is typically measuring the mass of each constituent element.
• Experimental data must be used to determine chemical formulas; they cannot be guessed or assumed without empirical evidence.

Percent Composition and Experimental Calculations

• Measurements taken during an experiment allow for the calculation of a compound’s percent composition.

• The fundamental formula for percent composition is: $\text{Percent Component} = \frac{\text{Mass of Element}}{\text{Mass of Compound}} \times 100\%$

• Scenario: Carbon and Hydrogen Sample

  • Total mass of sample: $10\,g$
  • Mass of Hydrogen ($H$): $2.5\,g$
  • Mass of Carbon ($C$): $7.5\,g$
  • Calculation for Hydrogen: $\frac{2.5\,g}{10\,g} \times 100\% = 25\% \text{ by mass}$
  • Calculation for Carbon: $\frac{7.5\,g}{10\,g} \times 100\% = 75\% \text{ by mass}$

Detailed Example: Analysis of a Multi-Element Liquid Compound

• Problem Statement: A $12.04\,g$ sample of a liquid compound contains $7.34\,g$ Carbon ($C$), $1.85\,g$ Hydrogen ($H$), and $2.85\,g$ Nitrogen ($N$). Determine the percent composition.
• Verification of Total Mass:$7.34\,g + 1.85\,g + 2.85\,g = 12.04\,g$
• Calculations:
  • Carbon Percentage:$\frac{7.34\,g}{12.04\,g} \times 100\% = 61\%$
  • Hydrogen Percentage:$\frac{1.85\,g}{12.04\,g} \times 100\% = 15.4\%$
  • Nitrogen Percentage:$\frac{2.85\,g}{12.04\,g} \times 100\% = 23.7\%$
• Check: The sum ($61\% + 15.4\% + 23.7\% = 100.1\%$) is approximately $100\%$. Small variations are normal due to rounding.

Determining Percent Composition from Chemical Formulas

• Percent composition can also be derived from a known chemical formula rather than just experimental mass data.
• Different compounds containing the same element will have unique weight percentages of that element depending on the total formula.
• Example: Nitrogen Content in Fertilizers/Chemicals
  • Ammonia ($NH_3$): $\approx 82\% \text{ Nitrogen by mass}$
  • Ammonium Nitrate ($NH_4NO_3$): $\approx 35\% \text{ Nitrogen by mass}$
  • Urea ($CH_4N_2O$): $\approx 47\% \text{ Nitrogen by mass}$

Comprehensive Case Study: Aspirin ($C_9H_8O_4$)

• Purpose: To find the percent composition of each element ($C$, $H$, and $O$) using the molecular formula and molar masses.
• Step 1: Calculate Mass of each Element Component
  • Carbon: $9 \times 12.01\,g\,mol^{-1} = 108.09\,g\,mol^{-1}$
  • Hydrogen: $8 \times 1.008\,g\,mol^{-1} = 8.064\,g\,mol^{-1}$
  • Oxygen: $4 \times 16.00\,g\,mol^{-1} = 64.00\,g\,mol^{-1}$
• Step 2: Determine Total Molar Mass of Compound
  • Molar Mass ($M$): $108.09 + 8.064 + 64.00 = 180.154\,g\,mol^{-1}$
• Step 3: Calculate Individual Percentages
  • Percent Carbon ($C$):$\frac{108.09\,g\,mol^{-1}}{180.154\,g\,mol^{-1}} \times 100\% = 60\%$
  • Percent Hydrogen ($H$):$\frac{8.064\,g\,mol^{-1}}{180.154\,g\,mol^{-1}} \times 100\% = 4.476\%$
  • Percent Oxygen ($O$):$\frac{64.00\,g\,mol^{-1}}{180.154\,g\,mol^{-1}} \times 100\% = 35.53\%$
• Verification: The sum of percents ($60\% + 4.476\% + 35.53\%$) equals $100\%$.

Principles of Empirical Formula Determination

• Chemical formulas do not directly represent masses; subscripts represent the relative number of atoms/moles.
• Mass data obtained experimentally must be converted to moles to find formula ratios.
• The Empirical Formula represents the lowest whole-number ratio of atoms in a substance.
• Scenario: Carbon/Hydrogen Sample
  • Known masses: $1.71\,g\,C$ and $0.287\,g\,H$.
  • Moles Carbon: $1.71\,g \times \frac{1\,mol}{12.01\,g} = 0.142\,mol\,C$
  • Moles Hydrogen: $0.287\,g \times \frac{1\,mol}{1.008\,g} = 0.285\,mol\,H$
  • Writing as subscripts: $C_{0.142}H_{0.285}$.
  • Dividing by smallest ($0.142$):
  • Carbon: $\frac{0.142}{0.142} = 1$
  • Hydrogen: $\frac{0.285}{0.142} \approx 2$
  • Resulting Empirical Formula: $CH_2$

Addressing Non-Whole Number Ratios: Chlorine and Oxygen

• Problem: A sample has $5.31\,g$ Chlorine ($Cl$) and $8.4\,g$ Oxygen ($O$).
• Step 1: Convert to Moles
  • Moles $Cl = \frac{5.31\,g}{35.45\,g\,mol^{-1}} = 0.150\,mol$
  • Moles $O = \frac{8.4\,g}{16\,g\,mol^{-1}} = 0.525\,mol$
• Step 2: Divide by the smallest molar amount
  • $Cl: \frac{0.150}{0.150} = 1$
  • $O: \frac{0.525}{0.150} = 3.5$
• Step 3: Factor Multiplication
  • Because $3.5$ is not a whole number, multiply both subscripts by the same factor (in this case, $2$) until both are whole numbers.
  • $1 \times 2 = 2$
  • $3.5 \times 2 = 7$
• Final Empirical Formula: $Cl_2O_7$

Systematic Procedural Steps

1. Derive Moles: Calculate the molar amount of each element from its mass (using atomic mass from periodic table).
2. Tentative Formula: Divide all molar amounts by the smallest molar amount calculated in Step 1 to generate subscripts.
3. Whole Number Ratios: If any subscripts are not whole numbers, multiply all subscripts by the smallest possible integer that yields whole numbers for all elements.

Determining the Empirical Formula of Hematite

• Problem: Hematite contains $34.97\,g$ Iron ($Fe$) and $15.03\,g$ Oxygen ($O$).
• Step 1: Molar Conversion$\text{Moles } Fe = \frac{34.97\,g}{55.85\,g\,mol^{-1}} = 0.6261\,mol$$\text{Moles } O = \frac{15.03\,g}{16.00\,g\,mol^{-1}} = 0.9394\,mol$
• Step 2: Divide by smallest ($0.6261$)
  • $Fe = 1$
  • $O = 1.5$
• Step 3: Multiplication Factor
  • Multiply by $2$ to clear the $0.5$ decimal.
  • $Fe: 1 \times 2 = 2$
  • $O: 1.5 \times 2 = 3$
• Resulting Empirical Formula: $Fe_2O_3$

Using Percent Composition as Mass Data

• If percent composition is provided instead of individual masses, assume a total sample size of $100\,g$.
• This allows the percentage value to be used directly as mass in grams.
• Problem: Ethanol Fermentation Gas
  • Carbon: $27.29\%$
  • Oxygen: $72.71\%$
• Step 1: Convert to Moles (Assuming $100\,g$ sample)
  • Carbon: $\frac{27.29\,g}{12.01\,g\,mol^{-1}} = 2.272\,mol$
  • Oxygen: $\frac{72.71\,g}{16.00\,g\,mol^{-1}} = 4.544\,mol$
• Step 2: Ratio Calculation
  • Divide by $2.272$:
  • Carbon = $1$
  • Oxygen = $2$
• Empirical Formula: $CO_2$

Transitioning from Empirical to Molecular Formula

• Differences:
  • Empirical Formula: Shows the relative number of elements (lowest whole number ratio).
  • Molecular Formula: Shows the absolute number of atoms in a single molecule.
• Requirements: To find a molecular formula, you must first know the molecular (molar) mass of the compound, typically determined via separate experiment.
• Relationship Factor ($n$):
  • The molecular formula is always a multiple relative to the empirical formula.
  • Calculate the integer $n$ using the ratio of masses: $n = \frac{\text{Molecular or Molar Mass}}{\text{Empirical Formula Mass}}$
• Molecular Formula Formulation:$\text{Molecular Formula} = n \times (\text{Empirical Formula})$

Example: Molecular Formula of a Covalent Compound

• Problem: A compound has an empirical formula of $CH_2O$. Its molecular mass is found to be $180\,amu$. What is the molecular formula?
• Step 1: Calculate Empirical Formula Mass
  • Carbon ($12.01$) + Hydrogen ($2 \times 1.008$) + Oxygen ($16.00$) $\approx 30\,amu$
• Step 2: Find the factor ($n$)$n = \frac{180\,amu}{30\,amu} = 6$
• Step 3: Apply the factor to the subscripts
  • $C_{(1 \times 6)} H_{(2 \times 6)} O_{(1 \times 6)}$
• Final Molecular Formula: $C_6H_{12}O_6$