Empirical Formula - Chapter 3

Introduction to Empirical Formulas and Combustion Analysis

• Focused on chapter 3 content for a general chemistry audience

• Topics covered: percent composition, empirical formulas, molecular formulas, and combustion analysis

Percent Composition

• Definition: Percent composition is the percentage by mass of each element in a compound.

Example 1: Chromium (III) Fluoride (CrF₃)

• Formula weight: 109.00 g/mol

• Find percent composition of Chromium (Cr) and Fluorine (F).

• Setup:

  • Number of atoms in formula:

    • Cr: 1

    • F: 3

  • Molar mass of Cr: 52.00 g/mol

  • Total molar mass calculation:

    • Molar mass of Cr = 52.00 g/mol

    • Molar mass of F = 19.00 g/mol

    • Total = 52.00 + (3 x 19.00) = 109.00 g/mol

  • Percent Cr calculation: $\%{Cr}=\frac{52.00Cr\frac{g}{mol}}{109.00\frac{g}{mol}}\cdot100=47.71\%$

    • Rounding based on significant figures:

    • 52.00 (4 sig figs)

    • 109.00 (5 sig figs)

    • Result limited to 4 significant figures. Round 4770.0 -> 4771.0.

  • Percent F calculation: $\%{F}={\frac{\left(3\right)19.00\frac{g}{mol}}{109.00\frac{g}{mol}}\cdot}100=52.29\%$

    • Rounded down from 52.29 rounding only the third decimal. Check for total (% Cr + % F = 100%).

Example 2: Sodium Phosphate (Na₃PO₄)

• Formula weight: 163.94 g/mol

• Find percent Na, P, and O.

• Setup:

  • Number of atoms in formula:

    • Na: 3

    • P: 1

    • O: 4

  • Atomic weights:

    • Na: 22.99 g/mol

    • P: 30.97 g/mol

    • O: 16.00 g/mol

  • Calculate the masses for Na, P, and O, and repeat similar calculations as in CrF₃.

Empirical Formulas

• Definition: Empirical formula represents the simplest whole-number ratio of the elements in a compound.

Example: Empirical Formula from Percent Composition

• Given:

  • 68.4% Cr

  • 31.6% O

• Conversion to grams:

  • Treat percentages as grams: 68.4 g Cr and 31.6 g O.

• Convert grams to moles:

  • $ext{Moles of Cr} = rac{68.4 g}{52.00 g/mol} = 1.31$

  • $ext{Moles of O} = rac{31.6 g}{16.00 g/mol} = 1.97$

• **Finding the simplest ratio:

  • Divide by the smallest value:**

    • $ext{Cr} = rac{1.31}{1.31} = 1$

    • $ext{O} = rac{1.97}{1.31} <br>ightarrow 1.5$

• Scale up to the nearest whole numbers:

  • Multiply by 2: C₂O₃.

  • To scale a fraction in an empirical formula calculation, you must multiply all the mole ratios by the smallest whole number that will turn the fractional value into a whole number. This step occurs after you have divided each mole value by the smallest number of moles in the set.

    Here are common multipliers used based on the decimal result:

    • $.50$: Multiply all ratios by $2$ (e.g., $1.5 \times 2 = 3$).

    • $.33$ or $.66$: Multiply all ratios by $3$ (e.g., $1.33 \times 3 = 4$ or $2.66 \times 3 = 8$).

    • $.25$ or $.75$: Multiply all ratios by $4$ (e.g., $1.25 \times 4 = 5$).

    • $.20$ or $.40$: Multiply all ratios by $5$ (e.g., $1.2 \times 5 = 6$).

    Example from the note:
    
    In the Chromium and Oxygen example, the ratio was found to be $1$ mole of $\text{Cr}$ to $1.5$ moles of $\text{O}$. To remove the $.5$ decimal, both values are multiplied by $2$, resulting in a final empirical formula of {Cr}2\{O}3.

Molecular Formulas

• Definition: Molecular formula gives the actual number of each type of atom in a molecule.

Example: Finding Molecular Formula from Empirical Formula

• Empirical formula: C₂O₃

• Calculation of molar mass of the empirical formula:

  • C₂O₃ = (2 x 12.01) + (3 x 16.00) = 30.03 g/mol

• Molecular weight provided in the problem: A specific value (180 g/mol) implies there is a multiple.

• $n = rac{180 g/mol}{30.03 g/mol} = 5.99 <br>ightarrow ext{rounds to 6}$

• Final molecular formula: C₁₂O₁₈.

Combustion Analysis

• Definition: Combustion analysis involves burning a sample to determine the composition of the reactants from the combustion products (CO₂ and H₂O).

Example: Combustion of Unknown Compound

• Given:

  • Sample mass = 4.28 mg
    d - Products: 6.21 mg CO₂ and 2.54 mg H₂O

• Assume complete combustion process and find amount of Carbon, Hydrogen through analysis:

• Calculate percent carbon and hydrogen as previously done, replacing values as necessary.

• Compute residual elements (often oxygen).

• Ensure total mass equals 100% to confirm steady comptime.

• Convert to moles and simplify ratios for empirical formula, scaling appropriately for molecular formula calculations based on molecular weight of the sample.

Conclusion

• Recap: Percent composition, empirical vs. molecular formulas, and the fundamental principles of combustion analysis provide vital tools in understanding chemical compounds.

• More examples and exercises available for practice on the learning platform (e.g., Canvas).

Final Note

• Stay consistent in methods to ensure comprehension. Rounding and significant figures are essential for accuracy.