# Study Notes: Analysis of Compounds ## Overview Chemical formulas provide critical information about the composition of compounds. They allow us to derive **stoichiometric ratios** and determine the relative amounts of elements in a compound. This section focuses on using chemical formulas to analyze compounds, calculate empirical formulas, and understand molecular structures. --- ## Key Concepts ### 1. **Stoichiometric Ratios from Chemical Formulas** Chemical formulas indicate the ratio of atoms in a compound. For example, the formula **HgBr₂** tells us: - For every 1 mole of mercury (Hg), there are 2 moles of bromine (Br). This can be expressed as: $$ S(\text{Br:Hg}) = \frac{2 \, \text{mol Br}}{1 \, \text{mol Hg}} $$ Other valid stoichiometric ratios for **HgBr₂** include: - $$ S(\text{Hg:HgBr₂}) = \frac{1 \, \text{mol Hg}}{1 \, \text{mol HgBr₂}} $$ - $$ S(\text{Br:HgBr₂}) = \frac{2 \, \text{mol Br}}{1 \, \text{mol HgBr₂}} $$ Stoichiometric ratios derived from chemical formulas are essential for determining empirical formulas and analyzing compound composition. --- ### 2. **Empirical Formula Determination** The **empirical formula** represents the simplest whole-number ratio of elements in a compound. It can be determined using: - **Combustion analysis**: A method where a compound is burned in oxygen, and the resulting products (e.g., CO₂ and H₂O) are measured. #### Combustion Train Setup - A sample is heated in a stream of dry oxygen. - **H** in the compound is converted to **H₂O**, which is absorbed in a tube. - **C** in the compound is converted to **CO₂**, which is absorbed in another tube. - The mass increase in each tube corresponds to the amount of H₂O and CO₂ produced. Combustion analysis assumes the compound contains only C, H, and O. If other elements (e.g., sulfur) are present, additional steps are required. --- ### 3. **Example: Empirical Formula Calculation** #### Problem: A 6.49 mg sample of ascorbic acid (vitamin C) is burned, producing: - 9.74 mg of CO₂ - 2.64 mg of H₂O Determine the empirical formula. --- #### Step 1: Calculate the moles of C and H From the masses of CO₂ and H₂O, use stoichiometric ratios to find the moles of C and H. 1. **Moles of Carbon (C):** $$ n_C = \frac{9.74 \, \text{mg CO₂}}{44.01 \, \text{g/mol CO₂}} \times \frac{1 \, \text{mol C}}{1 \, \text{mol CO₂}} $$ $$ n_C = 2.21 \times 10^{-4} \, \text{mol C} $$ 2. **Moles of Hydrogen (H):** $$ n_H = \frac{2.64 \, \text{mg H₂O}}{18.02 \, \text{g/mol H₂O}} \times \frac{2 \, \text{mol H}}{1 \, \text{mol H₂O}} $$ $$ n_H = 2.93 \times 10^{-4} \, \text{mol H} $$ --- #### Step 2: Calculate the mass of C and H Convert moles to mass for C and H: 1. **Mass of Carbon:** $$ m_C = n_C \times 12.01 \, \text{g/mol C} $$ $$ m_C = 2.65 \, \text{mg C} $$ 2. **Mass of Hydrogen:** $$ m_H = n_H \times 1.008 \, \text{g/mol H} $$ $$ m_H = 0.295 \, \text{mg H} $$ --- #### Step 3: Determine the mass of Oxygen (O) Subtract the masses of C and H from the total sample mass: $$ m_O = 6.49 \, \text{mg sample} - 2.65 \, \text{mg C} - 0.295 \, \text{mg H} $$ $$ m_O = 3.54 \, \text{mg O} $$ Convert to moles: $$ n_O = \frac{3.54 \, \text{mg O}}{16.00 \, \text{g/mol O}} $$ $$ n_O = 2.21 \times 10^{-4} \, \text{mol O} $$ --- #### Step 4: Determine the element ratios Find the simplest whole-number ratio of C, H, and O: 1. **C:H ratio:** $$ \frac{n_H}{n_C} = \frac{2.93 \times 10^{-4}}{2.21 \times 10^{-4}} = 1.33 \, \text{mol H/mol C} \approx \frac{4}{3} $$ 2. **C:O ratio:** $$ \frac{n_O}{n_C} = \frac{2.21 \times 10^{-4}}{2.21 \times 10^{-4}} = 1 \, \text{mol O/mol C} $$ Thus, the empirical formula is **C₃H₄O₃**. The molecular formula of ascorbic acid is **C₆H₈O₆**, which is exactly double the empirical formula. --- ## Key Takeaways 1. **Chemical formulas** provide stoichiometric ratios that are essential for analyzing compounds. 2. **Empirical formulas** represent the simplest ratio of elements in a compound and can be determined using combustion analysis. 3. **Molecular formulas** are multiples of empirical formulas and require additional information, such as molecular weight. 4. **Structural formulas** show how atoms are connected and arranged in 3D space, providing more detailed information about a molecule. When calculating the mass of oxygen, students often forget to subtract the masses of carbon and hydrogen from the total sample mass. --- ## Practice Problems 1. A 10.0 mg sample of a compound produces 14.7 mg of CO₂ and 6.0 mg of H₂O upon combustion. Determine the empirical formula of the compound. 2. A compound has an empirical formula of CH₂O and a molecular weight of 180 g/mol. What is its molecular formula? 3. Explain why the molecular formula of a compound might differ from its empirical formula. Always double-check your calculations for unit consistency and significant figures!