Determine the empirical and molecular formulas of each of the following substances: (a) Ibuprofen, a headache remedy, contains \(75.69 \% \mathrm{C}\), $8.80 \% \mathrm{H},\( and \)15.51 \% \mathrm{O}\( by mass and has a molar mass of \)206 \mathrm{~g} / \mathrm{mol}$. (b) Cadaverine, a foul-smelling substance produced by the action of bacteria on meat, contains \(58.55 \% \mathrm{C}\), \(13.81 \% \mathrm{H},\) and $27.40 \% \mathrm{~N}\( by mass; its molar mass is \)102.2 \mathrm{~g} / \mathrm{mol}$ (c) Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains \(59.0 \%\) C, \(7.1 \%\) H, \(26.2 \%\) O, and $7.7 \% \mathrm{~N}\( by mass; its molar mass is about \)180 \mathrm{u}$.

Step 1: Convert the percentage composition by mass to grams. Assume a 100 g sample. - C: 75.69 g - H: 8.80 g - O: 15.51 g Step 2: Convert the grams to moles using the atomic masses of each element. - C: \(\frac{75.69 \text{ g}}{12.01 \text{ g/mol}} = 6.306 \text{ mol}\) - H: \(\frac{8.80 \text{ g}}{1.008 \text{ g/mol}} = 8.730 \text{ mol}\) - O: \(\frac{15.51 \text{ g}}{16.00 \text{ g/mol}} = 0.9694 \text{ mol}\) Step 3: Calculate the mole ratio of elements in the compound. - \(\text{Mole ratio} = \frac{6.306 \text{ mol C}}{0.9694 \text{ mol O}} : \frac{8.730 \text{ mol H}}{0.9694 \text{ mol O}} : \frac{0.9694 \text{ mol O}}{0.9694 \text{ mol O}}\) - Mole ratio ≈ 6.50:9.00:1.00 Step 4: Find the empirical formula using the lowest whole number ratio of elements. - Empirical formula: C6H9O Step 5: Calculate the empirical formula mass. - Empirical formula mass: \((6\times 12.01) + (9 \times 1.008) + (1 \times 16.00) = 121.1 \text{ g/mol}\) Step 6: Determine the molecular formula using the molar mass provided (206 g/mol). - \(\text{Molecular formula} = \text{Empirical Formula} \times n\) - \(n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}\) - \(n = \frac{206 \text{ g/mol}}{121.1 \text{ g/mol}} ≈ 1.7\) - Thus, Molecular formula: C12H18O2

Follow the same steps as in substance (a): Step 1: Convert the percentage composition by mass to grams (assume a 100 g sample): - C: 58.55 g - H: 13.81 g - N: 27.40 g Step 2: Convert the grams to moles: - C: \(4.878 \text{ mol}\) - H: \(13.68 \text{ mol}\) - N: \(1.957 \text{ mol}\) Step 3: Calculate the mole ratio of elements in the compound. - Mole ratio ≈ 2.49:7.00:1.00 Step 4: Empirical formula: C5H14N Step 5: Empirical formula mass: 86.2 g/mol Step 6: Molar mass provided: 102.2 g/mol - \(n = \frac{102.2 \text{ g/mol}}{86.2 \text{ g/mol}} ≈ 1.2\) - Molecular formula: C5H14N (as n ≈ 1)

Follow the steps as in substance (a): Step 1: Convert the percentage composition by mass to grams (assume a 100 g sample): - C: 59 g - H: 7.1 g - O: 26.2 g - N: 7.7 g Step 2: Convert the grams to moles: - C: \(4.919 \text{ mol}\) - H: \(7.042 \text{ mol}\) - O: \(1.637 \text{ mol}\) - N: \(0.5495 \text{ mol}\) Step 3: Calculate the mole ratio of elements in the compound. - Mole ratio ≈ 9.00:12.80:3.00:1.00 Step 4: Empirical formula: C9H13NO3 Step 5: Empirical formula mass: 181.2 g/mol Step 6: Molar mass provided: 180 amu (u = unified atomic mass unit) Since the molar mass provided is slightly less (due to rounding errors) than the empirical formula mass, the empirical and molecular formulas are the same: - Molecular formula: C9H13NO3