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 \(59.0 \% \mathrm{C}\), \(7.1 \% \mathrm{H}, 26.2 \% \mathrm{O},\) and \(7.7 \% \mathrm{~N}\) by mass; its \(\mathrm{MW}\) is about 180 amu.

To find the moles of each element in the compound, divide the mass percentage by the molar mass of the respective element. Assume there are 100 grams of each compound, so the mass percentages of the elements can be taken as their actual mass in grams. (a) Ibuprofen: C: \( \frac{75.69 \ g}{12.01 \ g/mol} = 6.308 \ mol\) H: \( \frac{8.80 \ g}{1.008 \ g/mol} = 8.73 \ mol\) O: \( \frac{15.51 \ g}{16.00 \ g/mol} = 0.969 \ mol\) (b) Cadaverine: C: \( \frac{58.55 \ g}{12.01 \ g/mol} = 4.874 \ mol\) H: \( \frac{13.81 \ g}{1.008 \ g/mol} = 13.7 \ mol\) N: \( \frac{27.40 \ g}{14.01 \ g/mol} = 1.957 \ mol\) (c) Epinephrine: C: \( \frac{59.0 \ g}{12.01 \ g/mol} = 4.914 \ mol\) H: \( \frac{7.1 \ g}{1.008 \ g/mol} = 7.04 \ mol\) O: \( \frac{26.2 \ g}{16.00 \ g/mol} = 1.637 \ mol\) N: \( \frac{7.7 \ g}{14.01 \ g/mol} = 0.549 \ mol\)

To find the simplest whole-number ratio of elements, divide the moles of each element from Step 1 by the smallest number of moles among the elements. Round to the nearest whole number. (a) Ibuprofen: C: \(\frac{6.308}{0.969} = 6.51 ≈ 6\) H: \(\frac{8.73}{0.969} = 9.01 ≈ 9\) O: \(\frac{0.969}{0.969} = 1\) Empirical formula: C\( _6\)H\( _9\)O (b) Cadaverine: C: \(\frac{4.874}{1.957} = 2.49 ≈ 2\) H: \(\frac{13.7}{1.957} = 7.00 ≈ 7\) N: \(\frac{1.957}{1.957} = 1\) Empirical formula: C\( _2\)H\( _7\)N (c) Epinephrine: C: \(\frac{4.914}{0.549} = 8.95 ≈ 9\) H: \(\frac{7.04}{0.549} = 12.83 ≈ 13\) O: \(\frac{1.637}{0.549} = 2.98 ≈ 3\) N: \(\frac{0.549}{0.549} = 1\) Empirical formula: C\_9H[13]ON

Divide the molecular weight (molar mass) of the compound by the molar mass of the empirical formula. Multiply the empirical formula by the resulting value (rounded to the nearest whole number) to obtain the molecular formula. (a) Ibuprofen: Molar mass of the empirical formula: (6 × 12.01) + (9 × 1.008) + 16.00 = 107.582 Molecular weight ratio: \(\frac{206}{107.582} ≈ 1.913\) Molecular formula: C\( _{12}\)H\( _{18}\)O\( _2\) (b) Cadaverine: Molar mass of the empirical formula: (2 × 12.01) + (7 × 1.008) + 14.01 = 43.12 Molecular weight ratio: \(\frac{102.2}{43.12} ≈ 2.37\) Molecular formula: C\( _4\)H\( _{14}\)N\( _ {2}\) (c) Epinephrine: Molar mass of the empirical formula: (9 × 12.01) + (13 × 1.008) + 16.00 + 14.01 = 167.385 Molecular weight ratio: \(\frac{180}{167.385} ≈ 1.076\) Molecular formula: C\( _9\)H\( _{13}\)NO\( _3\) Final Results: (a) Ibuprofen - Empirical formula: C\( _6\)H\( _9\)O and Molecular formula: C\( _{12}\)H\( _{18}\)O\( _2\) (b) Cadaverine - Empirical formula: C\( _2\)H\( _7\)N and Molecular formula: C\( _4\)H\( _{14}\)N\( _ {2}\) (c) Epinephrine - Empirical formula: C\( _9\)H\( _{13}\)ON and Molecular formula: C\( _9\)H\( _{13}\)NO\( _3\)