Question

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 \(180 \mathrm{u}\).

Short answer

The empirical and molecular formulas for the given substances are as follows: (a) Ibuprofen: Empirical Formula: C6H9O Molecular Formula: C12H18O2 (b) Cadaverine: Empirical Formula: C5H14N Molecular Formula: C5H14N (c) Epinephrine (Adrenaline): Empirical Formula: C9H13NO3 Molecular Formula: C9H13NO3

Step-by-step solution

Substance (a) - Ibuprofen

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

Substance (b) - Cadaverine

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)

Substance (c) - Epinephrine

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

Key concepts

Chemical Composition Analysis

When we talk about chemical composition analysis, we're essentially trying to figure out the makeup of a compound in terms of its elements by percentage. This is a fundamental aspect of determining the **empirical formula**, which gives the simplest whole-number ratio of atoms in a compound.
The process generally starts by assuming a sample size, like 100 grams, to simplify percentage calculations into direct grams for each element, as seen in the examples.
By converting these mass percentages to grams first, it enables us to get a clear path to converting them into **moles**, which are crucial for deriving empirical formulas.

• Considering a 100 g sample makes calculations straightforward when dealing with percentages.
• Converting mass to moles using atomic masses is essential to find how much of each element is present at the atomic scale.

For instance, Ibuprofen with 75.69% carbon is converted to grams under the assumption of a 100-g sample, resulting in 75.69 grams of carbon. This conversion to grams is just an intermediate step, which smoothly transitions into **mole calculations** in the next section.

Mole Concept in Chemistry

The concept of a mole is essential to the understanding of chemical compositions and reactions. In chemistry, a mole is like a dozen—but much larger. One mole equals Avogadro's number, approximately \(6.022 \times 10^{23}\) entities, such as atoms, molecules, or ions, of a substance.
This makes the mole an excellent bridge between the atomic scale of elements and compounds and our macroscopic observations in the lab.
Converting mass to moles involves dividing the mass of each element by its **atomic mass** (from the periodic table). For instance, converting grams of hydrogen to moles involves dividing by its atomic mass, approximately 1.008 g/mol.

• Moles help in comparing amounts of different elements in a compound accurate to a molecular level.
• The mole ratio is then used to infer the simplest proportions of elements, important for deducing formulas.

Relating it to a practical example, Ibuprofen's 8.80 grams of hydrogen become about 8.73 moles. This number helps establish how hydrogen atoms relate to carbon and oxygen atoms in calculating both empirical and potentially molecular formulas.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions and is key in determining empirical and molecular formulas. It's about balancing the proportions of elements as they respond in chemical processes.
The empirical formula derived from mole ratios represents the simplest integer ratio of elements, while the molecular formula gives the actual number of each type of atom present in a compound.
To find the molecular formula, the molar mass of a compound is divided by its empirical formula mass to determine how many empirical units fit the molecular structure.

• The empirical formula is a stepping-stone in finding the molecular formula—derived only when the molar mass is known.
• The ratio \(n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}\) helps determine the molecular formula, often adjusting empirical numbers by multiplying.

Something like Cadaverine, whose empirical and molecular formulas turned out to be the same because \(n \approx 1\), shows how theoretical calculations meet practical observation when molar mass validations align exactly.