Problem 25 - Chemistry The Central Science In Si Units

Question

Calculate the percentage by mass of oxygen in the following compounds: (a) vanillin, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3} ;(\mathbf{b})\) isopropyl alcohol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\); (c) acetaminophen, \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2}\); (d) cyclopropanone, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O} ;\) (e) dioxin, \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2} ;(\mathbf{f})\) penicillin, \(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{4} \mathrm{~S}\).

Step-by-Step Solution

Answer

Vanillin: 31.56%, Isopropyl Alcohol: 26.62%, Acetaminophen: 21.17%, Cyclopropanone: 28.53%, Dioxin: 9.97%, Penicillin: 19.14%

Step 1: Understand the Formula for Mass Percentage

The formula for calculating the percentage by mass of an element in a compound is given by: \[ \%\text{ by mass of element} = \left( \frac{\text{Mass of the element in 1 mole of compound}}{\text{Molar mass of compound}} \right) \times 100 \]

Step 2: Calculate Molar Mass of Vanillin, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\)

Determine the molar masses of individual atoms: C (12.01 g/mol), H (1.01 g/mol), and O (16.00 g/mol). For vanillin, calculate: \[ \text{Molar mass} = 8(12.01) + 8(1.01) + 3(16.00) = 152.15 \text{ g/mol} \]

Step 3: Calculate Oxygen Percentage in Vanillin

Calculate the oxygen contribution: \[ \text{Mass of O} = 3 \times 16.00 = 48.00 \text{ g} \] Apply the formula: \[ \frac{48.00}{152.15} \times 100 \approx 31.56\% \]

Step 4: Calculate Molar Mass of Isopropyl Alcohol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\)

Calculate the molar mass: \[ \text{Molar mass} = 3(12.01) + 8(1.01) + 1(16.00) = 60.09 \text{ g/mol} \]

Step 5: Calculate Oxygen Percentage in Isopropyl Alcohol

Calculate the oxygen contribution: \[ \frac{16.00}{60.09} \times 100 \approx 26.62\% \]

Step 6: Calculate Molar Mass of Acetaminophen, \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2}\)

Calculate the molar mass: \[ 8(12.01) + 9(1.01) + 1(14.01) + 2(16.00) = 151.17 \text{ g/mol} \]

Step 7: Calculate Oxygen Percentage in Acetaminophen

Calculate the oxygen contribution: \[ 2 \times 16.00 = 32.00 \text{ g} \] \[ \frac{32.00}{151.17} \times 100 \approx 21.17\% \]

Step 8: Calculate Molar Mass of Cyclopropanone, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\)

Calculate the molar mass: \[ 3(12.01) + 4(1.01) + 1(16.00) = 56.07 \text{ g/mol} \]

Step 9: Calculate Oxygen Percentage in Cyclopropanone

Calculate the oxygen contribution: \[ \frac{16.00}{56.07} \times 100 \approx 28.53\% \]

Step 10: Calculate Molar Mass of Dioxin, \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2}\)

Calculate the molar mass: \[ 12(12.01) + 4(1.01) + 4(35.45) + 2(16.00) = 320.85 \text{ g/mol} \]

Step 11: Calculate Oxygen Percentage in Dioxin

Calculate the oxygen contribution: \[ 2 \times 16.00 = 32.00 \text{ g} \] \[ \frac{32.00}{320.85} \times 100 \approx 9.97\% \]

Step 12: Calculate Molar Mass of Penicillin, \(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{N}_{2} \mathrm{O}_{4} \mathrm{S}\)

Calculate the molar mass: \[ 16(12.01) + 18(1.01) + 2(14.01) + 4(16.00) + 1(32.07) = 334.41 \text{ g/mol} \]

Step 13: Calculate Oxygen Percentage in Penicillin

Calculate the oxygen contribution: \[ 4 \times 16.00 = 64.00 \text{ g} \] \[ \frac{64.00}{334.41} \times 100 \approx 19.14\% \]

Key Concepts

Percentage by Mass

The concept of 'percentage by mass' helps determine how much of a specific element is present in a compound. It's like cutting a cake and figuring out what share of it is chocolate – except with molecules! To calculate this, you'll use the formula:

• \[\text{Percentage by mass of element} = \left( \frac{\text{Mass of the element in 1 mole of compound}}{\text{Molar mass of compound}} \right) \times 100\]

Start by finding the total weight (or molar mass) of each compound, which is the sum of the weights of all its atoms. Then, determine the total weight of the specific element within that compound. Finally, plug these values into the formula! This gives you the percentage of the compound's weight that comes from that element. Understanding this allows chemists to grasp how much of each element contributes to the compound, a crucial step in both research and applications in chemical formulations.

Chemical Compounds

Chemical compounds are combinations of two or more elements chemically bonded together. These elements can appear in various forms, contributing to the properties of the compound. For a clearer view:

• Vanillin, for instance, is composed of carbon, hydrogen, and oxygen.
• Each compound has a distinct formula, like \( \mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3} \) for vanillin, denoting how many atoms of each element exists in it.
• The elements in these compounds are often held together by covalent or ionic bonds.

The molar mass mentioned in mass percentage calculations is pivotal here. By adding up the atomic weights based on the compound's formula, like in vanillin's calculation, you find the compound's molar mass. This allows for precise scientific applications, particularly in formulation of medicines and flavoring substances, where knowing the exact composition is crucial.

Oxygen Contribution

Oxygen's role in compounds is often substantial, influencing both reactions and properties. When calculating the percentage of oxygen by mass in a compound, you focus on how much its presence weighs in.For clarity, let's review how vanillin (\( \mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3} \)) incorporates oxygen:

• The compound has 3 oxygen atoms, each weighing about 16.00 g/mol.
• The collective weight of oxygen in the compound is thus \( 3 \times 16.00 \text{ g} = 48.00 \text{ g} \).
• This is then divided by the entire compound's molar mass (152.15 g/mol for vanillin) and multiplied by 100 to find the percentage.
• Thus, the oxygen contributes significantly, for example, 31.56% by mass for vanillin.

Acknowledging oxygen's role is critical because oxygen often dictates the reactivity and stability of compounds. It serves as an oxidizing agent and influences molecular architecture and biological activity. So, understanding its contribution is crucial in fields ranging from organic chemistry to environmental science.