Problem 47 - Chemistry The Central Science

Question

The organic anion is found in most detergents. Assume that the anion undergoes aerobic decomposition in the following manner: $$ \begin{array}{r} 2 \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-}(a q)+51 \mathrm{O}_{2}(a q) \longrightarrow \\ 36 \mathrm{CO}_{2}(a q)+28 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{H}^{+}(a q)+2 \mathrm{SO}_{4}^{2-}(a q) \end{array} $$ What is the total mass of \(\mathrm{O}_{2}\) required to biodegrade \(10.0 \mathrm{~g}\) of this substance?

Step-by-Step Solution

Answer

The total mass of \(\mathrm{O}_2\) required to biodegrade \(10.0 \space \mathrm{g}\) of the organic anion is \(22.7 \space \mathrm{g}\).

Step 1: Calculate the molar mass of the organic anion and \(\mathrm{O}_2\)

First, we need to find the molar mass of the given organic anion \(\mathrm{C}_{18}\mathrm{H}_{29}\mathrm{SO}_{3}^{-}\). To do this, we find the molar mass of each element in the anion and sum them up. The molar mass of C is 12.01 g/mol, H is 1.008 g/mol, S is 32.07 g/mol, and O is 16.00 g/mol. Molar mass of \(\mathrm{C}_{18}\mathrm{H}_{29}\mathrm{SO}_{3}^{-}\): \((18\times 12.01 \space g/mol) + (29 \times 1.008 \space g/mol) + (1\times 32.07 \space g/mol) + (3\times 16.00 \space g/mol) = 361.37 \space g/mol\) Molar mass of \(\mathrm{O}_2\): \((2\times 16.00 \space g/mol) = 32.00 \space g/mol\)

Step 2: Convert the mass of the organic anion to moles

Now, we convert the mass of the organic anion (\(10.0 \space g\)) to moles using the calculated molar mass: Moles of \(\mathrm{C}_{18}\mathrm{H}_{29}\mathrm{SO}_{3}^{-}\) = \(\frac{10.0 \space g}{361.37 \space g/mol} = 0.0277 \space mol\)

Step 3: Use stoichiometry to find the moles of \(\mathrm{O}_2\) needed

Given the balanced chemical equation, we can see that for every 2 moles of the organic anion, we require 51 moles of \(\mathrm{O}_2\). Using the ratio, we can calculate the moles of \(\mathrm{O}_2\) needed for the decomposition: Moles of \(\mathrm{O}_2\) = \(\frac{51 \,"\,\text{moles of } \mathrm{O}_2"}{2 \, "\,moles of }\mathrm{C}_{18}\mathrm{H}_{29}\mathrm{SO}_{3}^{-}\space "}\) \(\times 0.0277 \,"\,moles of }\mathrm{C}_{18}\mathrm{H}_{29}\mathrm{SO}_{3}^{-}\space "= 0.7094 \space mol\)

Step 4: Convert the moles of \(\mathrm{O}_2\) to mass

Finally, we multiply the moles of \(\mathrm{O}_2\) by the molar mass of \(\mathrm{O}_2\) to find the mass required for the biodegradation process: Mass of \(\mathrm{O}_2\) = \(0.7094 \space mol \times 32.00 \space g/mol = 22.7 \space g\) Therefore, the total mass of \(\mathrm{O}_2\) required to biodegrade \(10.0 \space \mathrm{g}\) of the organic anion is \(22.7 \space \mathrm{g}\).

Key Concepts

Molar Mass Calculation

The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Calculating the molar mass of a compound involves summing up the product of the atomic masses of each element within the compound and its respective number of atoms.

• To calculate the molar mass of an organic anion like \( \mathrm{C}_{18}\mathrm{H}_{29}\mathrm{SO}_{3}^{-} \), you consider each element in the anion: carbon (C), hydrogen (H), sulfur (S), and oxygen (O).
• The atomic masses are: C is 12.01 g/mol, H is 1.008 g/mol, S is 32.07 g/mol, and O is 16.00 g/mol.
• Multiplying the atomic mass by the number of atoms gives you: \((18 \times 12.01) + (29 \times 1.008) + (1 \times 32.07) + (3 \times 16.00) = 361.37 \space g/mol\).

The process is similar for diatomic oxygen (\( \mathrm{O}_2 \)), which involves multiplying the atomic mass of oxygen by two, yielding \(32.00 \space g/mol\). Understanding molar mass helps convert grams to moles and vice versa, crucial for stoichiometric calculations.

Stoichiometric Ratios

Stoichiometry involves determining the ratio in which chemical substances react. This ratio is derived from the coefficients in the balanced chemical equation, representing the number of moles of each substance needed or produced.

• In the given reaction of aerobic decomposition of the organic anion, \(2 \space moles\) of \( \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-} \) react with \(51 \space moles\) of \( \mathrm{O}_2 \).
• The stoichiometric ratio of \( \mathrm{O}_2 \) to \( \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-} \) is \(51:2\) or \(25.5:1\).

Knowing this ratio allows us to calculate how much oxygen is needed for the complete decomposition of a given amount of the anion. In our example, using the ratio \( \frac{51}{2} \), we calculate the moles of \( \mathrm{O}_2 \) required per mole of anion undergoing decomposition.

Aerobic Decomposition

Aerobic decomposition is a chemical reaction where organic substances break down in the presence of oxygen. This process involves microorganisms using oxygen to decompose organic matter into simpler compounds, producing carbon dioxide, water, and other substances.
Key points include:

• Oxygen is crucial for decomposition to occur aerobically, aiding in the oxidation of complex molecules.
• The balanced chemical equation shows this process, indicating the reactants (organic anion and \( \mathrm{O}_2 \)) and products (\( \mathrm{CO}_2 \), \( \mathrm{H}_2\), \( \mathrm{H}^+ \), and \( \mathrm{SO}_4^{2-} \)).

Aerobic decomposition is more efficient in energy release compared to anaerobic processes and is key in various applications, including wastewater treatment and organic waste composting.

Organic Anion

Organic anions are negatively charged ions derived from organic molecules. In many detergents, these anions play a crucial role in cleaning by breaking down oils and dirt.

• The organic anion in our exercise, \( \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-} \), consists of a long hydrocarbon chain and a sulfate group, giving it surfactant properties.
• Such anions can undergo aerobic decomposition, as shown in the chemical equation.

Understanding the structure and properties of organic anions helps us comprehend their role in environmental chemistry and biodegradation. Their negative charge allows them to interact with positively charged particles, aiding in cleaning and biochemical processes.

Chemical Equation Balancing

Balancing chemical equations ensures that the law of conservation of mass is observed, meaning the same number of each type of atom is present on both sides of the equation. In the given reaction:

• The reactants include \(2 \space \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-} \) and \(51 \space \mathrm{O}_2 \).
• The products are \(36 \space \mathrm{CO}_2\), \(28 \space \mathrm{H}_2\), \(2 \space \mathrm{H}^+ \) and \(2 \space \mathrm{SO}_4^{2-} \).

Each element's total is equal on both sides. Carbon, hydrogen, sulfur, and oxygen atoms are conservatively accounted for. This balancing ensures accurate stoichiometric calculations when determining reagent amounts. Being adept at chemical equation balancing is essential for anyone studying chemistry to predict reaction outcomes and perform precise laboratory experiments.