Problem 39
Question
The organic anion is found in most detergents. Assume that the anion undergoes aerobic decomposition in the following manner: $$ \begin{array}{r} 2 \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-}(a q)+51 \mathrm{O}_{2}(a q) \longrightarrow \\ 36 \mathrm{CO}_{2}(a q)+28 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{H}^{+}(a q)+2 \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}(a q) \end{array} $$ What is the total mass of \(\mathrm{O}_{2}\) required to biodegrade \(1.0 \mathrm{~g}\) of this substance?
Step-by-Step Solution
Verified Answer
The total mass of \(\mathrm{O}_{2}\) required to biodegrade 1.0 g of the given substance is 1.58 g.
1Step 1: Balance the given chemical reaction
It appears that the given chemical reaction is already balanced:
\[2 \mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}^{-}(a q)+51 \mathrm{O}_{2}(a
q) \longrightarrow 36 \mathrm{CO}_{2}(a q)+28 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{H}^{+}(a
q)+2 \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}(a q)\]
2Step 2: Calculate the amount (in moles) of the organic anion in 1 gram of the substance
The molecular formula of the given organic anion is \(\mathrm{C}_{18} \mathrm{H}_{29} \mathrm{SO}_{3}\). First, we need to calculate its molar mass. Using the atomic masses of each element, we get:
Molar mass = (18 × 12.01 g/mol C) + (29 × 1.01 g/mol H) + (1 × 32.07 g/mol S) + (3 × 16.00 g/mol O) = 515.36 g/mol
Now, we can calculate the amount (in moles) of the organic anion in 1 gram of the substance.
moles of organic anion = mass / molar mass
moles of organic anion = 1.0 g / 515.36 g/mol = \(1.940\times10^{-3}\text{ mol}\)
3Step 3: Apply the stoichiometry of the balanced reaction to find the amount (in moles) of \(\mathrm{O}_{2}\) needed for the decomposition.
According to the balanced equation, the ratio of moles of organic anion to moles of \(\mathrm{O}_{2}\) is 2:51. Now, using a proportion, we can find the moles of \(\mathrm{O}_{2}\) required:
\(moles\_O2 =\frac{51 \times 1.94\times10^{-3}}{2} = 4.93\times10^{-2} \text{mol}\)
4Step 4: Convert the moles of \(\mathrm{O}_{2}\) to mass using its molar mass.
The molar mass of \(\mathrm{O}_{2}\) = 2 × 16.00 g/mol = 32.00 g/mol
Now, we can convert the moles of \(\mathrm{O}_{2}\) to mass:
mass of \(\mathrm{O}_{2}\) = moles × molar mass
mass of \(\mathrm{O}_{2}\) = \(4.93\times10^{-2} \text{mol}\) × 32.00 g/mol = 1.58 g
Therefore, the total mass of \(\mathrm{O}_{2}\) required to biodegrade 1.0 g of the given substance is 1.58 g.
Key Concepts
Chemical Reaction BalancingMolar Mass CalculationMole-to-Mass Conversion
Chemical Reaction Balancing
Balancing chemical reactions is critical for understanding how different substances react with each other. It involves making sure that the number of atoms of each element is the same on both sides of the reaction equation. This is based on the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system.
The balanced chemical equation provides the ratios of molecules and moles that will react or be produced. In the exercise, the given organic anion reacts with oxygen, and we see that the molecules of each compound are carefully balanced. For every 2 molecules of the anion, 51 molecules of oxygen are required, and the products are formed accordingly. Without a balanced equation, it would be impossible to accurately predict product formation or reactant consumption, making it a foundational skill in stoichiometry.
The balanced chemical equation provides the ratios of molecules and moles that will react or be produced. In the exercise, the given organic anion reacts with oxygen, and we see that the molecules of each compound are carefully balanced. For every 2 molecules of the anion, 51 molecules of oxygen are required, and the products are formed accordingly. Without a balanced equation, it would be impossible to accurately predict product formation or reactant consumption, making it a foundational skill in stoichiometry.
Molar Mass Calculation
Knowing the molar mass of a substance is essential in stoichiometry as it links moles — a count of molecules — to mass, which is a measurable quantity in the lab. To find the molar mass of a compound, one must sum the molar masses of all the individual atoms in its formula. This is done by multiplying the atomic mass of each element from the periodic table by the number of atoms of that element in the compound and adding together all of the contributions.
For instance, the organic anion in our exercise has a molar mass determined by adding the weighted contributions from carbon, hydrogen, sulfur, and oxygen. This step is vital in the following mole-to-mass or mass-to-mole conversions and for understanding the amount of each substance involved in the chemical reaction.
For instance, the organic anion in our exercise has a molar mass determined by adding the weighted contributions from carbon, hydrogen, sulfur, and oxygen. This step is vital in the following mole-to-mass or mass-to-mole conversions and for understanding the amount of each substance involved in the chemical reaction.
Mole-to-Mass Conversion
After determining the number of moles of a substance, we often need to find its mass to carry out practical work like measuring out chemicals for a reaction. This process is made straightforward by using the molar mass, which acts as a conversion factor between moles and grams.
By multiplying the number of moles by the molar mass, we obtain the substance's mass in grams. For example, the final step of the exercise involves converting moles of oxygen to grams by using oxygen's molar mass. This conversion is fundamental in chemistry, as it allows scientists and students alike to measure out precise amounts of substances for reactions based on stoichiometric calculations.
By multiplying the number of moles by the molar mass, we obtain the substance's mass in grams. For example, the final step of the exercise involves converting moles of oxygen to grams by using oxygen's molar mass. This conversion is fundamental in chemistry, as it allows scientists and students alike to measure out precise amounts of substances for reactions based on stoichiometric calculations.
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