Calculate the amount (volume) of 6.25% (wt./vol.) NaOCl solution (commercial bleach)
required to oxidize 150 mg of 9-fluorenol to 9-fluorenone. Whenever appropriate, use
balanced chemical equations as a part of your calculation
To calculate the volume of 6.25% NaOCl solution required to oxidize 150 mg of 9-fluorenol to 9-fluorenone, we need to follow a few steps:
1. Convert the mass of 9-fluorenol from milligrams to grams. Since there are 1000 milligrams in a gram, 150 mg is equal to 0.15 grams.
2. Write the balanced chemical equation for the oxidation of 9-fluorenol to 9-fluorenone using NaOCl as the oxidizing agent. The balanced equation is as follows:
C13H10O + NaOCl -> C13H8O + NaCl + H2O
This equation tells us that for every 1 mole of 9-fluorenol, we need 1 mole of NaOCl.
3. Determine the molecular weight of 9-fluorenol (C13H10O) and calculate the number of moles. The molecular weight of 9-fluorenol can be found by adding up the atomic weights of each element: 13 carbon atoms x 12.01 g/mol + 10 hydrogen atoms x 1.01 g/mol + 1 oxygen atom x 16.00 g/mol = 180.23 g/mol.
Now, calculate the number of moles of 9-fluorenol:
0.15 g / 180.23 g/mol = 0.000831 moles.
4. Since the balanced equation tells us that 1 mole of 9-fluorenol requires 1 mole of NaOCl, the number of moles of NaOCl needed will also be 0.000831 moles.
5. Calculate the required volume of 6.25% NaOCl solution using the equation:
Volume (L) = moles / concentration (mol/L).
The concentration of 6.25% NaOCl solution can be converted to mol/L by dividing the weight percent by the molar mass (per liter). The molar mass of NaOCl is calculated as:
(22.99 g/mol for Na) + (16.00 g/mol for O) + (35.45 g/mol for Cl) = 74.44 g/mol.
The concentration can be calculated as:
6.25 g/100 mL x (1 L/1000 mL) x (1 mol/74.44 g) = 0.00836 mol/L.
Now, substitute the values into the equation:
Volume (L) = 0.000831 moles / 0.00836 mol/L.
6. Calculate the volume:
Volume (L) = 0.0996 L or 99.6 mL.
Therefore, to oxidize 150 mg of 9-fluorenol to 9-fluorenone, you would need approximately 99.6 mL of 6.25% NaOCl solution.