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.
Any help is appreciatede.
Do you go to UCI by any chance? Anyway, this is how the TA helped me:
6.25% w/vol
1000 g/L
62.5 g/L
150 mg/ 182.2 = 0.82 mmol x 74 (the NaOCl MW) = 60.92 mg
62.5 mg/mL x (x)mL = 60.92 mg
60.92 mg/ 62.5 = 0.97 mL which is roughly 1
and we know that our calculation is correct because the procedure asks for 1.2 mL, so it's fairly close!
Hope that helps!!
I go to UCI, and this helped a lot. Thank you!
To calculate the amount (volume) of NaOCl solution required to oxidize 150 mg of 9-fluorenol to 9-fluorenone, we need to follow a step-by-step approach.
Step 1: Write the balanced chemical equation
The chemical equation for the oxidation of 9-fluorenol to 9-fluorenone using NaOCl can be represented as follows:
C13H10O + NaOCl → C13H8O + NaOH
Step 2: Determine the molar mass of 9-fluorenol
The molar mass of 9-fluorenol (C13H10O) can be calculated by adding the atomic masses of its constituent elements:
(12.01 g/mol x 13) + (1.01 g/mol x 10) + (16.00 g/mol x 1) = 180.2 g/mol
Step 3: Convert the given mass of 9-fluorenol to moles
We can use the molar mass of 9-fluorenol to convert the given mass to moles:
150 mg x (1 g / 1000 mg) x (1 mol / 180.2 g) = 0.000833 mol
Step 4: Determine the stoichiometric ratio between 9-fluorenol and NaOCl
From the balanced chemical equation, we can see that the stoichiometric ratio between 9-fluorenol and NaOCl is 1:1. This means that 1 mol of 9-fluorenol reacts with 1 mol of NaOCl.
Step 5: Calculate the moles of NaOCl required
Since the stoichiometric ratio is 1:1, the moles of NaOCl required will also be equal to 0.000833 mol.
Step 6: Determine the concentration of NaOCl solution
The given solution is a 6.25% (wt./vol.) NaOCl solution. This means that 6.25 g of NaOCl is present in 100 mL (or 0.100 L) of solution.
Step 7: Convert the concentration of NaOCl solution to moles
To convert the concentration from wt./vol. to moles, we need to divide the mass of NaOCl by its molar mass:
6.25 g x (1 mol / 74.44 g) = 0.0839 mol
Step 8: Calculate the volume of NaOCl solution required
Using the stoichiometric ratio, we can set up the following equation:
0.0839 mol NaOCl / X L = 0.000833 mol NaOCl / 1
Solving for X gives us:
X = (0.000833 mol NaOCl) / (0.0839 mol/L) = 0.00992 L or 9.92 mL
Therefore, the amount (volume) of 6.25% NaOCl solution required to oxidize 150 mg of 9-fluorenol to 9-fluorenone is approximately 9.92 mL.