Calculate the amount (volume) of 5.25% (wt./vol.) NaOCl solution (commercial bleach) required to oxidize 100 mg of 9-fluorenol to 9-fluorenone. Whenever appropriate, use balanced chemical equations as a part of your calculation.

I know that I need to start off with the balanced chemical equation. I was wondering if this was the correct chemical equation:

2 9-fluorenol + 2 NaOCl ==> 2 9-fluorenone + 2 NaOH + Cl2

I think if you draw out the fluorenol and fluorenone the H atoms in your equation will not balance.

Try

9-fluorenol + NaOCl ==> 9-fluorenone + NaCl + H2O

Thank you for the formula

I was wondering for the answer I got 0.778 ml bleach. I was wondering if that was correct?

The given balanced chemical equation for the oxidation of 9-fluorenol to 9-fluorenone using NaOCl is not correct. The correct balanced equation is:

9-fluorenol + NaOCl → 9-fluorenone + NaCl + H2O

To calculate the amount (volume) of 5.25% NaOCl solution needed to oxidize 100 mg of 9-fluorenol, we need to follow these steps:

Step 1: Convert the given mass of 9-fluorenol to moles.

Molar mass of 9-fluorenol = 182.23 g/mol
Mass of 9-fluorenol = 100 mg = 0.1 g

Number of moles of 9-fluorenol = (0.1 g) / (182.23 g/mol)
≈ 0.000548 mol

Step 2: Use the stoichiometry from the balanced equation to find the molar ratio between NaOCl and 9-fluorenol.

From the balanced equation:
1 mole of NaOCl reacts with 1 mole of 9-fluorenol

Step 3: Calculate the number of moles of NaOCl required.

Number of moles of NaOCl = Number of moles of 9-fluorenol
= 0.000548 mol

Step 4: Convert the moles of NaOCl to grams.

Molar mass of NaOCl = 74.44 g/mol

Mass of NaOCl = (0.000548 mol) * (74.44 g/mol)
≈ 0.0408 g

Step 5: Calculate the volume of 5.25% NaOCl solution.

Since the concentration is given in weight/volume (% wt./vol.), we need to convert grams to milliliters.

Volume of NaOCl solution (mL) = (Mass of NaOCl) / (Density)
= (0.0408 g) / (0.0525 g/mL)
≈ 0.778 mL

Therefore, approximately 0.778 mL of a 5.25% NaOCl solution would be required to oxidize 100 mg of 9-fluorenol to 9-fluorenone.