What volume of 6.43% (wt.vol.) NaOCl solution is required to oxidize 221 mg of 9-fluorenol to 9-fluorenone?
I'm not to sure of where to start with this problem. Please help!
ur a noob
mega n00b
To solve this problem, we need to use the concept of stoichiometry and the given information about the concentration of NaOCl solution and the mass of 9-fluorenol.
Here are the steps to calculate the required volume:
Step 1: Determine the moles of 9-fluorenol.
To calculate the moles of 9-fluorenol, we can use its molar mass. The molar mass of 9-fluorenol (C13H10OH) is calculated as follows:
C (13 atoms) = 13 * atomic mass of carbon (12.01 g/mol) = 156.13 g/mol
H (10 atoms) = 10 * atomic mass of hydrogen (1.01 g/mol) = 10.10 g/mol
O (1 atom) = atomic mass of oxygen (16.00 g/mol) = 16.00 g/mol
Adding all the masses together, we get:
156.13 g/mol + 10.10 g/mol + 16.00 g/mol = 182.23 g/mol
Now, divide the given mass (221 mg) by the molar mass to get the moles of 9-fluorenol:
221 mg = 0.221 g
Moles of 9-fluorenol = 0.221 g / 182.23 g/mol = 0.001213 mol
Step 2: Write the balanced chemical equation.
The balanced chemical equation for the oxidation of 9-fluorenol to 9-fluorenone using NaOCl can be written as follows:
C13H10OH + NaOCl -> C13H8O + NaCl + H2O
From the balanced equation, we can see that it requires 1 mole of NaOCl to oxidize 1 mole of 9-fluorenol.
Step 3: Calculate the moles of NaOCl required.
Since the molar ratio between NaOCl and 9-fluorenol is 1:1 from the balanced equation, the moles of NaOCl required will be the same as the moles of 9-fluorenol.
So, the moles of NaOCl required = 0.001213 mol.
Step 4: Convert moles to volume of the NaOCl solution.
To convert moles to volume, we need to know the concentration of the NaOCl solution.
From the given information, the concentration of the NaOCl solution is 6.43% (wt.vol.). This means that there are 6.43 grams of NaOCl in 100 mL (or 6.43 g/100 mL) of solution.
To calculate the volume of NaOCl solution required, we can use the following formula:
Volume (mL) = (moles of NaOCl required / concentration of NaOCl solution in mol/mL) * 1000
To determine the concentration of the NaOCl solution in mol/mL, we need to convert the weight percent of NaOCl to molarity (mol/L). The molar mass of NaOCl is:
Na (22.99 g/mol) + O (16.00 g/mol) + Cl (35.45 g/mol) = 74.44 g/mol
The molarity is calculated as follows:
6.43 g/100 mL = 0.0643 g/mL
Molarity of NaOCl = (0.0643 g/mL / 74.44 g/mol) * 1000 = 0.8654 mol/L
Now, let's substitute the values into the formula:
Volume (mL) = (0.001213 mol / 0.8654 mol/mL) * 1000 = 1.40 mL
Therefore, approximately 1.40 mL of the 6.43% (wt.vol.) NaOCl solution is required to oxidize 221 mg of 9-fluorenol to 9-fluorenone.