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.
This question has come up on here a few times in the past, see for example http://www.jiskha.com/display.cgi?id=1261460213.
You need to start with the babalnced equation.
RCH-OH + NaOCl ==> RC=O + NaCl + H2O
so one mole of NaOCl reacts with one mole of fluorenol.
1. calculate the number of moles of fluorenol you have.
0.150 g/molecular mass of fluorenol
this is the number of moles of NaOCl
2. hence calculate the mass of NaOCl
3. then use the percentage concentration to calculate the volume needed.
can you please explain to me how to use the percentage concentration to calculate volume
To calculate the volume of 6.25% NaOCl solution required to oxidize 150 mg of 9-fluorenol to 9-fluorenone, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction.
The reaction involves the oxidation of 9-fluorenol (C13H10O) to 9-fluorenone (C13H8O). The balanced equation is as follows:
2 C13H10O + NaOCl → 2 C13H8O + NaCl + H2O
Step 2: Determine the molar mass of 9-fluorenol.
Using the periodic table, you can find the molar mass of each element in 9-fluorenol (C13H10O).
C: 12.01 g/mol x 13 = 156.13 g/mol
H: 1.01 g/mol x 10 = 10.10 g/mol
O: 16.00 g/mol x 1 = 16.00 g/mol
Total molar mass of 9-fluorenol = 156.13 g/mol + 10.10 g/mol + 16.00 g/mol = 182.23 g/mol
Step 3: Calculate the number of moles of 9-fluorenol.
Number of moles = mass / molar mass
Number of moles of 9-fluorenol = 150 mg / 182.23 g/mol
Note: Convert the mass to grams to ensure consistency in units.
Number of moles of 9-fluorenol = 0.822 mol (rounded to three decimal places)
Step 4: Use stoichiometry to determine the number of moles of NaOCl required.
From the balanced equation, we see that 2 moles of 9-fluorenol react with 1 mole of NaOCl.
Therefore, the number of moles of NaOCl required = (0.822 mol x 1 mol NaOCl) / 2 mol 9-fluorenol
Number of moles of NaOCl required = 0.411 mol (rounded to three decimal places)
Step 5: Calculate the volume of 6.25% NaOCl solution needed.
The concentration of the NaOCl solution is given as 6.25% (wt./vol.), which means there are 6.25 g of NaOCl per 100 mL of solution.
To convert to moles, we need to know the molar mass of NaOCl. Sodium (Na) has atomic mass 22.99 g/mol, Oxygen (O) has atomic mass 16.00 g/mol, and Chlorine (Cl) has atomic mass 35.45 g/mol.
Na: 22.99 g/mol x 1 = 22.99 g/mol
O: 16.00 g/mol x 1 = 16.00 g/mol
Cl: 35.45 g/mol x 1 = 35.45 g/mol
Total molar mass of NaOCl = 22.99 g/mol + 16.00 g/mol + 35.45 g/mol = 74.44 g/mol
Now we can calculate the number of moles of NaOCl in the solution:
Number of moles of NaOCl = (0.411 mol NaOCl / 74.44 g/mol) x (1000 mg/g) / (6.25 g/100 mL)
Note: Convert mg to g and mL to L to ensure consistent units.
Number of moles of NaOCl = 0.874 mol (rounded to three decimal places)
Finally, using the density of the NaOCl solution, we can calculate the volume of the solution needed:
Density of NaOCl solution = 6.25 g/100 mL
Volume of NaOCl solution = (0.874 mol NaOCl / 6.25 g/mL) x 1000 mL
Volume of NaOCl solution = 139.84 mL (rounded to two decimal places)
Therefore, approximately 139.84 mL of the 6.25% NaOCl solution is required to oxidize 150 mg of 9-fluorenol to 9-fluorenone.