Question

Calculate the millimoles of sodium hypochlorite used in the reaction: 5 mL of a 5.25% solution of Clorox bleach (show detailed calculations). Compare this value with the quantity of diphenylmethanol (millimoles) to be used in the reaction. Would you describe the oxidant as being present in slight excess or large excess?

Answer 1

1. MW of NaOCl = 74.44 g/mol = 74.44 mg/mmol
2. MW of 9-fluorenol = 182.22 g/mol = 182.22 mg/mmol
3. NaOCl in bleach = 5.25% (w/v) = 5.25 g NaOCl / 100 ml bleach = 5.25 mg NaOCl = 100 ul bleach

If you wondering how much to use on a mole-for-mole basis, then first write out the equation and make sure it balances

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

So there is a 1 mole-for-1 mole reaction between 9-fluorenol (9F-OH) and NaOCl.

So we set up the product as follows:

(100 mg 9F-OH) * (1 mmol 9F-OH / 182.22 mg 9F-OH) (1 mmol NaOCl / 1 mmol 9F-OH) * (74.44 mg NaOCl / 1 mmol NaOCl) * (100 ul bleach / 5.25 mg NaOCl) =

Note how all dimensions cancel and we are left with a value in microliters of bleach. You should always set up the expression as a product in this way. We are left with the numbers now:

(100) * (1 / 182.22) * (1 / 1) * (74.44) * (100 / 5.25) = 778.127956 ul bleach

Assuming 3 significant digits, this is 778 ul or 0.778 ml bleach

Normally you would use EXCESS NaOCl, since you don't want it to be the limiting reagent. You might use anywhere from 2 - 10 molar excess, so your real volume might be 1.5 - 7 ml bleach (approx. 2-10 molar excess).