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I'd like to determine the water content in a sample without buying an expensive machine (like Karl Fischer titrator). So I'd like to measure the water volumetrically. I've chosen formamide to dissolve the sample because of its high polarity.

I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.

1) Let's assume there is 0,5 ml water dissolved in the formamide fraction, is it possible to add 30 ml solvent B and get so a fraction of 30,5 ml (30 ml + 0,5 ml)?

2) Instead of formamide, do you know another solvent A which dissolves the sample including the water (and the watercontent of approx. 20% in the glucose syrup)) and a solvent B which extracts the water phase from the solvent A?

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    $\begingroup$ You can use Dean-Stark azeotropic removal of water with toluene or benzene $\endgroup$ – K_P Jan 20 '16 at 20:23
  • $\begingroup$ Can't do you a brine wash? Prepare a saturated aqueous solution of salt, then do a liquid-liquid partition and collect the aqueous layer. The volumes of brine and water will not be additive, but the masses will be. It's not a perfect separation, but certainly among the easiest options. $\endgroup$ – Nicolau Saker Neto Jan 22 '16 at 21:48
  • $\begingroup$ Maybe try silica gel? $\endgroup$ – G-man Jan 23 '16 at 7:29
  • $\begingroup$ @K_P Thank you K_P for your useful advice! Is it possible to heat my solution of formamide up to 105°C without distilling also the glycerol? (I think glycerol (because it's an alcohol) will make an azeotrop with water and therefore I doubt if this method is suitable) $\endgroup$ – laminin Jan 23 '16 at 9:35
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    $\begingroup$ I don't think you will have any problem with glycerol since it has a boiling point of $\ce{260 ^{\circ} C }$ Also I wouldn't use formamide at all. Just mix your sample with enough toluene with good stirring and let it reflux using the Dean-Stark and collect and measure the amount of water. Formamide usually contains some water so this can affect your calculations depending what accuracy you need $\endgroup$ – K_P Jan 23 '16 at 10:07
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I'd like to determine the water content in a sample without buying an expensive machine (like [a] Karl Fischer titrator).

Rather than try to extract the water and measure it volumetrically, I think you could try to measure its activity, or else perform your own Karl-Fischer titration, which does not necessarily require any equipment.

Water activity measurement

Practically this means hygrometers. That may sounds like another expensive machine, but E-bay tells me that capacitive hygrometers are available for under $20. If your sample consists of sucrose, glucose, and glycerol, those substances are unlikely to interfere with capacitive hygrometry measurements of water activity.

You might even get very crude accuracy by incubating a mirror at a known (chilled) temperature, and then waving it above your water/sucrose/glucose/glycerol sample, seeing if dew forms, and then repeating the process at several different temperatures.

Karl Fischer titration on your own

As an alternative, couldn't you just set up your own Karl Fischer titration? Commercial titrators often use coulometric detection for better accuracy, but setting up a volumetric titration is pretty easy. You need:

  1. Methanol (or some other alcohol)
  2. Imidazole (or some other amine base)
  3. Sulfur dioxide
  4. Iodine

Imidazole and sulfur dioxide should be combined in a 1:1 molar ratio in an excess of methanol. Iodine should be dissolved in methanol.

A 2012 paper has a nice description of how to do KF with visual end-point determination. You just need to buy a KF titration solution, or make your own:

Commercially available [...] Test Kit from [...] was used for reference moisture determination. All the equipments needed were provided with the kit. It consisted of two-component ethanol-based reagents: [a] working medium containing [a mixture of] imidazole, sulphur dioxide and diethanolamine, and [separately a titrant solution] containing iodine. According to manufacturer's instructions for use, 20 mL of [working medium] was added to the titration vessel, which was then tightly sealed with a septum. The titrant was delivered through the septum with a syringe to react with the present water, until a color change from colorless to yellow occurred. A known amount of sample was added by a syringe to the titration vessel via septum, and a newly filled titration syringe was used for a new titration until the color change from colorless to yellow occurred. Consumption volume of the titrant was read off the titration syringe. The water content, in percent by volume, was calculated from the consumption of the titrant and the sample volume.

If you've diluted with formamide

I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.

The nice part about both water activity measurement and Karl Fischer titration is that they are both sensitive and accurate enough to still work on your sample even after dilution or dissolution in formamide.

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  • $\begingroup$ Thank you Curt F. for your ideas! (We already use a device which gives a water activity of let's say 60%. Unfortunately it's not possible to calculate back to the water content.) There are three questions: 1) How can I bring in SO2 in methanol most easily? 2) If I don't know the water content exactly, will it works if I use twice the molar amount of SO2/imidazol as stochiometrical needed? 3) Is it possible to see the colorchange from colorless to yellow? $\endgroup$ – laminin Jan 28 '16 at 20:24
  • $\begingroup$ I'm not sure about your question (1). The answers to (2) and (3) are both "yes". $\endgroup$ – Curt F. Feb 8 '16 at 15:05
  • $\begingroup$ The answer to #1 depends a lot on the equipment you have. I suppose it's possible to use sodium sulfite as the $\cf{SO2}$ source and then ion exchange the resulting solution with an imidazolium-based ion exchange resin. Bubbling in $\cf{SO2}$ gas to a solution of imidazole free base also probably works. $\endgroup$ – Curt F. Feb 8 '16 at 15:08

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