Author Topic: Amount of maltodextrin for easiest Colloidal Gold chemical reduction  (Read 3003 times)

Offline PeterXXL

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Kephra, in your quickstart guide for colloidal gold you wrote... "1/2 teaspoon maltodextrin" for a 40 ppm of 250 ml colloidal gold based on 1 ml gold chloride and 10 drops (0.5 ml) 1 mol sodium carbonate.

But how much, in milligram, is actually required to make the reduction (well aware that in order to be on the safe side to make sure that everything is reduced, more than the minimum of reduction agent is needed)?


Offline kephra

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You can calculate that, (almost).
Assume maltodextrin is a chain of 17 glucose molecules.  (3 to 17 is the standard definition of maltodexrin)
Assume each maltodextrin molecule can donate 2 electrons
Calculate how many moles of gold you want to reduce.
Since gold chloride needs  3 electrons, calculate the ration of maltodextrin molecules to gold atoms
Calculate the moles of maltodextrin required.
Convert moles of maltodextrin to grams.

Why did I say almost?  Because the formula for maltodextrin is not exact.  It may be any chain length between 3 and 17 glucose molecules, or a mixture of different chain lengths (most likely).  Since that is unknown, the exact molecular weight is also unknown, so the exact weight needed cannot be calculated exactly. 
Colloidal Silver is only a bargain if you make it yourself.

Offline Dean

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #2 on: December 20, 2016, 08:38:26 AM »
Quick question,

Per the above, would this likely be the case for silver also?
(Raising the grams of maltodextrin based on the intended ppm of silver / gold)

Never used maltodextrin until today (turning up hopefully via amazon).
In light of my learnings that Karo appears to have a limit on suitability based on ppm, I was going to trial using maltodextrin as my default go to reducing agent For some trial batches of everything.

Learning that reducing is based on nano particles and not the water volume I'm assuming that my use of maltodextrin will be as PeterXXL's question of "weight of maltodextrin per concentration of nano.

If this be the case, are we talking a "weighable" amount when dealing with ppm's as low as 20 or would such a thought be only realistic from say 40 upwards.

I may just end up using Karo for 20 and maltodextrin for everything else.

I do like my cinnamon though too :-)

I'll be doing some 250ml test batches with that as well I think but understand that this will also always cap to a degree and would like the option to have something that I feel is genuinely uncapped for certain purposes.

Much to read :-)

Offline kephra

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #3 on: December 20, 2016, 01:51:56 PM »
Maltodextrin is not  a standardized product. 
Maltodextrin is a chain of glucose molecules, and that chain can be anywhere from 3 to 17 glucose molecules long.  Only 1 end of the chain is a reducing molecule so the safe thing is to assume worst case of a 17 glucose chain.

To calculate the required maltodextrin, the steps are:
Calculate the weight of silver in the solution.
Calculate the silver in moles by dividing the weight by the atomic weight of silver
Multiply the moles of silver by the molecular weight of glucose.
Multiply that number by 17

In simplified form, the worst case weight of mailtordextrin is:  ppm * ml * .028 0.14
Ex for  20 ppm, 500 ml:  20 * 500 * .014 = 140 mg of maltodextrin.
Ex for 80 ppm, 250 ml:  80 * 250 * .014 = 280 mg
Ex for 100 ppm, 1000ml: : 100 * 1000 * .014 = 1400 mg

For gold, the process is almost the same, except the multiplier is different since gold's atomic weight is a lot more, and it takes 3 times as much because gold is missing 3 electrons instead of 1.

For gold, the constant works out to 0.051  0.025
Ex: 50 ppm 500ml : 50 * 500 * .025 = 625 mg
Ex: 40 ppm 1000ml: 40 * 10000 * 0.025 = 1000 mg

More does not hurt. 
« Last Edit: December 20, 2016, 10:29:07 PM by kephra »
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Offline Dean

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #4 on: December 20, 2016, 02:19:06 PM »
Hi Kephra,

Extremely helpful references. I thank you so much for your patience.
You are after all trying to teach a gardener how to do brain surgery! (metaphorically)

In anticipation of the malto turning up I'm reading as much as poss and have a question.
When we use these products the way we do (Malto, glucose, corn syrup, ........)
do they stay as Malto, glucose, corn syrup or do they become chemically different, inert even due to the
loss of electrons to the silver / gold.

I ask because it would seem that Malto, appears to be one to be avoided if one has diabetes due to the
high glycemic index. I hadn't thought about consequences like that if someone might be super intolerant
(bit like giving a peanut to someone with a nut allergy).

But, if it "gives up" being malto the second it reduces the silver or gold, then I guess there is not question (or worry) to have.
regarding our choice of reducer.




Offline kephra

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #5 on: December 20, 2016, 02:34:03 PM »
For one substance to be reduced (gain electrons), another substance must be oxidized (lose electrons)
So when using a glucose based reducing agent, a glucose molecule is oxidized to gluconic acid.
In the case of maltodextrin, only one of its glucose molecules becomes oxidized.
Since the body breaks down maltodextrin (as it does starches)  most of the molecules become simple glucose.

So yes, it would affect blood sugar.

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Offline Dean

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #6 on: December 20, 2016, 05:01:25 PM »
That's an interesting detail actually. but I guess that all of our reducing agents are similar being sugars.
I'm not sure that the amount would ever be an issue but I think now it would be wise to always let the
user know that there is glucose in there. Just in case.

Would the exception be cinnamon (if someone has issues with sugar at all?)

Offline kephra

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #7 on: December 20, 2016, 07:15:49 PM »
Cinnamon extract reduces by its phenol content, not glucose, so shouldn't upset any diabetic condition.
Also, don't forget that the numbers for maltodextrin are worst case.  In practice, you should be able to at least halve those calculated values because its highly unlikely that all of the molecules have 17 glucose molecules. 
Colloidal Silver is only a bargain if you make it yourself.

Offline Dean

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #8 on: December 20, 2016, 08:45:41 PM »
Thank you Kephra,

Great information as usual. Really appreciate it.

(Up to page 9 by the way but having to focus on work a little at the mo. I hope that's ok  8)

Offline kephra

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #9 on: December 20, 2016, 10:24:23 PM »
I want to correct a statement I made.  I have the constant wrong for silver by a factor of two.  I forgot this morning that 1 glucose molecule reduces 2 silver ions according to the formula:
Ag2O + C6H12O6  --> 2Ag + C6H12O7
So the formula constant should be 0.014 instead of 0.028 for silver.

Likewise for Gold, it should be 0.025 instead of 0.051

Sorry, I should not do these calcs off the top of my head.
« Last Edit: December 20, 2016, 10:31:28 PM by kephra »
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Offline Dean

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #10 on: December 21, 2016, 07:24:34 AM »
I see it thank you  ;D

And I see you corrected the figure in earlier post.
Will be trying the maltodextrin with 250ml silver batches as soon as I have some minutes to myself.

This dissolves "as is" (without dilution) correct ?

Offline PeterXXL

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #11 on: December 21, 2016, 11:46:24 AM »
Does this mean that a molecule of ALL "reducing sugars" (glucose, fructose, maltose, lactose, maltodextrin, etc) will reduce 2 silver ions?


I want to correct a statement I made.  I have the constant wrong for silver by a factor of two.  I forgot this morning that 1 glucose molecule reduces 2 silver ions according to the formula:
Ag2O + C6H12O6  --> 2Ag + C6H12O7
So the formula constant should be 0.014 instead of 0.028 for silver.

Likewise for Gold, it should be 0.025 instead of 0.051

Sorry, I should not do these calcs off the top of my head.

Offline kephra

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Re: Amount of maltodextrin for easiest Colloidal Gold chemical reduction
« Reply #12 on: December 21, 2016, 12:03:08 PM »
Yes it does for silver.  For gold it takes 3 molecules of glucose to reduce 2 gold ions.
« Last Edit: December 21, 2016, 12:50:43 PM by kephra »
Colloidal Silver is only a bargain if you make it yourself.