Gold powder to Metallic form

[autumnwillow]

Just maybe somebody has an answer already so I won't have to sacrifice a fresh crucible.

Does one gram of gold powder yields to the same amount in metallic form? eg. 1 gold powder = 1 g when melted? Assuming no losses in the crucible?

Silver crystals yields the same for me.

 

[Lou]

Depends on the precision of your scale but in our imperfect world, the answer is no.

 

[anachronism]

Depends on the precision of your scale but in our imperfect world, the answer is no.

Am I missing something here Lou because if I take 100g of perfectly dry powder and melt it I get extremely close to 100g of bar. When I say extremely close I mean 99.9+ g

 

[upcyclist]

Agreed--in that perfect imaginary world where all substances are pure and all techniques are perfect, 1 ozt of gold powder = 1 ozt of gold button/bar/whatever. So your theoretical hunch is correct, just like it was with silver.

Theoretically, yes, they're the same. Practically, they're close but not exactly the same. Yes, some could be trapped in the flux. Some could not actually be gold, and something else happens when you apply heat (moisture, etc.). Some might blow away or go up in smoke. Some might get dropped or lost with poor transfer to the crucible.

If your scale were to have no decimal points, you'll probably have 100g left if that's what you started with. If it has 4 decimal points, you'll see how much you lost to that degree of accuracy.

In the end, we're having one of those discussions where some of us are talking about a theoretic or pure scientific point of view, others are talking from a practitioner's perspective, and others are somewhere inbetween.

Edited to clarify.

 

[Lou]

"In theory there is no difference between practice in theory, but in practice there is"

Basically, you will have melt losses, it's part of the fundamental process yield

what you get / what you actually had.

Say someone sent me 100 fine ounces contained in 14 K scrap. While I might get 99.995 ounces troy from it, some will still be lost...be it suspended fines in nitric from part, or gold residuals that won't wash off silver chloride that ultimately report in my silver cell, or volatilization of the gold during the melt that collects on the walls of the crucible, and on the list goes.

Loss mechanisms. Lot of my job as a consultant in this stuff is helping people understand and mitigate them.

Look at all the people doing stone removals with ammonia or sodium thiosulfate to remove AgCl...they're throwing some of the baby (gold, either suspended or dissolved) out with the bath water (solublilzed silver chloride complex).


and upcyclist...it's a degree of precision, how many zero's you have on the balance (readability really), not necessarily accuracy. You may have very precise measurements with horrible accuracy and very accurate measurements with horrible precision. In fact, they're probably related in some fashion

https://en.wikipedia.org/wiki/Accuracy_and_precision

 

[Barren Realms 007]

The end result also depends largely on the patience that one has in processing the material and paying attention to detail during processing to aquire to accurate final results.

 

[4metals]

Even at the mint where they make .9999 fine kilo bars, there is an accounting figure called the "give back". That give back is a pinch of gold powder to make up for any melt loss.

 

[Lou]

Yep. Never had anyone, ahem, "complain" about weights being heavy, come in light and get ready to reap the whirlwind with some customers.

 

[g_axelsson]

Every surface in this environment we live in has a thin cover of water molecules, even when dry. It can be minimized by drying at elevated temperature but never totally removed. Fine gold powder has a much larger surface per gram than silver crystals so the effect is more pronounced for fine powder.

When melting powder the water on the surface is boiled off, which can lead to a measurable weight loss.

High vacuum systems that I have worked with is running into this effect when the pressure goes down below a certain level and when a system has been opened to the atmosphere it takes a longer time to pump down as there are a lot of water vapors to pump out that comes from the walls. For ultra vacuum systems the apparatus is usually baked at elevated temperature or flooded with UV-light to knock the water from the surface.

Göran

 

[anachronism]

Ok so we're back theorycrafting guys? Yes of course there's going to be a tiny loss but it can be really tiny if as Barren pointed out you take care in processing your material to the point of pure powder that's bone dry.

I accept that if you don't dry properly then your "melt losses" will be greater. That's a given, but when we delve into the realms of water molecules coating the surface of properly dried gold powder then it moves into the realms of daftness (sorry Goran old friend.) Incidentally the purer your gold the larger the particles hence the less surface area.

Can we agree that from a practical perspective - given that the gold is pure, and has been dried properly the melt losses are tiny as a percentage of the gold that the average hobby refiner produces?

 

[upcyclist]

Hey, he asked

Sent from my SAMSUNG-SM-G891A using Tapatalk

 

[autumnwillow]

Giving out theories is good, its the whole concept of having a discussion.

Let's sum it up.

As long as you are able to create a perfect environment avoiding crucible losses, flue losses, flux losses, etc. Then, the answer is yes.
But in reality you cannot create that perfect environment so the answer is no.

The reason I asked this question is because I am about to do another experiment which involves dissolving gold and I'd rather not go thru the melting process and rolling it thin because what my experiment requires is only a large surface area.

Thank you guys!

 

[goldsilverpro]

Hello, Metallic pigments are utilized in all state-of-the-art printing processes in order to impart high quality gold or silver metallic effects.
Thank you

Actually, the initial question was faulty. Gold powder is already in metallic form.