Inquartation

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johnpotts2nd

Member
Joined
Feb 12, 2020
Messages
7
How much silver do I need to add to inquart 331.39 grams of 14K gold jewelry?


Sent from my iPhone using Tapatalk
 
You have 193,31 gr gold and 138.08 gr of silver and need to add 596,5gr to come to 6K
I have an excel sheet for that.
It keeps in account a factor of 0,95 because of karat gold being 1/2 karat lower than the stamp. According a video of streetips, if remember correct.
Its in dutch, but i'll translate and share it here.
Martijn.
 
Ok, Here it is, just translated it to English.
I sure hope some experienced members (kadriver maybe : wink:) will test it and correct me if i'm wrong.

I made this sheet just two days ago because a good friend of ours gave me a broken 14K necklace of 1.7 gram to refine because she liked that I was so excited over my new 10 gram button. That data is still in the sheet :)

EDITED: forgot to substract the weight of the gold present from the total target weight. It gave too much silver to add this way. This file is a right one.

It was my first attempt at inquarting jewelry!! went well, 1 gram of gold, and 0,7 gram of silver, what the sheet predicted.
I've entered your data too and the numbers are there.

Fill in the weight and karat of your gold, multiple different karat batches can be tossed on it, so you can throw it all in one pile and inquart, after weighing and entering the data.
The sheet calculates the weight of the gold and silver, their value, and the amount of silver needed to add to reach 6K.
I've taken into account the 0.95 factor Streetips talks about in his inquarting video, compensating for low carat gold and therefore more silver already being present than calculated. If you want to use the full 100% factor for silver to get to 6K or slightly below, change the orange field to 1 instead of 0.95, and use points, not comma's to enter numbers like 331.39
If more entries are made, the total is added up on the bottom.

Keep in mind that the price calculated is still based upon 50 euro's per gram of gold, please enter your currency and price per gram in the yellow fields. The rest of the data will change with it.
You can also change the word in the yellow fields from gram to ounces or whatever you want, but then you will have to recalculate the amount of HNO3 needed to your unit and fill it in in the green field. The rest of the results will be fine.

As aid, the sheet also calculates the amount of Nitric needed to consume the total amount of Silver.
The number I got from a thread here on GFR, and it was 1.44ml of HN03 at 69-70% needed to dissolve one gram of silver. That amounts to 1.008 gram of pure HNO3 (theoretically) and divided by the concentration of your nitric should give the right number. if you type in 70%, it gives 1.44ml/gr.

Type in the concentration of your HNO3 and it will calculate the right amount. It is a weight based calculation.
My assumption here is: 70% HNO3 is 70 gram of pure nitric in 100ml solution.
Please correct me if i'm wrong here!!

Then at last, the sheet calculates your efficiency, or the real percentage of the expected gold that was actually present. Which ever explanation you feel more comfortable with... :lol: :lol:
Just enter the amount you recovered in the yellow field(s) (and refined of course, it might show a higher number than 100 :shock: :wink: )

The sheet takes no responsibility for incorrect application of chemical processes, just providing the raw data.

Feel free to share, alter, or whatever with anyone.
Hope many of you can put this to good use. Have fun! Feedback is much appreciated.

The total amount of nitric may not be needed, or might not be quite enough. Add a part of it to start with and add slowly when the reaction slows.
Please add +\-50% nitric acid in small amounts and stay safe!

I was thinking of adding an AR calculator sheet. Need to look up some data first.

Martijn.
edit: The file in this post had faults in it. So here is a simplified version, but these calculations are correct. View attachment Inquarting xx-K calculator.xlsx
 
Thanks for sharing your work with everyone! I do have some comments:

In cell D2 your label is "Weight of silver". It assumes that all the non-gold weight is silver. It would be better labeled "Weight of Base Metals".

I see we can enter a value for low karat gold in cell A24, but you don't use it in calculating the gold values in column E. I see you use it in column H to calculate the amount of silver needed, but the calculation is wrong because it uses A24. Put in a value of 2 grams of 12K gold. 12K is half gold and half base metals. You show that correctly in columns C and D, as well as the target weight in column I, but because you use A24 in the calculation in column H your result of adding 1.90 grams is incorrect.

There are several problems with column J which calculates the amount of nitric needed to dissolve the base metals. First, as mentioned above, not all the non-gold weight will be silver. Copper and other base metals, which usually represent a greater percentage of the base metals in a piece of jewelry than silver, need more nitric than silver. So you could calculate the nitric used for the added silver, and the nitric needed for the original base metals, then add them together.

I'm not sure I understand your calculations in A21 and B21. Maybe it's just too late for my brain to work.

I think I'm confusing myself now, so I'll hope others comment as well. Hope my comments help.

Dave
 
Thanks for your reply Dave, I've changed somethings in the spreadsheet.
The 'expected gold value' in column is now corrected with the factor in A24. That value is now used to calculate the expected amount of base metals present, which are both subtracted from 'four times the expected gold weight'. This should give a true 6K, if the factor of 95% is correct. Am I right there?
You can still set the factor to 1.

The mention of only silver as a base metal, was kind of deliberate because base metals use up more nitric, and I wanted to give an estimate to what is needed, from another thread i got different values; General Reaction List
That's why you can set your own ml/gram basemetal to the desired number.
You could note the actual amounts as you go through different batches, and then adjust your number to your process and habits which will differ from person to person, giving different efficiency of nitric use.

The factor of 1.44 ml HNO3 was per gram of base metal from a post i can not find anymore, and was probably based on a higher silver content of the base metals present or based on a closed reaction vessel, reusing the nitric to some extent. Discussion on the actual number has been going on a long time, i know, and opinions differ. One thing is sure: never throw in everything at once. So use the calculation and adjust the number if needed. Love to know what your numbers are, (Without getting in to a fight over who is right 😉)

The calculations in A21 and B21 are based on these assumptions: (don't shoot me) 1.44ml @ 70% gives 70 gram of "pure' nitric per 100ml acid, which would result in (1.44*0.7)=1.008 gram of pure nitric per gram base metals, but only to calculate with.
My 53% nitric has 53 gram of nitric per 100ml... so the 1.008 gram divided by 0.53 gives the number in B21.
Probably a way to simple approach, so please correct me if i'm wrong.

If these calculations or numbers are off, let me know what it should be.

Filling the numbers of actual return of each batch, could provide some nice statistics to improve and fine tune your processes. And it could show the actual percentage of silver in the base metals of your different batches. if you weigh the cemented silver and know how much went in. For that you will have to inquart per batch.

Here is the new spreadsheet; if we're done discussing and some level of consensus has been reached, I will replace the first file with the final version.
Again thanks for the feedback.

View attachment Inquarting xx-K calculator.xlsx

Martijn.

and i have to apologize to johnpotts2nd, the amount of silver i said in my first reply was a bit toooo much. It went down from 596,5gr to 441.85, if you leave the factor at 1, and if you set the factor to 0.95, it will be: 403.19gr. Provided that this calculation is correct...
I hope you haven't inquarted yet :shock: :oops: :oops: that would waste some nitric.

To know how much gold you have in karat gold: the weight times the karat divided by 24.
The silver to add: Subtract that from the actual weight and you have your base metals, four times the weight of the gold is your target weight, and you already have your original weight, so subtract that from the target weight to know how much to add.
 
Martijn said:
The calculations in A21 and B21 are based on these assumptions: (don't shoot me) 1.44ml @ 70% gives 70 gram of "pure' nitric per 100ml acid, which would result in (1.44*0.7)=1.008 gram of pure nitric per gram base metals, but only to calculate with.
My 53% nitric has 53 gram of nitric per 100ml... so the 1.008 gram divided by 0.53 gives the number in B21.
Probably a way to simple approach, so please correct me if i'm wrong.

If these calculations or numbers are off, let me know what it should be.

I knew I forgot something in my last post.

The nitric percentage is calculated on a weight to weight basis. So 70% nitric acid would contain 70 grams of nitric per 100 grams of solution. The density of concentrated nitric (68 to 70%) is about 1.42 grams per ml. So 100 ml of concentrated nitric would weigh about 142 grams. 70% of that would be about 99.4 grams, not 70 grams.

Dave
 
Martijn.
If I'm not mistaken this is the post you mentioned and the number is wrong,
it is supposed to be 1.22ml 70% pr gram.

It was a post from GSP, may he rest in peace
https://goldrefiningforum.com/phpBB3/viewtopic.php?t=18598#p187310

Im playing a bit with the calculator, it seemed a bit disorganized.
If I'll come up with something, I'll PM you a copy and its up to you if you will share it.
 
Great. Looking forward to your input. I just put this together in a couple of hours.
One can always put in their own number. As long as the rest of the calculations are correct. Which they are now, I'm sure.
Take the info you need from it. Hide columns or remove. Play around with it and share it here.

@johnpotts2nd: didn't want to hijack your thread.
Let us know how you are doing.
Martijn.
 
Goldenscrap.com has an inquartation calculator you can put your values into and choose the karat type and it will give you the amount of silver you need to get it to 6k.

I always double check my written math with that calculator then multiply the silver amount by .95 since noticing that alot, not all, karat jewelry scrap can run a little lower than what's claimed on the stamp, if it even has a stamp after testing.
 
DylanDownright84. Can you send me the link to the gold inquartation calculator?

Appreciate it


Sent from my iPhone using Tapatalk
 
I think he's talking about this one.

Gold Inquarting Calculator

Just compared multiple carats at multiple weights on the site and the sheet (with the correction on 1), and the one from goldenscrap is 0.001 % lower on most.
That's less than most scales have for accuracy.
So the sheet is fine, the site is fine witch ever you prefer.
I'll replace the first file in this post.

Martijn.
 
You have 193,31 gr gold and 138.08 gr of silver and need to add 596,5gr to come to 6K
I have an excel sheet for that.
It keeps in account a factor of 0,95 because of karat gold being 1/2 karat lower than the stamp. According a video of streetips, if remember correct.
Its in dutch, but i'll translate and share it here.
Martijn.
Hello Martijn!
I learned from an old school jeweler a very exact formula to calculate the amount of silver/copper to add to get 6k gold.

1 Determine the karats of the gold to be quartered.
2 Subtract 6 from that number, the number 6 is always a constant.
3 The result of that subtraction is multiplied by the weight of your gold to be quartered.
4 The result of that multiplication is divided by 6.

So following the example of johnpotts2nd, we do the following:
14 - 6 = 8
8 * 331.39 = 2651.12
2651.12 / 6 = 441.85

441.85 are the grams of silver or copper you need to add to bring that 14k gold to 6k gold.

You can confirm that after melting you will have 773.24 grams of 6k gold, or what is the same, 193.31 grams of pure gold.
 
Hello Martijn!
I learned from an old school jeweler a very exact formula to calculate the amount of silver/copper to add to get 6k gold.

1 Determine the karats of the gold to be quartered.
2 Subtract 6 from that number, the number 6 is always a constant.
3 The result of that subtraction is multiplied by the weight of your gold to be quartered.
4 The result of that multiplication is divided by 6.

So following the example of johnpotts2nd, we do the following:
14 - 6 = 8
8 * 331.39 = 2651.12
2651.12 / 6 = 441.85

441.85 are the grams of silver or copper you need to add to bring that 14k gold to 6k gold.

You can confirm that after melting you will have 773.24 grams of 6k gold, or what is the same, 193.31 grams of pure gold.
As additional information, this formula works with any karat, even decimals. You can work perfectly by melting all the jewels you have and scanning by XRF the button or bar to know exactly its pure gold content. The % of gold or "law" is multiplied by 24, to obtain the karat, with decimals.
 
Hello Martijn!
I learned from an old school jeweler a very exact formula to calculate the amount of silver/copper to add to get 6k gold.

1 Determine the karats of the gold to be quartered.
2 Subtract 6 from that number, the number 6 is always a constant.
3 The result of that subtraction is multiplied by the weight of your gold to be quartered.
4 The result of that multiplication is divided by 6.

So following the example of johnpotts2nd, we do the following:
14 - 6 = 8
8 * 331.39 = 2651.12
2651.12 / 6 = 441.85

441.85 are the grams of silver or copper you need to add to bring that 14k gold to 6k gold.

You can confirm that after melting you will have 773.24 grams of 6k gold, or what is the same, 193.31 grams of pure gold.
That is a nice how to-step plan to calculate it. Just like my spread sheet or any calculator for that matter, but just like with chemical reactions, I like to understand what I'm doing so I like to logically approach a mathematical equation. My first reply in this thread had a mistake. I forgot what was already in the alloy. i made the spread sheet for those large volumes of different carat jewelry I was expecting to magically arrive at my doorstep. :rolleyes:

It's a ratio by weight, 24K equals 100%Au.
6K, or 25%Au is what you need to have a nitric-digestible button and still leave a gold sponge in stead of powder.
Some carat gold has up to 5% less gold (meaning more silver) in it than it should, so correct that if you want to. if you don't, you'll have a bit lower carat than 6, still no big problem. It can leave some fine powder after the nitric bath.

The rest is math.
I believe our numbers are the same.
 
I think he's talking about this one.

Gold Inquarting Calculator

Just compared multiple carats at multiple weights on the site and the sheet (with the correction on 1), and the one from goldenscrap is 0.001 % lower on most.
That's less than most scales have for accuracy.
So the sheet is fine, the site is fine witch ever you prefer.
I'll replace the first file in this post.

Martijn.
So karat gold inquarting will be the same as gf items too?
 
And yes I've used that website before. That's reason I asked. It has a calculator for " karat gold". Is this calculator the same for gf " gold filled" items?
As said in previous post, yes after the base metals are removed.
 

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