Plz Help Through Air weight & Water weight how to calculate

[sohailag]

Dear Brother,

Please Help me,

I have gold bar how I calculate actual pure gold in this bar.

Air weight & water weight.

Any brother give me any formula. I search the web found the specfic gravity gold test method but its not right. Its give the wrong result.

Please Help me in this .

Thanks in advance.

Best Regards,

Sohail

 

[butcher]

how could it give you wrong results? it is a simple formula, and proceedure.

is it that it does not give you the result's you want to see.

is it possible that your metal is not as heavy as you wish?

what are you trying to get the specific gravity of rock, metal bar?

maybe post your proceedure, with full details,and maybe we can see what your doing wrong or if there is a better way to test your material.

 

[sohailag]

This is the Formula I get from different sources;

Note the weight of the specimen (wet weight).
Now you have the dry weight and the wet weight of the specimen. Use the following calculation to calculate the amount of gold in your specimen.
- Subtract the wet weight from the dry weight and note the difference (D).

- Divide the dry weight by (D) and write down the result (SG) which is the specific gravity of the specimen.

- Subtract the specific gravity of quartz (2.60 to 2.65) from (SG) to get the result (R).
- The specific gravity of quartz varies from 2.60 to 2.65 because some quartz is more dense than others. This variation does not affect the result greatly, but it does show that the calculation, though close, will never be 100% accurate.

- Multiply the result (R) by (D) to get a result (R2).

- Divide (R2) by 25.97, which will give you the contained amount of gold in the specimen in ounces.

- Multiply the contained amount of gold in ounces by 31.103 to get the contained amount of gold in grams.

Below is an example using the formula above:

Specimen dry weight: 35.8 Grams 
Specimen wet weight: 27.9 Grams 
35.8 minus 27.9 equals 7.9 Difference (D) 
35.8 divided by 7.9 equals 4.53, the specific gravity of the specimen (SG) 
4.53 minus 2.63, the average specific gravity of quartz, equals 1.9 (R) 
1.9 (R) multiplied by 7.9 (D) equals 15.01 (R2) 
15.01 divided by the constant 25.97 equals 0.577 ounces of contained gold 
0.577 ounces multiplied by 31.103 (the number of grams in a troy ounce) equals 
17.94 grams of contained gold in the specimen.

I also make this formula in excel & upload it for users testing & correcting error mistake.

Please Help me My main purpose is this when I am melting mixed old products of gold of purities after making a shape of Bar before Refining I get the idea of much pure gold in this bar.

Thanks in advance.

Best Regards,

Sohail

 

[eeTHr]

Since you have gold bars, why would the specific gravity of quartz be needed?

 

[sohailag]

Dear brother,

I tell you again these gold bars are not pure I melt a old Jewellery
Of different purifies like in my country 21kt, 19kt, 18kt after melting
All together & make a shape of bar before refining how I calculate
How much pure gold in this mixed Jewellery melted bar.

I hope you understand.

Thanks for reply. Still waiting for solution.

Best regards,

Sohail.

 

[Palladium]

You just have to love this.

http://www.midstatesrecycling.com/karat_kalc

 

[eeTHr]

sohailag---

Since the formula you posted involves quartz in it, it appears to be for use on ore.

My point was that since you have karat gold, and not ore, including the figures for quartz may be where your error is coming from.

Getting the specific gravity for a bar is not that complicated.

There are two ways to do it.

1.) The specific gravity of a solid is the ratio of its weight in air to the difference between its weight in air and its weight immersed in water.

If you have a hanging spring scale, where you hold or hang the scale up, and it has a hook or recepticle to weigh the bar on, hanging from the bottom, then use this method.

2.) If you don't have a hanging spring scale, use the method for liquids, but determine the volume by the displacement method, below.

The volume of a solid with uneven surfaces is difficult to measure, so water can be used to measure it's volume, by displacement.

Either use a graduated container, or mark your own graduations, as follows. Put enough water into your empty container to easily cover the bar when it is placed into the container with the water later. Mark the water level on your container, then add a known amount of water, like 100 ml, and mark the new level on the container. Keep doing this until you have enough graduated marks that you can see will be more than the volume of the bar, as the water will rise that amount when you put the bar in.

Then remove the water you added, down to the first mark on the container.

If you have a container that is already graduated, put enough water in it to easily cover the bar, and come up to major graduation mark.

Set your bar on the bottom of the container with the water, and see where the new water level is. The difference between the water level without the bar, and the new level with the bar, equals the volume of the bar. Then go directly to step #3 in the above link, and substitute "bar" for "liquid," and do only steps #3 and #4.

 

[sohailag]

Dear Brother,

I get the formula from web but I mistake about quartz I realize.

You are right I check the links you send me. These links are useful but I am confused Sorry I am NEW in this please explain me I found this formula useful.

Specific gravity is a comparison of a substance's density to that of water. Water's density is 1 gram per cubic centimeter (1 g/cm[3]). The specific gravity of a substance is determined by dividing its density by that of water. Because specific gravity is determined by dividing two densities, the units cancel each other out, and the final value is given with just a number. Specific gravity can also be calculated by dividing the weight or mass of an object by the weight or apparent mass of the same object when it is submerged in water. Because each substance has a unique specific gravity, this test can determine the composition of an object. (I test this one)
Difficulty: Moderate
Instructions

Things You'll Need
Spring scale
Measuring cup
Water
◦ 1
Weigh your solid by hanging it from the spring scale.

◦ 2
Fill the measuring cup with enough water in which to submerge your object.

◦ 3
Hang the object from the spring scale and lower it into the water until it is completely submerged. Record the weight of the object when in water.

◦ 4
Divide the weight of the dry solid by the weight of the solid when it was submerged in water. The result is the specific gravity of the solid.


(I have the weight scale for dry & water weight)
Now for example.
I have a Dry weight 65.560
I have water Weight 61.800
I divided weight of the dry to weight of the air
65.560 / 61.800 = 1.060 Answer specific gravity of solid

Now What's Next can I divided 65.560 to 19.3 - 65.560 = 62.163 is pure gold I am right or wrong?

Please Help I am waiting Brother.

Thanks in advance.

Best Regards,

Sohail

 

[eeTHr]

sohailag---

Specific gravity is the ratio of the weight of a sample (in this case, your bar), to an equal volume of water.

The density ratio of gold:water is 19.32:1. This represents a Specific Gravity of 19.32.

If the specific gravity of your bar is only 1.06, then it is only 0.31 % of the density of gold. It would be a non-metallic substance, so I think your weights are wrong. It looks like you added an extra zero between the decimal point and the six.

The formula should go like this, using 1 Troy ounce of pure gold, as an example---

1.) Dry weight of sample is 31.10348 grams.
2.) Weight of sample suspended in water is 1.60991 grams (this is actually the weight of the displaced water).
3.) 31.10348 divided by 1.60991 = 19.32 Specific Gravity.

The technique of suspending the bar in water, gives the weight of the displaced water, not the weight of the bar. This allows for the division step to calculate the ratio of the weight of the bar to the weight of the equal (displaced) volume of water, and thus arrive at the Specific Gravity figure. The bar cannot touch the bottom or sides of the container.

You can also get the weight of the displaced water by suspending a bar into a container of water, on a regular scale, as shown here. This link also has calculators for figuring the amount of gold, when it is in combination with another mineral, but only if the other material in it is known.

If you want to determine for your own reference, approximately how much gold your bars contain, you could take a small sample, weigh it, then dissolve it and selectively precipitate just the gold, then weigh the resultant gold to figure a percentage by weight. The purity of the resultant gold would determine the accuracy of your calculations, and the extent of your refining techniques would determine how pure you could be getting the precipitated gold powder.

But to be most accurate you would need an assay.



Edited for correction---

If suspending from a scale, you get the weight of the bar, less the weight of the displaced water, so you need to subtract the "wet" weight from the "dry" weight, to get the weight of the equal volume of water which is displaced.

If you have the water container sitting on the scale, the difference in weight, with and without the bar suspended in it, is the weight of the equal volume of displaced water.

 

[eeTHr]

sohailag---

One more thing.

The standard for the water is 39 degrees F (4 degrees C) at 1 atmosphere of pressure, so water at that temperature and pressure would have a specific gravity of 1.

ref: Ways to Measure Specific Gravity.

Edit: As pointed out in samuel-a's link, below, the bar should be at the same temperature as the water, too.

 

[samuel-a]

http://www.goldrefiningforum.com/phpBB3/viewtopic.php...

 

[sohailag]

Dear Brother,

Sorry for the late reply I am out of my city to for office work.

Now I am Back.

I got it What you say. But hot to get displaced water weight.

I see video at you Tube about how to get water weight here is the link.

http://www.youtube.com/watch?v=GF4NqIYh6zg&feature=player_embedded

I also see the other place about this How they calculate the formula like this.

Weight in air - Weight in water = Weight difference

Weight in Air / Weight Difference = Answer is the Density of the object.

I get this this formula from here.

http://www.platinumguild.com/files/pdf/V3N6W_practical_applications.pdf

But the result is not right.

In your formula can you Explain Brother how I get displaced water weight.
Sorry to disturb you again & again but I am new Your Help is appreciate me to go in this field further.

Thanks in advance.

Best Regards,

Sohail

 

[eeTHr]

sohailog---

Due to the buoyancy of the bar, the weight measured when suspending it in water is the weight of the displaced volume of water. No further conversion is needed.

Thus, when you divide the weight of the bar, by the weight of the displaced water, you arrive at a ratio of the weight of the bar to the weight of an equal volume of water. This ratio is the specific gravity (which is, by definition, a ratio).

Pure gold will have a weight ratio, to an equal volume of water, of 19.32-to-1.

If your bar is close to that, and you're sure there is no tungston, or other heavier than gold metals, in it---then you have relatively pure gold.

If less than 19.32 specific gravity, then you would need to know that there is only one other metal in it, and what that metal is, in order to caluculate exactly what percentage of it is gold.

If the specific gravity is more than 19.32, then you have platinum, or other heavier than gold metal, in your bar. And again, if you know for sure that there is only one other metal present in it, and you know for sure what that other metal is, then you can calculate the percentages of gold and that other metal.

If you use the method of hanging the bar from a scale, be sure to either reset the scale to zero (tare) with the hanging device attached, or subtract it from the final weight when the bar is suspended in water. If you use the method of sitting the water container on a regular scale, be sure to subtract the weight of the container and water from the final weight with the bar suspended in the water.


If you are wondering why the suspended bar weight is the weight of the displaced water, it is because when you suspend the bar in the water, you are not adding the weight of the bar to the water, you are only adding the volume of the bar to the water. It is the same as if you just added a volume of water equal to the volume of the bar. You would get the same weight reading on the scale, whether the bar was gold or aluminum, as long as both bars were the exact same size.

The weight of the bar itself, only counts when you weigh it "dry."



Edit for correction---

If suspending from a scale, you get the weight of the bar, less the weight of the displaced water, so you need to subtract the "wet" weight from the "dry" weight, to get the weight of the equal volume of water which is displaced.

If you have the water container sitting on the scale, the difference in weight, with and without the bar suspended in it, is the weight of the equal volume of displaced water.

 

[sohailag]

Dear eeTHR,

I understand the buoyancy of the bar.

But I have the bar with mix mixed impurities I melted mix used gold jewellery of different karrat like 21kt, 19kt, 18kt these all jewellery are mixed with siver & copper mixing
now after melting these old gold jewellery with shape of bar you no how I get the water weight as you see in youtube video & I have Air weight now how I calculate the weight of pure
gold in this bar. My 2 friends have the formula when I get my melted bar to them they just get the air weight & water weight & give me the idea how much pure gold in this bar but they don't tell me the formula.

So your help appreciate me to read different formulas & research its increase my knowledge but I am at still same place at very first day. I hope you understand me.

Its very thankful to you guide me how I Calculate the things.

Thanks in advance,

Best Regards

Sohail
:idea: :roll: :?:


Best Regards,

Sohail

 

[goldsilverpro]

Dear Brother,

I get the formula from web but I mistake about quartz I realize.

You are right I check the links you send me. These links are useful but I am confused Sorry I am NEW in this please explain me I found this formula useful.

Specific gravity is a comparison of a substance's density to that of water. Water's density is 1 gram per cubic centimeter (1 g/cm[3]). The specific gravity of a substance is determined by dividing its density by that of water. Because specific gravity is determined by dividing two densities, the units cancel each other out, and the final value is given with just a number. Specific gravity can also be calculated by dividing the weight or mass of an object by the weight or apparent mass of the same object when it is submerged in water. Because each substance has a unique specific gravity, this test can determine the composition of an object. (I test this one)
Difficulty: Moderate
Instructions

Things You'll Need
Spring scale
Measuring cup
Water
◦ 1
Weigh your solid by hanging it from the spring scale.

◦ 2
Fill the measuring cup with enough water in which to submerge your object.

◦ 3
Hang the object from the spring scale and lower it into the water until it is completely submerged. Record the weight of the object when in water.

◦ 4
Divide the weight of the dry solid by the weight of the solid when it was submerged in water. The result is the specific gravity of the solid.


(I have the weight scale for dry & water weight)
Now for example.
I have a Dry weight 65.560
I have water Weight 61.800
I divided weight of the dry to weight of the air
65.560 / 61.800 = 1.060 Answer specific gravity of solid

Now What's Next can I divided 65.560 to 19.3 - 65.560 = 62.163 is pure gold I am right or wrong?

Please Help I am waiting Brother.

Thanks in advance.

Best Regards,

Sohail

In your example, the SG = (weight in air)/(weight in air - weight in water) = (65.560)/(65.560 - 61.800) = 65.560/3.76 = 17.4

Since the various percentages of the non-gold alloying ingredients (copper, silver, zinc - all of which contribute differently to the overall SG) in karat golds can vary considerably, depending on the exact color desired by the manufacturer, the SG method can, at best, only give an approximation of the gold content.

Also, to get decent accuracy when weighing, the scale you use should provide at least 3, and preferably 4, significant digits.

 

[eeTHr]

GSP is right.

I misstated part of my previous post, in that the weight is figured differently, depending on whether you are suspending the bar from a scale, or if the water container is sitting on a scale and you are suspending the bar into it.

If suspending from a scale, you get the weight of the bar, less the weight of the displaced water, so you need to subtract the "wet" weight from the "dry" weight, to get the weight of the equal volume of water which is displaced.

If you have the water container sitting on the scale, the difference in weight, with and without the bar suspended in it, is the weight of the equal volume of displaced water.

It sounds like your friends are using an estimated average of the specific gravities of the various possible alloys of added base metals. One reason for learning to refine is to make sure you are getting paid correctly.

 

[goldsilverpro]

sohailag,

Go to this link advertising the book I wrote and click on the 1st pdf attachment in the 1st post on the thread. At the bottom of page 41 you will see a chart of several karat gold alloys in various colors. Please note the variations in the percentages of the alloying metals and the resulting variations in the densities (specific gravities). For example, some typical alloy compositions of 18K yellow gold are: Copper - 5% to 15%; Silver - 10% to 20%; Zinc - 1% to 3%. The range of the densities of these variations is from 15.1 to 15.7, although they all contain 75% gold and all are yellow. Therefore, it is impossible to calculate the exact gold percentage from specific gravity measurements, unless you know the exact percentages of the alloying ingredients. However, if you knew these percentages, you would also know the gold percentage, by difference, and the measurement of the specific gravity would therefore be a moot point. Another thing that could produce error would be when the bar contains gold solder.

http://www.goldrefiningforum.com/phpBB3/viewtopic.php?f=84&t=5810

If you really want to know the gold content of the bar, either sample it correctly and run a triplicate fire assay or, refine it.

 

[sohailag]

Dear Gold & Silver Pro & eeTHR,

Sorry for late reply I just do the practical that's why I am late.

Your formula is right I get the accurate result. But for accuracy you are right I must need the balance with 3 or 4 decimal.

But one thing more I get the plain tap water. Its right to get plain tap water or get distolled water & one thing more for accurate water weight
I get any special bowl like plastic, metal or glass & big or small one an suggestions are appreciate.

Once Again very thank full to all brother who support me & other users. Thanks to team of Gold Refining forum & 2 my best supporting friends
Gold & Silver Pro and eeTHR.

Thanking you,

Best Regards,

Sohail

 

[goldsilverpro]

Dear Gold & Silver Pro & eeTHR,

Sorry for late reply I just do the practical that's why I am late.

Your formula is right I get the accurate result. But for accuracy you are right I must need the balance with 3 or 4 decimal.

But one thing more I get the plain tap water. Its right to get plain tap water or get distolled water & one thing more for accurate water weight
I get any special bowl like plastic, metal or glass & big or small one an suggestions are appreciate.

Once Again very thank full to all brother who support me & other users. Thanks to team of Gold Refining forum & 2 my best supporting friends
Gold & Silver Pro and eeTHR.



Thanking you,

Best Regards,

Sohail

For accurate work, tap water should not be used. Also, the temperature of the distilled water you use can affect its specific gravity and, for greatest accuracy, this would enter into your calculations. However, since the alloys can vary, the accuracy will be unreliable, even if your procedure is perfect. There are much better methods available. A simple scratch/acid test on a touchstone is more reliable.

 

[danib]

Dear All,
I have quite a similar problem that you answered so many times But Still I find no other option but asking for your help. I have been asked to develop a software to test gold purity. We don't know anything about possible metals and their combination alloyed with that particular gold bar. But still they are using this software for testing gold purity commercially in Gold test laboratories. I am sending you few calculations.
First Calculation:
First Weight (Dry Weight) 7.808gm
Second Weight(Suspended spring scale in water) 7.384gm
Total Impurity 0.3368gm
Total Pure Gold 7.4712gm

Second Calculation:
First Weight (Dry Weight) 10.3340gm
Second Weight(Suspended spring scale in water) 9.783gm
Total Impurity 0.3365gm
Total Pure Gold 9.9975gm
Carats: 23.2185K

Any help will be greatly appreciated.

Regards
Danib