help with simple gold Buttons Assaying

[samuel-a]

well, i'm using water and a scale to measure the volume of thing (like Archimedes )
but i'm not sure about the last calculation step.

so, i have a hypothetical 1 cubic centimeter gold button.
if it was 100% pure gold, it would have weight 19.32 grams.

let's say my button weight exactly 19 grams, meaning i have a 0.32 grams lighter then it should be.

with this data, how do i calculate the weight of the gold and the weight of unknown contaminates, or in other words, the purity of the button?



now, i have a guess but i'm not sure, and i can't find confirmation on google.
it goes like this:

i find the general ratio from early findings
0.32g / 19.32g = 0.0165
or
98.35% ( = 100 - 1.65 )

so it mean the button contain 18.6865 grams of gold (98.35% of 19 g) and 0.3135 grams of other elements.

Does it correct, or i'm missing somthing?
or sould i have to know the other elements content?

sorry for the dumb question, but i never had any formal education on physics & chemistry so i'm learning in the process.

Thanks
SAMUEL

 

[goldsilverpro]

The math isn't that simple. To predict the purity of the gold by measuring the density, you need to know what other metals are present and in what proportions. For example, the density of an alloy of 95Au/5Cu would be lower than an alloy of 95Au/5Ag, since the density of Cu is lower than the density of Ag.

 

[chemist]

Samuel -- I don't want to discourage you, but you have chosen a difficult task. It will probably be simpler to send a sample to GSP for assay.
If you want to proceed, it is going to be difficult. Theoretically it can be done, but in practice the errors will pile up on you and make your final values dubious.
Getting the weight is easy, it's the volume measurement that is difficult.
How are you measuring the volume of the gold sample? Measuring volume very precisely is very difficult for small volume samples. If you have something as large as a few Troy ounces, you may be able to get an accurate volume with a small graduated cylinder, but smaller volumes will require the use of a laboratory volumetric pipette or something similar.
The next problem will be the contaminants. What are they? If you know that your sample is a simple mixture of copper and gold, then we can calculate the percentage of copper fairly easily once you know the weight and the volume. But, if the contaminants are unknown or the percentages are unknown, then your calculation will probably be way off.
Good luck!

 

[samuel-a]

thank you guys

i assumed that the content of the other element will be the main problem...

well... better let professional to do this...


anyway, about volume mensuration it's actually not that complicated, here is a sketch :

1- placing a cup with distilled water on a scale.
2 - placing a mounting device to hang a plate inside the water.
3 - zeroing the scale.
4 - placing the nugget on the plate
5 - the change in grams is the volume of the object in cubic centimeters

 

[Noxx]

The specific gravity method will only work for assaying if gold is only alloyed with only one other element. Accuracy is generally poor, ranging from 0.5% to 2%.

 

[goldsilverpro]

Samuel,

I may be dense, but I can't see your setup working. The method is to isolate the water and container from the scale. Here's the proper setup using an Ohaus Cent-O-Gram balance, which has a built-in platform for doing this. Almost any other scale can be jury-rigged to do this, with a little ingenuity.
http://www.lusitaniamedal.com/specific_gravity.htm

You can also get fairly good measurements by using a container with a small overflow tube near the top. You fill it to the top with water until it overflows. You then hang the object in the water and collect the overflow water and weigh it.

In my book, there's a chapter on the ball-park evaluation of karat gold using specific gravity measurements. In the attachment below, one of the page samples shows the setup and a chart for the different karats. I don't know why this pdf is so jerky.

 

[samuel-a]

GSP, i think your missing somthing, look,
once i put the the container (filled with water) and hanger on the scale and zeroing, it does not matter to isolate the components, cous, when inserting the nugget, it will repulse an equal amount of water to the nugget's volume, that will represented by the change of weight in grams.

try it on a known dimensions object, it's working wonderfully

 

[goldsilverpro]

Samuel

With further thought, what you've stumbled onto is very interesting. You may have greatly simplified the measurement of specific gravity. I do think you've come to the wrong conclusions, though.

I'm not at work, so I can't measure this on a good balance. I'll try doing this later on. In the meantime, I tried it on a cheap kitchen scale that measures in ounces. I found a steel ax head that weighs 15.3 av.oz. (434 grams), in air, on this scale. I hung it by a string in water and the weight added to the scale was about 2 av.oz. (56.8 grams). If you divide 434 by 56.8, you get 7.6, which is a very good possibility for the density of steel. According to this site, the density of steel is between 7.47 - 8.03, depending on the alloy.
http://hypertextbook.com/facts/2004/KarenSutherland.shtml

I've been trying to think this out. As a thought experiment, if I hung my ax head on a spring scale and suspended it in the water while, at the same time, weighed the tared (zeroed) water and container with the ax-head hanging in it, I'm making 2 different weighings and here's what would happen. Assuming the density of the ax-head is 7.6 and it's weight in air is 15.3 oz, the spring scale would read 13.3 oz, as calculated from the standard density formula of Density = (weight in air) divided by (weight in air - weight in water). This 13.3 oz would be the weight of the ax-head in water. The 2.0 oz weighing would be the difference between the weight in air and the weight in water. The weight in air of 15.3 oz equals 13.3 + 2.0, the weight in water plus the weight lost by the buoyancy of the water. Makes sense to me.

In other words, it seems that the density can be simply measured by making a weighing using Samuel's setup and dividing this weight into the weight in air. I think his apparatus directly measures the weight in air minus the weight in water, without having to use the awkward standard density measuring setup.

At work, I'll do this more accurately. I'm excited.

Although congratulations may be premature - great job, Samuel!

We need some more brain-power on this.

Chris

 

[samuel-a]

coming from you, it mean a lot to me.

thank you for having interest in my idea. i just used simple logic to masure 'simply' the volume and apply it to density of an object.

i would love to hear your finall conclusion after experimenting at your work place.

 

[goldsilverpro]

Samuel,

I didn't think so, at first, but you are exactly right. It is the volume you're directly measuring in your setup. In the experiment I did, the ax-head weighed 434 grams. The volume is therefore 434/7.6 = 57 cc. The increase of weight with the ax-head was 56.8 grams (57 grams, rounded off). One cc of water weighs 1 gram. The increased weight with the ax-head in the water, therefore, equals the weight of the water displaced, in grams, which equals the volume of the ax-head, in cc.

You probably don't need the mounting stand. For most all but small things, you could probably just tie a very thin thread (or, polyester fishing line) around the object and hang it in the water by holding it - that's what I did with the ax-head. Of course, with the little object tray, like in your drawing, the tray is tared along with the container and water. This would be more accurate than a thread, since the thread could contribute to the weight. I would suggest holes or gaps in the bottom of the tray to allow drainage and prevent bubbles from being trapped between the object and the tray.

Like the standard density measurements, the water and the object should be at the same temperature - I think the ideal temperature is 39.2F (4C). For greater accuracy at other temperatures, there is a formula and charts for making minor corrections based on the actual temperature of the water. You need to tap the object with a rod to get any bubbles off of it - a drop of dishwashing soap also helps to prevent the bubbles from sticking to the object. You can't have any cavities in the object that won't be completely filled with water. And, of course, the object can't touch the container.

I have made density measurements all my life, using all sorts of methods, and I have never heard of or read of anyone using this method. I now have no doubt it will work. It's all in the math.

Chris

 

[Palladium]

See, you can teach an old dog new tricks.
I haven't seen Chris this excited about something in awhile.
That man has fun with a calculator.

 

[samuel-a]

GSP, i'm happy i could contribute somthing to this fine forum and people here.

i thought this a bit more deeply, it's a little flatulent but.... meybe this can be developt in to some new ball park measurement.
meaning, how far am i, in percentage from the desired density (in this case of 19.32 gold) of my final product.

this can be applied on any other element naturally .

for example, which is very common here:

3.4 grams gold button came from fingers.
then we know that our perfect volume for this button will be exactly 0.17598343685300207039337474120083 cubic centimeters, which will represent a density of 19.32 for a 3.4g button

now lets say that the nugget's actual volume is 0.178 cc
this lead us to understanding that the nugget density is 19.1011 (3.4g / 0.178) meaning, the nugget is far just 1.13% ( 1 - [actual SG 19.1011/ gold SG 19.32] ) from the desired density, and this percentage should displayed negatively.

a negative figure will indicate a low SG contaminates like copper,lead,tin, iron etc'... (not knowing their actual weight within the nugget of course)
and a positive figure will indicate a high SG contaminates platinum, osmium, iridium, tungsten

meaning that our nugget i pretty pure and that % figure will give us an estimation (a pretty good one) of how pure is our product so that we don't get fulled by the buyer that is assying our nugget.

we can call this masurmant - SPS (samuel's purity scale) :mrgreen:

anyway... this meant to be just a ball park process for the un pro's out there, and will not reaplace the professional assaying.

Edited: according to GSP's figurs which of course is right

 

[g_axelsson]

I did the same trick two months ago in a second hand store when buying a poorly hallmarked golden chain.

Pocket scale (up to 300g 0.1 g precision) with a small glass of water. Tared the scale to show zero with the glass on.
Then I submerged the chain in the water hanging in a thread and I could read the volume in cubic centimeters right off the scale. The tricky part was to get the chain curled up in a small ball so nothing would touch the bottom.
... yeah, and don't forget to weigh the gold first before it gets wet.

When calculating the carat I assumed the alloy was mostly gold and copper as it was red gold. The result was close enough to 18k that I would believe in the only stamp on it saying 750.
I bought the chain and sold it later on with a nice profit.

I've used this method before to measure density of rocks for the last ten years but this was the first time I used it on gold.

There are a few sources of errors in the calculation of density this way.
1. error and precision of the scale
2. the density of the water is changing with temperature
3. air bubbles could stick to your object
4. surface tension on the wire
5. volume of your wire used to suspend the object
I'm not going to do the calculation on how much each factor contributes, I leave that as an exercise to someone else.

:lol:

/Göran

 

[goldsilverpro]

Samuel,

You're a decimal place off.

The volume of the 3.4 grams of pure gold should actually be stated as .18 cc, rather than 1.759834-------. Same with the 1.78 - it would also be .18. The reason for this is that the number of digits (significant figures) in the answer can't exceed the lowest number of significant figures in any of the measured numbers used to get the answer. Since the 3.4 only contains 2 significant figures, you would have to round the answer off to 2 significant figures. Therefore, the correct answer would be .18, instead of .176, or the long string of digits you posted. Had you instead used a 2 decimal point scale and the gold weighed 3.40 grams (3 significant figures), you could state that the volume was .176 cc (3 significant figures). I know it is intuitive to think that the long string of digits you gave was quite accurate. However, in this case, .18 is more accurate. Modern calculators, with their long string of digits, can get us in trouble from a technical standpoint. Using the correct number of significant figures keeps you honest.

g_axelsson,

There are a few sources of errors in the calculation of density this way.
1. error and precision of the scale
2. the density of the water is changing with temperature
3. air bubbles could stick to your object
4. surface tension on the wire
5. volume of your wire used to suspend the object

All of these are also sources of error when using the standard method of weighing the object in air and then in water. About the only difference I can think of is that you should tare the system (or make a new measurement) immediately before adding the object, due to the continuing evaporation of the water. In the standard method, evaporation wouldn't affect the result.

Searching further, I did find a couple of references that state that, when suspending an object in water, the added weight equals the volume of the object. I can't believe I never considered this. It's really pretty simple. It seems like the more you know (or think you know), the harder it is to see the forest for the trees. On this forum, being humble and open-minded and admitting that you don't know it all are virtues. Without them, you'll definitely end up eating a lot of crow. Been there. Done that. And, like the saying goes, got the T-shirt.



In case anyone is interested, I just posted this on another forum: The question concerned the specific gravity of a Mexican 50 peso gold bullion coin - 90% gold and 10% copper.

The correct theoretical formula for predicting the specific gravity of a 2 metal alloy is given in this pdf (page 40 - middle column).
http://www.platinumguild.com/files/pdf/V3N6W_practical_applications.pdf

S.G. = 1/((fraction of A/S.G of A) + (fraction of B/S.G.of B))

For more than 2 metals, the denominator is extended to include them. For example, the formula for a 3 metal alloy is thus (I am using Fr. as an abbreviation for fraction):

S.G. = 1/((Fr. of A/S.G. of A) + (Fr. of B/S.G. of B) + (Fr. of C/S.G. of C))

In the case of a 90/10, Au/Cu alloy (The S.G.s of the 2 metals were taken from the CRC Handbook of Chemistry and Physics):

S.G. = 1/(( .90/19.3) + (.10/8.92)) = 1/(.0466 + .0112) = 1/.0578 = 17.3

Chris

 

[Palladium]

Hydrostatic weighing.