HEy guys....saw the posting on the pics thread about a quartz cobbler, and that based on SG deduced a few ounces of gold in that stone. Help me out here....how did you calculate that. I have scarce yellow quartz 'sand' that settles with my black sand, and I was wondering if it did so because gold flour was adhered to the quartz. Would be nice to find a stone like that.
Speciffic gravity of quartz???
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Jim in Idaho
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Solid quartz is about 2.6. Magnetite is 3.2 - 5.0. Hematite is about 5.0- 5.2. So, the black sand is roughly 50-100% heavier than the quartz.
Jim
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Per the chart I use, the specific gravity of quartz sand is 1.2
Solid quartz, as stated by Jim in Idaho above, has a specific gravity of 2.64 and lump quartz has a specific gravity of 1.55
Here is a very useful chart showing the specific gravity of many items - bookmark it: Specific Gravity Weights Of Materials from READE
Solid quartz, as stated by Jim in Idaho above, has a specific gravity of 2.64 and lump quartz has a specific gravity of 1.55
Here is a very useful chart showing the specific gravity of many items - bookmark it: Specific Gravity Weights Of Materials from READE
Well the some of the quartz I found in and settled out of my black sand is heavy. It has a tan color. I would say about 6-7. It is consistent size. Tiny. Minus 70. Just what I have found here in arkanss. In three river/ creeks.
http://www.reade.com/Particle_Briefings/spec_gra.html
http://www.reade.com/Particle_Briefings/spec_gra.html
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Jim in Idaho
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Jim in Idaho
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I think those guys at that link made a typo. I've never seen quartz listed above 2.6 or 2.8 at the highest.
Jim
Jim in Idaho
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it's possible that tan material is some form of silver, too, though I've never seen metal "coat" quartz, like paint...usually the metal is in veins, or flakes in the quartz. You might look at the grains with a microsope to see what's going on.
Jim
You can derive the equation you seek from understanding specific gravity. I'll give it a go.
Specific Gravity is defined as the ratio of the density of the material to the density of water. To express it mathematically:
SG = p(solid)/p(water); where p is as close as I can get to the Greek rho which is the symbol for density.
Density, p, is the ratio of the mass of the solid to its volume, or:
p(solid) = m(solid)/v(solid)
From this, the formula for SG becomes,
SG = [m(solid) * v(water)] / [v(solid) * m(water)]
Old Archimedes discovered the pricipal of buoyancy, which says that a solid, when placed in a fluid, becomes lighter by an amount equal to the weight of the fluid it displaces. So, if you can measure the weight of the object dry, then the weight when it is "floating" in water, you can calculate the weight of the water it displaces by taking the difference between the dry weight and the wet weight. (I won't get into a discussion about the differences between weight and mass, just say that they are essentially numerically equivelent when the weight is measured on the surface of the Earth.)
In other words, W(water) = W(solid,dry) - W(solid,wet)
This makes the SG equation:
SG = [W(solid) * v(water)] / [[W(solid,dry) - W(solid,wet)] * v(solid)]
You can simplify this further because (unless ther is a big bubble in your solid) the volume of the solid equals the volume of water when submerged, so;
SG = W(solid,dry) /[W(solid,dry) - W(solid,wet)]
Getting the weight wet is a little problematic, but think of using one of those old grocery store scales with the hanging pans and the big dial. From the first pan, hang another pan that is submerged in water. Re-zero the scale after the second pan is added. Place the sample in the top pan and get the dry weight then move the sample to the bottom pan, making sure the sample is completely covered in water and there are no clinging air bubbles, to get the wet weight.
To arrange this sort of setup with your home scale may take a little ingenuity and some home grown trapeze rig to support the scale above the container of water and suspend another pan in it, but you get the idea.
Finally, we're on to the real question, how to estimate the amount of gold included in a sample of gold and quartz.
Here's the derivation. Assume a lump of something that is a mixture of two (pure) minerals, both of whose SG is known. Call them SG1 and SG2.
You can measure the SG of the whole lump from the method above, and call it SGt.
Now, from inspection the following is known:
The mass (weight) of the lump is the sum of the masses (weights) of the two components and the volume of the lump is the sum of the volumes of the two components. This gives us two equations:
Wt = W1+W2, and
Vt = V1+V2.
The definition of SG (the ratio of the density of the object to the density of water), the fact that the density of water is, numerically, very very close to 1, and the weight on earth is numerically the same as the mass of an object, allows us to simplify the SG equation, numerically, to
SG = W / V. (The units don't work out, but the numbers are essentially correct.)
This means we can substitute the SG relation into the Volume equation and get two equations with two variables:
Wt = W1 + W2, and
(Wt / SGt) = (W1 / SG1) + (W2 / SG2).
I won't bore you with solving the equations, but assume that W1 is the weight of interest, then:
W1 = ((Wt * ((1/SGt)-(1/SG2))) / ((1/SG1) - (1/SG2))
With an application of the derivation of SGt from above, and a bunch of work, this, "simplifies" to:
W1 = (Wt*(SG1/(SG1-SG2))) - ((Wt-Wwet)*(SG2*SG1)/(SG1-SG2))
This is a generic formula that will work for any two materials, as long as you can accurately get the SG of the various components.
Now, if you assume your sample consists of pure gold (SG1 = 19.32), and quartz (SG2 = 2.65), then this simplifies to:
Wg = (3.071 * Wwet) - (1.91 * Wt), which is the equation that is often quoted.
There is a problem with this often quoted equation, natural gold is seldom, as in NEVER, pure so it will always have a lower specific gravity then what was assumed above. Some would think that this doesn't effect the calculations much, but it can put in an error of over 10%.
It would be best to actually measure the SG of some of the gold from your area to get a better estimate for SG1. Get a nugget, or a good quantity of fines, and do the same SG measurement that you did on the rock with the wet and dry weights, and then use these actual measurements to calculate the estimated SG1, for gold, from the SG formula we derived first.
As an example, take a quantity of fines from your area, say 10gm and measure the wet weight as 9.43g to get the SG of the "representative" gold from your area as SG1 = 17.5. This isn't an unusual number for "natural" gold.
Now, take your sample of local gold in quartz that weighs 65gm dry (Wdry) and 52 gm wet (Wwet). Assuming the host rock is quartz with a SG2 = 2.65, you can use the W1 formula to calculate that your weight of (impure) gold is about 36gm.
You can use the M1 formula again to approximate the Karat of your local gold by using 19.32 for SG1, and assuming, quite naturally, that the majority of impurity of the gold is silver (SG = 10.5) for SG2, and the wet and dry weights of 9.42 and 10. If you do this you get a Karat of about 21K, so you have about 31.7g of fine gold in your gold in quartz sample.
If you had used the simplified formula, you would have calculated 35.5g of gold. An overestimation of about 12%.
Simple, right?
JayeLK
Specific Gravity is defined as the ratio of the density of the material to the density of water. To express it mathematically:
SG = p(solid)/p(water); where p is as close as I can get to the Greek rho which is the symbol for density.
Density, p, is the ratio of the mass of the solid to its volume, or:
p(solid) = m(solid)/v(solid)
From this, the formula for SG becomes,
SG = [m(solid) * v(water)] / [v(solid) * m(water)]
Old Archimedes discovered the pricipal of buoyancy, which says that a solid, when placed in a fluid, becomes lighter by an amount equal to the weight of the fluid it displaces. So, if you can measure the weight of the object dry, then the weight when it is "floating" in water, you can calculate the weight of the water it displaces by taking the difference between the dry weight and the wet weight. (I won't get into a discussion about the differences between weight and mass, just say that they are essentially numerically equivelent when the weight is measured on the surface of the Earth.)
In other words, W(water) = W(solid,dry) - W(solid,wet)
This makes the SG equation:
SG = [W(solid) * v(water)] / [[W(solid,dry) - W(solid,wet)] * v(solid)]
You can simplify this further because (unless ther is a big bubble in your solid) the volume of the solid equals the volume of water when submerged, so;
SG = W(solid,dry) /[W(solid,dry) - W(solid,wet)]
Getting the weight wet is a little problematic, but think of using one of those old grocery store scales with the hanging pans and the big dial. From the first pan, hang another pan that is submerged in water. Re-zero the scale after the second pan is added. Place the sample in the top pan and get the dry weight then move the sample to the bottom pan, making sure the sample is completely covered in water and there are no clinging air bubbles, to get the wet weight.
To arrange this sort of setup with your home scale may take a little ingenuity and some home grown trapeze rig to support the scale above the container of water and suspend another pan in it, but you get the idea.
Finally, we're on to the real question, how to estimate the amount of gold included in a sample of gold and quartz.
Here's the derivation. Assume a lump of something that is a mixture of two (pure) minerals, both of whose SG is known. Call them SG1 and SG2.
You can measure the SG of the whole lump from the method above, and call it SGt.
Now, from inspection the following is known:
The mass (weight) of the lump is the sum of the masses (weights) of the two components and the volume of the lump is the sum of the volumes of the two components. This gives us two equations:
Wt = W1+W2, and
Vt = V1+V2.
The definition of SG (the ratio of the density of the object to the density of water), the fact that the density of water is, numerically, very very close to 1, and the weight on earth is numerically the same as the mass of an object, allows us to simplify the SG equation, numerically, to
SG = W / V. (The units don't work out, but the numbers are essentially correct.)
This means we can substitute the SG relation into the Volume equation and get two equations with two variables:
Wt = W1 + W2, and
(Wt / SGt) = (W1 / SG1) + (W2 / SG2).
I won't bore you with solving the equations, but assume that W1 is the weight of interest, then:
W1 = ((Wt * ((1/SGt)-(1/SG2))) / ((1/SG1) - (1/SG2))
With an application of the derivation of SGt from above, and a bunch of work, this, "simplifies" to:
W1 = (Wt*(SG1/(SG1-SG2))) - ((Wt-Wwet)*(SG2*SG1)/(SG1-SG2))
This is a generic formula that will work for any two materials, as long as you can accurately get the SG of the various components.
Now, if you assume your sample consists of pure gold (SG1 = 19.32), and quartz (SG2 = 2.65), then this simplifies to:
Wg = (3.071 * Wwet) - (1.91 * Wt), which is the equation that is often quoted.
There is a problem with this often quoted equation, natural gold is seldom, as in NEVER, pure so it will always have a lower specific gravity then what was assumed above. Some would think that this doesn't effect the calculations much, but it can put in an error of over 10%.
It would be best to actually measure the SG of some of the gold from your area to get a better estimate for SG1. Get a nugget, or a good quantity of fines, and do the same SG measurement that you did on the rock with the wet and dry weights, and then use these actual measurements to calculate the estimated SG1, for gold, from the SG formula we derived first.
As an example, take a quantity of fines from your area, say 10gm and measure the wet weight as 9.43g to get the SG of the "representative" gold from your area as SG1 = 17.5. This isn't an unusual number for "natural" gold.
Now, take your sample of local gold in quartz that weighs 65gm dry (Wdry) and 52 gm wet (Wwet). Assuming the host rock is quartz with a SG2 = 2.65, you can use the W1 formula to calculate that your weight of (impure) gold is about 36gm.
You can use the M1 formula again to approximate the Karat of your local gold by using 19.32 for SG1, and assuming, quite naturally, that the majority of impurity of the gold is silver (SG = 10.5) for SG2, and the wet and dry weights of 9.42 and 10. If you do this you get a Karat of about 21K, so you have about 31.7g of fine gold in your gold in quartz sample.
If you had used the simplified formula, you would have calculated 35.5g of gold. An overestimation of about 12%.
Simple, right?
JayeLK
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Nice chemistry/physics lesson. I remembered that it had to do with mass and volume, but just could not figure out how that fella knew that it was gold making his quartz heavier and not lead/tungsten, etc. I appreciate your time in this....not being facetious. I really do. thanks!
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