Hello:
Can anyone help out with some calculations from an Assay, if would be greatly appreciated.
As per a previous post re: fine gold recovery. Lots of good advice have just built classifier system to get all material under 1/8 of an inch with ability to go through 10 tons of material a day. End Result will be 3 tons of classified material from which we can expect 1000 pounds of black sand. We took a sample of the black sand which will be the final material run in a ball mill. So, from the results below can anyone extrapolate as to how much gold we can expect per 1000 pounds of black sand based on assay results? Or if anyone else has any ideas on how to make a determination if this is a worthwhile project to pursue that would be great too! This is new territory for me and all advice and ideas are encouraged and appreciated. Here are assay results,
Summary:
Sample preparation:
ICP-OES analysis:
ā¢ The calibration was done with a 5 ppm and 1 ppm mixed CMS-2 standard from Inorganic Ventures, Inc. These standards were dilutions from an original 10 ppm standard (Lot#D2-MEB324017).
ā¢ Weight started with: 5.4757 gm black sands; Aqua Regia used: 37.1841 gm.
ā¢ Sample (as received) was digested at 80ā°C for 1 hour.
ā¢ 2 mL of sample digest, 0.14 mL internal standard solution (Y/Sc) were diluted to 14 mL with deionized water. Thus, the dilution is 7X, with this taken into account.
ā¢ Where ever it is noted in the concentration as INTF, this means there was interference from other non-precious metals.
ā¢ Where it is noted as ND, this means ānot detectedā
ā¢ A different CMS-2 standard was tested at the end having the concentration of 7 ppm as a check.
ā¢ Qualifying Statement: Given the fact there is obviously a lot of other metals in this sample, the best we can do here is to note where there is clearly interference from other elements (i.e. where there is a large difference in the signal at two different emission wavelengths). If the response is similar at both wavelengths we have a good chance that the signal is probably real, but there is still a possibility that there is equal interference at both wavelengths, which is rare.
Results:
1. Black Sand Sample
a. Au (267.959 nm) = 7.54 ppm (0.000754 wt%)
i. Calibration: R2=0.999996 (5 ppm, 1 ppm, thru 0)
b. Au (242.795 nm) = 10.71 ppm (0.00107 wt%)
i. Calibration: R2=1.00000 (5 ppm, 1 ppm, thru 0)
2. Alternate CMS-2 Standard (7 ppm)
a. Au (267.959 nm) = 6.31 ppm (90.1% recovery)
b. Au (242.795 nm) = 6.31 ppm (90.1% recovery)
Note: 1 ppm = 0.0001 wt% or 1000 ppm = 0.1 wt%
Great care was taken in the experiment to minimize āfalse positiveā results. This is why two different wavelengths were used for each element and a ācheckā sample was used. However, it is still possible that some interference from elements in the unknown sample contribute to some of the signal observed.
The following was from the lab owner,
Buckyboy,
Given this was an acid digestion (not hydrofluoric acid) of black sands, the sample probably has at least 50% more gold in it. Aqua regia isn't going to break up or dissolve the sand (silica) so there could be gold encapsulated in the sand grains. Given that, you have to do some calculations to see if it is worth it.
Regards,
So if anyone can help break down this info I will look forward to your feed back. Thanks Mucho, Buckyboy!
Can anyone help out with some calculations from an Assay, if would be greatly appreciated.
As per a previous post re: fine gold recovery. Lots of good advice have just built classifier system to get all material under 1/8 of an inch with ability to go through 10 tons of material a day. End Result will be 3 tons of classified material from which we can expect 1000 pounds of black sand. We took a sample of the black sand which will be the final material run in a ball mill. So, from the results below can anyone extrapolate as to how much gold we can expect per 1000 pounds of black sand based on assay results? Or if anyone else has any ideas on how to make a determination if this is a worthwhile project to pursue that would be great too! This is new territory for me and all advice and ideas are encouraged and appreciated. Here are assay results,
Summary:
Sample preparation:
ICP-OES analysis:
ā¢ The calibration was done with a 5 ppm and 1 ppm mixed CMS-2 standard from Inorganic Ventures, Inc. These standards were dilutions from an original 10 ppm standard (Lot#D2-MEB324017).
ā¢ Weight started with: 5.4757 gm black sands; Aqua Regia used: 37.1841 gm.
ā¢ Sample (as received) was digested at 80ā°C for 1 hour.
ā¢ 2 mL of sample digest, 0.14 mL internal standard solution (Y/Sc) were diluted to 14 mL with deionized water. Thus, the dilution is 7X, with this taken into account.
ā¢ Where ever it is noted in the concentration as INTF, this means there was interference from other non-precious metals.
ā¢ Where it is noted as ND, this means ānot detectedā
ā¢ A different CMS-2 standard was tested at the end having the concentration of 7 ppm as a check.
ā¢ Qualifying Statement: Given the fact there is obviously a lot of other metals in this sample, the best we can do here is to note where there is clearly interference from other elements (i.e. where there is a large difference in the signal at two different emission wavelengths). If the response is similar at both wavelengths we have a good chance that the signal is probably real, but there is still a possibility that there is equal interference at both wavelengths, which is rare.
Results:
1. Black Sand Sample
a. Au (267.959 nm) = 7.54 ppm (0.000754 wt%)
i. Calibration: R2=0.999996 (5 ppm, 1 ppm, thru 0)
b. Au (242.795 nm) = 10.71 ppm (0.00107 wt%)
i. Calibration: R2=1.00000 (5 ppm, 1 ppm, thru 0)
2. Alternate CMS-2 Standard (7 ppm)
a. Au (267.959 nm) = 6.31 ppm (90.1% recovery)
b. Au (242.795 nm) = 6.31 ppm (90.1% recovery)
Note: 1 ppm = 0.0001 wt% or 1000 ppm = 0.1 wt%
Great care was taken in the experiment to minimize āfalse positiveā results. This is why two different wavelengths were used for each element and a ācheckā sample was used. However, it is still possible that some interference from elements in the unknown sample contribute to some of the signal observed.
The following was from the lab owner,
Buckyboy,
Given this was an acid digestion (not hydrofluoric acid) of black sands, the sample probably has at least 50% more gold in it. Aqua regia isn't going to break up or dissolve the sand (silica) so there could be gold encapsulated in the sand grains. Given that, you have to do some calculations to see if it is worth it.
Regards,
So if anyone can help break down this info I will look forward to your feed back. Thanks Mucho, Buckyboy!
Sorry to tell ya this but all that work and money thrown in the toilet. #1 rule of a ASSAY is -NEVER HIGRADE SAMPLES- all you get is extrapolated bs and not a comprehensive analysis. No not never,multiple ROUGH samples only-John 
