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Assuming the product is what it says it is: (1 gram Sodium Gold Chloride per ounce):
Sodium Gold Chloride is Gold that has been dissolved in Aqua Regia, then acidity neutralized with sodium hydroxide.
Formula NaAuCl4
The formula weight is then 23 + 197 + 4 * 35.5 == 362
1 Gram of this would then contain 197/362 grams of gold (0.544 grams pure gold)
Since an ounce is approx 30 ml, then each ml of solution would contain about 18 milligrams of gold
So .55ml of this sodium gold chloride would equal 1 ml of Salt Lake Metals gold chloride.
This is about 11 drops from a medicine dropper.
The formulas given should work fine, but a little less sodium carbonate is needed because the acid is already neutralized. It might not need any at all.
Thank you for giving it a thought.
Although, correct me if i am wrong but i think they say depending on the bottle it is...
CHLORIDE OF GOLD SODIUM 1gr/oz, 1/2 oz
GOLD SODIUM CHLORIDE, 1gr/oz, 3 oz
GOLD SODIUM CHLORIDE, 2gr/oz, 3 oz
So GOLD SODIUM CHLORIDE, 1gr/oz, 3 oz meaning that one gram of gold chloride solution on 3 ounces right ( 3 ounces 85 ml )?
Anyway still lets assume it is like the calculations you made i still cannot yet crasp how to then turn it into colloidal gold.
Meaning what amounts do i need to use with what chemicals.
Do i still use the same amounts as in described below topic link...?
http://www.cgcsforum.org/index.php/topic,1148.msg8854.html#msg8854Also which of the 3 bottles is best to use for making colloidal gold?
CHLORIDE OF GOLD SODIUM 1gr/oz, 1/2 oz
GOLD SODIUM CHLORIDE, 1gr/oz, 3 oz
GOLD SODIUM CHLORIDE, 2gr/oz, 3 oz
Hope this makes sense.
Thank you