Roy,
Well, I think that I can help... I'm a retired chemistry professor!
I think that your supposition that this is a 1 g vial is correct. This is generally how small amounts of precious metal salts are packaged... 1 g sealed in a glass ampule. Although, looking at the age of the package it could actually be a 15 grain vial. (There are 15.4 grains in a gram so the difference is unimportant here.)
And yes "1%" most probably indicates a 1g/100 mL solution. This should most correctly be written as "1%(w/v)" to distinguish among the the other possibilities, but I would assume in most cases where the "(w/v)" is omitted that we are talking a weight/volume (w/v) solution.
However, there is another "wrinkle" to take into account. It looks to me like you have the monohydrate. (That is what the "x H2O ion the label indicates.) You will may need to take that water of hydration into account when you make your solution depending on exactly what your recipe calls for.. more on this in a bit.
(Also, I doubt that the extra proton ("H" in the formula) of your "chloroauric acid" compared to simple "gold chloride" to will matter for making gold toner.)
The formula weight of gold chloride is 303.33. The formula weight of water is 18.02. Thus the formula weight of monohydrate is the sum of these, or 321.35. (Note that I am ignoring the "H", with a formula weight of 1, it is inconsequential here.) What all of this means is that your monohydrate has a bit less gold per weight that the anhydrous (without water) compound.
The "correction factor" is 303.33/32135 or 0.94. You may need to make this correction if your recipe calls for anhydrous gold chloride. If your recipe calls for monohydrate no correction is needed. If your recipe does not say (and many don't) then your guess is as good as mine!
To make a 1% (w/v) solution entails dissolving 1 g in 100 mL of solution. The most careful way to do this is to dissolve the solid in about 75% of the final volume and then to bring the solution up to the final volume. This allows you to account for the volume the solid takes up in the final solution.
However, for a very dilute solution like this you are safe just adding the solid to 100 mL and dissolving.
If your recipe calls for anhydrous you would apply the above correction fact by making only 94 mL of solution instead of 100 mL.
I hope this help...
