Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: bigblock on 30/09/2012 20:11:22
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What if I took a gram of pure gold and dissolved it into gold chloride. Next I removed the acid I used to dissolve the gold and was left with pure gold chloride. Would it still weigh a gram?
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No it would be heavier.
There's some ambiguity about what "gold chloride" is so the actual weight is not clear.
For true AuCl3
196.96 g of gold would give 303.325g of the chloride
so 1 gram would give about 1.5g
Dissolving gold in aqua regia then evaporation off the excess acid gives Chloroauric acid
Your 196.96 grams of gold would give you between 340 and 412 grams depending on how carefully you dried off the water.
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However,
The density of Gold (Au) is about 19.3 g/cc
The density of AuCl3 is about 4.7 g/cc
So, not only would you have more total product (by mass) of the AuCl3 than the quantity of pure gold that you began with (assuming a lossless reaction), but at about 1/4 the density, you would get about 6 times the quantity by volume.
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To explain BC's figures above, the usual form of chlorauric acid crystals is a tetrahydrate: HAuCl4.4H2O, molar mass 412 g/mol. If you tried to dry these nice looking yellow crystals off by heating, you might in theory get chlorauric acid with no water of crystallisation (molar mass 340 g/mol, but you would more likely drive off both water and HCl to obtain a mixture of the hydrated crystals, the anhydrous material, and the reddish brown gold(III) chloride.
By heating the gold(III) chloride above 250°C you can apparently get gold(I) chloride and chlorine gas (according to the textbook -- I have never tried that bit or the next), and above about 450°C you drive off the rest of the chlorine gas to return to gold