The Edinburgh Etch calculations

So, the calculations for the solution i’ve just created. I am the first to admit that the following may be absolute bollocks; although I was quite good at chemistry, its been a while since school…

Anyway, the proportions for the Edinburgh Etch are:
4/5 Saturated FeCl3 solution at (40% weight/volume)
1/5 Citric Acid solution at (25% weight/volume)

The complication is that since I don’t have giant drums of pre-mixed FeCl3 (40% w/v) solution about the place, I’m using the commonly available FeCl3’6H2O nuggets, or "Iron IIII Chloride Hexahydrate". This adds complication because you have six extra H2O molecules for every molecule of FeCl3 to take into account.

The molecular weight of anhydrous FeCl3 is 162.2.
The molecular weight of FeCl3’6H2O is 270.29.
Which implies that 60% of FeCl3’6H2O is FeCl3, and 40% is H2O.

This first time, I followed the instructions on the packet of hexahydrate, so I mixed 600ml of H2O with 500g of FeCl3’6H2O. Using the percentages above, this yields a solution with:
500 * 0.60 == 300g of FeCl3
600 + (500*0.40) == 800ml of H2O (since 1g of H2O == 1ml)

So I have ended up with… ((300/800)*100), or 37.5% weight/volume FeCl3 solution.

Unfortunately I’ve miscalculated the citric acid slightly, since  my calculation assumed 1000ml FeCl3 solution; it should still work, but etching times may be affected. My main problem is not knowing the exact volume of the resulting FeCl3 solution: I don’t have anything suitable for the measurement of the volume of nasty chemical solutions.

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