kevin_hulmes@uk.ibm.com on wed 8 jan 97
On 5th January 97, Bonnie Hellman asked about red and black copper oxides
and copper carbonate, and relative costs.....
In summary (and there's an explanation below if you want it) ......
To change from black copper oxide to red or to copper carbonate, my
calculations yield ......
For each part of black copper oxide, you should use either 0.9 parts of
red copper oxide or 1.4 parts of copper carbonate.
(Whichever book you got the conversion factor of 2 from, is wrong! Hamer
states 1.5, but that's based on an assumption of pure copper carbonate.
See below.)
For each dollar you spend on black copper oxide, you would spend 0.9
dollars on red copper oxide or 1.1 dollars on copper carbonate. (Using
the prices you supplied.)
EXPLANATION
Do people want this level of detail, or not? It takes a long time to
compile, so I'd like to know whether it pleases or annoys! Let me know if
you feel strongly either way.
(For those who hate chemistry .......
Please don't switch off when you see the words "molecular weight", it's
just a relative weight. So when you see that water is H2O, which tells
you that it has twice as many hydrogen atoms as oxygen, you can't know
that it has eight times as much oxygen by weight as it has hydrogen
unless you know that hydrogen weighs 1, and oxygen weighs 16, giving
a molecular weight for H2O of 18)
Black Copper Oxide is CuO (molecular weight 63.5 + 16 = 79.5)
Red Copper Oxide is Cu2O (mol. wt. 2x63.5 + 16 = 143)
Copper Carbonate is more difficult. It isn't pure carbonate, but rather
a mixture of two forms of basic carbonate. The content of any
particular sample will depend on the original source, but usually
(I believe) approximates to the malachite form.
Pure Copper Carbonate is CuCO3 (mol. wt. 63.5 + 12 + 3x16 = 123.5)
Malachite is CuCO3.Cu(OH)2 (mol. wt. 2x63.5 + 12 + 5x16 + 2x1 = 221)
Azurite is 2CuCO3.Cu(OH)2 (mol. wt. 3x63.5 + 2x12+ 6x16 + 2x1 = 312.5)
If you then divide the molecular weights by the number of Cu atoms, you
get the relative weights which contain the SAME quantity of Copper.
CuO mol wt = 79.5 mol wt/1 = 79.5
Cu2O mol wt = 143 mol wt/2 = 71.5
CuCO3 mol wt = 123.5 mol wt/1 = 123.5
CuCO3.Cu(OH)2 mol wt = 221 mol wt/2 = 110.5
2CuCO3.Cu(OH)2 mol wt = 312.5 mol wt/3 = 104.2
If you want to convert a recipe from one copper source to another, you
can do that by dividing each number in the last column by the number for
your current source. This will give you the number of parts, kilos,
lbs, or whatever you use, and your original source will become 1.
So, if your original source was Black Copper Oxide, divide all by 79.5
CuO mol wt/1 = 79.5 Relative weight is 1
Cu2O mol wt/2 = 71.5 " 0.9
CuCO3 mol wt/1 = 123.5 " 1.55
CuCO3.Cu(OH)2 mol wt/2 = 110.5 " 1.4
2CuCO3.Cu(OH)2 mol wt/3 = 104.2 " 1.3
So you could use 0.9 parts of red copper oxide for each part of black.
Or, if your sample of copper carbonate approximates to malachite, you
could use 1.4 parts of copper carbonate for each part of black copper
oxide.
You can then work out the equivalent costs by multiplying relative
weights by the price per kilo, lb, or whatever (must all be the same).
In your case Bonnie ......
CuO Rel. weight is 1 USD 4.75 Rel. cost is 4.75
Cu2O " 0.9 USD 4.75 " 4.27
CuCO3.Cu(OH)2 " 1.4 USD 3.75 " 5.21
So, the red oxide is the cheapest source for you.
Note that this ignores other factors such as particle size, which may
change the balance of benefit. Also, the level of accuracy in the above
numbers is spurious. Consider the variable purity of different samples,
and the fact that black iron oxide is hygroscopic, and will therefore
absorb water if not kept in an airtight container. Also, red copper oxide
is regarded as less stable than black, but I don't know if that's
significant in the way it's stored and used by a potter.
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