Let the glass formula be (1-y)[x AO2 - (1-x)BO2] - y C2O3. where A, B & C are the precursors and x & y are the compositions in mol%.
Tell me the procedure of how to calculate the amount of AO2, BO2 & C2O3 in grams (individually) in a particular batch.
Hmmmmm
you are a research scholar plz do some mathematics taking mole concept in account.
You can get good result only when u will do some hard work by ur own..
Determine the mol% on the basis of your formula, and then simply use the conversion of mol (n) to wt (m) with using the molar mass M (calculated using the elements table), m = n*M, do not forget to normalize and you will finally have the oxyde wt%. This works for ternary, quaternary and more complex glass/rock systems, chemical systems... Look on the internet for more details.
@ Charles Le Losq,
Sir, I know the conversion of mol% to Wt.% for binary glasses. But for trinary glasses i have some doubt. If possible plz help me, As per the example i have mentioned above, in order to to calculate the amount of AO2 & BO2, the following steps i followed,
(1-y)[x AO2 - (1-x)BO2] - y C2O3
let y = 0.3 & x = 0.8 mol%
let the batch amount be 25gm
0.7[0.8 AO2 - 0.2 BO2] - 0.3 C2O3
let the mol. wt. of [0.8 AO2 - 0.2 BO2] is M and Mol. wt. of C2O3 is N
therefore,
0.7 x M - 0.3 x N
let the sum of above is P
therefore,
(0.7xM)/ P will give the wt.% of 0.7[0.8 AO2 - 0.2 BO2], then on multiplying this wt% value with the amount of batch we can obtain the amount of 0.7[0.8 AO2 - 0.2 BO2], in grams. ---------> (1)
similar steps applies for 0.3 C2O3. ---------> (2)
Now in order to know the particular amount of AO2 and BO2 in the above glass system, we should multiply 0.7 with both terms in the bracket & should take its above amount in grams as batch amount and follow the same steps???
To avoid confusion just convert directly your (1-y)[x AO2 - (1-x)BO2] - y C2O3 formula in x mol% AO2, y mol% BO2, z mol% C2O3 (x+y+z = 100), then convert it into wt% with molar mass of AO2, BO2 and C2O3. I think it may be simpler like that...
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