%B = (6*atomic mass B)/(molar mass Ca2B6O11.5H2O)]*100 = ?
The way I see it you don't have two problems. The above calculates the percent B for a pure sample of the mineral. Something like 15.78% B. I wouldn't buy it.
Homework question?
Colmanite [Ca2B6O11 * 5 H20] is a salt whose boron content gives glass high clarity. You are a buyer for an optical fiber manufacturer and want pure material. You have just received a Certificate of Analysis for a lot that is 12.8 +- 1.2% (i.e;11.6 to 14.0%) by weight of boron. Would you buy it? (Show work of both problems)
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