Ca Anhydrous vs Dihydrate math question

marke · Feb 4, 2019

Some have argued that anhydrous chemicals are cleaner or have better purity due to the fact that it has less water. Is this water in the solution the cause of this impurity? Is one really better than the other? Does the dihydrate powder need more in terms of weight to meet the same concentration. Ie: 1 gram per lite of ca anhy=x ppm ca, and the dihydrate at 1 gram/lt=y ppm ca. Please help solve for x and y and explain difference if any exist? Thanks in advance!!

JimWelsh · Feb 4, 2019

X = 1 / MW of CaCl2 (anhydrous) * atomic mass of Ca.
Y = 1 / MW of CaCl2 (dihydrate) * atomic mass of Ca.
Difference is MW of anhydrous vs MW of dihydrate.

marke · Feb 4, 2019

keeping me in suspense----what is the difference in mw or better atomic mass? Complete the formula----please! How about the quality issue?

mborn · Feb 4, 2019

There is no quality issue. The water in the dihydrate is not an impurity for this application. The only quality issue would be the grade of chemical you use. You need to add ~1.3247 times more of the dihydrate.
The molecular weight of calcium chloride dihydrate is 147.01 g/mol: Dissolve 147.01 grams in water to a final volume of 1 liter for a 1 Molar solution.
The molecular weight of calcium chloride anhydrous is 110.98 g/mol: Dissolve 110.98 grams in water to a final volume of 1 liter for a 1 Molar solution.