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Using NaH2PO4-2H2O instead of NaH2PO4. How to calculate?

Posted: Mon Jul 09, 2012 12:48 pm
by Lumo
I have a recipe that calls for 3.2g of NaH2PO4 (anhydrous). We have NaH2PO4-2H2O (dihydrate) in stock and I have been told I can use that after working out how much to use.

I know that the FW of the anhydrous is 119.98 and of the dihydrate it is 156.01.

BUT, how oh how do I work out how much of the dihydrate to use instead?

Any help or insight would be much appreciated.

Calculate by Ratio: Using NaH2PO4-2H2O instead of NaH2PO4.

Posted: Mon Jul 09, 2012 1:25 pm
by GrahamKemp
N = W/M; molecule amount equals weight divided by molecular mass.

You want identical amounts of sodium dihydrogen phosphate, thus: NH=NA

.: WH = WAMH/MA
= 3.2 g * 156.01/119.98
≈ 4.1(6) g

4.1(6) g sodium dihydrogen phosphate dihydrate will be 3.2(0) g sodium dihydrogen phosphate and 0.9(6) g water of crystalization.

Posted: Tue Jul 17, 2012 10:47 am
by Lumo
Great explanation. Many thanks!