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

[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.

[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.

[Lumo]

Great explanation. Many thanks!