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In a gravimetric determination of phosphorus an aqueous solution of dihydrogen phosphate ion is treated with the mixture of ammonium and magnesium ions to precipitate magnesium Ammonium Phosphate. This is heated and decomposed to magnesium pyrophosphate which is weighed. A solution of $\ce{NaH2PO4}$ yielded 1.054 gram of $\ce{Mg2P2O7}$. What weight of $\ce{NaH2PO4}$ was originally present?

I attempted the question and my answer was coming out to be 1.32 gm where as the real answer is 1.14 gm

P.S: I wanted to post the pic of my solution so that i could be corrected , but just cant figure out how to upload the photo.

And also can somebody give the answer without using POAC

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  • $\begingroup$ Maybe this is a case of hydrate vs anhydrous salt... $\endgroup$ – Mithoron Apr 11 '19 at 19:02
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1.054 gram of $\ce{Mg2P2O7}$ equals $1.054/222.57 = 0.004736$ moles

But there are two atoms of $\ce{P}$ in $\ce{Mg2P2O7}$.

grams of $\ce{NaH2PO4} = 2\times0.004736\times 120.0 = 1.137$

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