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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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    $\begingroup$ "And also can somebody give the answer without using POAC". The stoichiometric ratios have to come from a balanced equation, so there is no way to avoid atomic conservation. $\endgroup$ – Karsten Theis Jun 6 at 3:23
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The amount of $\ce{Mg2P2O7}$ in 1.054 gram of a pure sample is: $$ n = \frac{\pu{1.054g}}{\pu{222.57g mol-1}} = \pu{0.004736 mol}$$

There are two atoms of $\ce{P}$ in $\ce{Mg2P2O7}$, but only one in $\ce{NaH2PO4}$. Without considering the two reactions done in detail, we can infer a 2:1 ratio:

$$\ce{2 NaH2PO4 + ... -> Mg2P2O7 + ...}$$

Thus, the mass of $\ce{NaH2PO4}$ is $$ m_\ce{NaH2PO4} = 2 \cdot \pu{0.004736 mol} \cdot \pu{120.0 g mol-1}= \pu{1.137g} $$

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    $\begingroup$ If you use $\ce{NaHPO4\cdot H2O}$ instead, you get the other answer, like @Mithoron mentioned in the comment to the question. $\endgroup$ – Karsten Theis Jun 6 at 3:19
  • $\begingroup$ As @KarstenTheis mentioned, the OP must have calculated based on the salt being $\ce{NaH2PO4\cdot H2O}$ for which the mass would be: $$ m_\ce{NaH2PO4\cdot H2O} = 2 \times \pu{0.004736 mol} \times \pu{137.99 g mol-1}= \pu{1.307 g} $$ $\endgroup$ – MaxW Jun 6 at 3:46

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