# Why does monoammonium phosphate (MAP) have 52% of phosphorus (P₂O₅)?

The formula of monoammonium phosphate (MAP) is $$\ce{NH4H2PO4}$$. Manufacturers say that MAP has 10% of nitrogen (N) and 52% of phosphorus (P). However, its molecular mass is 115 u and the atomic mass of P is 31 u $$(31/115 × 100\% = 27\%)$$. How is there 52% of P?

• They don't say there is 52% of P. They say there is 52% of P2O5. – Ivan Neretin Mar 19 at 17:17
• The gist is that phosphorus can be added to fertilizers in several different forms. To have a standard way of reporting the fertilizer analysis, phosphorus is reported as being $\ce{P2O5}$ rather than whatever form it actually is. – MaxW Mar 19 at 18:07
• Thanks for answering, but there is only one phosphate anion ($H_2PO_{4}^{-}$) in the original formula. How can it be defined in terms of phosphorus pentoxide $P_2O_5$? – Kelvin Mar 19 at 20:19
• Sorry! I understood after reading this wikipedia article. The molecular mass of $P_2O_5$ is 142 u so that P is 43% of the compound ($2*31/142 \approx 0.4366$). When I need to represent P in terms of $P_2O_5$, I must multiply the molecular mass of P by the inverse of that ratio ($1/0.4366 \approx 2.29$). Then, in this case, I have $31/115*2.29 = 0.61$, which is 61% of $P_2O_5$. This value - 61% - is acceptable for purified MAP, but I don't know why "basic" MAP has only 52% of phosphorus pentoxide having the same formula. – Kelvin Mar 19 at 22:23
• Apparently because it is not pure. Look, the content of N is also lower than expected. – Ivan Neretin Mar 20 at 10:17

When it comes to fertilizers, usually what is advertised by general name monoammonium phosphate 10-52-0, in reality is ammophos, which is a mix of monoammonium and diammonium phosphates, $$\ce{NH4H2PO4}$$ and $$\ce{(NH4)HPO4}$$ with some water, so the brutto formula is

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}$$

where $$0 ≤ x ≤ 1$$. To find out the range for $$y$$, let's find the average molecular mass from the given nitrogen ($$\ce{N}$$) and phosphorous ($$\ce{P2O5}$$) content $$α_i$$:

$$\bar{M} = \frac{1}{2}\cdot\left(\frac{M(\ce{N})}{α(\ce{N})} + \frac{M(\ce{P2O5})}{2α(\ce{P2O5})}\right) = \frac{1}{2}\cdot\left(\frac{\pu{14 g mol-1}}{0.10} + \frac{\pu{142 g mol-1}}{2\cdot 0.52}\right) \approx \pu{138 g mol-1}$$

Using the expression for molecular mass of brutto formula and its average value:

$$x = 1.35 - 1.06y$$

Since $$0 ≤ x ≤ 1$$, the following holds true for $$y$$:

$$0.33 ≤ y ≤ 1.27$$

resulting in formula with the following ranges for the variables:

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}, \quad x \in [0; 1], y\in [0.33; 1.27]$$

Since monoammonium phosphate usually prevails in this mix, there is a substantial amount of water; e.g. one of the possible brutto-formulas would be the following monohydrate:

$$\ce{(NH4)_{1.3}H_{1.7}PO4 * H2O}$$

Obviously, all the above is valid in the absence of third-party contaminations, e.g. if pure phosphoric acid and ammonia were used for the synthesis.