${\mathrm{M}}_{{\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4}$ = 149
One molecule of (NH4)3PO4 contains 3×14 = 42 amu nitrogen atom
Mass percent of Nitrogen = $\frac{\mathrm{weight}\mathrm{of}\mathrm{nitrogen}}{\mathrm{Molecular}\mathrm{weight}\mathrm{of}{\left({\mathrm{NH}}_4\right)}_3{\mathrm{PO}}_4}$ × 100
                =$\frac{42}{149}\times 100$ = 28.2%