# How many ions are there in 4.5 g of (NH_4)_3PO_4?

Approx. $0.121 \cdot m o l$ of phosphate and ammonium ions.
The formula mass of ammonium phosphate is $149.09 \cdot g \cdot m o {l}^{-} 1$. A mass of $4.5 \cdot g$ corresponds to a molar quantity of $\frac{4.5 \cdot g}{149.09 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.02 \times {10}^{- 2} \cdot m o l$ with respect to ${\left(N {H}_{4}\right)}_{3} P {O}_{4}$, and thus $3.02 \times {10}^{- 2} \times 4 = 0.121 \cdot m o l$ of ions, given that there are 4 individual ion per formula unit of ammonium phosphate.