# What is the mass of hydrogen in 49.2 g ammonium hydrogen phosphite?

If you mean ammonium hydrogen phosphate, ${\left(N {H}_{4}\right)}_{2} H P {O}_{4} \text{ (ammonium biphosphate)}$, then there are $3.35$ $m o l$ $\text{H atoms}$.
$\text{Moles of ammonium hydrogen phosphate}$ $\frac{49.2 \cdot g}{132.06 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.373 \cdot m o l$.
Clearly, there are $9$ $m o l$ $H$ per $m o l$ ${\left(N {H}_{4}\right)}_{2} H P {O}_{4}$, thus $3.35 \cdot m o l$ $\text{hydrogen atoms}$.