# How many phosphorus atoms are present in a 25.0*g mass of "phosphorus pentoxide"?

$\text{0.352 moles of phosphorus atoms are present in this quantity.}$
Number of moles of ${P}_{2} {O}_{5}$ $=$ $\frac{25.0 \cdot g}{141.95 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.176 \cdot m o l$, with respect to ${P}_{2} {O}_{5}$.
There are thus $2 \times 0.176 \cdot m o l$ of $\text{phosphorus}$.
${P}_{2} {O}_{5}$ is an empirical formula. The actual molecule is ${P}_{4} {O}_{10}$, which is an extremely hygroscopic white powder. Do the different formulations make a difference to the number of phosphorus atoms present?