# How many molecules are in 75.0 g of diphosphorus pentoxide P_2O_5?

Mar 2, 2016

Molar quantity $=$ $\text{Mass"/"Molar Mass}$ $=$ $\frac{75.0 \cdot \cancel{g}}{141.94 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.528$ $m o l$. ${P}_{2} {O}_{5}$ has, of course, a molar mass of $141.94 \cdot g$. (Of course, the molecule is actually ${P}_{4} {O}_{10}$, but it does not really matter for this treatment.)
Now it is a fact that $1$ $m o l$ of stuff contains $6.022 \times {10}^{23}$ individual items of that stuff.
So, $0.528$ $m o l$ $\times$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $=$ ??? $\text{individual "P_2O_5" units}$. How many moles of phosphorus atom are in such a quantity; how many oxygen atoms?