What mass of phosphorus is contained in a 15.5*g mass of P_2O_5?

Mar 10, 2017

$\text{Mass of phosphorus} \cong 7 \cdot g$.

Explanation:

We need (i) to assess the molar quantity of ${P}_{4} {O}_{10}$ (which of course is the same as ${P}_{2} {O}_{5}$, and which formula we can use without loss of generality).

$\text{Moles of}$ ${P}_{4} {O}_{10}$ $=$ $\frac{15.5 \cdot g}{283.89 \cdot g \cdot m o {l}^{-} 1} = 0.0546 \cdot m o l$.

Now in one mole of ${P}_{4} {O}_{10}$, CLEARLY there are 4 moles of phosphorus. So (ii) we need to multiply the given molar quantity by $4 \times 30.9737 \cdot g \cdot m o {l}^{-} 1$, where $30.9737 \cdot g \cdot m o {l}^{-} 1$ is the atomic mass of phosphorus.

$\text{Mass of phosphorus}$ $=$ $0.0546 \cdot m o l \times 4 \times 30.9737 \cdot g \cdot m o {l}^{-} 1$

$\cong 7 \cdot g$.