How many total atoms are in .660 g of P_2O_5?

May 18, 2016

$14 \times {10}^{21}$ atoms

Explanation:

Let us first calculate the number of moles of Phosphorous ($V$) Pentoxide i.e. ${P}_{2} {O}_{5}$

mass of ${P}_{2}$ ${O}_{5}$ = $0.660$ $g$

molar mass of ${P}_{2}$ ${O}_{5}$ = $142$ $g$ $m o {l}^{-} 1$

$\text{number of moles of}$ ${P}_{2}$ ${O}_{5}$ = $\text{mass" /"molar mass}$

$n = \frac{0.660}{142}$ $g$ $m o {l}^{-} 1$

$n = 0.0046$ $m o l$ of ${P}_{2}$ ${O}_{5}$

one mole of ${P}_{2}$ ${O}_{5}$ has 5 moles of Oxygen atoms.

using this we can write two conversion factors

1 mol ${P}_{2}$ ${O}_{5}$/ 5 mol Oxygen or 5 mol Oxygen / 1 mol ${P}_{2}$ ${O}_{5}$

using conversion factor

$0.0046$ $m o l$ of ${P}_{2}$ ${O}_{5}$ x $\left[\left(5 \text{ mol " O_2 ) /( 1 " mol } {P}_{2} {O}_{5}\right)\right]$

= $0.023$ $m o l$ of Oxygen

$0.023$ $m o l \times 6.02 \times {10}^{23}$ $\text{atom}$ / $m o l$

= $14 \times {10}^{21}$ $\text{atoms}$