Question

How many total atoms are in $0.810{\text{ }}g$of${P_2}{O_5}$?

Answer 1

A mole is a unit measurement for the amount of substance in the international system of units i.e., SI unit. A mole of a particle or a mole of a substance is defined as $6.02214076 \times {10^{23}}$ of a chemical unit, that can be ions, atoms, molecules, etc. Originally it was defined as, the number of atoms in $12{\text{ }}g$of carbon-12.

Complete step-by-step answer:
The phosphorus pentoxide i.e., ${P_4}{O_{10}}$ and it is derived from the empirical formula ${P_2}{O_5}$. It is a white crystalline solid. It is also anhydride of phosphoric acid. It’s a powerful dehydrating agent and desiccant.
The $1\,g$ of an atom will be equal to $1$mole.
In equation;
$1\,g$of atom $= \,\,1\,$mole
Since, $1$mole contains Avogadro’s number (${N_A}$ ) of atoms
So, $1\,mole\, = \,6.022\, \times \,{10^{23}}\,atoms$
Therefore, for $1$ atom will be;
$1\,atom\, = \,\dfrac{1}{{6.022\, \times \,{{10}^{23}}}}\,g\,atoms$
So, the answer will be;
$1\,atom\, = 1.66\, \times \,{10^{ - 24}}\,g\,atoms$
$1$mole ${P_2}{O_5}$ contains Avogadro’s number (${N_A}$ ) of atoms
So,$1$ mole ${P_2}{O_5}$ contains $6.02214076 \times {10^{23}}$ atoms

Let’s calculate the number of moles of Phosphorus pentoxide i.e., ${P_2}{O_5}$
Mass of ${P_2}{O_5}$ = $0.810{\text{ }}g$
Molar mass of ${P_2}{O_5}$= $142\;g\;mo{l^{ - 1}}$
Number of moles of ${P_2}{O_5}$;
$\Rightarrow n\,\, = \,\,\dfrac{{mass}}{{molar\,mass}}$
$\Rightarrow n\,\, = \,\,\dfrac{{0.810\,g}}{{142\,gmo{l^{ - 1}}}}\,\,$
$\Rightarrow n\,\, = \,\,0.00570\,\,mol$ of ${P_2}{O_5}$

But here, the question says about total atoms are in $0.810{\text{ }}g$of${P_2}{O_5}$
$\Rightarrow n\,\, = \,\,0.00570\,\,mol$ of ${P_2}{O_5}$
$\Rightarrow 0.0057\,\,mol\,\, \times \,\,6.023\,\, \times \,\,{10^{23}}\,atom/mol$
In ${P_2}{O_5}$ we have $2$ phosphorus atoms and $5$ oxygen atoms.

So, $7$atoms.
$= \,7\, \times \,0.0057\,\,mol\,\, \times \,\,6.023\,\, \times \,\,{10^{23}}\,atom/mol$
$= \,24.0\,\, \times \,{10^{21}}\,atoms$

So, the answer is $= \,24.0\,\, \times \,{10^{21}}\,atoms$are there in $0.810{\text{ }}g$of${P_2}{O_5}$

Note: If they ask for individual compound,
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 ${P_2}{O_5}$
So, from the first conversion we get,
$\Rightarrow {\text{0}}{\text{.00570}}\,\,mol\,{\text{of}}\,{P_2}{O_5}\,{{ \times }}\,\dfrac{{5\,\,mol\,\,{\text{Oxygen}}}}{{1\,\,mol\,\,{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}}}$
$0.0285\,\,mol$ of oxygen
So, the total number of oxygen atoms,
$\Rightarrow 0.0285\,\,mol\,\, \times \,\,6.023\,\, \times \,\,{10^{23}}\,atom/mol$
$\Rightarrow 17\,\, \times \,\,{10^{21}}\,atoms$of oxygen
Since, in this equation we found the moles of oxygen in atoms.
One mole of ${P_2}{O_5}$ has $2$ moles of phosphorus atoms.

Let’s see from this
We get,
${\text{0}}{\text{.00570}}\,\,mol\,{\text{of}}\,{P_2}{O_5}\,{{ \times }}\,\dfrac{{2\,\,mol\,\,Phosphorous}}{{1\,\,mol\,\,{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}}}$
$0.0114\,\,mol$ of phosphorous
So, the total number of atoms,
$\Rightarrow 0.0114\,\,mol\,\, \times \,\,6.023\,\, \times \,\,{10^{23}}\,atom/mol$
$\Rightarrow 68\,\, \times \,\,{10^{21}}\,atoms$of phosphorous
Since, in this equation we found the moles of phosphorus in atoms.