Question

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

Answer 1

Hint: 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.