Question

How many total atoms are in $0.410$ g of ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$?

Answer 1

Hint: We will use the Avogadro number to determine the number of atoms in the given gram. For this first, we have to calculate the moles of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ in the given gram. For this, we will use the mole formula. Then by using the Avogadro number we will determine the numbers of atoms.

Formula used: ${\text{mole}}\,\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,\,{\text{mass}}}}$

Complete step by step answer:
We will use the mole formula to determine the number of moles of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
Molar mass of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is $141.94$ g/mol.
On substituting $0.410$ grams for mass and $141.94$ for molar mass of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$,
${\text{mole}}\,{\text{ = }}\,\dfrac{{0.410}}{{141.94}}$
${\text{mole}}\,{\text{ = }}\,2.88 \times {10^{ - 3}}$
So, the moles of the diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is $2.88 \times {10^{ - 3}}$.

According to the Avogadro number,
One mole of any substance =$\,6.02 \times {10^{23}}$ molecules
So, One mole =$\,6.02 \times {10^{23}}$ molecules of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$
One molecule of ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ = $7$ atoms
So, $\,6.02 \times {10^{23}}$ molecules or one mole of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ will have,
= $\,6.02 \times {10^{23}}\, \times 7$ atoms of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$
= $\,4.214 \times {10^{24}}$ atoms of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$

So, one mole of ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ contains$\,4.214 \times {10^{24}}$ atoms.
Then $2.88 \times {10^{ - 3}}$ moles of ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ will have, One mole ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ = $\,4.214 \times {10^{24}}$ atoms $2.88 \times {10^{ - 3}}$ mole${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ = $\,4.214 \times {10^{24}} \times \,2.88 \times {10^{ - 3}}$ atoms.
= $1.21 \times {10^{22}}$ atoms

So, $1.21 \times {10^{22}}$ atoms of diphosphorus pentoxide ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ molecules are there in $0.410$ g of ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$.

Therefore, $1.21 \times {10^{22}}$ atoms is the correct answer.

Note: The number of atoms present in $12\,{\text{g}}$ of carbon-12 is known as one mole. In case of monoatomic, one mole of substance contains Avogadro's number of atoms. The subscript after each atom represents the numbers of that atom. The superscript represents the charge of an ion not the number of that ion. In ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$, five oxygen atoms and two phosphorous atoms are present, so the total number of atoms is seven. If we have to determine the number of atoms of oxygen and phosphorus separate, then we will multiply the Avogadro number and moles of ${{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ with five four oxygen atoms and with two for phosphorus atoms.