Verified Answer :
We need a basis of 100 g mg2p2o7. We need to determine the amount of p2o5 from the amount given. We do as follows:
mass p2O5 = 100 g mg2p2o7 ( 1 mol / 364.4978 g ) ( 1 mol P2O5 / 1 mol Mg2P2O7 ) ( 283.89 g / mol ) = 77.89 g P2O5
Percent = 77.89 / 100 x 100 = 77.89%
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The percentage of \({{\text{P}}_{\text{2}}}{{\text{O}}_5}\) in \({\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\)is \(\boxed{63.78{\text{\%}}}\).
Further explanation:
Molar mass:
The molar mass of a compound is the summation of masses of all atoms or molecule present in the 1 mole of compound. In other words, Molar mass of a compound is the ratio of mass of compound in grams to the number of moles present of that compound. The S.I unit of molar mass is gram per mol or g/mol.
Step 1: The molar mass of \({\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\) is calculated as follows:
\({\text{Molar mass of}}{\mathbf{}}{\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_&{\text{2}}}{{\text{O}}_{\text{7}}}&=\left[\begin{aligned}&{\text{2}}\left({{\text{Atomic mass of Mg}}}\right)+\hfill\\&{\text{2}}\left({{\text{Atomic mass of P}}}\right)+\hfill\\&{\text{7}}\left({{\text{Atomic mass of O}}}\right)\hfill\\\end{aligned}\right]\) …… (1)
The atomic mass of Mg is 24.305 g.
The atomic mass of P is 30.974 g.
The atomic mass of O is 16 g.
Substitute these values in equation (1).
\(\begin{aligned}{\text{Molar mass of}}{\mathbf{}}{\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}&=\left[{{\text{2}}\left({{\text{24}}{\text{.305 g}}}\right)+{\text{2}}\left({{\text{30}}{\text{.974 g}}}\right)+{\text{7}}\left({{\text{16 g}}}\right)}\right]\\&=\left[{{\text{48}}{\text{.61 g}}+{\text{61}}{\text{.948 g}}+{\text{112 g}}}\right]\\&=222.558\;{\text{g/mol}}\\\end{aligned}\)
Step 2: The molar mass of \({{\text{P}}_{\text{2}}}{{\text{O}}_5}\) is calculated as follows:
\({\text{Molar mass of}}{\mathbf{}}{{\text{P}}_{\text{2}}}{{\text{O}}_5}=\left[\begin{aligned}&{\text{2}}\left({{\text{Atomic mass of P}}}\right)+\hfill\\&{\text{5}}\left({{\text{Atomic mass of O}}}\right)\hfill\\\end{aligned}\right]\) …… (2)
The atomic mass of P is 30.974 g.
The atomic mass of O is 16 g.
Substitute these values in equation (2).
\(\begin{aligned}{\text{Molar mass of}}{\mathbf{}}{{\text{P}}_{\text{2}}}{{\text{O}}_5}&=\left[{{\text{2}}\left({{\text{30}}{\text{.974 g}}}\right)+{\text{5}}\left({{\text{16 g}}}\right)}\right]\\&=\left[{{\text{61}}{\text{.948 g}}+80{\text{ g}}}\right]\\&=141.948\;{\text{g/mol}}\\\end{aligned}\)
The reaction of formation of \({\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\)is as follows:
\({\text{2MgO}}+{{\text{P}}_{\text{2}}}{{\text{O}}_5}\to{\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}\)
The percentage of \({{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{5}}}\)in\({\mathbf{M}}{{\mathbf{g}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{7}}}\)can be calculated as follows:
\(\begin{aligned}{\text{Percent of }}{{\text{P}}_{\text{2}}}{{\text{O}}_5}\left(\%\right)&=\frac{{{\text{Mass of 1 mole }}{{\text{P}}_{\text{2}}}{{\text{O}}_5}}}{{{\text{Mass of 1 mole M}}{{\text{g}}_2}{{\text{P}}_{\text{2}}}{{\text{O}}_7}}}\times100\\&=\frac{{141.948{\text{ g}}}}{{222.558{\text{ g}}}}\times100\\&=63.78{\text{}}\%\\\end{aligned}\)
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: Percentage, p2o5, mg2p2o7, molar mass of a p2o5, molar mass of a mg2p2o7, gram per mol, Mg, O, P, 63.78 %, atomic mass of Mg, atomic mass of P is 30.974 g, atomic mass of O is 16 g.