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Question: How many moles are present in \(198.34\,g\) of \(F{e_3}{\left( {P{O_4}} \right)_2}\,?\) (i) \(0.45...

How many moles are present in 198.34g198.34\,g of Fe3(PO4)2?F{e_3}{\left( {P{O_4}} \right)_2}\,?
(i) 0.455mol0.455\,mol
(ii) 0.555mol0.555\,mol
(iii) 0.655mol0.655\,mol
(iv) 0.755mol0.755\,mol

Explanation

Solution

First multiply the atomic weight of the individual atoms with the number of individual atoms present in the molecule. Add them up to obtain the molecular weight of the given molecule. Then use the equation, number of moles present in a molecule =GivenWeightMolecularWeight = \,\dfrac{{Given\,Weight}}{{Molecular\,Weight}} to calculate the number of moles present in the molecule.

Complete step by step solution:
In the given molecule Fe3(PO4)2F{e_3}{\left( {P{O_4}} \right)_2}, the individual atoms are Fe,PFe,\,P and OO.
Atomic weight of Fe=55.845gmol1Fe\,\, = \,\,55.845\,g\,mo{l^{ - 1}}
Number of FeFe atoms present in the molecule Fe3(PO4)2=3F{e_3}{\left( {P{O_4}} \right)_2}\,\, = \,\,3
Atomic weight of P=30.974gmol1P\,\, = \,\,30.974\,g\,mo{l^{ - 1}}
Number of PP atoms present in the molecule Fe3(PO4)2=2F{e_3}{\left( {P{O_4}} \right)_2}\,\, = \,\,2
Atomic weight of O=15.999gmol1O\,\, = \,\,15.999\,g\,mo{l^{ - 1}}
Number of OO atoms present in the molecule Fe3(PO4)2=8F{e_3}{\left( {P{O_4}} \right)_2}\,\, = \,\,8
Therefore, the molecular weight of Fe3(PO4)2F{e_3}{\left( {P{O_4}} \right)_2}
=(= \,(Atomic weight of Fe×Fe\, \times Number of FeFe atoms present in the molecule)+()\, + \,( Atomic weight of P×P\, \times Number of PP atoms present in the molecule)+()\, + \,( Atomic weight of O×O\, \times Number of OO atoms present in the molecule))
= \,\left\\{ {\left( {55.845 \times 3} \right) + \left( {30.974 \times 2} \right) + \left( {15.999 \times 8} \right)} \right\\}\,g\,mo{l^{ - 1}}
=(167.535+61.948+127.992)gmol1= \,\left( {167.535 + 61.948 + 127.992} \right)\,g\,mo{l^{ - 1}}
=357.475gmol1= \,357.475\,g\,mo{l^{ - 1}}
Now, we know Number moles present in a molecule=GivenWeightMolecularWeight.........(1) = \,\dfrac{{Given\,Weight}}{{Molecular\,Weight}}.........\left( 1 \right).
In this case the given weight =198.34g = \,198.34\,g
Therefore using equation (1)\left( 1 \right) we get
Number moles present in Fe3(PO4)2F{e_3}{\left( {P{O_4}} \right)_2} =198.34g357.475gmol1=0.555mol = \,\dfrac{{198.34\,g}}{{357.475\,g\,mo{l^{ - 1}}}}\,\, = \,\,0.555\,mol.
Hence the correct answer is (ii) 0.555mol0.555\,mol.

Additional Information: Mole is the unit of measurement for the amount of substance in SI units. A mole of a substance or a mole of particles is defined as containing exactly 6.022×10236.022 \times {10^{23}} particles, which may be atoms, molecules, ions, or electrons. The current definition was adopted in November 20182018 as one of the seven SI base units, and is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in 12g12\,gof 12C^{12}C.

Note: The most important step in this question is the calculation of the molecular weight of the given molecule. Calculate the molecular weight with proper units and in a stepwise manner. If the molecular weight comes out wrong then there will be error in the calculation of number of moles of the molecule.