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Question: One moles of iron III sulfate, \[F{e_2}{\left( {S{O_4}} \right)_3}\] , contains how many moles of \[...

One moles of iron III sulfate, Fe2(SO4)3F{e_2}{\left( {S{O_4}} \right)_3} , contains how many moles of SO42SO_4^{ - 2} ions? How many moles of OO atoms?

Explanation

Solution

As we know that, 11 mole of any atom or molecule or ion contains Avogadro’s number (NA{N_A} ) of molecule
So, 1mole=6.022×1023atoms1\,mole\, = \,6.022\, \times \,{10^{23}}\,atoms
Therefore, for 11 molecule will be;
1atoms=16.022×1023moles1\,\,atoms\, = \,\dfrac{1}{{6.022\, \times \,{{10}^{23}}}}\,moles
So, the answer will be;
1atoms=1.66×1024moles1\,atoms\, = 1.66\, \times \,{10^{ - 24}}\,moles

Complete step-by-step answer: First, we need to calculate the number of moles of sulfate ion and then we need to calculate the moles of oxygen;
As we know that, the number of moles will be equal to the given mass divided by the molecular mass a compound.
Hence, the formula will be as follows,
n=massmolarmassn\,\, = \,\,\dfrac{{mass}}{{molar\,mass}}
Where,
n=n\, = the amount in moles (mol)(mol)
Mass will be in the terms of   (g)\;(g)
Molar mass will be in the terms of   (g/mol)\;(g/mol)
As we know the molar mass of each element is;
Fe=56Fe\, = \,56
S=32S\, = \,32
O=16O\, = \,16
But we need to calculate the moles of sulphate ions first, so for that we need to calculate the molar mass of the compound and then we need to calculate the molar mass of sulphate ions.
So, let’s calculate the molar mass of the iron III sulfate, Fe2(SO4)3F{e_2}{\left( {S{O_4}} \right)_3};
 = 56×2+3×32+3×4×16{\text{ = }}\,56\, \times \,2\, + \,3\, \times \,32\, + \,3\, \times \,4\, \times \,16
=400 g/mol= \,400{\text{ }}g/mol
The molar mass of the sulphate ions in the compound is;
 = 3×32+3×4×16{\text{ = }}\,3\, \times \,32\, + \,3\, \times \,4\, \times \,16
=288g/mol= \,288\,g/mol
Moles of sulphate ions in the compound is;
n=288400×1n\,\, = \,\,\dfrac{{\,288\,}}{{400}}\, \times \,1
=0.72mol= \,0.72\,mol
Next, we need to calculate the moles of oxygen;
The molar mass of the oxygen in the compound is;
 = 3×4×16{\text{ = }}\,3\, \times \,4\, \times \,16
=192g/mol= \,192\,g/mol
Moles of sulphate ions in the compound is;
n=192400×1n\,\, = \,\,\dfrac{{\,192\,}}{{400}}\, \times \,1
=0.48moles= \,0.48\,moles

Note: 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 6.02214076×10236.02214076 \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 g12{\text{ }}g of carbon-12.