How many grams of iron are in 21.6 g of iron (III) oxide?
Answer
599.4k+ views
Hint: This question can be solved using the mole concept. Mole is the SI unit of measurement and is used to determine the amount of a substance. The molecular formula of iron (III) oxide is $F{{e}_{2}}{{O}_{3}}$.
Complete answer:
The mass of one mole of a substance is known as the molar mass of that substance. Its SI base unit is kg/mol but it is usually expressed in g/mol. It is a bulk property of a substance, not a molecular property.
First, we need to determine the molecular mass of iron (III) oxide.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
We know that the atomic mass of iron (${{M}_{Fe}}$) is 55.847 g/mol and the atomic mass of oxygen (${{M}_{O}}$) is 15.999 g/mol
So, the molecular mass of one mole of $F{{e}_{2}}{{O}_{3}}$ (\[{{M}_{F{{e}_{2}}{{O}_{3}}}}\]) is
\[\begin{align}
& {{M}_{F{{e}_{2}}{{O}_{3}}}}=2\times {{M}_{Fe}}+3\times {{M}_{O}} \\
& {{M}_{F{{e}_{2}}{{O}_{3}}}}=2\times 55.847+3\times 15.999 \\
& {{M}_{F{{e}_{2}}{{O}_{3}}}}=159.691g/mol \\
\end{align}\]
It is given to us that the mass of the sample substance is 21.6 grams.
Now, the number of moles in a given sample can be given by
\[n=\frac{w}{M}\]
Where n is the number of moles,
w is the mass of the substance given (in grams), and
M is the molar mass of the substance (in g/mol).
So, the number of moles of the substance in the sample is
\[\begin{align}
& n=\frac{21.6g}{159.691g/mol} \\
& n\cong 0.135mol \\
\end{align}\]
We know that 1 mole of $F{{e}_{2}}{{O}_{3}}$ contains 2 moles of Fe and 1.5 moles of ${{O}_{2}}$.
\[F{{e}_{2}}{{O}_{3}}\to 2Fe+\frac{3}{2}{{O}_{2}}\]
So, 0.135 moles of $F{{e}_{2}}{{O}_{3}}$ will contain
\[\begin{align}
& {{n}_{Fe}}=2\times 0.135mol \\
& {{n}_{Fe}}=0.27mol \\
\end{align}\]
And the mass of iron in $F{{e}_{2}}{{O}_{3}}$ will be
\[\begin{align}
& {{w}_{Fe}}={{n}_{Fe}}\times {{M}_{Fe}} \\
& {{w}_{Fe}}=0.27\times 55.847 \\
& {{w}_{Fe}}=40.21g \\
\end{align}\]
Hence, there are 40.21 grams of iron in 21.6 g of iron (III) oxide.
Note:
It should be noted that if one mole of a substance is present, it has exactly the Avogadro number (${{N}_{A}}=6.022\times {{10}^{23}}$) of particles and the mass of one mole of a compound is equivalent to the sum of the mass of all the particles contained in one mole of a substance. This can also be used to define the molar mass of a substance.
Complete answer:
The mass of one mole of a substance is known as the molar mass of that substance. Its SI base unit is kg/mol but it is usually expressed in g/mol. It is a bulk property of a substance, not a molecular property.
First, we need to determine the molecular mass of iron (III) oxide.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
We know that the atomic mass of iron (${{M}_{Fe}}$) is 55.847 g/mol and the atomic mass of oxygen (${{M}_{O}}$) is 15.999 g/mol
So, the molecular mass of one mole of $F{{e}_{2}}{{O}_{3}}$ (\[{{M}_{F{{e}_{2}}{{O}_{3}}}}\]) is
\[\begin{align}
& {{M}_{F{{e}_{2}}{{O}_{3}}}}=2\times {{M}_{Fe}}+3\times {{M}_{O}} \\
& {{M}_{F{{e}_{2}}{{O}_{3}}}}=2\times 55.847+3\times 15.999 \\
& {{M}_{F{{e}_{2}}{{O}_{3}}}}=159.691g/mol \\
\end{align}\]
It is given to us that the mass of the sample substance is 21.6 grams.
Now, the number of moles in a given sample can be given by
\[n=\frac{w}{M}\]
Where n is the number of moles,
w is the mass of the substance given (in grams), and
M is the molar mass of the substance (in g/mol).
So, the number of moles of the substance in the sample is
\[\begin{align}
& n=\frac{21.6g}{159.691g/mol} \\
& n\cong 0.135mol \\
\end{align}\]
We know that 1 mole of $F{{e}_{2}}{{O}_{3}}$ contains 2 moles of Fe and 1.5 moles of ${{O}_{2}}$.
\[F{{e}_{2}}{{O}_{3}}\to 2Fe+\frac{3}{2}{{O}_{2}}\]
So, 0.135 moles of $F{{e}_{2}}{{O}_{3}}$ will contain
\[\begin{align}
& {{n}_{Fe}}=2\times 0.135mol \\
& {{n}_{Fe}}=0.27mol \\
\end{align}\]
And the mass of iron in $F{{e}_{2}}{{O}_{3}}$ will be
\[\begin{align}
& {{w}_{Fe}}={{n}_{Fe}}\times {{M}_{Fe}} \\
& {{w}_{Fe}}=0.27\times 55.847 \\
& {{w}_{Fe}}=40.21g \\
\end{align}\]
Hence, there are 40.21 grams of iron in 21.6 g of iron (III) oxide.
Note:
It should be noted that if one mole of a substance is present, it has exactly the Avogadro number (${{N}_{A}}=6.022\times {{10}^{23}}$) of particles and the mass of one mole of a compound is equivalent to the sum of the mass of all the particles contained in one mole of a substance. This can also be used to define the molar mass of a substance.
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