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Question: The number of atoms of Cr and O are \(4.8\; \times \;{10^{10}}\) and \(9.6 \times {10^{10}}\) respec...

The number of atoms of Cr and O are 4.8  ×  10104.8\; \times \;{10^{10}} and 9.6×10109.6 \times {10^{10}} respectively. Its empirical formula is?
A) Cr2O3C{r_2}{O_3}
B) CrO2Cr{O_2}
C) Cr2O4C{r_2}{O_4}
D) None of these

Explanation

Solution

We know that the empirical formula is the simplest whole-number ratio of atoms of elements constituting the compound. It gives a relative number of atoms in a molecule or a relative number of moles of in one mole of the molecule.

Complete step by step answer:
We know that 1 mole of atoms contains 6.022×10236.022 \times {10^{23}} number of atoms.
It is given that the number of atoms of Cr and O 4.8  ×  10104.8\; \times \;{10^{10}} are and 9.6×10109.6 \times {10^{10}} respectively.
So 4.8  ×  10104.8\; \times \;{10^{10}} number of atoms of Cr contain 4.8×10106.022×1023  moles\dfrac{{4.8 \times {{10}^{10}}}}{{6.022 \times {{10}^{23}}}}\;moles
i.e., it contains 7.97×10−14  moles7.97 \times {10^{ - 14}}\;moles
Now, In 9.6×10109.6 \times {10^{10}} number of atoms of O contain 9.6×10106.022×1023  moles\dfrac{{9.6 \times {{10}^{10}}}}{{6.022 \times {{10}^{23}}}}\;moles
i.e., it contains 1.59×10−13  moles1.59 \times {10^{ - 13\;}}moles
Now dividing them by the smallest value we get
Cr  =7.97×10−141.59×10−13=0.5Cr\; = \dfrac{{7.97 \times {{10}^{ - 14}}}}{{1.59 \times {{10}^{ - 13}}}} = 0.5
O  =1.5×10−131.5×10−13=1O\; = \dfrac{{1.5 \times {{10}^{ - 13}}}}{{1.5 \times {{10}^{ - 13}}}} = 1
Multiplying both by 2 we get Cr = 1 and O = 2
Therefore the empirical formula becomes CrO2Cr{O_2}

So option (b) is correct.

Additional Information: We should also keep this in mind that,
Molecular formula mass is the whole number multiple of the empirical formula mass for a given compound.
Molecular  formula  massEmpirical  formula  mass=n\dfrac{{Molecular\;formula\;mass}}{{Empirical\;formula\;mass}} = n
Empirical formula and molecular formula are also related to this whole number, nn as :
Molecular  formula  =  (Empirical  formula)nMolecular\;formula\; = \;{\left( {Empirical\;formula} \right)_n}

Note: There is also a shortcut method to find the empirical formula.
From the definition of empirical formula, we understood that it is the ratio of the number of atoms constituting the compound. It is given that
The number of an atom of Cr  =4.8×1010Cr\; = 4.8 \times {10^{10}} and
The number of atoms of O  =  9.6×1010O\; = \;9.6 \times {10^{10}}

The ratio between them =  Number  of  atoms  of  CrNumber  of  atoms  of  O  =  1:2 = \dfrac{{\;Number\;of\;atoms\;of\;Cr}}{{Number\;of\;atoms\;of\;O}}\; = \;1:2
So the Cr has a value of 1.
And the O has a value of 2.
Therefore, the empirical formula becomes CrO2Cr{O_2}