Question

The number of atoms of Cr and 0 in a compound is \[4.8 \times {10^{10}}\] and \[9.6 \times {10^{10}}\] respectively. Its empirical formula is:
A. \[C{r_2}{0_3}\]
B. \[Cr{0_2}\]
C. \[C{r_2}{0_4}\]
D. \[Cr{0_5}\]

Answer 1

Hint: Empirical formula gives the lowest value of ration of the elements present in a molecule. But from this formula, the actual number of the elements present in the molecule cannot be defined. The chemical formula is a way to write a substance using the chemical symbol and number subscripts to mention the number of the atoms in that chemical compound.

Complete step by step answer:
To calculate empirical formula from a given percentage of the constitutional atoms, there are some steps to follow.
 First, let the compound be 100 grams in weight. Therefore, this percentage will be masses of those constitutional atoms of the molecule.
Now change those masses into moles by dividing with their molar masses.
After this, calculate the lowest ratio of those constitutional atoms in that molecule. By dividing each number of moles value with the lowest number to get an integer value.
Write the empirical formula using the ratio of constitutional atoms of the molecule.
In this question directly the number of atoms are given. so, only calculate the ratio of the atoms to find out the empirical formula. The process is shown below.
The ratio of the atoms of Cr and oxygen is,
 \[
  \dfrac{{4.8 \times {{10}^{10}}}}{{9.6 \times {{10}^{10}}}} \\
   = \dfrac{1}{2} \\
 \]
Therefore \[Cr:O = 1:2\] .
So, the empirical formula should be, \[Cr{0_2}\]

Therefore, the correct option is B.

Additional information:
Now, if there are some given percentages of atoms of a compound as follows. The empirical formula would be,
 \[C = 38.71\% \] , \[H = 9.67\% \] and \[0 = 51.62\% \]
Now when the weight of the molecule is 100 grams the masses of carbon, hydrogen, oxygen is 38.71 grams, \[H = 9.67\] grams, and \[0 = 51.62\] grams.
Now their number of moles are,
 \[
  C = \dfrac{{38.71}}{{12}} \\
  C = 3.22 \\
 \]
 \[\begin{array}{*{20}{l}}
  {H = \dfrac{{9.67}}{1}} \\
  {\;H = 9.67}
\end{array}\]
 \[
  O = \dfrac{{51.62}}{{16}} \\
  O = 3.22 \\
 \]
Now, the lowest ratio of constitutional atoms of the molecule. Is
 \[
  C:H:O \\
   \Rightarrow 3.22:9.67:3.22 \\
   \Rightarrow \dfrac{{3.22}}{{3.22}}:\dfrac{{9.67}}{{3.22}}:\dfrac{{3.22}}{{3.22}} \\
   \Rightarrow 1:2:1 \\
 \]
The empirical formula is, \[C{H_2}O\]

Note: The molecular formula of a compound is with the actual number of the elements present in a compound. To calculate the molecular formula of that particular molecule, the relation between the molecular formula and empirical formula should be known. The relation is,
 \[molecular\,formula = {\left( {Cr{O_2}} \right)_n}\] Where \[n = \dfrac{{Molecular\,weight}}{{Empirical\,weight}}\] .