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# An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C = 38.71% and H, 9.67% and 0 = 51.62%. The empirical formula of the compound would be:A. CHOB.  $C{H_4}O$ C.  $C{H_3}O$ D.  $C{H_2}O$

Last updated date: 19th Apr 2024
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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. Chemical formula is a way to write a substance using the chemical symbol and number subscripts to mention the number of the atoms into that chemical compound.

Complete step by step solution:
To calculate an empirical formula from a given percentage of the constitutional atoms ,there are some steps to follow.
(i) At first the compound is 100 gram in weight . Therefore, this percentage will be masses of those constitutional atoms of the molecule.
(ii) Now change those masses into moles by dividing them with their molar masses.
(iii) 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.
(iv) Write the empirical formula using that ratio of constitutional atoms of the molecule.
Now, given the percentages in the question as follows,
C = 38.71% , H= 9.67% and 0 = 51.62%
Now when the weight of a molecule is 100 grams the masses of carbon, hydrogen, oxygen are, 38.71 grams, H= 9.67 grams and 0 = 51.62 grams.
Now their number of moles are,
$C{\text{ }} = {\text{ }}\dfrac{{38.71}}{{12}}$
$C = 3.22$
$H = \dfrac{9.67}{1}$
$H = 9.67$
$O = \dfrac{51.62}{16}$
$O = 3.22$
Now ,the lowest ratio constitutes atoms of the molecule is
$\text{C:H:O}$
$\Rightarrow 3.22:9.67:3.22$
$\Rightarrow \dfrac{{3.22}}{{3.22}}:\dfrac{{9.67}}{{3.22}}:\dfrac{{3.22}}{{3.22}}$
$\Rightarrow 1:3:1$

The empirical formula is,  $C{H_3}O$. So, the correct option is (C).

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 , ${\text{ molecular formula = }}{\left( {{\text{C}}{{\text{H}}_{\text{2}}}{{\text{0}}_{\text{2}}}} \right)_{\text{n}}}$  where  ${\text{n = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{Empirical weight}}}}$.