Question

A compound gave the following data: C$=57\cdot 82%$, O$=38\cdot 58%$ and the rest hydrogen. Its vapor density is 83. Find its empirical and molecular formula. [C=12. O=16 and H=1]

Answer 1

Hint: The molar mass or molecular formula weight of a compound is double times its vapor density. Empirical formula tells us about the simplest ratio of elements in the compound whereas molecular formula tells us about the total no. of atoms of an element present in the compound. The ratio for any element present in the compound is found by dividing its given percentage in that compound to its atomic mass.
Formulae used: $Molecular\,formula\,weight=2\times vapor\,density$
                        $Formula\,ratio\,=\,\dfrac{Molecular\,formula\,weight}{Empirical\,formula\,weight}$
$Molecular\,formula\,=\,formula\,ratio\times empirical\,formula$

Com[plete Step By Step Solution:
The percentage of C=57.82% and O=38.58%, then % of hydrogen present is-$\begin{align}
&\implies 100-57\cdot 82-38\cdot 58 \\
&\implies 100-96\cdot 4 \\
&\implies 3\cdot 6 \\
\end{align}$
Percentage ratio of carbon in the compound is given by its given percentage by its atomic mass i.e. $\dfrac{57\cdot 82}{12}=4\cdot 82$
The atomic mass of oxygen is 16 so its percentage ratio is $\dfrac{38\cdot 58}{16}=2\cdot 41$
Similarly atomic mass of hydrogen is 1 and its percentage ratio is 3.6
Converting these ratios into simple ratios to find the empirical formula, we will divide all the three ratios by 2.41 because it is the smallest of the all. Therefore we get,
Ratio of C $=\dfrac{4\cdot 82}{2\cdot 41}=2$
Ratio of O$=\dfrac{2\cdot 41}{2\cdot 41}=1$
Ratio of H$=\dfrac{3\cdot 6}{2\cdot 41}=1\cdot 5$
We need whole no. values to find empirical formulas so will convert these simple ratios into whole no. values by multiplying them by 2 because we need to convert$1\cdot 5$ into the nearest whole no. So, the ratios will become
For carbon = 4
For oxygen = 2
For hydrogen = 3
So, the required empirical formula will be ${{C}_{4}}{{H}_{3}}{{O}_{2}}$.
To find molecular formula we need formula ratio and for that we need molecular formula weight. To find molecular formula weight we will use vapor density and the relation-
Molecular formula weight $=2\times $vapor density
Molecular formula weight$=2\times 83$
$\,\,=166$
We can find empirical formula weight by using empirical formula for compound which will give us,
$\begin{align}
&\implies 12\times 4+3\times 1+16\times 2 \\
&\implies 48+2+32 \\
&\implies 83 \\
\end{align}$
So now obtaining formula ratio which is given by,
$\text{Formula}\,\text{ratio}\,=\dfrac{\text{Molecular}\,\text{formula}\,\text{weight}}{\text{Empirical}\,\text{formula}\,\text{weight}}$
$\text{Formula}\,\text{ratio}\,=\,\dfrac{166}{83}\,=\,2$
So, $Molecular\,formula\,=\,formula\,ratio\times empirical\,formula$
$\begin{align}
&\implies 2\times {{C}_{4}}{{H}_{3}}{{O}_{2}} \\
&\implies {{C}_{8}}{{H}_{6}}{{O}_{4}} \\
\end{align}$

Note: The empirical formula for two compounds can be similar as it is the simplest ratio of its constituent elements but two different compounds can never have the same molecular formula. The empirical formula should always be in the whole number values. Molecular formula provides more information of a compound than empirical formula because it tells us about the exact no. of atoms of an element present in the compound. To convert the ratios in simpler ratios always divide them by the smallest ratio.