Question

a organic compound contain 57.14% of carbon ,6.16% of hydrogen, 9.52% of nitrogen, 27.18% of oxygen. Calculate the empirical formula and molecular formula. If its molecular mass is 294.3g/mol

Answer 1

sir how to came 9 and 2.5?

Answer 2

Given:

The percentage composition C=57.14%, N=9.52%, 0=27.18%, H= 6.16%.

To Find:

The empirical formula and a molecular formula of the compound.

Solution:

To find the empirical formula we will follow the following steps-

As we know,

The molecular weight of carbon, oxygen, nitrogen and hydrogen is 12,16, 14 and 1 respectively.

Now,

$Number of moles \: of carbon = \frac{57.14}{12} = 4.76, Number of moles \: of oxygen = \frac{27 .18}{16} = 1.69, Number of moles \: of nitrogen = \frac{9.52}{14} = 0.68, Number of moles \: of hydrogen = \frac{6.16}{1} = 6.16$

Diving the number of moles with the smallest containing moles we get carbons, oxygen and hydrogen in the empirical formula.

$Number of carbon = \frac{4.76}{0.68} = 7, Number of oxygen= \frac{1.69}{0.68} = 2.5, Number of nitrogen= \frac{0.68}{0.68} = 1, Number of hydrogen= \frac{6.16}{0.68} =9$

Putting the number of carbon, hydrogen, oxygen and nitrogen atoms in the subscript of respective atoms. we will get the empirical formula of the compound.

The empirical formula of the compound is C7H902.5N1.

Mass of C7H4.502.5N1 = 146.0

Molecular formula =

$\frac{empirical \: mass}{molecular \: mass}$

$= \frac{294.6}{146}$

= 2.

Here, the molecular formula is two times the empirical formula.

So, the molecular formula is C14H9O5N2.

Henceforth, the empirical formula of the compound is C7H902.5N1 and the molecular formula is C14H9O5N2.