Question

An organic monobasic acid contains $18.6\% {\text{ }} carbon, {\text{ }}1.55\% {\text{ }} Hydrogen, {\text{ }} and {\text{ }}55.4\% {\text{ }}Chlorine$ and its molecular mass is 129. Calculate its molecular formula.

Answer 1

By calculating empirical formula and empirical mass, dividing empirical mass by given molecular mass will give value of n which will ultimately give molecular formula.

Step by step explanation:
In this question, we need to find the molecular formula of the given composition which would be done by finding out the empirical formula of the compound.
Empirical formula is the simplest positive integer ratio of all the atoms present in the compound. For example, ${H_2}{O_2}$ will have empirical formula as $HO$ (since all are present in$2:2{\text{ }} or {\text{ }}1:1$).
In the given question, step by step answer would be obtained and the steps are explained as follows: -
Calculate $\%$ composition of all the elements present i.e. Carbon, Hydrogen, Oxygen and Chlorine
Divide the $\%$ composition by molecular mass of respective atoms to get in moles.
Divide the moles of all by a number such that it gives the simplest ratio of compound.
This simplest compound is empirical formula and calculate its molecular mass
Divide the empirical mass by molecular mass which will give an integer value
Multiply this integer value by empirical formula which will give finally molecular formula
So, Given Composition is: -
Carbon =$18.6\%$, Hydrogen =$1.55\%$, Chlorine = $55.4\%$
So, Total = $18.6 + 1.55 + 55.4{\text{ }} = {\text{ }}75.19\%$
Therefore, the rest present is oxygen
So, Oxygen = $100 - 75.19{\text{ }} = {\text{ }}24.81\%$
2. Dividing $\%$ composition by respective molecular mass
Carbon = $\dfrac{{18.6}}{{12}} = 1.55$
Hydrogen = $\dfrac{{1.55}}{1} = 1.55$
Chlorine = $\dfrac{{55.4}}{{35.5}} = 1.55$
Oxygen = $\dfrac{{24.81}}{{16}} = 1.55$
3. Therefore, the ratio in which these are present is $C:{\text{ }}H:{\text{ }}O:{\text{ }}Cl{\text{ }} = {\text{ }}1:{\text{ }}1:{\text{ }}1:{\text{ }}1$
4. So, empirical formula = $CHOCl$
Empirical mass = $12{\text{ }} + {\text{ }}1{\text{ }} + {\text{ }}35.5{\text{ }} + {\text{ }}16{\text{ }} = 64.5$
5. ${\text{ n = }}\dfrac{{Molar{\text{ }}mass}}{{empirical{\text{ }}mass}}{\text{ = }}\dfrac{{129}}{{64.5}} = 2$
So, the molecular formula of the compound becomes = $2 \times CHOCl$
= ${C_2}{H_2}{O_2}C{l_2}\;$
and it is given that it is organic monobasic acid so means it contains 1 acidic group and therefore, formula is $CHC{l_2}COOH$

Note: To find empirical formula or molecular formula just calculate the simple ratio in which they are present and from the empirical formula rest things can be easily calculated. You must be well familiar with atoms and their atomic number and masses which would really help in these types of questions.