Question

A compound contains 4.07% hydrogen,

24.27% carbon and 71.65% chlorine. Its

molar mass is 98.96 g. What are its

empirical and molecular formulas?​

Answer 1

Step 1. Conversion of mass per cent to grams.

Since we are having mass per cent, it is

convenient to use 100 g of the compound

as the starting material. Thus, in the

100 g sample of the above compound,

4.07g hydrogen, 24.27g carbon and

71.65g chlorine are present.

Step 2. Convert into number moles of each element.

Divide the masses obtained above by

respective atomic masses of various

elements. This gives the number of moles

of constituent elements in the compound

$\small \bold{Moles \: of \: hydrogen = \frac{4.07g}{1.008g} = 4.07}$

$\small \bold{Moles \: of \: carbon = \frac{24.27g}{1.008g} = 2.021}$

$\small \bold{Moles \: of \: chlorine = \frac{71.65g}{35.453g} = 2.021}$

Step 3. Divide each of the mole values

amongst them

Since 2.021 is smallest value, division by

it gives a ratio of 2:1:1 for H:C:Cl .

In case the ratios are not whole numbers, then

they may be converted into whole number by

multiplying by the suitable coefficient.

Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements

$CH_2Cl$ is, thus, the empirical formula of

the above compound.

Step 5. Writing molecular formula(a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula.

For $CH_2Cl$, empirical formula mass is

12.01 + (2 × 1.008) + 35.453

= 49.48 g

(b) Divide Molar mass by empirical

formula mass

$\bold{ \frac{Molar \: mass }{Empirical \:formula \: mass} = \frac{98.96g}{49.48g} = 2 =( n )}$

(c) Multiply empirical formula by n

obtained above to get the molecular

formula

Empirical formula = $CH_2Cl, n = 2$. Hence

molecular formula is $C_2H_4Cl_2$.