Question

Dichloroethane, a compound that is open used for dry cleaning contains carbon, hydrogen, and chlorine. It has a molar

mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen what is its molecular
formula?

Answer 1

Answer: The empirical and molecular formula for the given organic compound is
`CH_2Cl` and
`C_2H_(4)Cl_2` respectively.

Step-by-step explanation:

We are given:

Percentage of C = 24.3 %

Percentage of H = 4.1 %

Percentage of Cl = (100 - 24.3 - 4.1) = 71.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 24.3 g

Mass of H = 4.1 g

Mass of Cl = 71.6 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =
`\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(24.3g)/(12g/mole)=2.025moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(4.1g)/(1g/mole)=4.1moles`

Moles of Chlorine =
`\frac{\text{Given mass of chlorine}}{\text{Molar mass of chlorine}}=(71.6g)/(35.5g/mole)=2.017moles`

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.017 moles.

For Carbon =
`(2.025)/(2.017)=1.003\approx 1`

For Hydrogen =
`(4.1)/(2.017)=2.032\approx 2`

For Chlorine =
`(2.017)/(2.017)=1`

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : Cl = 1 : 2 : 1

Hence, the empirical formula for the given compound is
`C_1H_(2)Cl_1=CH_2Cl`

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

• The equation used to calculate the valency is :

`n=\frac{\text{Molecular mass}}{\text{Empirical mass}}`

We are given:

Mass of molecular formula = 99 g/mol

Mass of empirical formula = 49.5 g/mol

Putting values in above equation, we get:

`n=(99g/mol)/(49.5g/mol)=2`

Multiplying this valency by the subscript of every element of empirical formula, we get:

`C_((1* 2))H_((2* 2))Cl_((1* 2))=C_2H_(4)Cl_2`

Thus, the empirical and molecular formula for the given organic compound is
`CH_2Cl` and
`C_2H_(4)Cl_2` respectively.