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AnnyKZ [126]
2 years ago
15

Dichloroethane, a compound that is often 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?
Chemistry
1 answer:
OverLord2011 [107]2 years ago
3 0

Answer: The molecular formula for the given compound is C_2H_4Cl_2

Explanation : 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}}=\frac{24.3g}{12g/mole}=2.025moles

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

Moles of Chlorine = \frac{\text{Given mass of chlorine}}{\text{Molar mass of chlorine}}=\frac{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 = \frac{2.025}{2.017}=1.00\approx 1

For Hydrogen  = \frac{4.1}{2.017}=2.03\approx 2

For Chlorine = \frac{2.017}{2.017}=1

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is C_1H_2Cl_1=CH_2Cl

Mass of empirical formula = CH_2Cl  = 1(12) + 2(1) + 35.5 = 49.5 g/eq.

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=\frac{99g/mol}{49.5g/mol}=2

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

CH_2Cl=(CH_2Cl)_n=(CH_2Cl)_2=C_2H_4Cl_2

Thus, the molecular formula for the given compound is C_2H_4Cl_2

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