Question

5.An organic compound contains C=61% ,H=11.8% and O=27.12%.Calculate the empirical formula and molecular
formula of the compound. The vapour density of the compound is 59.
(molecular mass=2Xvapour density)

Answer 1

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Answer 2

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_{14}O_2$ respectively.

Explanation:  

We are given:

Percentage of C = 61 %

Percentage of H = 11.8 %

Percentage of O = 27.12 %

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

Mass of C = 61 g

Mass of H = 11.8 g

Mass of O = 27.12 g

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

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{61g}{12g/mole}=5.08moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.8g}{1g/mole}=11.8moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.12g}{16g/mole}=1.695moles$

• 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 1.695 moles.

For Carbon = $\frac{5.08}{1.695}=2.99\approx 3$

For Hydrogen  = $\frac{11.8}{1.695}=6.96\approx 7$

For Oxygen  = $\frac{1.695}{1.695}=1$

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

The ratio of C : H : O = 3 : 7 : 1

Hence, the empirical formula for the given compound is $C_3H_{7}O_1=C_3H_7O$

• To calculate the molecular mass, we use the equation:

$\text{Molecular mass}=2\times \text{Vapor density}$

We are given:

Vapor density = 59

Putting all the values in above equation, we get:

$\text{Molecular mass}=2\times 59=118g/mol$

• 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 = 118 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

$n=\frac{118g/mol}{59g/mol}=2$

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

$C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_7O$ and $C_6H_{14}O_2$ respectively.