Question

determine the empirical and molecular formula of a compound composed of 18.24 g carbon, 0.51 g hydrogen, and 16.91 g fluorine and has a molar mass 562.0 g/mol.

Answer 1

Answer: The empirical formula for the given compound is $C_3HF_2$  and molecular formula for the given compound is $C_{24}H_8F_{16}$

Explanation : Given,

Mass of C = 18.24 g

Mass of H = 0.51 g

Mass of F = 16.91 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{18.24g}{12g/mole}=1.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.51g}{1g/mole}=0.51moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{16.91g}{19g/mole}=0.89moles$

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 0.51 moles.

For Carbon = $\frac{1.52}{0.51}=2.98\approx 3$

For Hydrogen  = $\frac{0.51}{0.51}=1$

For Fluorine = $\frac{0.89}{0.51}=1.74\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 3 : 1 : 2

The empirical formula for the given compound is $C_3H_1F_2=C_3HF_2$

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:

Molar mass  = 562.0 g/mol

Mass of empirical formula = 3(12) + 1(1) + 2(19) = 75 g/eq

Putting values in above equation, we get:

$n=\frac{562.0}{75}=7.49\approx 8$

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

$C_3HF_2=(C_3HF_2)_n=(C_3HF_2)_8=C_{24}H_8F_{16}$

Thus, the molecular formula for the given compound is $C_{24}H_8F_{16}$