Question

A compound contains 10.13% C and 89.87% Cl (by mass). Determine both the empirical formula and the molecular formula of the compound given that the molar mass is 237 g/mol.
CCl3
C2Cl
CCl

Answer 1

Answer:

The empirical formula is = $CCl_3$

The molecular formula = $C_2Cl_6$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 10.13

Molar mass of C = 12.0107 g/mol

% moles of C = 10.13 / 12.0107 = 0.8434

% of Cl = 89.87

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 89.87 / 35.453 = 2.5349

Taking the simplest ratio for C and Cl as:

0.8434 : 2.5349

= 1 : 3

The empirical formula is = $CCl_3$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12*1 + 3*35.5 = 118.5 g/mol

Molar mass = 237 g/mol

So,  

Molecular mass = n × Empirical mass

237 = n × 118.5

⇒ n ≅ 2

The molecular formula = $C_2Cl_6$