A 2.85-g sample of an unknown chlorofluorocarbon is decomposed and produces 564 mL of chlorine gas (Cl2) at a pressure of 752 mmHg and a temperature of 298 K. What is the percent chlorine (by mass) in the unknown chlorofluorocarbon

0.028%

Explanation:

A general formular for a chlorofluorocarbon (CFC) can be expressed as:

R₂CFCL

This is, an organic compund that has CL-C and F-C bonds (where R is a carbon chain)

For the mentioned decomposition, reaction is:

R₂CFCL → Cl₂ ...

According to the information:

P = 752 mm Hg

T = 298 K

V = 564 ml = 0.564 Lt of Cl₂

If we consider that chlorine gas is an ideal gas, we may use the formula:

PV=nRT = (m/M) RT

m = (PVM) / (RT)

M=molar mass of Cl₂ = 70 g/mol

R = gas constant = 62.36 mm Hg*Lt/mol*K

m = mass of Cl₂

Then:

m Cl₂ = (752x0.564x70) / (62.36x298)

m Cl₂ = 0.0016 g

If we consider (according to above reaction from CFC) that 2 moles of Cl₂ comes from 1 mol of Cl (from R₂CClF):

mass of Cl on CFC = (0.0016) / 2 g = 0.0008 g

So finally, the percent chlorine in the unknow chlorofluorocarbon is:

% Cl = [ (0.0008) / 2.85) ]x100 = 0.0028%