Answer:
a) Percentage by mass of carbon: 18.3%
Percentage by mass of hydrogen: 0.77%
b) Percentage by mass of chlorine: 80.37%
c) Molecular formula: 
Explanation:
Firstly, the mass of carbon must be determined by using a conversion factor:

The same process is used to calculate the amount of hydrogen:

The percentage by mass of carbon and hydrogen are calculated as follows:
%C
= 18.3%
%H
=0.77%
From the precipation data it is possible obtain the amount of chlorine present in the compound:
= 0.43g AgCl
Let's calculate the percentage by mass of chlorine:
%Cl=
= 80.37%
Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:



Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be
≈ 2

≈3
The empirical formula for the compound is:

The mass of this empirical formula is:
mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g
This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.