*Chemistry (empirical formula)

29.72 mg of a compound containing carbon, hydrogen, chlorine, and oxygen isburned to give 41.27 mg CO2 and 5.63 mg H2O. A seperate experiment using 25.31mg of the compound produced 57.13 mg AgCl. What is the empirical formula ofthis compound?

Solution :
Mass of C in the compound = Mass of C in CO2
Mass of H in the compound = Mass of H in H2O
Mass of Cl in the compound = Mass of Cl in AgCl

In the compound:
% by mass of C = {41.27 x [12/(12 + 16x2)] / 29.72} x 100% = 37.87%
% by mass of H = {5.63 x [(1x2)/(1x2 + 16)] / 29.72} x 100% = 2.10%
% by mass of Cl = {57.13 x [35.5/(35.5 + 108)] / 25.31} x 100% = 55.84%
% by mass of O = 100% - (37.87% + 2.10% + 55.84%) = 4.19%

Mole ratio C : H : Cl : O
= (37.87/12) : (2.10/1) : (55.84/35.5) : (4.19/16)
= 3.16 : 2.10 : 1.57 : 0.262
= 12 : 8 : 6 : 1

Hence, the empirical formula = C12H8OCl6