Question

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670 g of H2O. The molar mass is 532.7 g/mol. What is the molecular formula of the compound?

Answer 1

m = 0.6 g

CHO

m = 1.043 g of CO2

m = 0.567 g of H2O

MW = 532.7

first, find empirical

mol of water = mass/MW = 0.5670 /18 = 0.0315

mol of CO2 = mas/MW =1.043 /44 = 0.023704

mol of H = 2*0.0315 = 0.063

mol of O (From H2O) = 0.0315

mol of O (from CO2) = 2*0.023704 = 0.047408

total mol of O = 0.047408+0.0315 = 0.078908

mol of C = 0.023704

Then

ratios:

C:O = 0.078908/0.023704 = 3.32

H:C = 0.063 /0.023704 = 2.657

C H2.65 O3.32

Total mol = 0.0011

MW = mass/mol

MW= 0.6/0.0011 = 545.4 g/mol