Question

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 36.86g CO2 and 10.06g H2O. The molar mass of estriol is 288.38g/mol .

Find the molecular formula of the unknown compound

Answer 1

we use the mass of carbon dioxide to work out the mass of carbon
molar mass of CO2 = 44 g/ mole
36.86 g of CO2 has 36.860 /44 = 0.83773 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2
so moles of C in the compound = 0.83773 moles
mass of C = 0.83773 x 12 = 10.06697 g

we use the mass of water to find out the mass of hydrogen
molar mass of H2O = 18 g/ mole
10.06 g of H2O has 10.0600 / 18 = 0.55842 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 1.11683 moles
mass of H = 1.11683 x 1.0079 = 1.1257 g

we calculate the O by difference from the mass of sample once we know the mass of C and H
mass of H + C = 11.19263 g
mass of sample = 13.420 g
mass of O by difference = 2.22737 g
moles of O = 2.22737 /16 = 0.13921 moles

we now work out the molar ratio of all the elements in the compound
molar ratio of C : H : O = 0.83773 : 1.11683 : 0.13921
smallest number 0.13921
divide the ratio by the smallest number we get
molar ratio of C : H : O = 6.0 : 8.0 : 1.0
empirical formula is C6H8O
this has a formula mass of 72+8+16 = 96g
to find the number of empirical units in a mole of compound we divide the molar mass by the formula mass

= 288.38 / 96 = 3.0

so the molecular formula of estriol is 3 x the empirical formula or C18H24O3