Question

A 1.143 gram sample of an unknown substance isolated from the liver of a polar bear is shown to contain only C, H, and O. After complete oxidation in an environment with excess oxygen, the sample produced 1.674 grams of CO₂ and 0.6858 grams of H₂O. Is the empirical formula of this substance consistent with its being a carbohydrate?

Answer 1

let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2
= 1.674/44
= 0.038045

Number of moles of H2O = mass of H2O / molar mass H2O
= 0.6858/18
= 0.0381

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.038045
so, x = 0.038045

Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*0.0381 = 0.0762

mass O = total mass - mass of C and H
= 1.143 - 0.038045*12 - 0.0762*1
= 0.610255

number of mol of O = mass of O / molar mass of O
= 0.610255/16
= 0.038141
so, z = 0.038141
Divide by smallest to get simplest whole number ratio:
C: 0.038045/0.038045 = 1
H: 0.0762/0.038045 = 2
O: 0.038141/0.038045 = 1


So empirical formula is:CH2O