Question

An unknown compound has the formula C_xH_yO_z.You burn 0.2115 g of the compound and isolate 0.5164 g of CO₂ and 0.2114 g of H₂O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula?

Answer 1

moles of CO2 = 0.5164 / 44 = 0.01174

moles of C= 0.01174

mass of Carbon = 12 x 0.01174 = 0.1408

moles of H2O = 0.2114 / 18 = 0.01174

moles of H = 2 x 0.01174 = 0.02349

mass of hydroegen = 0.02349

Oxygen mass = 0.2115 - (0.1408 + 0.02349 ) = 0.04721g

moles of O = 0.04721 /16 = 0.00295

C                O               H

0.01174     0.00295     0.02349

4                 1                    8

C4H8O   -----------------------> empirical formula

empirical formula mass = 72 g/mol

n = molar mass / empirical formula mass

    = 72.1 / 72

     = 1 (nearly)

molecular formula = n x empirical formula

                             = C4H8O