Question

An unknown compound has a formula of CxHyOz. You burn .1523g of the compound and isolate .3718g of CO2 and .1522 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

total Carbon of the compound converted as CO2

total Hydrogen converted as H2O

44gm. of CO2 is having 12 gm of C

0.3718 gm. of CO2 ............?

    = 0.3718*12/44 = 0.1014 gm of C

18 gm of H2O contains 2 gm , of Hydrogen

0.1522 gm. of H2O contains............?

      = 0.1522*2/18 = 0.0169 gm.of H

weight of Oxygen = weight of sample -[weight of carbon + weight of H]

                         = 0.1523 - [0.1014+0.0169]

                         = 0.034 gm.

now we have to calculate no. of moles of individual element

no.of moles of C = weight/atomic.wt   = 0.1014/12 = 0.00845

no.of moles of H    = 0.0169/1 = 0.0169

no. of moles of O = 0.034/16 = 0.002125

in the next step

divide the no.of moles of each element with smallest no.of mole among them

C .......................................0.00845/0.002125 = 4

H .........................................0.0169/0.002125 = 8

O ........................................0.002125/0.002125 = 1

Assign these values at the subscripts of each element respectively

C4H8O (this is the emperical formula)

moleculer formula = n*(emperical formula)

n = moleculer weight/emperical weight = 72.1/72 = 1

so moleculerformula = emperical formula = C4H8O