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What is the Empirical and Molecular Formula of Caffeine?

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the Empirical and Molecular Formula of caffeine, given these information:

    0.376g caffeine would yield to 0.682 CO2 , 0.174g H2O and 0.110g N. The molecular weight if caffeine is 194 g/mole.

    2. Relevant equations


    3. The attempt at a solution

    0.682 CO2 [tex]\rightarrow[/tex] gC = 0.186 gC
    0.174g H2O[tex]\rightarrow[/tex] gH = 9.735 x 10-03 gH
    0.110gN [tex]\rightarrow [/tex]0.110 gN

    0.376g sample = gC + gH + gN + gO
    0.376g sample - 0.186 gC - 9.735 x 10-03 gH - 0.110gN = gO
    gO = 0.070

    C= [tex]0.186/12.01[/tex] = [tex]0.15/.004375[/tex] = 3.42 [tex]\approx[/tex] 3

    H= [tex].009735/1.008[/tex] = [tex].009735/.004375[/tex] = 2

    N= [tex]0.110/14.01[/tex] = [tex].00785/.004375[/tex] = 1.7 [tex]\approx[/tex] 2

    O= [tex]0.070/16.00[/tex] = [tex].004375/.004375[/tex] = 1

    EF: C3H2N2O

    n= [tex]\frac{MW}{EFW}[/tex]

    n= [tex]\frac{194}{82.066}[/tex]


    MF= n(EF)
    MF= 2(C3H2N2O)

    MF= C6H4N4O2

    Is this correct? :/ I think It's not but I hope it is! :)
  2. jcsd
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