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What is the Empirical and Molecular Formula of Caffeine?

  • Chemistry
  • Thread starter haengbon
  • Start date
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Homework Statement



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.

Homework Equations



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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]

n=2

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

MF= C6H4N4O2

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

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