- #1

- 38

- 0

## Homework Statement

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

0.376g caffeine would yield to 0.682 CO

_{2}, 0.174g H

_{2}O and 0.110g N. The molecular weight if caffeine is 194 g/mole.

## Homework Equations

---

## The Attempt at a Solution

0.682 CO

_{2}[tex]\rightarrow[/tex] gC = 0.186 gC

0.174g H

_{2}O[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: C

_{3}H

_{2}N

_{2}O

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

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

n=2

MF= n(EF)

MF= 2(C

_{3}H

_{2}N

_{2}O)

MF= C

_{6}H

_{4}N

_{4}O

_{2}

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