(adsbygoogle = window.adsbygoogle || []).push({}); 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 CO_{2}, 0.174g H_{2}O and 0.110g N. The molecular weight if caffeine is 194 g/mole.

2. Relevant equations

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3. 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! :)

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# Homework Help: What is the Empirical and Molecular Formula of Caffeine?

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